What are the limits of size differences in a host–guest complex?

The size of the cavity is one of the many factors collaborating to host-guest chemistry with cyclodextrin (CD). For example, depending on the size and the properties of the guest molecule, CD may form complexes with either the entire molecule or only a part of the molecule (The diameter and the volume of the cavity of β-CD are around $\pu{7.8 Å}$ ($\pu{0.78 nm}$) and $\pu{262 Å^3}$, respectively, and β-CD complexes well with aromatic rings; Ref.1). It is accepted in general (Ref.2) that the binding forces involved in complex (host-guest) formation are:

Van der Waals type interactions (or hydrophobic interactions) between the hydrophobic units of the guest molecules and the CD cavity.
Hydrogen bond between the polar functional groups of the guest molecules and the hydroxyl groups of the CD rims.
Release of high energy water molecules from the cavity during the complex formation process.
Release of strain energy into the ring structure system of the CD.

Note that the role of hydrogen bonding is not universal since stable complexes are formed with hosts and hydrophobic guests such as benzene derivatives (Ref.3), which do not form hydrogen bonds.

Also, regardless of which types of stabilizing forces are involved, the most important factors in determining the stability of the inclusion of the complexes are: (i) The geometric capability (which is the OP's concern); (ii) the polarity of the guest molecule; (iii) medium; and (iv) the temperature. The reference 2 states that:

Geometric, rather than the chemical factors, are critical in determining the type of guest molecules that can penetrate into the cavity. If the guest is too small, it passes easily through the cavity and the bond will be weak or will not occur. The formation of complexes with the molecules significantly larger than the cavity is also possible, but only some limited groups or the side chains penetrate into the cavity.

It is also important to note that cavity size of CD seems to depend on the method used by the researcher (Ref.4). For instance, Ref.2 reported that inner diameters of $\pu{5.7 Å}$ for $\alpha$-cyclodextrin, $\pu{7.8 Å}$ for $\beta$-cyclodextrin, and $\pu{9.5 Å}$ for $\gamma$-cyclodextrin, but did not verify each was the upper or the lower or middle diameter. Meantime, Ref.3 reported that top and bottom inner-diameters of $5.3$ and $\pu{4.7 Å}$ for $\alpha$-cyclodextrin, $6.5$ and $\pu{6.0 Å}$ for $\beta$-cyclodextrin, and $8.3$ and $\pu{7.5 Å}$ for $\gamma$-cyclodextrin. Therefore, it is hard to predict which molecule would bind to appropriate CD without knowing exact parameters of their cavity. Regardless of the reported cavity sizes, Ref.3 shows that the grafting of $\beta$-cyclodextrin to silica gel conveniently removed water soluble organic molecules with different sizes such as p-nitrophenol, p-nitroaniline, m-nitrophenol, p-chlorophenol, and phenol.
It is also evident that cyclodextrin-catalyzed organic synthesis can be achieved using $\beta$-cyclodextrin as the catalyst. In these reactions, you may find part of starting materials with different sizes making stable host-guest complexes with $\beta$-CD (Ref.5).
On the other hand, an identical molecule can be bound to all three CDs ($\alpha$-, $\beta$-, and $\gamma$-cyclodextrin) grafted to a biomolecule with different impact (Ref.5). For example, $\mathrm{^1H \ NMR}$ studies suggested that methyl-$\beta$-CD forms inclusion complexes with the lipid guest molecules such as cholesterol, linoleic acid, and $\beta$-sitosterol within the CD cavity regardless of their sizes (Ref.6):

Note: Although it is not relevant to the question in hand, it is worth noting that  the chiral nature of the cavity in these molecules makes them useful for enantiomeric separation applications (Ref.7).

References:

Vijaykumar Parmar, Gayatri Patel, Nedal Y. Abu-Thabit, “Chapter 20:  Responsive cyclodextrins as polymeric carriers for drug delivery applications,” In Stimuli Responsive Polymeric Nanocarriers for Drug Delivery Applications, Volume 1: Types and Triggers (Woodhead Publishing Series in Biometerials); Abdel Salam Hamdy Makhlouf, Nedal Y. Abu-Thabit, Eds.; Woodhead Publishing, an imprint of Elsevier: Duxford, United Kingdom, 2018, pp. 555-580 (ISBN: 978-0-08-101977-9).
Julia Martin, Enrique Jacobo Díaz-Montaña, Agustín G. Asuero, “Chapter 1:  Cyclodextrins: Past and Present,” In Cyclodextrin: A Versatile Ingredient; Poonam Arora, Neelima Dhingra, Eds.; IntechOpen: London, United Kingdom, 2018, pp. 3-44 (ISBN: 978-1-78923-068-0).
A. Bibby, L. Mercier, "Adsorption and separation of water-soluble aromatic molecules by cyclodextrin-functionalized mesoporous silica," Green Chem. 2003, 5, 15-19 (DOI: 10.1039/B209251B).
Avilasha A. Sandilya, Upendra Natarajan, M. Hamsa Priya, "Molecular View into the Cyclodextrin Cavity: Structure and Hydration," ACS Omega 2020, 5(40), 25655–25667 (DOI: https://doi.org/10.1021/acsomega.0c02760).
Chang Cai Bai, Bing Ren Tian, Tian Zhao, Qing Huang, Zhi Zhong Wang, "Cyclodextrin-Catalyzed Organic Synthesis: Reactions, Mechanisms, and Applications," Molecules 2017, 22(9), 1475 (17 pages) (DOI: https://doi.org/10.3390/molecules22091475).
Lajos Szente, Ashutosh Singhal, Andras Domokos, Byeongwoon Song, "Cyclodextrins: Assessing the Impact of Cavity Size, Occupancy, and Substitutions on Cytotoxicity and Cholesterol Homeostasis," Molecules 2018, 23(5), 1228 (15 pages) (DOI:  https://doi.org/10.3390/molecules23051228).
Mikhail V. Rekharsky, Yoshihisa Inoue, "Complexation Thermodynamics of Cyclodextrins," Chem. Rev. 1998, 98(5), 1875-1918 (DOI: https://doi.org/10.1021/cr970015o).


Solubility comparison of hexan-1-ol and cyclohexanol in water

In general is very difficult to predict solvatation some experimental models are used such as Abraham solvation equation.
I suppose that the mistake you have done is the fact that you haven't consider one of the most important factor: the molecule structure.

In this case you are comparing a cyclic compound with a linear one, if you could have a look to the potential surface of the two molecules you will see something like this:

What makes the alchol soluble is the hydrophilic hydroxyl group while the lipophilic chain is not. The  ratio between idrophilic and lipophilic area is greater in the cyclohexanol compared to 1-hexanol.


Is there a systematic way to determine the expected cubic structure of metals?

There is no systematic way (besides brute force - that is a system, right?) to predict the crystal structure of the metals.  Even sticking just to the transition metals, as you have noted, there are quite a variety.  The primary ones are as you have noted $fcc$, $bcc$, and $hcp$ (ignoring the rhombohedral Hg).

Several issues arise here. First, there is little difference between $fcc$ and $hcp$ - they are both close packed structures with the only the stacking order changing (ABCABC for $fcc$ and ABABAB for $hcp$). An estimate of the free energy difference between these two crystal structures can be obtained 'SGTE Data for Pure Elements', A.T. Dinsdale, CALPHAD 15(4) 317-425 (1991), the compilation of elemental thermodynamic data. Here one finds that for Ag ($fcc$), the $hcp$ phase is only 300J/mol higher free energy (at T=0K).  Similarly, for Co ($hcp$), the $fcc$ phase is only 427 J/mol higher. So, the free energy differences aren't necessarily that much, making an $a\ priori$ differtiation impossible. (I'll note here that Ru and Os are $hcp$ crystals, not simple cubic as you stated).

The second factor that is difficult to determine is the impact of magnetic free energy on the equilibrium crystal structure (as included in the above reference).  Iron is the most obvious candidate to consider this, and it does not disappoint.  As observed, the equilibrium crystal structure at room temperature is $bcc$.  Heating leads to a change to $fcc$ at $910C$, then back to $bcc$ at $1400C$.  If the magnetic contributions to the free energy are ignored, the observed crystal structures would instead be $hcp$ at low temperatures, a transition to $fcc$ at around $200C$, and a brief appearance of $bcc$ right before melting. I see no way that one could predict these particular changes in phases with temperature.


From PubChem

So the molecule is sort of boat shaped.


When a piston cylinder system that evolves isothermally and quasistatically is irreversible?

If it is done quasi statically with friction involved and, if the system is in contact with a constant temperature bath at the initial system temperature T, then, for the same volume change, the change in entropy of the system will be the same.  That means that all the entropy generated as a result of frictional irreversibility will be transferred to the surroundings (the constant temperature bath).  This will be equal to the frictional work divided by the temperature T.  So the entropy of the universe will increase by this amount.


Do I use the formula $\mathrm{p}K_\mathrm{a}= 2 \mathrm{pH} + \log(c (\mathrm{acid}))$ ?

No, for the pure acide use:

$$K_\mathrm{a} = \frac{([\ce{H+}])^2}{[\ce{HA}]_\mathrm{initial}-[\ce{H+}]}$$

Then, if you are still having problems, see this question and its answer:

Difference in calculated pH and the real pH of a phosphate buffer?


It is NOT the amount of light (number of photons), it is the energy of each photon (its frequency).  When the energy of a photon exceeds the vacuum work function of the surface, an electron is unbound.  There obtains a charged surface and a free electron.  Unless the surface is part of a circuit (e.g., grounded or connected to the cathode), the next electron out must overcome the vacuum work function plus the electrical potential.

The electron is emitted from the conduction band of the bulk metal surface.  The positive charge is delocalized across the surface if it is of constant radius of curvature.   Surface electric field divergence is inversely proportional to radius of curvature.  Unless the surface is very smooth you can get cold cathode emission or electrical breakdown of the encapsulating medium.


How was it concluded that the H3O+ rather than H+ is the "acid" ion?

Actually, the initial theories before Lewis suggested that $\ce{H+}$ is the cause of acidity. However, it soon turned up that an ion as small as the nucleus of hydrogen (you may simply call it a proton) can't be created in low energy reactions due to its high polarising power. So, $\ce{H+}$ is though the cause of acidic nature in aqueous solutions, $\ce{H+}$ never exists as $\ce{H+}$ but as $\ce{H3O+}$, $\ce{H9O4+}$, etc. It is because the proton is heavily hydrated. Though we are not very sure of what is the actual hydrated form, we usually refer it to as $\ce{H3O+}$ or $\ce{H+ (aq)}$.


How to calculate the heat of dissolution from a calorimeter experiment?

You have multiplied the mass of the sample, 1.50g, by temperature change and heat capacity.

However, the water provides most of the heat for the reaction.

The total mass of the solution is 1.50g + 35.0g = 36.5g.

You should be multiplying  36.5g by the temperature change and heat capacity.

Then, you need to consider how many moles 1.50g KCl is.
Divide the change in enthalpy of the solution by the number of moles of KCl to determine the molar heat of solution of KCl.


In this case, the absolute configuration remains same (as R) for $\mathrm{S_N1}$ and changes (into S) for $\mathrm{S_N2}.$ This is because the priority order of non substituted groups (wrt $\ce{C2})$ remains the same before and after substitution of $\ce{Br}$ by $\ce{OH}.$

Let the initial mole fraction of (R)-2-bromobutane be 1. After the reaction, let the mole fractions of R- and S-2-bromobutane be $x$ and $1-x$, respectively.

According to question condition, as $\mathrm{S_N1}$ proceeds twice as fast as $\mathrm{S_N2},$ we get

$$x = 2(1 - x) \quad\implies\quad x = 2/3$$

So we have excess of R-enantiomer, and the specific rotation is

$$\frac{2}{3}\cdot (+13.5) + \frac{1}{3}\cdot (-13.5) = +4.5$$


Is *rinsing* (after washing) with cold water is as efficient as rinsing with hot water?

Depends on carry-over & solubility

Generally, when you have two cleaning fluids, one of which is less effective than the other (the "rinse"), you need to use the good solvent to get the contaminant concentration below the solubility limit of the poor solvent, before switching solvents. Otherwise, you will need a heck of a lot of the poor solvent (albeit that in this case it is only cold water.) How do you do that with only 2 L of good solvent (soapy hot water)?

This sort of question is formally studied in both industrial cleaning plants, and decontamination processes. Determining the ideal process is not exactly rocket science but it is fairly complicated.

It depends on the solubility of the contaminant in the washing fluid -- usually, the dynamic solubility, since processes like these can't sit around waiting for a long soak. However, even if the contaminant is perfectly miscible in the cleaning fluid, it still also depends on "carry-over": the fraction of cleaning fluid that remains on or in the cleaned object between rinses.

In the case where the contaminant is miscible and the carry-over fraction is very small, then you actually get the best results (theoretically, at least) from a very large number of very small rinses. But that won't apply in your case: a wet t-shirt is capable of holding quite a large fraction of 2 litres.

For more practical size of problem like yours, the optimal solution is challenging to find analytically, but straightforward to find numerically. For example, you can set up a simple simulation in (say) MS Excel and run the Solver to get lowest residual concentration against an input parameter of "fraction of hot water used for each wash."

Oh, and it simplifies things considerably to know: you get best result when each titre is the same. That is, you don't need to consider using, say, 2/3 of the hot water on a first wash and 1/3 on a second. If you have n washes, just use 1/n each time.

Incidentally, for many practical parameters, the solution is so sensitive to carry-over fraction that often the most economical approach is to minimise carry-over; in your case, give the t-shirt a really thorough wringing out between rinses!


Betaine HCl is usually delivered in capsules which pass through the mouth and esophagus intact.
Regarding the webmd article that says " Even if it were true, betaine hydrochloride wouldn’t help. It only delivers hydrochloric acid but does not itself alter stomach acidity."
This statement seems contradictory. Indeed, I found article the tests stomach ph after betaine hcl usage and confirms that indeed it reduces it:
https://www.ncbi.nlm.nih.gov/pubmed/23980906


Should β-mercaptoethanol and anything that touches it be handled only under fume hoods for safety aspects?

β-Mercaptoethanol is not that dangerous a chemical. Yes, you should not drink it, you shouldn’t dip your bare fingers in it, you shouldn’t inhale tons of it and you shouldn’t use it to rub yourself down. But other than that it is hardly more dangerous than most other thiols. Simple methanol is probably more toxic and is also routinely handled.

The thing about β-mercaptoethanol is that it  stinks like hell.  And it has the tendency to stick around stinking for quite some time. Like with most thio-compounds, you can smell it long before it becomes seriously dangerous; your sense of smell is a warning against being too risky. You know yourself that the gloves stink if you used the bottle, that the pipette tips stink after pipetting it and that any flasks that contained it tend to stink for quite a while afterwards.

And that is the reason why some labs rigorously segregate it from other chemicals; have dedicated wastes for pipette tips and gloves all kept under a fume it. It is all simply meant to stop the stench as early as possible, to keep it confined in a single fume hood and keep most of the lab air smelling better. The safety aspect is secondary. If you develop a routine of changing equipment every time after handling β-mercaptoethanol, you are less likely to forget to take off your glove and the lab has less smell to deal with.

Therefore, you would be doing your lab colleagues a favour by keeping the bottle confined within its own small plastic bag; but it’s only an olfactory favour, not that much a health favour.


What reagents or series of reactions would allow for selective hydroxylation of an allylic methyl group outside of a ring?

As has been stated by @Beerhunter those benzylic hydrogens are a serious concern . Nonetheless, here is a different offering. To solubilize carboxylic acid 1 both the phenolic group and the acid should be protected. Here, both of the groups are protected as methyl groups. [An ester group and silylated (TBS?) phenol would also suffice.] The double bond of 2 is epoxided to 3. The critical step is the isomerization of the epoxide to an exo-methylene allylic alcohol. There are several reagents that do this transformation. Yamamoto1 has shown that methyl groups are more reactive than methylene groups with the epoxide rearranging reagent diethyl aluminum 2,2,6,6-tetramethyl piperidide although no benzylic methylene examples were provided.
Acetylation of the allylic alcohol affords allylic acetate 4. Grieco2 has reported that bis-acetonitrile PdCl2 isomerizes allylic acetates to their tautomer in what is an equilibrium reaction favoring the more stable isomer. In this instance, the trisubstituted, endocyclic double bond compound 5 is expected. The remaining steps involve removal of the protecting groups of choice to afford the target 6.

1) S. Tanaka, A. Yasuda, H. Yamamoto and H. Nozaki, J. Am. Chem. Soc., 1975, 97, 3252.

2) P. A. Grieco, T. Takigawa, S. L. Bongers and H. Tanaka, J. Am. Chem. Soc., 1980, 102, 7587.


Lignin and cellulose in cotton, work together like the epoxy resin and  the glass fibers in  fiberglass composite materials (a fiberglass boat, for example). So, the cellulose is the fibrous component and is the primary load-bearing element.  While lignin is the polymeric matrix that provides stiffness and rigidity.

The more cellulose is in cotton, the higher is the tensile strength.


That's actually a good question. I was lucky enough that Dietrich Gudat, who wrote a long chapter on this type of chemistry in the Riedel - Moderne Anorganische Chemie was my professor in inorganic molecular chemistry and we discussed these things in detail back then but it's been a while.

I think it's due to how you want your final structure to look like or VSEPR. You have bascially three ligands, two rests and the lone-pair on the carbon. And by having it sp² you get quite some distance with the 120° angles. This splits your possible energy states in one which is called a1, that is basically the sp² and an b1 (the p-orbital). As the a1 has more s-character it is lower in energy and thus you can, depending on how big the gap is, either get a triplett or a singulett carbene.

This will change however if you move to the isolobal and isoelectronic nitrenes. Here we have no sp² but an sp which is automatically filled with two electrons. The other two electrons occupy the degenerate sets of px and py and thus you always end up with a triplett.


Do hard drive platters contain toxic chemicals?

I never opened a newer 2.5" hard disk, but I've read that these may contain glass platters.

However, I've seen and touched stacks of platters from old 3.5" hard disks - a friend of mine makes clocks from hard disk cases.

These platters have an aluminium core, on which another metallic layer, that serves as a kind of primer, and the magnetic material are applied by sputtering. The surface is covered by a protective polymer layer.

Platters have been drilled or cut in a lot of electronics projects, e.g. to make hdd clocks based on the persistence of vision effect. Apart from the usual safety rules for metal works (protect your eyes, don't touch the hot drill, etc.) one only has to make sure that the rotating platter is properly mounted and doesn't start flying around!

When the platters aren't mechanically altered and only painted, there is even less danger. The platters are not toxic and do not even have sharp corners - it is safe to keep them around in the living room.

On a side note, the magnets from these hard disks are more dangerous. These magnets are pretty strong and when playing with them, you can easily jam your fingers between them ;)


Yes, there are specific rules regarding the products of periodic acid cleavage of 1,2-diols (Malaprade reaction). As to the mechanism, it is illustrated here. It is similar to the mechanism for KMnO4 oxidation of 1,2-diols. The structures of the products of 1,2-cleavage of carbohydrates is a diagnostic tool used as an aid in determining their structures. The oxidative cleavage of ethylene glycol (1, Fig. 1) affords two equivalents of formaldehyde. Primary alcohols give formaldehyde. Cleavage of the first single bond in glycerol 2 necessarily affords formaldehyde and α-hydroxyacetaldehyde. Aldehydes are readily hydrated in water. As far as periodic acid is concerned, the hydrate is a 1,2-diol. Cleavage of the second C-C bond produces not only another equivalent of formaldehyde but also an equivalent of formic acid. Secondary alcohols produce formic acid via aldehydes. (continued)

From the foregoing information, D-glyceraldehyde (3, Fig. 2) yields formaldehyde from the primary alcohol and 2 equivalents of formic acid, one from the secondary alcohol and one from the aldehyde group. Cleavage of one C-C bond of 1,3-dihydroxyacetone (4) yields formaldehyde and glycolic acid, which in turn provides formaldehyde and CO2. Ketones, and ultimately carboxyl groups, produce CO2. The oxidative transformations are summarized in the red box in Fig. 2.

The expected products from periodic acid cleavage of D-glucose, D-mannitol and D-fructose are predicted below based upon the formulated rules.


The L and D nomenclature is directly transform from its origin, D- and L-phenylalanine. The following scheme summarize the transformation:

The blue box in right-hand corner explains how L- and D-$\alpha$-amino acids get their names using Fischer projection (this is similar to D- and L-monosaccharides, replacing $\ce{OH}$ group by $\ce{NH2}$ group). If you reduced the carboxylate group of amino acid to a methyl group, the optical rotation of $\alpha$-carbon change from $(S)$ to $(R)$ because second highest priority of $\ce{COOH}$ group to third highest ($\ce{CH3}$ group) based on the size of the $\ce{R}$ group (see general formula of amino acid in top right corner). For example, when $\ce{R=CH2Ph}$ group, the amino acid is phenylalanine, which has $(S)$-rotation, but reduction of caboxylate group to methyl make the rotation $(R)$. However, the product $(R)$-amphetamine still carries the original L-nomenclature (see examples in bottom red box).
Note: The conversion of phenylalanine to amphetamine in several steps were achieved in beginning of 21st century (Ref.1).This is the ultimate enanthimerically pure version.
Reference:

Dominick A. Quagliato, Patrick M. Andrae, Edward M. Matelan, “Efficient Procedure for the Reduction of α-Amino Acids to Enantiomerically Pure α-Methylamines,” J. Org. Chem. 2000, 65(16), 5037–5042 (https://doi.org/10.1021/jo000242h).


What are solvolytic displacement reactions?

A book could be written on this subject!

A solvolysis reaction is an $\ce{S_{N}1}$ reaction where the solvent acts as the nucleophile and replaces the leaving group.  The following reaction is a solvolysis reaction run in ethanol.

I suspect there are some industrial uses of solvolysis reactions and they have been used in labs and by drug companies as a step in a synthetic sequence to prepare some complex molecules.

But I suspect the biggest contribution from the study of solvolysis reactions is the impact they have had on our understanding of carbocation chemistry.  From our study of solvolysis reactions we have learned how and why some carbocations rearrange before being captured by solvent.

While studying such rearrangements the solvolysis of the norbornyl system was examined.

It turns out that

norbornyl compound 1 solvolyzes several hundred times faster than compound 3 - why?
if compound 1 is prepared optically active, the products are racemic - why?
If compound 1 is radiolabeled, the label is scrambled in the products - why

It took decades to answer these and other questions.

These investigations led to a whole new field of carbocation chemistry: the classical - non-classical carbocation debate (see the norbornyl link above for a discussion of this topic).  These investigations added a whole new section to physical organic chemistry textbooks.  What was learned from these solvolysis studies completely changed our thinking about carbocation chemistry.


Comparison of C-Cl bond length in CH3Cl and CF3Cl

The paragraph you cite does not define Bent's rule (see What is Bent's rule?, Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?).

If you compare $\ce{H3CCl}$ and $\ce{F3CCl}$, then the $\ce{C-H}$ bond has less carbon-p character than an idealised sp³ orbital, while the $\ce{C-F}$ bond will have more carbon-p character than the idealised sp³ orbital. This in turn means that in $\ce{H3CCl}$ the $\ce{C-Cl}$ bond will have more carbon-p character as an idealised sp³ orbital and is therefore longer, while in $\ce{F3CCl}$ the $\ce{C-Cl}$ bond will have less carbon-p character and will therefore be shorter.

Note of caution: I have read many questions about the concepts introduced in J. D. Lee's Concise Inorganic Chemistry; many of which revolve around outdated theories. I do not have a copy of that book, but from what I have experienced so far, I do not think it is a good book. Alas, in this instance it appears to be correct though.


Symmetry representations for acetonitrile (CH3CN)

I am afraid I disagree with Geoff's assignment of acetonitrile to the point group $C_{\infty v}$ and "approximation" with $C_{2v}$, so I will make another attempt to answer this question. The way I have structured this answer, the resources I used, and the level of detail included follow a similar question on this site.

Molecule structure

First you need to know the geometrical structure of your compound. This can be found easily from Wikipedia. The image is reproduced here.

Assigning the point group

Then you need to find the point group of acetonitrile. This can be done by following flow charts, one of which is available here. While it may be a little difficult at the beginning, with some practise you can look at a molecule and determine its point group accurately, so it is worth working through the flow chart for a range of molecules until you are confident with this. There are other arrangements of this flow chart, but all flow charts should give you the same result, so use the one provided in your course if possible.

With the flow chart I have linked to, we go through these steps (if you have a flow chart organised differently, you will encounter the same steps, but likely in a slightly different order - check this is the case!):

molecule is not linear
molecule contains only one $C_3$ axis
molecule DOES contain a proper rotation axis which is $C_3$ (rotation over methyl group)
the highest rotation axis is $C_3$ but there are no perpendicular $C_2$ axes
the molecule does NOT contain a horizontal reflection plane ($\sigma_n$). This can sometimes be confusing -- in general, a horizontal plane is perpendicular to the highest rotation axis.
the molecules DOES contain 3 vertical reflection planes ($\sigma_v$).
therefore the point group is $C_{3v}$.

"Orbitals"

In your question, you said that you need to "to assign the symmetry labels to orbitals of acetonitrile". This is a little unclear, because you can mean that you want to assign symmetry labels to atomic orbitals of acetonitrile, or molecular orbitals of acetonitrile. I am going to assume that you need the molecular orbitals here. But how do we find the molecular orbitals?

Molecular orbitals are constructed by making combinations of atomic orbitals. Which means that, a logical next step is to consider all the valence atomic orbitals (VAOs) on all the atoms of acetonitrile.

Atomic orbitals and symmetry sets

To help us talk about VAOs, I have put numbers on all the atoms.

Carbon, atoms 1 and 2: $2s$, $2p_x$, $2p_y$, $2p_z$

Nitrogen, atom 3: $2s$, $2p_x$, $2p_y$, $2p_z$

Hydrogen, atoms 4, 5 and 6: $1s$

When we try to tackle these types of problems, it is common to construct symmetry sets of atomic orbitals that would transform in the same way under the operations of the point group of the molecule. You will also notice that you cannot use any of the operations of the group to transform certain atomic orbitals on the same atom into each other (cannot transform $2s$ to $2p$, cannot transform $2p_x$ to $2p_z$, etc), so these should be considered as separate sets. The $p_x$ and $p_y$ orbitals are mixed by the rotations and reflections of the group, so we should consider them as one symmetry set.

In acetonitrile, there is only one nitrogen, so we have to consider as separate sets:

$2s$
$2p_z$
$2p_x$, $2p_y$

Carbon 1 and carbon 2, though both carbons, cannot be transformed from one to another, and your chemical intuition should also tell you they are different carbons because they have different bonds attached to them, so they are also in separate sets of their own. The three hydrogen atoms, on the other hand, are related by rotation (a $C_{3v}$ rotation on the principal axis, so they can form one set of 3 hydrogen 1s orbitals.

Assigning symmetry labels to each set of valence atomic orbitals

The usual trick at this point is to simply read off the orbitals that have corresponding symmetry labels, e.g. the $p_x$ orbital would transform as $x$ on the character table. This definitely works for any atom that lies on the origin point of the point group, but it is not completely obvious where you would take the origin point for putting a (right handed) axis system on acetonitrile. The "middle" of this molecule would appear to be somewhere between the two carbons, and that may seem a little awkward.

Remember that in all cases, the choice of origin and axis system is actually arbitrary, however you should pick the most convenient one. Now, for acetonitrile you can pick any point on the principal rotation axis ($C_{3}$) and it would work, because that is an inherent feature of the $C_{3v}$ point group. If this is not obvious, I suggest looking at the answer I referenced previously, and also working through any other examples that you can find.

Luckily for us, $C_{3v}$ has quite a small character table (i.e. it does not have many symmetry elements), so we can work through it and check. The character table of the point group $C_{3v}$ is shown below and can also be found here. The character tables for many common point groups can be found here.

$s$ orbitals that lie on the principal axis

Now let's consider the $2s$ orbital or carbon atom 1. It should be fairly obvious that it transforms back to itself under all the operations of the point group (unchanged under identity $E$, any of the $C_3$ rotations, any of the $\sigma_v$ reflections), and so it transforms as $A_1$. It should also become very obvious that this is also true for the $2s$ orbitals of carbon atom 2 and nitrogen atom 3, even though these atoms are not symmetry-related, their atomic orbitals can have the same symmetry under the point group. This will be important when we come to construct molecular orbitals.

$p$ orbitals that lie on the principal axis

For p orbitals, we need to impose an axis system. Conventionally, the principal axis is taken as the $z$ direction. You have to follow these conventions when doing symmetry analysis.

Let's consider the $2p_z$ orbital or carbon atom 1, which lies along the principal axis. This again transforms as $A_1$.

The $2p_x$ and $2p_y$ orbitals are a little tricky. They remain identical under the identity operator $E$, giving a character of 2. For the reflection operator $\sigma_v$, one of the three $\sigma_v$ planes would keep one identical while inverting the other one (giving zero), and the other two $\sigma_v$ planes give characters of zero, so the overall character for $3 \sigma_v$ is zero. The $C_{3v}$ is the difficult one, where you have to put vectors on the axis, rotate them, work out the geometry with trigonometry, form the $2 \times 2 $ matrix and work out the trace (i.e. the character) is -1. Or you can use the fact that the character tables show (x,y) transform as the irreducible representation $E$.

Again, the same symmetry labels apply for all the atomic orbitals for all atoms that lie on the principal axis.

Set of hydrogen $1s$ orbitals that do not lie on the principal axis

For these hydrogen $1s$ orbitals we examine explicitly how they behave under all operations of the point group.

$E$: 3  (all three $1s$ remain unchanged from themselves)
$C_v$: 0 (all three change and do not remain at the same position)
$\sigma_v$: 1 (for any one operation, only 1 of the orbitals remain the same, the rest change)

This is a reducible representation. You need to reduce it to find the symmetry labels for the hydrogen $1s$ orbitals.

Symmetry labels for all valence atomic orbitals

For the whole molecule of acetonitrile, the valence atomic orbitals give the irreducible representation $4 A_1 \oplus 4 E$. From the discussion above you should be able to deduce the exact symmetry label for any atomic orbital on any atom of the molecule.

Molecular orbitals

You know that you need to generate as many molecular orbitals as atomic orbitals, and only the atomic orbitals of the same symmetry can mix. Molecular orbitals use lower case letters, so you expect two types of symmetry labels based on the irreducible representations of the VAOs. You should be able to say how many molecular orbitals of each type you expect, remembering that the irreducible representation $E$ is two-dimensional and indicates a degenerate pair of VAOs.

Symmetry analysis cannot give you the energy of these molecular orbitals, however, so you cannot know how they are ordered relative to one another.


One thing I see:  when you work out the MW (or AW) of X, you assume 4 moles of X based on the chemical formula.  But, properly, you have twice as many moles of X as Na ($\ce{Na_2X_4}$), and you found just 0.5 mole of Na.  So instead of 4 moles of X you have just one mole of X making the atomic weight if X 10.8 g/mol not 2.7.  Hmmm, isn't there an element in the second row of the Periodic Table with that atomic weight?


The two forms of amino acids that your question asks about are called zwitterions (technically IUPAC says they should be called dipolar ions but the term zwitteron is pervasive).

Since amino acids contain both acidic (the carboxylic acid) and basic (the amine) moieties the molecule is able to undergo what almost appears to be an intramolecular acid-base reaction in which the acidic proton of the carboxylic acid protonates the amine, this is what gives rise to the species on the right of the image below.

Its worth pointing out the this property of the amino acids is valid at physiological pH for the naturally occurring amino acids, generally in aqueous medium (which living organisms have. Also as Jan pointed out in the comments, other polar solvents would also allow the amino acid to show this characteristic).

At other pH values, the molecules become singly ionised, favouring either the cationic or anionic forms. The range of pH values at which the amino acid is zwitterionic will also vary depending on which amino acid you're looking at (the pKa values are different for every amino acid, even if only subtly)


I'm trying to verify the integrity of my cheap "24k gold" leaf

EDIT: as pointed out by Stian Yttervik, my previous answer (below) is probably too dangerous for someone without prior experience with handling hazardous chemicals.

I would therefore recommend that you use nitric acid: simply add a few drops of clear fuming nitric acid to a test tube containing your gold leaf and stir with a glass rod.

The solution will then go green/blue if it contains copper. Only do this with very small quantities of gold leaf and when outside as one of the products of the reaction with copper is nitrogen dioxide - an extremely toxic brown gas.

I would recommend concentrated $\ce{H2SO4}$, as it is easily available and produces a brightly coloured salt when reacted with copper, thus making it easy to tell whether your gold leaf is 24K or not without any expensive weighing equipment.

Make sure you do this outside or in a very well-ventilated area, as it will give off small amounts of $\ce{SO2}$, a toxic gas. Wear thick gloves and eye protection and make sure you have a water supply nearby in case of an accident.

Pour 10 ml of concentrated sulfuric acid (98%) into a boiling tube (or any other narrow Pyrex container you might have) and heat the liquid with a Bunsen burner until it starts to boil. Avoid the fumes as these are corrosive. Without removing the boiling tube from the flame, add a few small sheets of gold leaf and stir the solution with a glass rod for a few minutes.

If you find that the solution is completely colourless, there certainly was no copper in the gold leaf. However, even the slightest tinge of blue/green would indicate copper sulfate - formed by reacting copper with sulfuric acid.

When you're done, make sure to dispose of the sulfuric acid by emptying it into a litre or so of water (acid in water, not the other way around) and then neutralising this with bicarbonate of soda


What is the role of size in formation of complex compounds?

The 3d electrons are attracted strongly to the nucleus (positive charge) and this overalls 'pulls' the electron cloud closer to nucleus making the atomic radii much smaller, which is why there is a general trend of decreasing atomic radii across the period.

Since in transition elements series the 4s orbitals are filled first before 3d orbit, this is turn makes the transition element very reactive and readily form compound as the 4s electrons are released easily, (in addition 3d orbitals provide a screening effect) and excited to a higher energy level during typical reaction to form  complex salts.


What is the relationship between something magnetic and unpaired electrons?

There are five forms of magnetic behavior. Ferromagnetism is when a material is attracted to a magnetic field and can spontaneously hold a magnetic field alone (the typical form of magnetism).  Paramagnetism is when a material is attracted to a magnetic field but cannot spontaneously hold a magnetic field alone (the attraction may be too weak to observe practically).  Diamagnetism is a material that expels magnetic fields (essentially non-magnetic). there is also ferrimagnetic and anti-ferromagnetic but though these are interesting they not incredibly relevant to the answer.

Electrons form the magnetic fields of materials.  If a metal's atom have unpaired electrons the metal can attract a magnetic field.  If all of the electrons are paired in opposite spin orbitals then the spins cancel each other out and no response to a magnetic field is observed.

Cobalt has unpaired electrons making it at minimum demonstrate paramagnetic behavior. It is however able to hold a magnetic field making it ferromagnetic.  Most metals have unpaired electron and are paramagnetic (for example: aluminum, zirconium, titanium, magnesium and potassium)  Ferromagnetic metals become paramagnetic at elevated temperatures and inversely many paramagnetic metals become ferromagnetic at low temperatures.


Why assume a neutral amino acid is given for acid-base reaction?

There are a few observations to make.

The amino acid only has two $\mathrm{p}K_\mathrm{a}$ values. Thus, it cannot have a side chain that is in any way acidic or basic. We only have three different possibilities: $\ce{H2AA+, HAA, AA-}$.
It was crystalline when added. Therefore, it must have been in its neutral $\ce{HAA}$ state. Otherwise, it would have been given as ‘the $\ce{HCl}$ salt …’ or ‘the $\ce{NaOH}$ salt …’

If we had any excess base in there, it would have to have been given in the question as the sodium hydroxide crystal adduct or something.


How to obtain hydrophobic Doxorubicin from its hydrochloride salt?

Add a base to an aqueous solution of the compound.  The base could be ammonia water or ammonium hydroxide.  Make sure you add enough base to raise the pH to 12 or higher.
Add a water-immiscible organic solvent with a low boiling point to the solution and mix well.  A good solvent might be dichloromethane.
Let the phases separate and carefully remove the organic phase.  (If you use dichloromethane, the organic phase will be the bottom phase, but this is not true of all solvents.)
Repeat step 2 several times, each time combining the newest organic extract phase with the extract from the previous steps.
Evaporate the solvent under air, or, if you have it, an inert gas such as nitrogen.


What actually happens in strong acid- weak base reactions?

The strategy of major species

You correctly describe the strategy. First, let reactions go to completion (in the direction that makes sense, i.e. weak base and strong acid forms weak acid and spectator ion - never the other way around). Then, check what the major species are and estimate the pH. Finally (and this step is often omitted), see if equilibria involving minor species (such as hydroxide for acidic solutions) need adjustment.

Why does the strategy of major species work?

For the example, the OP already calculated the estimated amount of species: 0.0075 mol of ammonium, and 0.013 mol of ammonia. With a total volume of 65 mL, that comes out as:

$$c_\ce{NH4+} = \pu{0.115M}$$
$$c_\ce{NH3} = \pu{0.20M}$$

From this, we can estimate the pH to be a bit more basic than the $\mathrm{p}K_\mathrm{a}$, i.e. 9.49. Now we can think about the minor species, hydroxide and hydronium. At pH = 9.49, there should be an excess of hydroxide over hydronium, by about 0.00003 M. Dissociation of water does not give an excess, so it has to come from ammonia turning into ammonium:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

However, when this reaction makes hydroxide, it also changes the ratio of ammonium to ammonia, changing the pH. The only reason we don't keep running in circles, adjusting one equilibrium and then the other, is that this adjustment is tiny compared to the concentration of the major species. It amounts to decreasing the ammonia concentration by 0.00003 M and increasing the ammonium concentration by the same amount. The shift in pH is so small that it makes no difference when the pH is written with appropriate number of significant figures.

Alternate strategy

For the calculators we use, this is a good strategy to deal with multiple equilibria (in your case, that of the weak acid/base pair and autodissociation of water).

If you had an analog computer that adjusts the concentrations based on a "pH slider" and reports out whether all the constraints are met, there would be no reason to do this in multiple steps. Instead, you would just push your pH slider from one extreme to the other and stop when all equations are satisfied.

In your case, you could calculate the ratio of ammonia to ammonium from the pH and the $\mathrm{p}K_\mathrm{a}$, and the hydroxide concentration from the autodissociation constant of water. Then, you vary the pH until the charge balance (chloride plus hydroxide has to match hydronium plus ammonium) is achieved. It is a fun exercise to program this into a spread sheet.

What actually happens?

All reactions (any of the acids - water, hydronium or ammonium - reacting with any of the bases - water, ammonia or hydroxide) go on at the same time. The further from equilibrium a reaction is, the faster is the net change. Once all the reactions are at equilibrium, there is no more net change. The details depend on the kinetics but are not relevant to the equilibrium state that is finally reached.


If i keep adding sulfuric acid into water, will it keep dissolving forever?

If you remove the last two or so percent of water from concentrated $\ce{H_2SO_4}$, it starts fuming, giving off SO3 until it again contains enough water to be stable.

$\ce{H_2SO_4 <-> SO_3 + H_2O}$

Pure sulfuric acid does not exist, so there is no clear yes/no answer to your question. Also $\ce{H_2SO_4}$ (like all other mineral acids) does not just mix with water, but strongly reacts. So it's clear there can be no miscibility gap like with water/ether or similar pairings.


What happens to the structure of tetracyclines when they expire?

Apparently the culprits are anhydro-tetracycline (ATC) and 4-epianhydrotetracycline (EATC) - see Determination of tetracycline and its major degradation products by liquid chromatography with fluorescence detection. J. Pharm. and Biomed. Anal. 18 (1998), 839-845 for structural information and references to toxicity of those byproducts as known at the time of publication.

It is worth noting that renal tubular acidosis (RTA) - the condition that was reported in 1963 to have been precipitated by ingestion of expired tetracycline - is not presently considered as much of an issue as it was at the time (see this review). Finally, this piece in the New York Times from 2014 (entitled Do some drugs become dangerous after expiration?) has more information and references concerning tetracycline and RTA.


Why does pH of a buffer solution change according to Henderson–Hasselbalch equation?

Suppose you have a $\pu{1.0 L}$ of acetate buffer solution made by $\pu{500.0 mL}$ of $\pu{1.0 M}$ $\ce{CH3COONa}$ solution adding to $\pu{500.0 mL}$ of $\pu{1.0 M}$ $\ce{CH3COOH}$ solution. Thus, $[\ce{CH3COOH}] = [\ce{CH3COONa}] = \pu{0.5 M}$. Therefore, according to Henderson–hasselbalch equation, $\mathrm{pH} = \mathrm{p}K_\mathrm{a} = 4.75$ ($\mathrm{p}K_\mathrm{a} = 4.75$ for acetic acid).

Now you add a little amount of acid ($\pu{0.01M}, \; \pu{5 mL}$). It disturbs the equilibrium and change both $[\ce{CH3COONa}]$ and $[\ce{CH3COOH}]$, according to the reaction:

$$\ce{CH3COONa + HCl -> CH3COOH + H2O}$$

Now, the amount of $[\ce{CH3COONa}]$ is reduced by $\pu{0.01M} \times \pu{0.005 L} = \pu{0.00005 mol}$ and, similarly, $[\ce{CH3COOH}]$ is increased by $\pu{0.00005 mol}$.

Hence, new $[\ce{CH3COONa}]$ is $\pu{0.49995 mol/L}$ and $[\ce{CH3COOH}]$ is $\pu{0.50005 mol/L}$ (neglecting the volume change for convenience). Thus, new $\mathrm{pH}$:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \left(\frac{[\ce{CH3COONa}]}{[\ce{CH3COOH}]}\right)= 4.75 + \log \left(\frac{\pu{0.49995 mol/L}}{\pu{0.50005 mol/L}}\right) = 4.7499 \approx 4.75$$

You see, it is a very minimal change. It is true that if you add more acid or base, change could be significant. However, all buffers have buffer capacity (usually, $\mathrm{p}K_\mathrm{a} \pm 1$). Nobody ever said $\mathrm{pH}$ of buffer is constant. It fluctuate withing the capacity. If you add too much acid or base, its $\mathrm{pH}$ would drastically change.


Outermost electronic configuration of f block elements

As you said, if you look at the electron configurations of Lanthanides and Actinides, Thorium seems to be the only exception from this empirical rule. You could expand the rule to also include the case of Thorium but I would say it is more useful to have the rule describe the majority of cases and remember $\ce{Th}$ as an exception.


Why do s-block elements impart color to a flame?

All elements colour the flame somehow. The salt that is held into the flame dissociates homogenously (i.e. you’re not creating unpaired ions but rather single metal atoms and their counter-radicals), the heat energy excites the electrons and when they fall back down they emit photons at certain wavelengths, corresponding to the energy differences between the orbital an electron was excited to and the one it falls down to.

The further away from the nucleus the electrons are, the easier they are to excite. Thus valence electrons are most likely to be excited. The more valence electrons you have (i.e. the further right you go in a given period) the harder it gets to excite one of them because the whole atom contracts and the energy levels of the outer shells are lowered. Therefore, it is a lot easier for electrons of elements on the left half of the periodic table to be excited.

Furthermore, the more electrons you can excite, the more complicated the resulting spectrum will be. For only one or two electrons, you usually expect (and observe) distinct lines of definite wavelengths; the more you have, the larger the possibility of broader bands.

Therefore, the easiest element to recognise would be hydrogen, which only creates very few single, distinct lines. Alkaline metals, having few valence electrons, will still create easy spectra with distinct lines (albeit more). It is similar for alkaline earth metals, although the spectra tend to get more complicated (see barium for an example of something with broad bands). For many, if not all other metals, the spectra just get too difficult to interpret and consistently assign very quickly. They will all seem white-ish to some degree with hardly any possibility of distinguishing characteristic lines easily. Thus, in the typical inorganic chemistry ion lottery lab course, typically only alkalines and alkaline earths are detected via flame colouring.

I also remember from that course that copper halides seem to provide a more characteristic green colour (that I didn’t even try recognising — I relied on other methods to detect copper). I think that there is a mulitude of combined effects in action to give this combination its characteristic colour. In any case, I don’t know what exactly is going on there.


Why does microwave heat up things so much more quickly than visible light?

Microwaves don't have more energy, they just resonate at the frequency that causes molecular bonds to rotate. This specifically applies to dielectric molecules, molecules like water that have electric dipoles.
From Wikipedia:

Water, fat, and other substances in the food absorb energy from the microwaves in a process called dielectric heating. Many molecules (such as those of water) are electric dipoles, meaning that they have a partial positive charge at one end and a partial negative charge at the other, and therefore rotate as they try to align themselves with the alternating electric field of the microwaves. Rotating molecules hit other molecules and put them into motion, thus dispersing energy. This energy, when dispersed as molecular vibration in solids and liquids (i.e., as both potential energy and kinetic energy of atoms), is heat. Sometimes, microwave heating is explained as a resonance of water molecules, but this is incorrect such resonances occur only at above 1 terahertz

For reference, energy is related to wavelength by the formula:
$$E=\frac{hc}{\lambda}$$
and
$$\lambda=\frac{c}{f}$$
where

$\lambda$ is wavelength
$c$ is the speed of light
$f$ is frequency
$h$ is Plank's constant
$E$ is energy

You can see by the first equation that as $\lambda$ gets smaller $E$ gets larger.


I would say the answer on this website is pretty near perfect.

To summarize, you need 2 things:

Enough friction between your nose and the spoon
The center of mass of the spoon should be on a line vertically downwards THROUGH the tip of your nose as illustrated in this image from the same website:

The friction is highest if both your skin and the spoon are free of greasy `lubricants'. Breathing on the spoon will add more friction by depositing a very thin (only a few atomic layers) film of water which, on this scale, behaves solid-like due to ordering effects as shown in this paper.


Which has higher priority (Cahn-Ingold-Prelog)?

As mentioned by Yomen , the double bond counts for two atoms with a single bond between them . They are called ghost atoms . This makes aldehyde group at higher priority than the ether .

Wikedia has a very nice explanation on it :

  If an atom A is double-bonded to an atom B, A is treated as being singly bonded to two atoms: B and a ghost atom that has the same atomic number as B but is not attached to anything except A. In turn, when B is replaced with a list of attached atoms, A itself is excluded in accordance with the general principle of not doubling back along a bond that has just been followed, but a ghost atom for A is included so that the double bond is properly represented from both ends.

A triple bond is handled in a similar manner .

Wikipedia link : Cahn–Ingold–Prelog priority rules


Protonation of Carbonyl oxygen in Fischer Esterification

There are a few reason why the carbonyl oxygen gets protonated -

1) The carbonyl oxygen is sp2 hybridized which is more electronegative than the sp3 oxygen. So, it has more tendency to give electrons since it is more negative.

2) The lone pairs of carbonyl carbon are more available since they are not in conjugation whereas the lone pairs of the other oxygen are not much available since they are in conjugation with carbonyl carbon.

3) When carbonyl oxygen donates its electron the resulting structure is more stabilized (because of conjugation) than the structure formed when the other oxygen atom donates.

So, that's why carbonyl oxygen has far more tendency of donating electrons than the -OH grp.


Oxidation of an aldehyde - how is this a loss of electrons?

OILRIG always holds, but you need to be careful at what you are actually looking at. When we say a compound gets oxidised, that is technically not true. In this framework, only atoms get oxidised or reduced.
In the reaction
$$\ce{R-CHO + [O] -> R-COOH}$$
we usually omit the reduction part, because it's always the same: the oxygen.

To identify the whole redox process, use oxidation numbers. Keep in mind that these are simply a bookkeeping tool and the partial charges of the atoms involved might be quite different. Setting $\ce{R {=} CH3}$ or the above reaction it boils down to this:
$$\ce{
\overset{+1}{H}_3\overset{-3}{C}-\overset{\color{\red}{+1}}{C}\overset{+1}{H}\overset{-2}{O}
+ [\overset{\color{\green}{0}}{O}]
->
\overset{+1}{H}_3\overset{-3}{C}-\overset{\color{\red}{+3}}{C}\overset{-2}{O}\overset{\color{\green}{-2}}{O}\overset{+1}{H}
}$$

Now you can see that the carbonyl carbon increases its oxidation state from $\color{\red}{+1}$ to $\color{\red}{+3}$, so formally it lost $\pu{\color{\red}{2} e^-}$. The oxidising agaent, indicated with $\ce{[O]}$, at the same time decreases its oxidation state from $\color{\green}{0}$ to $\color{\green}{-2}$, therefore formally gaining $\pu{\color{\green}{2} e^-}$. The redox reaction is complete.

Let's look at a real world example, the Baeyer-Villiger oxidation:

As you can see, the ketone gets oxidised, more precisely the carbonyl carbon gets oxidised as increases its oxidation state. On the other side, the peroxide oxygens get reduced as they decrease their oxidation state.


The vicinal-glycols can be cleaved to corresponding aldehydes and/or ketones in high yield by the action of periodic acid ($\ce{HIO4}$) or lead tetraacetate ($\ce{Pb(OAc)4}$). This oxidative cleavage of a carbon-carbon single bond provides a two-step involving cyclic intermediate reaction mechanism with high-yield. A generally accepted equation for these oxidations (A) is shown below:

The reaction using $\ce{HIO4}$ is called Malaprade Reaction (Ref.1), while that involving $\ce{Pb(OAc)4}$ is called Criegee Reaction. Malaprade Reaction is described as:

  The oxidation of adjacent diols with periodic acid or its salt in aqueous solution is generally known as the Malaprade reaction. Several solvents have been used to increase the solubility of organic substrates and the reaction proceeds faster under acidic conditions. This reaction has been further extended to the cleavage of $\alpha$‐hydroxy carbonyl compounds, 1,2‐dicarbonyl compounds, $\alpha$‐amino alcohols, $\alpha$‐amino acids, and polyhydroxy alcohols, and successfully applied for structural analysis.

Both cis- and trans-glycols (e.g., cis- and trans-cyclohexane-1,2-diols) undergo the oxidation with $\ce{HIO4}$ or its salts. As a rule, cis-glycols react more rapidly than trans-glycols and there is evidence for the presence of heterocyclic intermediates as shown in the diagram (B). For example, the publication by Buist, et al. (ref.2) revealed that cis-cyclohexane-1,2-diol cleaves about 20-times faster than trans-cyclohexane-1,2-diol. However, the equilibrium between the diol and the cyclic periodate intermediate (faster first step) slightly favors the trans-isomer (the ratio of cis/trans equilibrium constants is $400:1000$; Ref.2) while the rates of cleavage of cis-cyclohexane-1,2-diol to products is much faster compared to its trans-isomer (the ratio of cis/trans rate constants is $3300:165$; Ref.2). Thus, theoretically your believe of only cis-vicinal-diols would get oxidized with $\ce{HIO4}$ is completely wrong.

Therefore, theoretically, each available $\ce{-C(OH)-C(OH) -}$ function would get oxidized ($\alpha$-methyl-D-glucopyranoside has 4 such functions), costing one $\ce{HIO4}$ molecule per function (Ref.3). However, rate of oxidation is depend on conditions, steric effects, $\mathrm{pH}$, etc. For example, the oxidation of $\alpha$-methyl-D-glucopyranoside by a solution of $\ce{HIO4}$ in pyridine has cleaved only the $\ce{C}$3$-\ce{C}$4 bond while the reaction of 6-O-trityl-$\alpha$-methyl-D-glucopyranoside is less selective under same condition and is directed to a greater degree to the $\ce{C}$2$-\ce{C}$3 bond (Ref.4).

References:

Zerong Wang, "Malaprade Reaction (Malaprade Oxidation)," In Comprehensive Organic Name Reactions and Reagents; John Wiley & Sons, Inc.: New York, NY, 2010 (https://doi.org/10.1002/9780470638859.conrr406). ISBN: 9780471704508.
G. J. Buist, C. A. Bunton, J. H. Miles, “149. The mechanism of oxidation of α-glycols by periodic acid. Part V. Cyclohexane-1 : 2-diols,” J. Chem. Soc. 1959, 743-748 (https://doi.org/10.1039/JR9590000743).
Glen Dryhurst, In Periodate Oxidation of Diol and Other Functional Groups: Analytical and Structural Applications; First Edition, Pergamon Press: Oxford, United Kingdom, 1970. ISBN-10: 1483128385.
R. G. Krylova, S. N. Ryadovskaya, L. I. Kostelian, A. I. Usov, “Periodate oxidation of α-methyl-D-glucopyranoside, 6-O-trityl-α-methyl-D-glucopyranoside and 6-O-tritylcellulose in pyridine,” Bulletin of the Academy of Sciences of the USSR, Division of chemical science 1972, 21, 2005-2009 (https://doi.org/10.1007/BF00854627).


I finally found the answer.

In temporary hardness, we have to remove $\ce{Mg^2+}$ ions by precipitating it. In $\ce{Mg(OH)2}$, $\ce{Mg^2+}$ ions in water are present in the concentration that is cube root of the solubility product, but in $\ce{Mg(CO3)}$ it is the square root of the solubility product. So, despite $\mathrm{K_{sp}}$ of $\ce{Mg(OH)2}$ being higher than $\ce{Mg(CO3)}$, $\ce{Mg^2+}$ ions in $\ce{Mg(OH)2}$ are comparatively less than $\ce{Mg^2+}$ ions in $\ce{Mg(CO3)}$.


Proving 1 = 2 by POAC( Principle of Atom Conservation )

As said by many others ,  The equation 1 given above is simply put very wrong  . As it can be applied only when a balanced chemical equation is given to us ,  otherwise we will reach at all sorts of contradictions as in the question above .

However many non- rigorous teachers , textbooks often write equation 1 in POAC (which is always true in a balanced chemical equation ) and then say now equation 2 is obtained from equation 1(which is partially incorrect) . It is true equation 2 comes naturally from equation 1 , but only for a balanced chemical equation.

So then how come we apply equation 2 in POAC in all sorts of question ? And still come up with the right answer ?

The answer is that equation 2 is true in itself always whether the chemical equation given ( in POAC ) is balanced or not . So it is not related to equation  1 for its validity .

How is equation 2 always valid ?

As pointed out by @iammax in the comments actually we (sort of )  essentially balanced the chemical equation given to us in the equation 2 itself .

For example in the $KClO_3$ question linked above , the moles of $O_2$ , moles of $KClO_3$ given to us are actually according to the stoichiometric coefficients , which enable us to use POAC without balancing the reaction (even though we have done that indirectly by using No. of moles of $O_2$ , $KClO_3$).

To conclude we can say after using No. Of moles of reactant , product we have indirectly balanced the chemical equation as the Ratio of their number of moles will be
Equal  to Ratio of their respective stoichiometric coefficients . After which we just use Conservation of Atoms( as the name says it) and multiply the number of moles of the molecule with the number of Atoms present  on which POAC is being applied ( For example we multiply no. Of moles of $O_2$ by 2 when we apply POAC on oxygen atom ).


What would the following NGP reaction yield?

Reaction-1 is a simple example of an $\mathrm{S_N2}$ reaction, where the acetate group ($\ce{AcO-}$) substitutes the tosyl group ($\ce{OTs}$) forming the product (1) as follows:

But for Reaction-2, the lone pairs present on the $\ce{NMe2}$ provide anchimeric assistance in the removal of the tosyl group, leading to the formation of a bicyclic intermediate (both rings are not coplanar though). The acetate anion can now open the smaller ring in two ways and hence, lead to two different products labelled (1) and (2) as follows:

Finally,to deduce the enantiomeric relation between (1) and (2), the following transformations can be done:


Acetic acid odor while testing CO2 sensor; can it be easily removed, is there a better way?

You could use a less volatile acid, like oxalic. But I think the reaction with NaHCO3 probably gives off a mist as the reaction bubbles, and that might be why the odor is so strong.

How about using seltzer water (or Perrier? or Diet Coke!) as the source of CO2? I estimate that the pressure of CO2 is about 45 psi in the unopened bottle, and if you attach a balloon to a freshly opened bottle and let it outgas slowly, the balloon will fill up and be essentially pure CO2. If a latex balloon is unsuitable or too hard to handle, a mylar balloon ($1 at the Dollar Store) will work just as well. The gas will be saturated with water vapor.


how to identify reaction medium by looking at the equation ( for h2o2)?

Is there really a way of recognizing the medium by looking at reaction equations??

Yes and no...

For the particular reaction, you have to know some chemistry. In acid solutions hydrogen peroxide acts as an oxidizing agent: $$\ce{H2O2 + 2H^+ +2e^- -> 2H2O}$$

In neutral and basic solutions hydrogen peroxide acts as an reducing agent: $$\ce{H2O2 -> 2H^+ +2e^- + O2(g)}$$

The silver oxide is being reduced so the hydrogen peroxide must be acting as a reducing agent.

The real answer is that the balanced reaction isn't really enough information to duplicate the reaction. What temperature? How long for the reaction to take place? Since reaction would use excess hydrogen peroxide, how much was in excess and so on...

PS - Sometimes you have to consult the oracle to know the laboratory conditions. I asked Carnac the magnificent for more information.

This reaction was done by a red-headed girl. She was wearing a white lab coat over a red blouse and a blue skirt. She did have safety glasses which is a really sexy look over her hazel eyes in the chem lab. She was also wearing white leather sneakers.


Does hydrogen peroxide - urea become more active when illuminated?

I hate to answer my own question, but after some digging, I believe that the use of a light source is in fact a gimmick, at any concentration of the reagent, contradicting the only other answer so far (But I'm happy to be corrected, if I'm wrong)

This meta-study https://www.ncbi.nlm.nih.gov/pubmed/22525016 concluded that at high concentrations of the reagent (typical of in-office use), the light is useless; and there was insufficient or no data for lower concentrations (as of 2012):

  Light increases the risk of tooth sensitivity during in-office
  bleaching, and light may not improve the bleaching effect when high
  concentrations of HP (25-35%) are employed. Therefore, dentists should
  use the light-activated system with great caution or avoid its use
  altogether. Further rigorous studies are, however, needed to explore
  the advantages of this light-activated system when lower
  concentrations of HP (15-20%) are used.

A slightly more recent study came to the same conclusion https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4178330/ :

  The in-office bleaching treatment of vital teeth did not show
  improvement with the use of light activator sources for the purpose of
  accelerating the process of the bleaching gel and achieving better
  results.

I'm inclined to think that if light were useful at all, they'd see this at higher concentrations of HP too, but I haven't seen any recent studies that address the lower concentrations specifically.


Karl's comment (see above) prompt me to suggest the following rough experiment which can be done in the classroom with minimum equipment:
Suppose you are required to find out the specific heat capacity of water. Weigh $m_1 \ \pu{g}$ of warm water at $T_1 \ \pu{^\circ C}$ to a Styrofoam cup with a lid. Also weigh $m_2 \ \pu{g}$ of ice at $T_2 \ \pu{^\circ C}$ to another Styrofoam cup. Then add ice to warm water (Note: Measure the temperature of warm water just before addition to make sure it is still at $T_1 \ \pu{^\circ C}$). Stir the mixture carefully while the coup is covered by the lid. Measure the temperature of the mixture until reading is steady. Suppose the steady temperature is $T_3 \ \pu{^\circ C}$. Ensure this time no ice is visible (meaning all dissolved). Now we are ready for the calculations:
Suppose the specific heat capacity of water is $c_{(\ell)}$ and the specific heat capacity of water is $c_{(s)}$. For novices, we can assume that $c_{(\ell)} \approx c_{(s)} = c$. Thus, heat lost by warm water at equilibrium:
$$Q_\text{lost} = m_1 \times c \times (T_1 - T_3) \tag1$$
For ice, if $T_2 \lt 0$, ice would gain heat first to become ice at $\pu{0 ^\circ C}$, then $\pu{0 ^\circ C}$ ice melts to $\pu{0 ^\circ C}$ water, and warm up to $T_3 \ \pu{^\circ C}$ water. If heat of fusion of ice is $L \ \pu{Jg-1}$, total heat gained by ice at equilibrium:
$$Q_\text{gained} = m_2 \times c \times (0 - T_2) + m_2 \times L + m_2 \times c \times (T_3 - 0) \tag2$$
At equilibrium, $Q_\text{gained} =  Q_\text{lost}$. Therefore, equations $(1)$ and $(2)$:
$$m_1 \times c \times (T_1 - T_3) = m_2 \times c \times (0 - T_2) + m_2 \times L + m_2 \times c \times (T_3 - 0) tag3$$
Only unknown in the equation $(3)$ is $c$, you can solve the equation $(3)$ for $c$.
Also note that you can avoid $m_2 \times c \times (0 - T_2)$ part by choosing $\pu{0 ^\circ C}$ ice, because it is easy to get $\pu{-20 ^\circ C}$ from freezer, it it kept in constant temperature.


Radial wave-functions and spherical harmonics of p-orbitals in a certain hypothesis

In "real life", the hydrogen atom has its only electron in the s orbital. And unless it gets to an excitated state, that's the only thing that matters (whether you're studying the energy of the atom or the bounds it can make in terms of energy or symmetry).
Nevertheless, the H atom is likely to have a spherical symmetry, so any of its p-orbitals will be orthogonal to the others, so their linear combination is likely to keep the overall spherical(-ish) symmetry of the atom.
Here is a representation of such orbitals :

(Reference)
Hope this helps!


Finding equation of line of the P-T graph for a gas dissociation reaction

The overall notion is that gases A, B and C are all behaving as an ideal gases which can be represented by the mythical ideal gas X. Then the moles of X are equal to moles of A+B+C. Then we can use the ideal gas equation:

$$pV=nRT$$

which simplifies to:

$$p = \dfrac{nR}{V}\times T$$

Given that the degree of dissociation of A is 0.4 and we started with 1 moles of A, then 0.6 moles of A remain and 0.4 moles got converted.

$$\ce{2A -> 3B + 2C}$$

or

$$\ce{A -> }\dfrac{3}{2}\ce{B + C}$$

let B and C equal mythical gas X and

$$\ce{A -> }\dfrac{5}{2}\ce{X}$$

but since A = 0.6 moles of X, the total moles are

$$n = 0.6 + 0.4\times\dfrac{5}{2} = 1.6\text{ moles}$$

We know that n = 1.6 moles and that V = 0.16 L so:

$$p =\left(\dfrac{n}{V}\right)RT = \dfrac{1.6\text{ mole}}{0.16\text{liter}}RT = 10RT\dfrac{\text{moles}}{\text{liter}}$$

Now the problem doesn't really specify what value of the ideal gas constant, $R$, to use nor what units the temperature, $T$, should be.

The temperature must be an absolute scale since the line passes through the origin. Thus of the main choices of Kelvin and Rankine, using Kelvin seems reasonable for temperature.

but there are still options for the pressure units of R

(a) $8.314\text{ L}\cdot\text{kPa}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$
(b) $8.314\times10^{−2} \text{ L}\cdot\text{bar}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$
(c) $62.36 \text{ L}\cdot\text{torr}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$
(d) $8.206\times10^{−2}\text{ L}\cdot\text{atm}\cdot\text{K}^{−1}\cdot\text{mol}^{−1}$

For $10R$ to be between 0.4 and 1.1 the only answers that fit are (b) and (d). Thus the pressure could be either in atm or bar. Presumably the OP's book would have stated a preference for R.


Is an imine considered a secondary amine?

No. Imines cannot be considered an example of secondary amines any more than pyridine can be considered a tertiary amine.
The easiest way to think about this is with oxidation state: imines are at a higher oxidation level than amines, just as aldehydes are at a higher oxidation level than alcohols.
Aside from this 'intuitive' explanation, the IUPAC gold book (a source of chemical nomenclature) defines both amines and imines, clearly differentiating the two:

amines. Compounds formally derived from ammonia by replacing one, two or three hydrogen atoms by hydrocarbyl groups, and having the general structures $\ce{RNH2}$ (primary amines), $\ce{R2NH}$ (secondary amines), $\ce{R3N}$ (tertiary amines).
imines. Compounds having the structure $\ce{RN=CR2}$ (R = H, hydrocarbyl). Thus analogues of aldehydes or ketones, having NR doubly bonded to carbon; aldimines have the structure RCH=NR, ketimines have the structure $\ce{R'2C=NR}$ (R' ≠ H). Imines include azomethines and Schiff bases. Imine is used as a suffix in systematic nomenclature to denote the C=NH group excluding the carbon atom.


Reaction of glucose acetal with acetic anhydride

I think there's nothing to discussed about this further as all comments directing to the correct answer: $x = 4, \ y = 6,$ and $ z = 5.$ I think it's better show in the structures:

Glucose is naturally occurring in pyranose hemiacetal form. When subjected to react with $\ce{H+/R-OH}$, the remaining acetal hydroxy group (anomeric hydroxyl group) will undergoes alkylation. Thus, the first scheme is showing alkylation of this anomeric hydroxyl group by methanol (Ref.1). The resultant methylglucose has four free hydroxy group, hence, $x = 4$.
When subjected to reduction with suitable conditions (e.g., $\ce{Ni/H2}$, $\ce{Ru/C/H2}$, etc.), the hemiacetal group would be reduced to a $\ce{-CH2OH}$ group (second scheme in the image). The resultant product is sorbitol (Ref.2), which has six free hydroxy groups, hence, $y = 6$.
Finally, pyranose glucose structure has five free hydroxy groups, hence, $z = 5$.

References:

B. Helferich, W. Schäfer (Checked by H. T. Clarke and M. R. Brethen), "$\alpha$-methyl d-Glucoside," Org. Synth. 1926, 6, 64 (DOI: 10.15227/orgsyn.006.0064).
J. Wildschut, J. Arentz, C.B. Rasrendra, R.H. Venderbosch, H.J. Heeres, "Catalytic Hydrotreatment of Fast Pyrolysis Oil: Model Studies on Reaction Pathways for the Carbohydrate Fraction," Environmental Progress & Sustainable Energy 2009, 28(3), 450-460 (https://doi.org/10.1002/ep.10390).


How do orbitals overlap in difluorosilane?

A simple model for bonding in this molecule is taking $\ce {Si}$ to adopt $\ce {sp^3}$ hybridisation, as suggested from the tetrahedral geometry of the molecule. Subsequently, we can refine our model by applying Bent's rule (see more about it here), which tells us that $\ce {p}$ orbital character concentrates towards more electronegative substituents. Since the electronegativity of the elements has the following order: $\ce {F}$ > $\ce {H}$ > $\ce {Si}$ with Pauling electronegativity values of $\ce {3.98}$, $\ce {2.20}$ and $\ce {1.90}$ respectively, the $\ce {3p}$ character of $\ce {Si}$ would concentrate towards $\ce {F}$ while $\ce {3s}$ character concentrates towards $\ce {H}$. Thus, we would have $\ce {Si}$ using $\ce {sp^{3+x}}$ orbitals to form the bonds with $\ce {F}$ and $\ce {sp^{3-x}}$ to form the bonds with $\ce {H}$, where $x$ is some small value between $\ce {0}$ and $\ce {1}$.


Van der Waals interaction and magnetic dipole dipole interaction

Magnetostatic interactions are much smaller than electrostatic interactions. For atomic and molecular systems, the magnitude of the magnetic potential energy is roughly $\alpha^2 \approx 5\times 10^{-5}$ times the magnitude of the electrostatic potential energy. Therefore magnetic dipole (and higher orders) are usually neglected.

The multipole expansions for the magnetostatic long-range potentials are formally equivalent to the usual electrostatic long-range, with $\mu_0/4\pi$ replacing the $1/4\pi\epsilon_0$ prefactor, where $\mu_0$ is the vacuum permeability.

References:

M. Lepers and O. Dulieu,  arXiv:1703.02833v2 [quant-ph] https://arxiv.org/abs/1703.02833v2
1 Z.C. Yan, J.F. Babb, A. Dalgarno, and G.W.F. Drake, Phys. Rev. A - At. Mol. Opt. Phys. 54, 2824 (1996).


Is iron (III) oxide-hydroxide the same as iron (III) hydroxide?

Start with my answer here: What is chemical formula of Bauxite? On oxides and hydroxides of aluminium - that's a quick explainer on notation.

Now, to your questions:

  where I can learn more about such unusual compounds

Easy. Here: The Iron Oxides: Structure, Properties, Reactions, Occurrences, and Uses by R. M. Cornell and U. Schwertmann. ISBN: 3-527-30274-3. PDF link

The problem with iron hydroxides is that there are several polymorphs (same composition, different crystal structures), and they are often mixed together in one sample. Add to that that not all iron is fully oxidised, that these compounds are hygroscopic (they adsorb water moisture on their surfaces), and they can contain impurities of whatever was alloyed with the iron (nickel, carbon, chromium, etc etc) forming even more compounds.

A single polymorph rare exists by itself (but when they do, they are spectacular; examples: 1 2 3). Pure iron hydroxide is usually yellow, and pure iron oxide is bright red. Anything in between is going to be brown. When found in nature it's called "limonite", and when formed by rusting of iron, it's called "rust".


How fast does the diamond lose the atoms at sunlight?

Thats true, but even an UV lamp would take 10 bilion years to remove a microgram (one-million part of a gram). It is caused by carbon atoms oxidizing, forming CO2. This process is accelerated by the presence of UV light.


How to create different fluorescent colors using household items

If you dismantle highlighter pens, you can find the felt material inside. This material contains fluorescent chemicals, such as pyranine. This felt like material is usually encased in a plastic film, remove the film and soak the felt in water.

If you use too much water then it may become too dilute and you may start to encounter an issue with the ink/water mixture sticking to the flower.

Are you using a black light to view the fluorescent colour?

it might be worth trying to cut the stems and keep the flowers in the ink containing water. Then as they absorb the water they will absorb the fluorescent ink as well and may take on some of the colour (and viewable under black light).


Is the final temperature of a solution affected by the order of mixing components?

Temperature is a so-called state function:

  In thermodynamics, a state function or function of state is a function defined for a system relating several state variables or state quantities that depends only on the current equilibrium state of the system. State functions do not depend on the path by which the system arrived at its present state. A state function describes the equilibrium state of a system.

The final temperature of your system does not depend on how the system got there (order of mixing, for example).


Does copper react with HCl when current is present?

A reaction between copper metal and hydrochloric acid would go something like this: $$\ce{Cu + 2HCl -> CuCl2 + H2}$$
...which we can write as: $$\ce{Cu + 2H+ -> Cu^2+ + H2}$$
If we break this up into half reactions, we get: $$\ce{Cu -> 2e- + Cu} ~~~E°=-0.34~V$$ $$\ce{2H+ + 2e- -> H2} ~~~E°=0.00 ~V$$
If we add the voltages together, we get the cell potential of the combined reaction is $-0.34+0.00=-0.34~V$. Since the cell potential is negative, the reaction is not spontaneous, so the reaction is not going to occur without any artificial interference ($\Delta G°=-nFE°_{cell}$).

However, if you connect a battery (galvanic cell) to this reaction, you can supply the necessary electrons for this reaction to occur (electrolytic cell). Since you are connecting a 12 V battery, there is more than enough voltage to cause this reaction to occur.


Cr(II) and Mn(III) - their oxidizing and reducing properties?

Related question with same answer but in a different context of the 4f block: Why don't we see these lanthanide species?

You have a misconception regarding the stability of oxidation states. The factors you have listed are honestly not very important in determining the stability of a certain oxidation state. They will tip a delicate balance in favour of one oxidation state or another, but they are hardly the sole determining factors.

If you are going from $\ce{M+}$ to $\ce{M^2+}$, you need to consider two main things.

You need to ionise the second electron from $\ce{M+}$, denoted $I_2$. That represents an input of energy.
$\ce{M^2+}$ is smaller and more highly charged than $\ce{M+}$, so you can gain some energy back on the basis that the $\ce{M^2+}$ ion is more strongly solvated or forms stronger ionic/covalent bonds with anions.

Loosely speaking, if the energy you get out of (2) is enough to compensate for the energy you lose in (1), then $\ce{M+}$ will act as a reducing agent. And the converse is true as well. It is for this reason that one does not observe Mg(I) compounds (apart from some esoteric organometallics). The $I_2$ of Mg is so small that it can be recouped extremely easily. Likewise, the $I_2$ of K is so large that there's no way any stronger bonding can compensate for that.

For the cases you have described:

$\ce{Cr^2+}$ simply does not act as an oxidising agent because $I_2$ of Cr is rather small. The same is true of the whole 3d block. How many 3d compounds do you know of with the metal in a +1 oxidation state? Without resorting to organometallic compounds, all I can suggest is copper(I). That's precisely because their $I_2$ values are small enough.
Similarly, $\ce{Mn^3+}$ does not act as a reducing agent because $I_4$ is way too large. Remember that successive ionisation is extremely difficult.

What your textbook should have written is:

The $I_3$ of Mn is anomalously low because you are "losing the stability of a half-filled 3d subshell".
The hydration enthalpy of $\ce{Cr^3+}$ is anomalously large (for a 3+ ion) because there is a large ligand-field stabilisation energy associated with the $(t_{2g})^3$ configuration. (This is factor number 2, not factor number 1).

These factors are small compared to the raw size of the ionisation energies. Yes, the $d^3$ ion $\ce{Mn^4+}$ also enjoys a large LFSE and its hydration enthalpy would be larger than expected. However, that 4th ionisation energy is just too big and that simply cannot be compensated for by the comparatively tiny increase in LFSE.


How exactly does copper react with hydrogen peroxide, salt and citric acid?

The acids, hydrochloric or citric, have a common property : they produce $\ce{H+}$ in water. These ions are necessary to react with copper and hydrogen peroxide $\ce{H2O2}$. Indeed metallic copper reacts with hydrogen peroxide $\ce{H2O2}$ and an acid in the following way : $$\ce{Cu + H2O2 + 2 H+ -> Cu^{2+} + 2 H2O}$$ So whatever the acid chosen, hydrochloric or citric, the copper will be dissolved the same way. With citric acid, the dissolution should be slower. But it leads to the same final result : metallic copper will be destroyed. The only difference is the final composition of the liquid. With HCl, the final liquid is made of $\ce{Cu^{2+}}$ and $\ce{Cl-}$ ions, which may be partially recombined. With citric acid, the final liquid is made of citrate ions, which will make a complex with $\ce{Cu^{2+}}$. Citral ions from citric acid will make a small change of color, but has no other effect. Both solutions are equally dangerous, or equally non-dangerous if you prefer, because copper solutions are always toxic. You should take care of the obtained copper solutions in the same way, whatever their compositions.
Personally I would recommend using citric acid instead of hydrochloric acid. It is less corrosive, and has nearly the same effect on copper + $\ce{H2O2}$.


pKa values for protonated nitrogens in ketoconazole

The 6.4 pKa is close to the reported pKa of imidazole (6.95), so 6.4 is at the imidazole ring.

That leaves pKa 2.3 (or 3.64) for the other basic nitrogen. While that nitrogen is part of a piperazine, piperazine itself (pKa 9.8) isn't a good comparison. The nitrogen in question is also conjugated to an aromatic ring, so it would be better to look at aniline, with a pKa of 4.6. That's much closer to ketoconazole. Add in the inductive effect of the proximal amide, and a pKa in the 2-3 range makes sense.

Many common pKa's can be found on Prof. Dave Evans's pKa list.


On mixing 2 parts of $\ce{NH3}$ with 1 part of $\ce{CuCl2}$, following reaction will occur.

$\ce{CuCl2 + 2NH3 -> Cu(NH3)2^{2+} + 2Cl-}$

This will give you soluble deep-blue copper-ammonia complex: $\ce{Cu(NH3)2^{2+}}$

and also, $\ce{NH3}$ does not react with $\ce{Na2SO4}$.

And $\ce{H2SO4}$ with $\ce{Na2CO3}$ will follow the reaction below :

$\ce{H2SO4_{(aq)} + Na2CO3_{(s)} -> H2O + CO2_{(g)} + Na2SO4_{(aq)}}$

As $\ce{Na2SO4}$ solution is colorless, so you can perform any appropriate test to identify the evolved gas i.e. $\ce{CO2}$.

A simple method used to test $\ce{CO2}$ is :

  Bubble the gas through limewater (calcium hydroxide). If carbon dioxide is present, the limewater will turn cloudy and a milky white precipitate will form.


difference between absorption spectroscopy and extinction spectroscopy

There should not be any difference in experiment, because the extinction is something you have to calculate by measuring the transmittance/absorbance of a given frequency.

The transmittance $T$ of a material is the ratio between the received intensity of a certain frequency to the transmitted intensity of the same frequency. $$T=I/I_0$$
The absorbance $A$ is defined as the decadic logarithm of the inverse transmittance. $$A=\log_{10}\left(I_0/I\right)=\log_{10}\left(1/T\right)$$

By using Beer-Lambert’s law $$I=I_0\operatorname{e}^{-\varepsilon~c~d}$$ one can calculate the extinction $\varepsilon$ from the measured transmittance/absorbance by $$\varepsilon=\frac{A}{c~d}$$ where $c$ is the concentration of the absorbing material and $d$ is the distance that the light has to travel in that given material.


How to know the aluminium sulfate solubility as function of the pH?

If the question only is about the diagram and the reference:

  Rubin, A. J., & Hayden, P. L. (1973). Studies on the Hydrolysis and Precipitation of Aluminum (III). page 57.

But the interpretation is not clear for me, so here is a new question


Why is oxygen absorbed in a colorless alkaline solution of pyrogallol and why does it turn dark brown?

Pyrogallol is the common name for 1,2,3-trihydroxybenzene.  Hydroxy groups are strong activavtors of aromatic systems.  By the time 3 hydroxyl groups have been added to a benzene nucleus you have a very reactive molecule, reactive enough to react with oxygen in the air.  The reaction is very complex and free radicals are thought to be involved.  Free radical reactions involving activated aromatic rings often lead to colored, high molecular weight products that are not easy to characterize.  Here's a link to a paper that suggests the initial products are probably the ortho-quinone and 2-pyrone pictured below.  Certainly the quinone will be colored, and as it reacts further to produce the higher molecular weight products, they will be colored too.  Here's another link to a very old paper suggesting that some of the higher molecular weight products contain at least 20 carbon atoms.

The balanced equation describing the initial radical oxidation of pyrogallol to the quinone can be written as
$$\ce{2~ pyrogallol + O_2 -> 2 ~o-quinone + 2H2O}$$


This is really a plumbing issue much more than a chemical dissolution issue. The problem is uncertain: "seems" to be a latex rubber glove; complete blockage.
I'm trying to envision how a rubber glove completely blocks a drain. Perhaps by partially blocking the drain, then catching hair and other particles - but on what? Perhaps on something else sticking into the drain - a root? Some other partial blockage?
Whatever; let's imagine pouring some solvent into the drain. For now, let's forget the PVC pipe: it is much thicker than the glove and will not dissolve before the glove does. As soon as the glove dissolves, the built-up head of solvent should flush away and prolonged rinsing with water will pretty much clean off the PVC pipe.
Unless the glove is caught on some projection and merely softens, while the 5 gallons of solvent drips away over a few minutes, leaving the glove softened and intermixed with the other captured trash. Now the problem has intensified. As solvent dissipates, the glove will become the adhesive that binds the trash together and to the PVC. But then the solution will be obvious: discover where the blockage is (with your snake), then remove that portion of the pipe and replace it. And don't do that ever again.
But if you insist on a chemical solution, I suggest air: pump up the pipe with a tire pump or one of those air blasters for toilet blockages and push the blockage down to larger pipe. It would also be helpful to load the blocked pipe up to your drain opening with a solution of dish detergent for lubrication and to decrease the volume that you have to pressurize.
If you must get more aggressive, there are tables of glove resistance (https://www.augusta.edu/services/ehs/chemsafe/PDF%20files/gloveselechart.pdf) online, but it would be easier to do your own experimenting. Since you have suspected it is a rubber glove, you might have other rubber gloves available that you could put into various solutions for a few hours or days to see how the glove behaves. But I think the glove is only part of the problem, and a good commercial drain cleaner would attack what I suspect is the helper-blocker, and dissolve that, and alkali isn't too gentle with latex rubber anyway (although the chart suggests natural rubber withstands alkali fairly well - except for ammonia). Also bleach is tough on natural rubber. Of course, YOU MUST NOT EVEN THINK OF MIXING AMMONIA WITH BLEACH!!!


I am not an expert in food safety, so understand this is for information only.

Aluminum, when exposed to the atmosphere, will immediately oxidize.  Aluminum oxide is not very soluble an will likely pass straight out of the body when ingested.  Aluminum oxide is in certain toothpastes as an abrasive.

If aluminum is exposed to fluoride, chloride, or sulfate there could be an issue if it converts to a soluble form.  As long as grinders are cleaned according to the manufacturer's recommendations this should not be an issue.

The amount of aluminum oxide which wears off a grinder is related to the abrasiveness of the material being ground.  Stainless steel wears about six times less than aluminum.  A titanium grinder is even better.


How would Dalton's law be affected when there are two ideal gases in a container at different temperatures?

Determine total number of moles of air in bulbs:
$$n=\frac{PV}{RT}=\frac{(1)(0.4+0.2)}{(0.082)(293)}=0.025$$

Consider final state of system.  Let $P_F$ be the final equilibrium pressure, $n_1$ be the final number of moles in the larger bulb, and $n_2$ be the final number of moles in the smaller bulb:
$$n_1=\frac{(P_F)(0.4)}{(0.082)(373)}=0.0131P_F$$
$$n_2=\frac{(P_F)(0.2)}{(0.082)(273)}=0.0089P_F$$
$$n=n_1+n_2=0.022P_F=0.025$$
$$P_F=1.14\tag{atmospheres}$$


Graphene in its pristine form, a single sheet of pure sp2 hybridized benzenes, is (relatively speaking) one of the least chemically reactive materials known because of several reasons: one, its enormous size (we keep making larger and larger single "molecules", now thanks to IBM up to 4 inches!). Secondly, its physical hindrance: you can only do bottom/top down chemistry, granted this isn't that big of an issue but it does greatly reduce using large compounds. Last is it's stableness: this thing is crazy stable because of its gigantic p-butts and other effects.

The long story short is that you don't have any functional groups or easy points of entry. The way to think about it is "what does graphene look like and what makes those other molecules reactive?". The answer is...benzene and really any other hydrocarbon with no functional groups. The difference there is their size. But the same rules apply! You can still oxidize graphene, at the edges for example, much like you would benzene. And in the past years, we've seen even more radical methods, the most common being Hummer's Oxidizing, and some other clever ones like a Claisen Rearrangement directly onto the surface of graphene.

Hope that helps!

EDIT: after Greg's comment

Don't apologize! This'll get the discussion going, I was hoping to see more answers other than mine. I think you're right in terms of absolute stability but my statement is a phenomenon harder to describe than mere size. From experience, we noted that on very large graphene flakes, our functionalization of them would distort the sheets, but not break them, creating lots of hindrances and thereby decreasing its reactivity; it basically went from sheet shape to a jumbled up ball. However, on smaller (much smaller) sized flakes, the individual molecules (though hindered and flexed in a similar way) would, at the same concentrations of reactants, get functionalized more, ratio of functional groups to carbons on the sheet.

This is an indirect effect of the massive size of the graphene sheets. So I guess you're right in that size isn't exactly a stability factor but rather, again, it's all about hinderance. However, I wanted to point it out because the massive size of such a flat molecule is quite spectacular and does yield some interesting behavior we don't normally think of.


Water is (almost) everywhere on earth in copious amounts. It is liquid at earth conditions (H-bridges!). It's rather non-toxic (or organisms were able to evolve to tolerate water well). It is a good solvent for the molecules involved in energy metabolism and biostructure ana- and catabolism. Water has neutral pH, it happily donates and accepts protons which makes it ideal for participation in metabolism.

I'd broaden the question, asking for other molecules that could support life as we know it in a similar way like water does it and also check other liquid molecules which are available in large quantities in the universe (e.g. ammonia, nitrogen, hydrogen) if they could support other forms of life that depend on analogous chemical processes, or even go beyond that (silicon- and aluminum oxides, etc.)


Why does 2+ oxidation state become more stable relative to 3+ oxidation state for first row of transition metals?

It is due to the difference in their second and third ionisation enthalpies. Due to the large difference in the second and third ionisation enthalpies arising from the disturbal of the stable half filled electronic configuration, +2 oxidation state is more favourable.


Reductive amination in case of secondary amines

In modern organic synthesis, reductive amination of aldehydes and ketones is considered one of the important and principal ways to make secondary and tertiary amines. It is a powerful and reliable strategy for the formation of $\ce{C–N}$ bonds. One of the advantages of the reductive amination is it can avoid the problem of over-alkylation that is always the problem with direct alkylation of amines with alkyl halides.

Reductive amination can be achieved following either a one-step or a two-step procedure. In both procedures, an amine and a carbonyl compound condense to form an imine (with primary amine) or iminium ion (with secondary amine). This imine or iminium ion is then reduced in situ (one-step reaction) or separated and subsequently reduced (two-step reaction) to form an amine product. Following is the suggested mechanism for reductive amination:

Most common condition is sodium triacetoxyborohydride ($\ce{NaB(OAc)3H}$) in THF or dichloroethane (DCE) under slighly acidic conditions (common acid used is acetic acid)(Ref.1).

Reference:

Ahmed F. Abdel-Magid, Kenneth G. Carson, Bruce D. Harris, Cynthia A. Maryanoff, Rekha D. Shah, “Reductive Amination of Aldehydes and Ketones with Sodium Triacetoxyborohydride. Studies on Direct and Indirect Reductive Amination Procedures,” J. Org. Chem. 1996, 61(11), 3849-3862 (https://doi.org/10.1021/jo960057x).


Why can't antiaromatic compounds just escape planarity and become non-aromatic to avoid destabilization?

They do try to escape planarity if possible. As a simple example, consider cyclooctatetraene. It has 8 electrons, which is a non-Huckel number, considering it's a multiple of four, and would correspond to an antiaromatic system, if the ring was planar, and allowed delocalization. However, cyclooctatetraene isn't planar. It's rather "tub-shaped", as seen in this structure here:

Click the image for interactive 3-D model

The electrons aren't delocalised either. Cyclooctatetraene is known to add bromine, and generally behave like a cyclic polyalkene in many aspects.

The only cases where molecules don't escape planarity is when they cannot. This happens in the case of cyclobutadiene. Such a molecule is so small, it cannot be distorted to such extents. As a result, it remains planar, antiaromatic, and immensely unstable:

Click the image for interactive 3-D model


How is the yellow colour of nitric acid removed?

I think you can explain this using the equilibrium:

$\ce{NO_2(g) <=> NO_2(dissolved)}$

When you pass $CO_2$ through the warm acid, the partial pressure of $CO_2$ ($P_{CO_2}$) above the gas increases (Since partial pressure generally represents the amount of gas above the solution).

Since

total pressure $P = P_{NO_2} + P_{CO_2}$ (Dalton's Law)
And $P$ remains constant at atmospheric pressure (assuming that the system is not closed).

$P_{NO_2}$ will decrease.

According to Le Chatelier's Principle, the equilibrium proceeds backward to compensate the decrease in $P_{NO_2}$ and thus more $NO_2$ gas is released. The gas flows quickly along with the bubbled gas and this prevents the forward reaction.

This seems like quite a roundabout method but it works. I'm sure there could be a much simpler explanation though.

The oxidation mentioned is most probably the disproportion reaction of $NO_2$ when it comes in contact with water:

$\ce{3NO_2 + H_2O -> 2HNO_3 + NO}$


How to calculate the amount of substance of sodium that are in 10 mL of 30% sodium chloride?

You are on the right track and with a few assumptions I'd say your calculation is correct. The one of your teacher is not.
Let's back up a moment and understand what is going on.

First of all we should make sure, that we are asking the right question. As you have given it, it seems incomplete. It should probably rather be:

  What amount of substance of sodium (ions) can be found in 10 mL of a 30% aqueous solution of sodium chloride?

We have been given a mass fraction, which we need to convert to the amount of substance concentration first. The general formula for this is
$$w_i = \frac{\rho_i}{\rho} = \frac{c_i M_i}{\rho},$$
where $\rho$ is the density of the solution, $\rho_i$ is the mass concentration of component $i$, $M_i$ is the molar mass of component $i$, and $c_i$ is the amount of substance concentration of $i$.

In your calculation you made the assumption, that the density of a 30% sodium chloride solution is $\rho = 1~\mathrm{g/mL}$, which is not correct. You can look up a couple of values for brine solutions on wikipedia.
For the purpose of this example, let's assume $\rho(\ce{NaCl~30\%}) \approx 1.2~\mathrm{g/mL}$.

The molar mass of sodium chloride is $M(\ce{NaCl}) = 58.4~\mathrm{g/mol}$. Be careful with the units, as you used the incorrect ones.

The amount concentration is given through the formula
$$c_i = \frac{n_i}{V},$$
where $n_i$ is the amount of substance (not number of moles!) and $V$ is the volume of the solution.

You correctly identified that there is a one-to-one correlation between sodium ions and sodium chloride as expressed by
$$\ce{NaCl (s) ->C[H2O] Na+ (aq) + Cl- (aq)}.$$
Therefore we can write
$$n(\ce{Na+}) = n(\ce{NaCl}).$$

Putting everything together we arrive at the following formula:
\begin{align}
n(\ce{Na+}) &= \frac{w_{\ce{NaCl}}\cdot \rho(\ce{NaCl~30\%})}{M(\ce{NaCl})} \cdot V\\
            &= \frac{0.3\cdot 1.2~\mathrm{g/mL}}{58.4~\mathrm{g/mol}}\cdot 10~\mathrm{mL}\\
            &= 0.061~\mathrm{mol}
\end{align}

Basically you thought process was correct under the assumption of the inaccurate density. And it was off by $10^{-3}$ because you (and your teacher) used the wrong units.


Inductive effect of alkyl ammonium groups

The inductive effect is both distance-dependent and χ-dependent. N-H bonds are 1.01 Å, yet the χ of H is only 2.1. N-C bonds are 1.42 Å, yet the χ of C is 2.5. At the extremes, where there are only N-C or N-H bonds, the ordering of -I is easy to rationalize.

However, the two, short N-H bonds in -NH2R+ outweigh the higher χ of N-C bonds, giving -NH2R+ a higher -I than -NHR2+.


How to calculate the latent heat when there is a boiling temperature range?

Conceptually, it is possible to calculate the enthalpy of any mixture of these three compounds in the vapor phase and in the liquid phase.  So heat balances on pieces of equipment can thereby be properly carried out.  But, in the case of such a multi-component mixture, the concept of latent heat ceases to have realistic meaning and applicability in practice.