Introduction to Chemistry Textbook - Theory based

Chemistry by Zumdahl is very good. I teach General Chemistry I and this is the book we use. I still reference it frequently. You will want the instructor solutions to accompany it. I let students check this out for classroom use, but you should have it for yourself at least. (I saw for the first time this semester a version that is loose leaf; this is great since it is sort of big. We have a custom version that takes out parts not covered. There is also a nice ebook.) This book is lovely, except for electron affinity.
For introductory chemistry, Zumdahl and DeCoste wrote Introductory Chemistry: A Foundation. We use this for Chem 101. Also very good, but a little too elementary for my taste and there is not as many topics discussed. But this may be what you want. You are probably somewhere between these two.

The next two are probably more rigorous than you would like, but worth mentioning.

Principles of Instrumental Analysis by Douglas A. Skoog is outstanding, but probabbly more advanced than you would like.
Physical Chemistry - Atkins. This book is very dense and is definitly rigorous. This will take an undergrad 2-3 semesters to get through. This may be a little overkill. You will actually see many general chemistry textbooks take exercises from this book. Really need the solutions bad for this book, it is that hard. A great place to find exam questions.


In principle you could find a lot of equations, all leading to the same result. But these are only (half-)equations. Most of the times the reaction mechanism is much more complicated than just these reaction equations.

The decomposition of hydrogenperoxide is still an active field of research in various media with various catalysts.

Unfortunately with the massive overflow of the above, I was only able to find a quite old PhD thesis by Trice Walter Haas (Iowa State University, 1960)

Here is a small line-up of reaction equations that will play a role in the decomposition, at least to that part that I can imagine it. And I am just assuming that there are radicals involved, because there usually are.
\begin{align}
\ce{H2O2 &<=> H+ + HO2- }\\
\ce{HO2- + H2O2 &<=> HO2 + HO + OH-}\\
\ce{HO2 + OH &<=> H2O + O2}\\
\end{align}


In liquid or gas and even in solid hydrogen peroxide has a gauche structure. One reason for this is to reduce electronic repulsion and the other is to give a well packed structure in solid phase.also the interactions between different molecules leads to a gauche structure in liquid and gas.


Oxymercuration and demercuration mechanism

The mechanism you are looking for is probably not known and might well be never known. According to Brown et. al. there is a rapid and reversible equilibrium between associated and dissociated reactants.[1] They tested this for cyclohexene, norbonene and apobornylene, but I believe it is fair to generalise in this matter.
$$\ce{R2C=CR2 + Hg(OOCCF3)2 <=> (F3CCOO)R2C=CR2Hg(OOCCF3)}$$

Due to this, even the formation of a mercurinium ion seems to be sometimes not observed. For more information please have a look at the mechanistic studies of Pasto and Gontarz,[2] where they focussed on the stereo selectivity of substituted cyclohexenes, and references within.

Whatever molecule you might assume acting as the electrophile, it is quite obvious, that the bond mercury oxygen bond is highly ionic, with most of the electron density located in the acetyl moieties, leaving a rather substantial positive charge at the mercury itself. The following structures were optimised at DF-BP86/def2-SVP and the charges displayed are natural charges from the Natural Bond Orbital analysis to illustrate the above point.

The lower structure of $\ce{Hg(OAc)2}$ is slightly more stable.

Herbert Charles Brown, Min-Hon Rei, Kwang Ting Liu, J. Am. Chem. Soc., 1970, 92 (6), 1760–1761.
Daniel J. Pasto, John A. Gontarz, J. Am. Chem. Soc., 1971, 93 (25), 6902–6908.


How adding one more electron increases the ionization energy?

It's not just adding one more electron. Another proton is also added to the nucleus. The overall trend across a row is driven by shielding effects.

In general, as you go across a row, the shielding effect is fairly small compared to the effect of increasing nuclear charge.

As you note, there is a slight deviation in this trend between 5A (i.e., $\ce{N}$) and 6A (i.e., $\ce{O}$) because the extra electron in oxygen is paired. This increases the electron-electron repulsion, so the ionization energy decreases slightly.

As you note, $\ce{F}$ (7A) resumes the overall trend of increasing ionization energy driven by the attractive effects of the higher effective nuclear charge $Z_{eff}$.


why disproportionation of free radicals does not happen in kolbe's electrolysis if we use secondary or tertiary carboxylic salt as substrate?

Because of the free-radical-scavenging activity of the secondary/tertiary  carboxylic salts?

Hence, radicals that could possibly undergo disproportionation reactions resulting in alkene are removed.

As a possible example, see 'Evaluation of the Free-Radical-Scavenging Activity of Diclofenac Acid on the Free-Radical-Induced Haemolysis of Human Erythrocytes'.


What is the difference between the structure of nitrito-o and nitrito-n?

The last letter of nitrite-o and nitrite-n depicts the atom which is attached to the other atom. This phenomenon is called linkage isomerism and is an important concept in both organic and inorganic chemistry.

inorganic chemistry

This complex exhibits linkage isomerism which results in two possible ways of attachment of ligand to the central atom. First one, the $\ce{NO2}$ ligates with the central atom through the nitrogen atom and thus is called pentaammine nitrito-N-cobalt (III) chloride and the second one,  $\ce{NO2}$ ligates with the central atom through the oxygen atom and thus is called pentaammine nitrito-O-cobalt (III) chloride. In this case, $\ce{NO2}$ is called ambident ligant and is represented as nitrito-κ-N and nitrito-κ-O, replacing the old system of trivial names such as nitro and nitroso.

organic chemistry

Alkyl nitrite forms two isomers:- $\ce{RNO2}$ (called alkyl nitrite-N) and $\ce{R-O-N=O}$ (called alkyl nitrite-O). [For preparation of these two isomers, see this answer].


Why do the two cycloheptadiene isomers both show four distinct shifts in carbon-13 NMR?

There are indeed 4 carbon environments in both of these molecules, shown below;

A plane of symmetry can be drawn through each, with one environment having only one carbon nucleus. All other sites have 2 carbons.

I don't know that any words will make it any clearer than the picture, but if you aren't sure after looking at this, I can elaborate.


Light is emitted from flames by two primary mechanisms: one is small particles glowing incandescently because they are hot (the same mechanism that drives an incandescent light bulb); the other is from electronic transitions from specific energy levels in excited atoms in the flame produces as a by product of the combustion process (this is partly why flames can coloured specifically by elements present in the burning material).

But why do flames and their emissions appear so far from the primary materials that are burning? Clearly solids themselves can burn (eg charcoal in a barbecue is glowing from surface combustion of the carbon with atmospheric oxygen and the emission you see is because this makes the carbon hot enough to glow with heat).

But when things are that hot, plenty of other reactions can occur and this is particularly noticeable if the thing that is burning is less pure than charcoal (which is mostly carbon). Raw coal, for example, contains a lot of volatile impurities (smokeless coal is deliberately treated to reduce the volatile impurities). When the primary reaction of burning carbon happens in raw coal, those volatiles are often turned to gases and driven off the coal to combust as gases far from the solid coal. Of course this is happening at the same time as solid carbon in the coal is burning so the two effects are mixed up to give both glowing coal and gases burning some distance from the glowing coal. Old home chemistry sets used to contain experiments that allowed these effects to be separated. Raw coal is placed in a vessel connected to a tube allowing any emerging gas to be directed far away from the coal. The coal is heated with an external source (eg a Bunsen burner). The emerging volatile gases can then be lit without setting the coal on fire giving a flame a long distance away from the heated coal.

The same can be done with wood (which contains even more volatile and flammable components than coal).

So, when wood or coal burn in an uncontrolled way both the burning of the solid and the burning of the volatiles contained with the solid occur at the same time but the volatiles (as gases) can move far from the solids before they burn. This is why the position of the fame can be far from the burning solid.

Also worth noting

In addition to the volatiles being driven from the burning solid, partial combustion can also create gases like carbon monoxide which can then travel some distance away from the solids while burning (though CO tends to have not very bright blue flame). In fact a controlled version of this reaction (which sometimes also used steam to create hydrogen) was once the primary way of creating town gas (which was widely distributed to support gas lighting and cookers in cities.

If the primary goal is to create flames designed to illuminate rather than heat then burning can be designed to create a lot of small hydrocarbon particles in the flame. This is intentional for things like candles. Here the heat may come from burning the was but the light comes because that burning creates small incandescent soot particles which are heated enough to glow turning a lot of the energy into light.


Converting Chlorine Dioxide to Free Residual Chlorine

Some chemistry, first add H2O2 and NaOH:

$\ce{2 ClO2 + H2O2 + 2 NaOH -> 2 NaClO2 + O2 (g) + 2 H2O }$

Source: https://chemiday.com/en/reaction/3-1-0-1114
Continue adding H2O2, as:

$\ce{2 NaClO2 + 2 H2O2 ->  2 NaClO + 2 O2 (g) + 2 H2O }$
$\ce{2 NaClO + 2 H2O2 -> 2 NaCl + 2 O2 (g) + 2 H2O }$

which implies a net reaction of:

$\ce{2 ClO2 + 5 H2O2 + 2 NaOH -> 2 NaCl + 5 O2 (g) + 6 H2O }$

So, slowly add H2O2 until the bubbling ceases. Collect the volume of oxygen and convert to moles. Divide by 2.5 to determine moles of ClO2 and your chlorine therefrom.
Here is a related reference, Titrimetric and photometric methods for determination of hypochlorite in commercial bleaches, which cites the use of hydrogen peroxide and acidic permanganate.


Prove that a Change of Volume will not Change the Value of the Equilibrium Constant

In your workyou have changed the "equilibrium constant" which is a fixed while keeping the number of moles constant of all species in both instances. Rather its the opposite

Consider the reaction;$$\ce{AB <=> A + B }$$ with equilibrium constant $K_c$ at temperature $T$ and volume $V_0$

Initially let $n(AB) = x_0$ , $n(A) = y_0$ , $n(B) = z_0$ where $n()$ refers to the number of moles.

then $$K_c =  {\frac{{\frac {x_0}{V_0}}{\frac {y_0}{V_0}}}{{\frac {z_0}{V_0}}}}$$

Now the volume is reduuced from $\ce{V_0\bond{->}V}$
Hence all the concentrations increase.
Then the reaction quotient at the moment the volume is changed;$$Q_c =  {\frac{{\frac {x_0}{V}}{\frac {y_0}{V}}}{{\frac {z_0}{V}}}}$$

Now $Q_c > K_c$ hence equilibrium shifts to the right reducing the number of moles of $A$ and $B$ increasing the number of moles of $AB$.Due to this effect $Q_c$ will gradually reduce and at acertain point will becaome equal tp $K_c$ with different number of moles.

$$K_c =  {\frac{{\frac {x_0}{V_0}}{\frac {y_0}{V_0}}}{{\frac {z_0}{V_0}}}} = {\frac{{\frac {x}{V}}{\frac {y}{V}}}{{\frac {z}{V}}}} $$

The only parameter remaining constant(under constant temperature) is the Equilibrium constant is a certain parameter(i.e. Volume here) is changed, other parameters will vary keeping Equilibrium constant a constant.


How exactly does the oxidation of TEMPO by NaOCl/KBr work?

I would propose the following mechanism:

Note that the oxygen radical anion isn't that bad as a leaving group as the Cl-O bond is weak. It can participate in an electron transfer with chloride anion, or rather the bromide at the beginning of the reaction as it is more abundant (I have written the protonated form - hydroxyl radical as it is less likely that two anions will come close enough for electron transfer):

The bromine radical can in turn react with another TEMPO molecule. That is why two equivalents of TEMPO react with one equivalent of hypochlorite.

Keep in mind that there are countless possibilities and deviations from the mechanism I proposed.


Hybridization is determined by molecular geometry. Carbons participating in triple bonds like those in acetylene have two regions of electron density. Carbon dioxide is a linear molecule, and also has two regions of electron density. The fact that both have two regions of electron density are what make these carbons $\mathrm{sp}$ hybridized.

To reenforce hybridization is determined by molecular geometry, consider the benzyne intermediate. It has a triple bond, but is $\mathrm{sp^2}$ hybridized because it is too strained to become linear and only have two regions of electron density.


Is the colour of leaves due to d-d transitions in chlorophyll?

Let's rephrase your hypothesis and test it:

  The green colour of leaves results from a d-d transition of the central atom of metal complexes known as chlorophylls.

Green leaves contain two types of chlorophyll, Chl a and Chl b, both have $\ce{Mg}$ as the central atom and show two distinct bands in the UV-VIS absorption spectrum.

The stronger ($\epsilon \approx 100000\, \mathrm{cm}^{-1}\cdot \mathrm{M}^{-1}$) band in the 400 nm (blue) region is known as the Soret band, the weaker  $\epsilon \approx 30000\, \mathrm{cm}^{-1}\cdot \mathrm{M}^{-1}$) in the 660 nm (red) region is known as the Q band.

Now imagine that Chl a is treated with weak $\ce{HCl}$. In the resulting product, known as Pheophytin a, the central atom ($\ce{Mg^{2+}}$) is replaced with two protons.

In the UV-VIS absorption spectrum, however, we would still observe the two bands. They are slightly shifted and show somewhat different extinction coefficients, but the pattern is fundamentally the same.

We may conclude that our initial hypothesis is wrong. If it is not the absorption of the central atom in our complex, the ligand must be responsible.

Our new working hypothesis therefore is:

The colour of green leaves results from the absorption of the polycyclic pheophytin ligand and its extended $\pi$ system.

Let's simplify the Pheophytin a molecule: Removal of the E ring and replacing all the fancy substituents with ethyl groups leads to octaethylporphyrin as the core structure.

In the UV-VIS spectrum, we would still be able to see the strong Soret band, which corresponds to a $\pi,\pi^*$ transition from the ground state to the second excited singlet state ($S_0 \rightarrow S_2$).


Acid-Base Titration of ammonium dihydrogenphosphate?

The pH can be measured vs. the amount of $0.1$ M NaOH added to your solution with a pHmeter. Then two titrations would be visible : first the titration of dihydrogenphosphate ($p\ce{K_a = 7.21}$) and then the titration of ammonium ($p\ce{K_a = 9.22}$). If you don't have a pHmeter, you would probably use a colored indicator. But the end of the first titration will not be determined with precision, as the two $p\ce{K_a}$'s are not different enough. You will only be able to determine the end of the titration of the two acids together. To do this, a rather basic indicator is needed, like Alizarine yellow whose $p\ce{K_a = 10.5}$.


Comparing size of ions lithium, gallium ions

That is impressive.  What you may need to bear in mind is that as the nuclear charge increases, all the electron orbitals move in closer to the nucleus.  You may have been assuming, consciously or unconsciously, that the electron shells appear at a relatively fixed distance from the nucleus -- that is, that the n=1 shell is always at about the same distance from the nucleus, the n=2 shell is always at about the same additional distance, and so on.

Not the case!  For example, if you take a look at the quantum mechanical model of the hydrogen atom and solve it for a "hydrogen-like" atom with atomic number Z and one electron (so, He+, Li++, Be+3,....,U+92 ha ha) then you will find that the average orbital radius of the electron scales like 1/Z.  So if we ignored electron-electron repulsion entirely, the radius of each shell would decrease like 1/Z as Z increases.  That's quite a contraction.  It would mean that the n=1 shell in carbon, say, would be 1/6 the size of what it is in hydrogen.  In the case of Ga, Z=31, the n=1 shell would be 1/31 the size it is in H!  Each new shell would contract similarly.

So in the absence of electron-electron repulsion we would expect the size of atoms to shrink drastically as Z increases, going across the Periodic Table, and bumping up slightly at the beginning of each row with the addition of a new shell.  Of course, real atoms do have electron-electron repulsion, lots of it, so the effect is considerably muted.

Nevertheless, the size of atoms does shrink considerably going from left to right in the Periodic Table, and it's worth remembering this is because all of the electron shells shrink, not just the valence shell.


Relationship between molecularity and the reaction order

Molecularity applies to single elementary reaction steps and is seldom higher than 2.
The reaction order is the "effective molecularity" of the whole set of linked reactions, if it had been a single step reaction. It need not to be integer and has just indirect relation to molecularities of elementary steps.
The reaction order is either determined experimentally, either derived from the reaction schema via determined conditions for steady states of intermediate products.


In order for a process to happen, it has to increase the combined entropy of the system in which it happens and of the surrounding (2nd law of thermodynamics). As we will see in a bit, the more work the process does on the surroundings, the less the entropy of the surroundings increases, putting a limit on how much work the process can do.

What happens to the energy leaving a system?

When a process brings the system from a higher to a lower energy, the energy leaves the system in the form of work done on the surroundings or heat transferred to the surroundings (1st law of thermodynamics). For a given set of initial and final states, the amounts of work and heat can vary, but the sum has to be the same. For example, if our system is a charged battery, we can either run a motor with it, lifting up a rock (work done on the surrounding). Or we can short the battery, creating thermal energy that is transferred to the surrounding in the form of heat. We can even have a situation where heat and work have opposite signs (e.g. do more work and have heat transferred into the system to get the same energy balance).

How is heat and work related to entropy changes?

The change in entropy inside the system depends solely on the initial and final state. For the surrounding (with nothing else going on in the surrounding), the change in entropy depends on the heat transferred. The more heat is transferred to the surrounding, the more the entropy of the surrounding increases. To maximize the work, you want to minimize the heat transferred to the surroundings (or even transfer heat to the system), but only to the point the that total entropy still increases (otherwise the process could not run). So if you match the entropy change in the system with the opposite entropy change in the surrounding (via heat exchange), you get the maximum work out of your process.

Can I see a picture?

Sure. The picture shows that the sum of heat and work is equal to the enthalpy change of the system. Values shown with a downward arrow are negative and with an upward arrow positive. For heat and work, negative values mean energy is transferred out of the system. For example, work done by the system on the surrounding is a downward arrow, and exothermic reactions have a downward arrow for $\Delta H$.

The picture also indirectly shows the entropy (via the arrows labeled heat and $-T \Delta S$. If the arrow for $-T \Delta S$ is downwards, entropy in the system increases (scenarios labeled "more dispersed"). If the arrow for heat is downwards, entropy in the surrounding increases. To calculate the actual entropy, you would have to divide by the temperature, but the way it is shown everything can go into a single picture.

I have three scenarios, depending on whether the process is exothermic or endothermic and whether the entropy in the system increases or decreases. One scenario (endothermic reaction with a decrease of the system's entropy) is not shown (you would have to do work on the system to drive such a process, and we are trying to make the process do work for us instead).

In scenario 1, the process is exothermic. Some of that available energy is transferred in the form of heat (to make sure that the overall entropy does not decrease) and some of it is transferred in the form of work done by the system.

In scenario 2, the process is also exothermic, but different from scenario 1 the entropy of the system increases in this scenario. The work done by the system is (surprisingly, perhaps) larger than the enthalpy change, and the extra energy comes from heat transfer from the surroundings.

In scenario 3, the process is endothermic. You might think that this process can't do any work because - where would the energy come from? It comes from the surrounding, partially to provide the energy for the endothermic process and partially to be transferred back into the surrounding in the form of work. This can happen because just as in scenario 2, the entropy of the system increases.

  [...]the Gibbs free energy is the maximum amount of work (or useful work) that a system can do, whereas entropy is a measure of the non-available enthalpy.

This sort of describes scenario 1, which I pasted again below with two boxes for emphasis.

The pink box represents the 1st law of thermodynamics. There is some enthalpy available from the process, and this will leave the system either as work or heat or some combination, as long as it adds up to the enthalpy change in the system. The blue box represents the 2nd law of thermodynamics. The entropy increase of the surrounding (through transfer of heat to it) has to be larger or equal to the entropy loss in the system. If you combine the two laws, you get the maximum work the system can do. In this scenario, the change in entropy of the system times the temperature represents that portion of the enthalpy not available for work (labeled $-T \Delta S$ in the diagram). You can't, however, use that label in the other scenarios where the entropy of the system increases. In those scenarios, the entropy change in the system results in more work available than the enthalpy change would suggest.

This is all rather complex and might make you dizzy. The Gibbs energy helps us to summarize all this in a single quantity. It combines entropy and enthalpy changes in the system in a neat way that tells us about the maximum work. In the absence of work, it also tells whether the process will happen in that direction or not.

Disclaimer for those that have a physical chemistry background

This is for a closed system at constant pressure and temperature (i.e. P-V work is done against a constant pressure, and initial and final state have the same temperature). Whenever it says work, it refers to non-P-V work only. Labels $\Delta H$, $\Delta G$, $\Delta S$ should read $\Delta H_\text{sys}$, $\Delta G_\text{sys}$, $\Delta S_\text{sys}$.


Your best guess for the boiling point of any version of Coke would be 100 C, the boiling point of water.

Diet Coke is mostly water (the flavourings are a very small amount relative to the amount of water). The largest ingredient will be the sweetener but there will be only a fraction of a gram of that. It is unlikely you will notice any deviation from the properties of water.

Standard Coke has quite a lot of sugar in it. A standard can (~300ml) contains about 40g of sugar. To put it another way, the contents are more than 10% sugar by weight and the solution is about 1/3 mol/L of sucrose (other sugars will be slightly different). A standard calculation using the ebullioscopic constant for water suggests the elevation of the boiling point will be barely 0.2 C, so small you'd struggle to measure it without good instruments and a good experimental setup.


Is a ring intermediate formed in hydroboration?

That should be a transition state, not an intermediate.

Also, there should not be a positive charge on your boron in the final compound you drew.


Why is the acoustic absorption coefficient of hexane in air less than air?

Best hypothesis so far is that as gas density increases the transducer couples more efficiently to the medium, giving the impression of a decrease in absorption coefficient due to the increased amplitude at the receiver.


Why lattice constant (unit cell) is always a parallelepiped?

A unit cell is always a parallelepiped because this is the definition of a unit cell. In case when you don't have a repeating unit of this kind (which might indeed be the case in real life), you don't have a crystal, also by definition. You might be dealing with a quasicrystal or something.

3D periodicity is what defines a crystal. If you have it, this means you have three non-coplanar periodic vectors, and these inevitably define a parallelepiped.


Why do esters have to be distilled during purification?

According to Wikipedia ethyl formate is 9% soluble in water, that figure may be increased by any excess ethanol present, so you will be losing a significant portion in the proposed work-up. Is this enough error to invalidate the experiment?


Is it possible for a carbon atom with a negative formal charge to have ONE bond and ONE lone pair?

The hydrogen atoms are not shown, so this is really $\ce{CH3CH2CH2CH2-}$ where the rightmost carbon atom has a lone pair and a formal negative charge.

The species is the n-butyl anion:


Does a chiral centre have to have 3 or 4 different groups attached?

The answer is that all chiral centers are stereocenters, but not all stereocenters are chiral centers thus the words are not interchangeable.

A chiral center requires four or more  different molecular groups attached to a single center atom (asymetrical). In the case of pnictogens (nitrogen, phosphorus, etc...) or chalcogens (sulfer, selenium, etc...) an electron pair can count as one of the molecular groups.

An alkene is a stereocenter that requires 3 or 4 different molecular groups such that no two common groups are on the same side of the bond (must be able to determine an E/Z configuration). In the case of the blue stereocenter, the methyl groups are on the same side and thus switching their position would create an identical compound meaning the alkene is not a stereocenter.  For the red center switching the position of the bromine and methyl groups would create the Z-isomer from the E-isomer thus is a stereocenter, but not a chiral center.  If however, the hydrogen of the blue stereogroup were to switch positions with a methyl group then the blue alkene would be a stereocenter with three distinct molecular groups.


NMR: Most deshielded protons (electron withdrawal)

I don't believe there is anything really wrong with this question. Yes, both molecules have multiple signals, but one of them contains a signal which corresponds to it being the most deshielded. In fact, one of these, molecule F, contains the most deshielded signal AND the most shielded signal.

"In any case it's clear that the question is asking about which molecule has the highest degree of electron withdrawal ..." not necessarily so. The correlation between electron withdrawing groups and shielding/deshielding is a strong rule of thumb, but can become complicated with aromatic systems and competing electronegativities of substituents. In any case, you need to be looking at parts of the molecule; shielding is quite a localised effect, and as I said above, molecule F actually has the most shielded AND deshielded protons.

In any case, if you consider the electronegativity of oxygen; a proton attached to this will be strongly deshielded. Stick them on an aromatic ring para to a carbonyl group, and you will generate a resonance system like below:

In this case, the phenolic proton is by far the most deshielded of all protons in the molecules shown; I'd expect a chemical shift of over 10ppm. In response to some of the comments about -OH signals being over the place; it is important not to confuse raw chemical shift and averaged chemical shift resulting from chemical exchange with other labile protons, such as the (often observed) exchange of phenol protons with water. In these cases, the resulting chemical shift depends on the relative concentrations of the exchanging species, the exchange rate, and the original chemical shifts of the non-exchanging species.


What is happening at the various boundaries of a phase diagram?

I got that you are trying to visualize what is actually happening. I have not done the compression myself, but this is what I think will be observed. It will be good if someone with real experience can confirm the answer.

At the beginning you are in the region marked 'vapour'. When you are approaching the liquid-vapour boundary from below, before you touch the boundary, you will still observe only vapour, no liquid at all. You are still in the region marked 'vapour'. Nothing will happen.

When you just reached the boundary, you will see some liquid droplets formed. It should look just like some dew formed at the wall. Regarding the relative amount of liquid and vapour at equilibrium, the pressure of vapour present must be equal to pressure exerted by you (or else the piston will move!) and equal to pressure on the boundary (determined by temperature). Say if you try to push the piston further inside, the volume occupied by sample will decrease, and more liquid formed, but you will still stay on the boundary. You will notice that the pressure needed to push piston inside further does not change, because the pressure of vapour does not change (push in, vapour pressure should increase, but some vapour condensed, lower back the pressure). Only when all vapour has been converted into liquid, you can depart from the boundary, move into the region marked 'liquid'. However, before all vapour turned into liquid, if you stop pushing in and hold your piston still, the liquid and vapour amount won't change. So there is no unique relative amount of vapour and liquid at equilibrium as written in your question. Above liquid-vapour boundary you have only liquid and the piston surface is resting on the liquid surface.

That state is known as supercritical fluid.


What is reaction between isopropyl alcohol and acetone?

I've used acetone/ipa mix to flush chromatography columns. There is no reaction between them without another reagent present


Redox and acid-base reactions in planetary atmospheres

We most certainly can call atmospheres acidic or basic.  Put some quicklime, or even magnesia, in our air and you see that it undergoes an acid-base reaction with acidic materials (water vapor, carbon dioxide) in our atmosphere.  Over time, our weakly acidic atmosphere gradually dissolves in water and then reacts with alkali and alkaline earth metal silicates too, turning rock into clay.

Our atmosphere became acidic as it picked up oxygen, which is indeed an "acid former" through the volatile (thus atmospheric) compounds it forms with several nonmetals.  In contrast, the gas (or at least, clouds therein) on the giant planets contain some ammonia and would appear basic if we brought typical Earth chemicals there.


I guess that the IUPAC Gold Book rules for helicity apply here.

  The chirality of a helical, propeller or screw-shaped molecular entity. A right-handed helix is described as P (or plus), a left-handed one as M (or minus).

The above knotane is right-handed, so it should be P.

For more complex molecular knots such simple rules most likely will fail.


Drops or patches of water in your sink are basically water that is not evenly spread out. The surface tension of the outermost water molecules acts a bit like a "balloon" containing the inner molecules. To a small degree it is thus possible for the water molecules to be stacked on top of each other because the water's surface tension is stronger than the gravitational pull.

Toothpaste contains detergents like sodium dodecyl sulfate which act as a surfactant reducing the surface tension of water.
When you coat your finger in a mixture of water and a weak detergent like toothpaste or a stronger one like soap and touch the water drops in the sinks, the water mixes with the detergent. This lowers the surface tension and gravity spreads the drop out. The water is thus not really moving away from your finger, but the drops are being spread out under and around your finger.


Which is more basic, piperidine or pyrrolidine and why?

I think I may have a possible insight into the small difference. Wikipedia gets the pKa from J. Am. Chem. Soc. 1957, 79 (20), 5441–5444 and lists pyrrolidine at 11.27 and piperidine at 11.22.

Calculating the equilibrium structures (B3LYP 6-31g*) for both species in their neutral and protonated states, I found something interesting.

  For piperidine

Neutral and protonated species both achieve the optimal staggered conformation about all bonds.

  For pyrrolidine

For the neutral species, the C-C bonds have their hydrogens in an eclipsed conformation. In the protonated species, the ring gets a slight twist, which allows the C-C bonds to have their hydrogens in a staggered conformation.

  So...

The hydrogens being moved into a staggered conformation upon protonation of pyrrolidine could be the source of the extra stabilisation of the protonated form.

I'm also thinking if the cation form is favoured slightly more in pyrrolidine because its aliphatic portion is less, thus leading to a bigger release of solvation energy upon protonation (calculated to be 1 kJ/mol more favourable for pyrrolidine, but this could be within the error of the method).

If I had to pose arguments for this small difference in basicity, these would be the ones I'd choose.


how to determine the volatility of an organic compound using the physical properties of organic molecules

Vapor pressure is the physical property that expresses a compound's volatility at a given temperature, and the enthalpy of vaporization describes the temperature dependence of the vapor pressure.

For some compounds, like water, there are tables and formulae readily available for determining the vapor pressure at different temperatures. If you know the compound's vapor pressure at a given temperature and you also know the enthalpy of vaporization for the compound, you can then estimate the vapor pressure of the compound at another temperature.  This method assumes that the enthalpy of vaporization of the compound is constant with temperature, which is only a good approximation over a relatively small temperature range. You would need to know the heat capacity of the compound to correct for the temperature dependence of the enthalpy of vaporization.

Comparison of boiling points is often used to say "compound x is more volatile than compound y" because compound x has the lower boiling point. This statement is only true for the pressure at which the boiling point was given, often 1 atmosphere.  The vapor pressure of a polar compound like a low-molecular weight alcohol has a greater temperature dependence (enthalpy of vaporization) than a non-polar compound like an alkane.  So an alkane and an alcohol with the same boiling point at one atmosphere will not have the same boiling point at another pressure. Correspondingly, an alkane and an alcohol with similar vapor pressures at one temperature can have very different vapor pressures (volatilities) at another temperature.


Rutherford's Alpha Ray Scattering Experiment & Plum-Pudding Model

First of all, yes, your teacher is right for most of the part but it is not that they will go without any collision. Instead, they will penetrate through the foil like a bullet shot at paper.
It is not that the alpha rays will not be deviated but that if the charge was evenly spread throughout the atom, it would not be able to provide sufficient repulsive forces to wear out the kinetic energy already given to the alpha particles and slow them down. As a result, the alpha particles should have penetrated the atom like a bullet penetrates a sheet of paper.


Calibration Ammonia Sensor and Gas Concentration

The calibration gas will purge the air out of your chamber, so that it contains 50 ppm NH3.  This assumes that there are no reactions that change the mixing ratio of the ammonia (e.g. reaction with water in your chamber).  The length of the tube doesn't matter unless the gas reacts with your tubing, though length will decrease the pressure of the gas coming out of the tube.


How to know whether s-p mixing will happen in heteronuclear molecules?

I don't think you can know, a priori, whether or not s-p mixing occurs without first having some experimental data.  I'll use water, a triatomic molecule, as my case in point.

For over 50 years students have been told that water is roughly $\ce{sp^3}$ hybridized.  The general description is that there are two equivalent O-H $\ce{sp^3}$ sigma bonds and two equivalent lone pairs, also in $\ce{sp^3}$ orbitals.  The lone pair - lone pair repulsion is greater than the sigma bond - sigma bond repulsion, so the lone pair-O-lone pair angle opens up slightly and the $\ce{H-O-H}$ angle closes down to the observed 104.5 degrees.

With the advent of photoelectron spectroscopy it was found that the two lone pairs in water were not equivalent (2 signals were observed for the lone pairs).  Now, the hybridization of water is described as follows:

2 $\ce{sp^3}$ O-H sigma bonds
one lone pair in a p orbital
and the second lone pair in an $\ce{sp}$ orbital

If chemists knew how to predict s-p mixing, it wouldn't have taken them over 50 years to understand how they mix in water.


We can and do use volume fractions to express the level of a gas of vapour in air, it is volume ppm.  For example air to a first approximation is 20 % oxygen and 80 % nitrogen.

This means 20 % of the molecules in air are oxygen while 80 % are nitrogen. If we treat the gases as ideal gases then air is by volume 20 % oxygen.


There are six unpaired electrons, and we can swap any two of them, so we have:
$${6\choose2} = 15$$


How to dissolve a spider with household chemicals?

You're not the old lady that swallowed a fly and trying to avoid your lyrical fate, are you?

"Biological" washing powder might do the trick - plenty of proteolytic enzymes - probably take several days at least.  Then you'd have to wash out the resultant spider soup.

I hope you like coffee with a lot of body to it.


When you make assumption for guessing the solubility of the molecule in some solvent you should consider the whole molecule, his structure and his size.

We can start with the ether, in the below figure (from wikipedia the one at right ) is shown a diethylether molecule.

You can see that is quite a non-polar molecule (charge 0,green, in the most part of the surface) so this explains why has a low solubility in a polar solvent like water. Same thing for Vitamin A:

The most part of the molecule is not polar. Only one $-OH$ can't make the difference. And so is for 1-Octadecanol that has a solubility in water of $1.1  \times 10^{-3} mg/L$ and Nonadecyl alcohol (1-nonadecanol) seems something like $4.8  \times 10^{-3} mg/L$ both up to my references have very low solubility in water (where did you find that were not?). For the $CO2$ is true what you have stated it has not a very strong polarity (but in fact it has a little bit!) because is a linear molecule.

This last example let me introduce to an important factor that influence the solubility: molecule size.
Little molecule are more easily solvatated by solvent molecule, this bring us to the case of cellulose:

Cellulose is very hydrophilic and so should be soluble in water, but this doesn't happen because cellulose is a polymer (in figure is show his monomer) that can have a molecular weight very hight. Polymers may swell but can't be solubilize if their chains are bonded with chemicals bonds until the bonds are break.


Why does DMSO help facilitate halohydrin reactions?

Alkenes predominantly gives relevant 1,2-dihalo product when reacted with a halogen (e.g., $\ce{Br2}$) in a organic solvent such as chloroform or dichliromethane. If the reaction is performed in water, the product is the corresponding 1,2-halohydrin due to restriction applied by polar water molecules caging $\ce{Br-}$ ion. Organic molecules are sparingly soluble in water as solvent. Thus, the halohydrin reaction is often done in a mixture of organic solvent and water. For example, bromohydrin can be achieved by using N-bromosuccinimide as the electrophilic bromine source in $\ce{DMSO/H2O}$.

Use of $\ce{DMSO/H2O}$ as a solvent has few advantages:

The bromine source, N-bromosuccinimide (NBS) is readily soluble in this solvent system [Ref. 1]).
$\ce{DMSO}$ is also participated in reaction mechanism. Reference 1 has detailed labeling procedure using $\ce{DMSO^{18}}$ to show that isotopic $\ce{^{18}OH}$ was incorporated in $\ce{OH}$ group of bromohydrin [Ref. 1].
Evedence also suggest the participation of water in the medium, probably due to increasd basic nature of water in $\ce{DMSO}$ ($\mathrm{p}K_\mathrm{a}$ of water in $\ce{DMSO}$ is $31.4$ at $\pu{25 ^\circ C}$ [Ref. 2]).

Suggested mechanism for involvement of $\ce{DMSO}$ as follows:

References:

Bromohydrin Formation in Dimethyl Sulfoxide: D. R. Dalton, V. P. Dutta, D. C. Jones , J. Am. Chem. Soc., 1968, 90(20), 5498–5501 (https://pubs.acs.org/doi/abs/10.1021/ja01022a030).
Acidities of water and simple alcohols in dimethyl sulfoxide solution: W. N. Olmstead, Z. Margolin, F. G. Bordwell, J. Org. Chem., 1980, 45(16), 3295–3299 (https://pubs.acs.org/doi/abs/10.1021/jo01304a032).


Think of it this way.
$$dP = \left(\partial P\over\partial T\right)_VdT+ \left(\partial P\over\partial V\right)_TdV$$
Now, we want P to be constant (that is, $dP=0$). What does that entail? A certain relation between $dV$ and $dT$ (which, BTW, are now $\partial$ rather than $d$):
$$\left(\partial V\over\partial T\right)_P=-\left(\partial P\over\partial T\right)_V\left/\left(\partial P\over\partial V\right)_T\right.$$


Converting Chlorine Dioxide concentration to Chlorine

You can, in fact, calculate the chlorine concentration from the concentration of chlorine dioxide. Chlorine will normally be in some compound, so when concentration of chlorine is given, it is usually concentration of chlorine compounds.

The concentration of chlorine dioxide is 0.6 mg/L. First convert the mg to g. this results in .006 g. Divide that value by the molar mass of chlorine dioxide (67.45 grams/mole) to get .00008895 moles of chlorine dioxide, or .00008895 moles of chlorine. Multiply this by the molar mass of chlorine (35.45 grams/mole) to get .003153 grams of chlorine. Convert this to milligrams to get 3 mg (one significant figure), so the concentration of chlorine from chlorine dioxide is 3 mg/L.


Gas evolution in reaction between zinc and HCl

When HCl reacts with zinc or any other active metal such as aluminium, hydrogen is produced. This is what your textbook tells you. Whats is perplexing is the appearance of "fog" in the Erlenmeyer. In the beginning there is air above the liquid. When hydrogen evolution starts, the flask begins to fill with $\ce{H2}$ gas and creates enough pressure to displace the column of water in the buret. So that part of your observation is correct and you see a colorless gas displacing water column in the buret. What is in the conical flask that is causing the fog? If you were to touch the Erlenmeyer you could have noticed that it was hot, secondly there was a very vigorous evolution of hydrogen bubbles. There are several reasons, each contributing to the fogginess in the flask. That "fog" is just not hydrogen, air but vapors and very fine droplets of acid.

(i) creation of very fine droplets of acid by vigorous bubbling process (major contribution)
(ii) steam generated by the heat of the reaction
(iii) Hydrochloric acid has the tendency to produce fumes in moist air (seen with very concentrated acid). You did not mention the concentration of the acid used.


Oxygen is not strictly needed, it just happens to be very, very good at what it does!

Specifically, combustion requires three things:

1) Fuel: The thing that burns. This is is often a hydrocarbon, or other organic molecule. The simplest possible fuel is pure hydrogen gas.

2) Energy: What gets the reaction started, which is true of most chemical reactions.  Note that this so-called activation energy is usually much less than the energy ultimately released from combustion.  This is like rolling a boulder some distance in order to roll it down a hill, which releases much potential energy.

3) Oxidizer: The molecule that accepts electrons.  It turns out that combustion requires the fuel to be oxidized, that is, it donates electrons.  So we need something to accept the electrons, and that's the oxidizer, which is then reduced.

Oxygen is a great oxidizer because it is so electronegative, which means it really wants to accept electrons.  Only fluorine is more electronegative, and fluorine is a superb oxidizer: blow fluorine gas at nearly any substance and it bursts into flames.  For example, see this video: Fluorine.

Oxygen is in many ways a perfect oxidizer for supporting life: it doesn't set things on fire like fluorine, but otherwise can oxidize very many things.  When you eat, oxygen is oxidizing the food (fuel) you ingested, to generate energy you need to live.  So there is "combustion" going on inside you!


I have to say thank you to Maurice for his suggestion to use more concentrated HCl. I took the same pieces of floppy disk that I used before and increased the acid concentration by a factor of 10 and let them sit for another week.
At first I was disappointed, because there was no visible change in the plastic or the solutions. But when I took the pieces out of the acid and cleaned them off with an alcohol soaked kimwipe, the remaining Iron Oxide came right off. It's not perfectly clean, most likely because I had taped the pieces into my notebook after the first treatment, and I didn't clean the tape residue off before the second treatment. This is also probably why the remaining absorbance at 700 nm is so variable. But I think this method works as long as I don't get the plastic covered in tape goo. The acid treated plastic also might feel a little less stiff than the 0 M control, but this is subjective.


Why is the first ionization enthalpy of zinc higher than expected?

It is because the shell is complete.

Zinc's electronic configuration is $\rm [Ar] 3d^{10} 4s^2$, which is a complete configuration for a transition metal. The next element in the same period is $\ce{Ga}$, which starts filling $\rm 4p$ orbitals ($\rm [Ar] 3d^{10} 4s^2 4p^1$).

The zinc's electronic configuration causes a sort of special stability because of the filled electronic states. Also, the effective nuclear charge is greater than $\ce{Cu}$'s, so it's more difficult to ionize $\ce{Zn}$.


Can aqua regia/royal water dissolve lead?

Aqua regia doesn't actually dissolve lead metal. It converts to insoluble lead salt. From here:

When we are working with an acidic chloride solution, like HCl (with
an oxidizer) or aqua regia, most of the lead will become an insoluble
lead chloride which if it is left to settle can be removed by careful
decanting of the solution and filtering, what lead does carry through
the process with this solution can then be removed with a few drops of
sulfuric acid, and precipitated out of solution as a lead sulfate.

Here is a YouTube video for your reference.


Equilibrium constants for lewis acids such as Ca2+

You're confusing terms here. $\ce{K_{a}}$ is the (Brønsted) acid dissociation constant. This suggests that $\ce{K_a}$ refers to the degree that a Brønsted acid dissociates into its conjugate base and hydrogen protons.

$\ce{K_{a}}$ is inapplicable to Lewis acids because Lewis acids are not proton donors.

Brønsted acids are Lewis acids in that Brønsted acids such as the hydrogen proton can accept a pair of electrons. But the converse is not true.

Lewis acids are electron acceptors. How do we measure the strength of Lewis acids? There is no universally accepted measure as there is for Brønsted acids. We could look at the enthalpy of reaction (Lewis adduct formation) between a Lewis acid and a Lewis base. This is a common method.


OH- and HO- are both acceptable.  The charge is on the oxygen, of course.  One can argue cation then anion (salts) versus alphabetical (e.g., AuBe, which is brittle and weak rather than ductile, suggesting ionicity).

Organic formulas are C, H, then alphabetical.  Order may be altered to conform to structure (e.g., propane as $\ce{H3CCH2CH3}$) or convenience $\ce{H3C^{-}Li^{+}}$ (which is nowhere near the real structure of methyllithium, but is how it reacts).


Dynamic equilibrium in unsaturated solutions

Proper definitions of chemical equilibrium will not involve reaction rates whatsoever. Thermodynamics does not care about time. Chemical potential is the work required to form a molecule in solution, irregardless of the time it takes. Statistical Mechanics says chemical potential is the work required to move a molecule from infinitely far away (ideal gas since there are no interactions at infinity) into the solution.

Also, Free Energy is entirely independent of the equilibrium constant, but the equilibrium constant is not independent from Free Energy. At equilibrium, Free Energy will be at a minimum. Throw out definitions that use equilibrium constants. If you so wish, you can calculate the equilibrium constant from the free energy.

\begin{equation}
    K = \exp^{-\frac{\Delta G^0}{RT}}
\end{equation}

Where $\Delta G^0$ is the ideal gas free energy reaction. If you know the ideal gas free energy change of a reaction, you may calculate K. However, K is a useless thing in reality, or at least, leads to more confusion than it is worth.

For chemical equilibrium, this is the definition at global equilibrium

\begin{equation}
   \left( \frac{\partial G}{\partial \xi} \right)_{T,P} = 0
\end{equation}

Where $\xi$ is the extent of reaction. In your case the free energy was minimized by the solid dissolving rather than staying a solid, so your system came to equilibrium given the constraint that there was no more solid that could be dissolved. There was also no constraint saying that there needed to be solid present.

Equilibrium and solubility limit are two different things. All equilibrium points are constrained by definition. At the very least you are constraining T and P. When you are measuring solubility you are actually implementing an additional constraint: find the minimum free energy of the system with the constraint that a pure solid and an aqueous phase must coexist - this necessitates there initially being enough solid to be able to meet this constraint at equilibrium.

Equilibrium reactions do not require time, and I have no time for equilibrium constants :)


Boltzmann Distribution in Molecular Dynamics Simulation?

Within statistical mechanics (SM) a molecular property $X$ is computed by $$\left<{X}\right>_{SM}=\sum^{states}_i X_i p_i$$where $X_i$ is the value of $X$ for energy state $i$ and $p_i$ is the probability of being in energy state $i$ with energy $E_i$:$$p_i=\frac{e^{-E_i/kT}}{\sum_i e^{-E_i/kT}}$$Within molecular dynamics (MD) the corresponding property is computed by$$\left<{X}\right>_{MD}=\frac{1}{M}\sum^M_i X(t_i)$$where $M$ is the number of time-steps and $X(t_i)$ is the value of $X$ at time $t_i$.

You can also compute $\left<{X_{SM}}\right>$ from an MD by finding all the "unique" states (e.g. conformations) and computing their energy.  "Unique" means that even if a state is observed more than once in the MD you only include it once in the sum over states.

In principle $$\left<X\right>_{SM} = \lim_{t \to \infty} \left<{X}\right>_{MD}$$ but this assumes that the method used to calculate the energy and forces are identical in both calculations.

For example, for MD explicit solvent calculations with periodic boundary conditions it not really possible to calculate the energy of a state ($E_i$).  Also, vibrational effects must be included in $E_i$, which is typically done using the harmonic oscillator approximation, while the MD will include anharmonic effects. Similarly, the harmonic oscillator approximation is quantized (QM) while the vibrations in the MD simulation are not (classical).


Is it possible for a compound to be soluble in a mixture of solvents but not the individual solvents?

The short answer is: yes, this is possible. Unfortunately, solubility is a fairly complex phenomenon to explain simply. Let's start with some examples where solubility is higher in a binary mixture than either solvent alone.

For a solid-liquid-liquid example: phenanthrene-cyclohexane-diiodomethane.[1]
For a liquid-liquid-liquid example like the one you were asking about: trifluoromethane-tetrafluoromethane-ethane at low temperatures and high pressures.[2]

In the Hildebrand Theory of solutions, there is a quantity called the "internal pressure", loosely how much these compounds want to break apart. In this model, this is usually something like the molar energy of vaporization divided by the molar volume. If the internal pressure of the solid is between the internal pressures of the liquids (as in this case), the solid is more soluble in the mixture than in either one individually. Another way to asking your question is about phase separation in ternary mixtures. You can find many examples of curved lines in ternary phase diagrams which are due to the ternary mixture not being the sum of its parts.

Qualitatively, you could imagine that some solvent solvates some parts of the molecule well and other poorly and vice versa for the other solvent. Hence, together they solvate better than each individually.

Gordon, L. J.; Scott, R. L. J. Am. Chem. Soc. 1952, 74 (16), 4138–4140. DOI: 10.1021/ja01136a054
Paas, R.; Peter, K. H.; Schneider, G. M. J. Chem. Thermodyn. 1976, 8 (8), 741-747. DOI: 10.1016/0021-9614(76)90053-7


Is it possible to extract magnesium from epsom salt?

Supposed you are trying to obtain magnesium metal from Epsom salt, $\ce{MgSO4.7 H2O}$:
Is is worth the effort?
Magnesium metal in the form of chips (particle size 4-30 mesh) in decent purity (99.98%) is available from suppliers such as Sigma-Aldrich for about 150 Euro per kilogram.
Can I use electrolysis to generate $\ce{Mg}$ metal?
Yes, but you wouldn't want to do that at home. An older industrial process uses electrolysis of molten magnesium chloride, $\ce{MgCl2}$ (mp. is around 700 °C). While $\ce{Mg}$ metal is formed at the cathode, chlorine gas is released at the graphite anode. Did I mention that this isn't meant to be done at home?
Can I convert $\ce{MgSO4}$ to something more usable to make $\ce{Mg}$ metal?
Yes, in principle. A group of Chinese researchers examined the conversion of  $\ce{MgSO4}$ to  magnesium oxide, $\ce{MgO}$ by heating in the presence of natural gas at temperatures around 500 °C and outlined their results in
Geochemical Journal, 2011, 45, 97-108.
The initial pressure by natural gas at room temperature was 3 MPa, but upon heating to the final temperature, the pressure in the reactor was as high as 12-20 MPa! This can't be done at home and it shouldn't be.
What if I would have $\ce{MgO}$? What could I do with it?
Magnesium oxide, $\ce{MgO}$, is indeed the most important starting material in the production of magnesium metal in the Pidgeon process. Here, magnesium oxide is heated in the presence of silicon in an evacuated reactor. Silicon is oxidized to silicon dioxide, magnesium oxide is reduced to magnesium metal. The latter boils off (bp. around 1100 °C) and is condensed. This isn't kitchen chemistry either.
In summary, you're better off buying some magnesium metal from a supplier of your choice!


Are non-ionic compounds generally as dehydrating as ionic compounds in drinking water?

Actually, the salt in seawater doesn't kill you directly. It dehydrates you and you die of thirst. If you mixed 1 part of seawater with 2 parts of fresh water, it would not dehydrate you, but you would need to eliminate the excess salt so you would need to drink a lot of this salty water.

There are many living things in seawater that can kill you. Red Tide (excuse me, harmful algal bloom) is an algae common to warm water that produces a toxin. Very little of this toxin is needed to kill a person.


Why is the inductive effect for a protonated amino group greater than that of a nitro group?

The short answer is that $\ce{NH3+}$ is a cation, and any cation will always draw electrons towards it stronger than any neutral species.

The electronegativity of the species involved is only a secondary cause. However, one can safely say that $\ce{NH2}$, if nitrogen’s lone pair is (somehow) impeded from excersizing it’s $+\ce{M}$ effect, would have a lesser $-\ce{I}$ effect on the whatever than $\ce{NO2}$ would.


How diphenylcarbazide is used to detect Mercury II ions?

The Diphenylcarbazide Test for the presence of mercury was used as early as 1920s. However, since the test is very delicate (the test can detect $\pu{250 \mu g/L}$ mercury in solution), it has been used predominantly in neutral solutions, until it have been adapted to Quantitative Analysis Scheme by Scott in 1929 (Ref.1) to ditect in acidic medium:

  The mercury is precipitated in Group I is dissolved in aqua regia, that of Group II is dissoved in hydrochloric acid and sodium chlorate. In either case, the solution is evaporate to a volume of about $\pu{1 cc.}$, placed in a test tube, and diluted with $5$ to $\pu{6 cc.}$ of water. Four to eight drops of a saturated alcoholic solution of diphenylcarbazide is added and a large excess of solid sodium carbonate is gradually dropped into the solution. When the mercury is present, the form produced on neutralization assumes a blue tinge, and after the addition of an excess of carbonate the entire solution turns blue.

The author stated at the end that neither sodium acetate nor sodium bicarbonate can be successfully substituted for the sodium carbonate, indicating the sensitivity of the test for $\mathrm{pH}$. It is worth noting that the test outlined here (vide supra) has been successfully used in Quantitative Analysis Courses in University of Georgia, Athens, GA for two years before publication.

The major drawback of the test is the deterioration of stock solution of diphenylcarbazide in different solvents. Thus, it needs to be freshly prepared. To avoid this problem, a solution has been found in the publication by Urone by 1955 (Ref.2). Accordingly, non-aqueous ethyl acetate and acetone were the better solvents, diphenylcarbazide solutions of which are stable for months. Diphenylcarbazide solutions of Methyl ethyl ketone, methyl cellosolve (2-methoxyethanol), and 2-propanol are usable for 1-2 weeks. Aqueous solutions and solvents tending to be basic such as methanol and ethanol, and those containing traces of water and basic impurities, do not make good solvents for stock solutions of the colorimetric reagent (Ref.2).

Notes: (1) All solutions tested (diphenylcarbazide solutions in seven different solvents) in Ref.2 are $0.25\%$ in diphenylcarbazide. (2) Also, you don't have to synthesize 1,5-diphenylcarbazide; it is commercially available in Millipore-Sigma. (3) I don't think anybody can answer your question about "how are the test strips for heavy metals in drinking water made from this compound," because it should be a corporate secret, similar to $\ce{KI}$ papers for oxidant check.

References:

Alfred W. Scott, “Adaptation of the Diphenylcarbazide Test for Mercury to the Scheme of Qualitative Analysis,” J. Am. Chem. Soc. 1929, 51(11), 3351-3352 (https://doi.org/10.1021/ja01386a503).
P. F. Urone, “Stability of Colorimetric Reagent for Chromium, s-Diphenylcarbazide, in Various Solvents,” Anal. Chem. 1955, 27(8), 1354-1355 (https://doi.org/10.1021/ac60104a048).


Canonical Restricted Open-shell Hartree–Fock equation

The treatment found in Szabo-Ostlund, although perhaps it is not explicitly stated, is appropriate only for non-degenerate spin-singlet electronic states, which in the Hartree-Fock method correspond to closed-shell systems.
For open-shell systems one may use the restricted open-shell Hartree-Fock method (ROHF), which is somewhat more complicated. In particular, in the ROHF method six arbitrary constants have to be introduced. Many choices for these parameters have been suggested.
The total ROHF energy, as well as the overall ROHF wave function and therefore the expectation value of any operator, are independent of these six parameters, so that all these ROHF sub-variants are largely equivalent.
However, the six parameters affect the Hartree-Fock orbitals and their orbital energies. Also, approximate post-HF methods such as MP2 or CISD will depend on the particular choice of those six parameters.
Knowles et al suggested in 1991 a particular choice leading to called "semi-canonical orbitals" whose orbital energies do obey Koopmans theorem and which are also a good choice for MP2 calculations.
I suggest the following paper for an introduction to the problem (the paper by Knowles metioned above is ref. 18):
T. Tsuchimochi and G. E. Scuseria, ROHF Theory Made Simple, J. Chem. Phys. 133, 141102 (2010) https://arxiv.org/abs/1008.1607
The paper above shows that one can obtain semi-canonical ROHF orbitals by a minor modification of the unrestricted Hartree-Fock method and that this approach is numerically very stable.


Energy of photons emitted by de-excitation of electrons

When an electron of an atom returns from an excited state to the ground state, it emits energy in the form of a photon.

That is one of several pathways for deexcitation. Often, internal conversion, the radiationless deeexcitation happens instead.

  How does the change in energy level compare to the energy of the emitted photon?

It depends on the relaxed geometry of the excited state, as compared to the ground state.

Remember that the excitation happens on a femtosecond scale. That is amazingly fast, light travels a distance of just $\mathrm{0.3\,\mu m}$ in $\mathrm{1\,fs}$.

Since it takes longer than some femtoseconds to move the much heavier nuclei, the molecule in its electronically excited state still has the geometry of the ground state. This can be far from the lowest vibrational level (the relaxed geometry) of the excited state.

Before emission of a photon occurs, ectronically excited molecules relax and adopt their lowest vibrational state.

This means that the energy of the emitted photon often is (much) lower than the excitation energy. This effect is known as Stokes shift.


Rank the following radicals in order of decreasing stability

Here, 1 and 3 are aromatic since they follow Huckels rule(Or am I wrong here?

Yes you're wrong over here, for applying the Huckel's rule the compound must be a closed circuit of delocalized electrons. In both the cases the circuits are not closed

If you're not aware of Conjugation then give that a read .
Now
The most stable compound is 3 since the extent of conjugation is the greatest i.e 2 pi bonds.

3 is more stable than 4. Even though 4 is aromatic it is an aryl free radical the p orbital containing the radical is out of the plane and can't be stabilized through resonance/Hyperconjugation.
In fact I would say that option 4 is the least stable compound .

The second most stable compound is is 2 since it is stabilized via resonance which is more effective than Hyperconjugation; 1 is stabilized by Hyperconjugation.

3 is more stable than 2 since the extent of conjugation is greater.

The order of stability will be
$$\mathrm{3 > 2 > 1 > 4}$$


What metal is under the stainless layer when it wears of?

1 =no , 2= no, 3 Likely brass , 4= no. The buckle is likely a brass although the bottom edge where you polished it is copper color . That could result if you used some cleaning agent. It could be made from copper but modern mass production would use brass. It was plated with nickle and / or chromium. Nickel plating is usually put down first , under chromium plating.


Basicities of meta-nitrophenoxide and para-nitrophenoxide

Yes, it is. These resonance structures describe basicity in para-nitrophenoxide.


Wikipedia states that upon heating, copper(II) acetate reacts with metallic copper to make copper(I) acetate, which would be colorless.  While this is no guarantee of what happened with your solution, you can mix a sample of the solution with saltwater (chloride ions).  White precipitate = copper(I).

In light of a comment, I am compelled to expand on the possibility of copper(I) in solution.  According to the Wiley Online Library, this salt is soluble in water with hydrolysis to cuprous oxide.  At the same time we have the copper Pourbaix diagram reported from Ref. 1, indicating that the oxide is stable only above a pH of 4.

This means any significant leftover acetic acid could prevent the hydrolysis and keep the copper(I) in solution, apparently stabilized by complexing with the acetate ion given that disproportionation is not reported.  (Such a stabilization is not included in the above diagram but would contract the Cu2O region further.)

Since this is only indirect evidence, again testing for a copper(I) chloride precipitate by mixing a sample with saltwater is recommended.

Reference

1.
Oh, Youn-Jin & Park, Gyung-Soon & Chung, Chan-Hwa. (2006). Planarization of Copper Layer for Damascene Interconnection by Electrochemical Polishing in Alkali-Based Solution. Journal of The Electrochemical Society - J ELECTROCHEM SOC. 153. 10.1149/1.2200288.


Decomposition products of hexaurea-chromic-chloride-trihydrate

Using the article Klaus Warzecha linked to and some trial-and-error, a possible decomposition could look like this and is in accordance with the TG diagram in the question:

Decomposition of $\ce{3 H2O}$ and $\ce{2 CO(NH2)2}$:
$\ce{[Cr(CO(NH2)2)6]Cl3 $\cdot$ 3 H2O -> [Cr(CO(NH2)2)4]Cl3 + 3 H2O \uparrow + 2 CO(NH2)2 \uparrow}$
Decomposition of $\ce{2 CO(NH2)2}$ and $\ce{2 HCl}$:
$\ce{[Cr(CO(NH2)2)4]Cl3 -> CrH6C2N4ClO2 + 2 CO(NH2)2 \uparrow + 2 HCl \uparrow}$
Decomposition of $\ce{2 CO2}$, $\ce{N2}$, $\ce{3 H2}$ and $\ce{1/2 Cl2}$ and oxidation of $\ce{Cr}$ to form $\ce{CrN2}$:
$\ce{CrH6C2N4ClO2 -> CrN2 + 2 CO2 \uparrow + N2 \uparrow + 3 H2 \uparrow + 1/2 Cl2 \uparrow}$

Given the temperature range and pressure (probably standard pressure), the formation of $\ce{CrO2}$ is unlikely.


Yes, the mechanism is correct and is different from the normal mechanism because here the more stable tertiary carbocation formation.


How do I decide the electromeric effect on pent-2-ene?

Here what you should consider first is hyperconjugation, rather than inductive effect (a brief explanation by me).

You're right that $\ce{Nu-}$ will either attack on $\ce{C-3}$ or $\ce{C-2}$, and hence two carbocations would be formed. Now, we'll analyse the stability of these carbocations on the basis of hyperconjugation which is as follows,

carbocation at $\ce{C-2}$ will have a total of 5 HC structures
carbocation at $\ce{C-3}$ will have a total of 4 HC structures

Based on the above difference, it is easy to conclude that $\ce{Nu-}$ will attack on $\ce{C-2}$ carbon. Whereas based on inductive effect, those carbocations are nearly equally stable.

Note: HC is short for "hyperconjugative structures".


Why do atoms on the periodic table have the number of valence electrons that they do?

Atoms have protons (and neutrons) in the nucleus that define the element. For example, Fluorine always nine protons. The number of electrons must balance the number of protons in the atomic state. If fluorine had ten electrons, i.e. a full inner and outer shell, it would be a negative ion with net charge -1, so the electrostatic charge would repel the last electron.
It is only with the help of other atoms nearby that charges can be (almost) balanced. In $\ce{LiF}$, the fairly positively charged $\ce{Li+}$ ion attracts the comparatively negatively charged $\ce{F-}$ ion, so there is no net unbalance in a crystal of $\ce{LiF}$, though in each pair, the valence electron spends most time in the outer shell of fluorine.


Main reason is behind is the reaction mechanism involved. $\ce{HBr}$ or other halides usually react with alkenes through an electrophilic addition. Due to the weak $\ce{O-O}$ bond, peroxides act as a source of free radicals. Thus in the presence of peroxides, a free radical chain pathway is followed.

The other hydrogen halides are reluctant to this reaction, because the chain reaction can't proceed due to energy requirements for the generation of the corresponding halogen free radicals.


Why shouldn't lithium batteries be discharged below 50%?

There are two main effects that cause damage to a Li-ion battery due to "deep cycling":

During the charge/discharge cycle, lithium ions are intercalated within the electrodes. This causes physical stress to the electrode material and spalling of their coatings. Deeper discharge causes increased stress. This effect is made worse with the temperature range from heavy current use as in a laptop.
Lithium metal may accumulate on the anode, taking it out of play. This effect may be due more to over-charging than to deep-discharge, though.


Can an SN1 reaction occur in compounds with the leaving group attached to a primary carbon, which can later rearrange?

Your thinking is correct that they can rearrange and then form stable carbocation (though your first example is wrong as it would become anti-aromatic after rearrangement) ,
but the answer to your question is NO , they would be substituted by SN2 mechanism being
1 degree . The reason behind this is that the compound will not try to form a carbocation first where it is not that stable , you can think of it in this way that at the initial stage the compound doesn't know that it can rearrange later to form stable carbocation, this is something that we can see but not the compound and thus it will go by normal SN2 mechanism.


Methods to increase the speed of dissolving potassium sulfate in water apart from heating, grinding or shaking

Since you're really asking about speeding dissolution, not increasing solubility, use a mixer; either:

From above, e.g., a drill attachment for mixing paint, or
From below, e.g., a magnetic mixer/stirrer.

The links are just as examples; you could devise your own mixer.


Polluting agents generated by road traffic in a city

You might also want consider the metals which appear in road dust, I recall having seen some work on neutron activation anaylsis of dusts collected near roads. The rubber tyres are likely to contain some metals. Cars will be likely to release metals in other ways. I would consider the dust from the operation of disk brakes, the wear on tyres, fuel additives and also corrosion of bodywork and engine parts. I know that zinc and cadmium plated parts can release both ionic and particulate forms of these metals.

One paper on the subject was by Siddique et. al. im Radiochimica Acta, 2013, volume 101, pages 273-277.

It is interesting that lead levels in the air have fallen since leaded petrol was banned in the UK, but some workers have reported finding platinum group metals in the air near to roads.

Using ICPMS Simitchiev et. al, in TALANTA, 2008, volume 77, pages 889-896 found Pt, Rh and Pd in road dust.


What does the term "simple" in the phrase "simple whole number ratio" mean in Gay Lussac's law?

3/2 is simple. 45498334/9232493 is not.

The reason it works is because reactions of gases producing gases involve relatively simple compounds. If they were complicated, the molecular weight would likely be high enough that the substance would not be a gas.

This is akin to Loong's comment about the Law of Multiple Proportions, which also mentions "small whole number ratios." The difference is that for the Law of Multiple Proportions, "small" ratios doesn't apply in the case of something like palytoxin ($\ce{C129H223N3O54}$).

But palytoxin isn't a gas. For the reason mentioned above, you're almost certain to get simple coefficients for gas to gas reactions.


3rd period heteroatoms can stabilize adjacent carbanions charges via hyperconjugation.

Consider ylides used for Corey,-Chaykovsky, Peterson, and Wittig olefinations. Granted those have formally positive sulfur, silicon, and phosphorus, but you will not see similar easy deprotonation with tetraalkylammonium for example.

The reactions of silicon-stabilized carbanions are described more fully in Clayden 1st ed., pp 1304–5 (this material seems to have been cut from the 2nd edition).


Can a stoichiometric mixture of oxygen and methane exist as a liquid at standard pressure and some (low) temperature?

There's a NASA report that looks into this: "ON THE SOLUBILITIES AND RATES OF SOLUTION OF GASES IN LIQUID METHANE", Hibbard and Evans, 1968 and concludes that such mixtures are possible.

Starting on page 8:

  Figure 5(a) presents the curves for oxygen, argon, carbon monoxide,
  and nitrogen. Also shown are the two experimental values for nitrogen.
  Agreement is excellent at 99.83K and good at 110.9K. The curves for
  these gases show that solubility should decrease with increasing
  temperature and the nitrogen data confirm this. This figure shows the
  mole fraction solubility of oxygen to be 1.0 at 90K. This means that
  oxygen, which has a normal boiling temperature of 90.1K would
  continuously condense in, and be miscible in all proportions, with
  liquid methane at 90K. This is confirmed by reference 11 where, in a
  study of the solubility of methane in liquid oxygen, it was concluded
  that these formed a near-ideal solution at -297 F (90K)

(emphasis added) Reference 11 mentioned in there is "Hydrocarbon-Oxygen Systems Solubility", McKinley and Wang, 1960 (unfortunately paywalled) which also has interesting discussion of the stability (i.e. presence or absence of a tendency to explode) of various mixtures.

Figure 5 is reproduced below. Note how the solubility of oxygen rises rapidly as temperature drops.


Are all gases equal in their ability to carry with them lower volatility matter at a given temperature?

It depends on how a solvent is defined. Regardless of semantics, supercritical steam 217.75 atm, 373.946 °C is a very good solvent. It will dissolve fats, hydrocarbons or plastics. It also becomes very corrosive to metals.


Does sodium borohydride reduce C=N group?

Waylander's comments answer the title question as well as I could.  Sodium borohydride, and even modified versions where some of the hydride ligands are replaced by others like $\ce{NaBH3CN}$, can reduce imines.

But, if we have only a limited amount of borohydride we will likely see the carboxyl group primarily reduced, as it has a more polar pi bond and thus it is a more reactive electrophile.  So you may in principle get mostly 1 with limited sodium borohydride, or 4 if you have enough sodium borohydride to attack both the carboxyl and less reactive imine functions.

The carbon-carbon bond here, not conjugated to either electronegatove function, is interactive to borohydride and thus products 2 and 3 are not reached.


How do electrostatic analysers act as kinetic energy analysers?

The point is that all ions produced in the ion source will not have a kinetic energy outside of the acceleration region equal to $q V$. When produced the ions will most probably be produced at (slightly) different positions in the acceleration field, with a small $\Delta V$ potential deviation from the average acceleration potential, and with a thermal kinetic energy distribution ($E_{kin,thermal}$).

In these conditions, the real kinetic energy of the ions as they exit the ion source will be slightly different:

$$E_{kin,real} = q V + q \Delta V + E_{kin,thermal}$$

and the purpose of the electrostatic analyser is to separate the ions depending on their actual kinetic energy and thus compensate for the distribution of kinetic energies of the ions exiting the ion source.


Solving a quadratic equation for pressure of real gas with unknown volume

Assuming the Van der Waals equation with $b=0$, I agree with the solution to the equation:

$$pV_\mathrm{m}^2 - 24V_\mathrm{m} + 2 = 0\label{eqn:1}\tag{1}$$

I'll point out that I'm following the notation of the given solution, but $V_\mathrm{m}$ seems odd to me. I'd think that $V_\mathrm{m}$ would be reserved to mean the molar volume at STP.

But equation \eqref{eqn:1} has two unknowns, $p$ and $V_\mathrm{m}$. Letting $a' = p$, $b' = -24$, and $c' = 2$ then the appropriate quadratic equation is of course

\begin{alignat}
2V_\mathrm{m} &= \dfrac{-b' \pm\sqrt{b'^2 - 4a'c'}}{2a'}\tag{2}\\
V_\mathrm{m} &= \dfrac{-(-24) \pm\sqrt{(-24)^2 - 4×p×2}}{2p}\tag{3}\\
V_\mathrm{m} &= \dfrac{24 \pm\sqrt{(-24)^2 - 8p}}{2p}\tag{4}\\
V_\mathrm{m} &= \dfrac{12 \pm\sqrt{144 - 2p}}{p}\label{eqn:5}\tag{5}
\end{alignat}

So we have two unknowns and one equation. Thus the problem is unsolvable without additional information. Notice that would still be the case if the simple ideal gas equation, $PV=nRT$, had been used. The ideal gas equation could be solved only to $pV_\mathrm{m}=24$.

For equation \eqref{eqn:5} there are three cases for the sqrt term $\sqrt{144 - 2p}$:

The term is negative.

If $144 - 2p$ is negative, then both roots would be imaginary.
The term is equal to zero.

We have $\sqrt{144 - 2p} = 0\implies 144 - 2p = 0,$ so $p = 72$. Therefore, $V_\mathrm{m} = 12/72 = 0.167.$
The term is greater than zero.

Then $72 > p$ also, and mathematically $V_\mathrm{m}$ has two roots.

Now, again there are three possibilities.

(a) $\sqrt{144 - 2p} < 12$

This is impossible since $V_\mathrm{m}$ would have two valid values.

(b) $\sqrt{144 - 2p} = 12$

This is impossible since $V_\mathrm{m} = 0$ and the gas volume can't be $0$.

(c) $\sqrt{144 - 2p} > 12$

This would yield one negative value and one positive value for $V_\mathrm{m}$, which is OK for $V_\mathrm{m}$. However it also means that $p < 0,$ which is nonsensical.

Thus the only reasonable solution is if the square root term is equal to zero.

P.S. The comment by user Poutnik made me look at this again, and I now understand the book's solution…


How are these two organic resonance structures different?

Resonance structures do not describe the conformation (=energy state) of a molecule, but are didactic. They show how you can just flip schemactic $\pi$ electrons around the structure, without breaking any connection. The bond lenghts and angles are constant in the actual molecule, at values somewhere between those indicated by the resonance structures.

Between resonance structures, you don't look at a possible symmetry of the location of atoms, but regard them as unique and fixed (see also https://en.wikipedia.org/wiki/Born-Oppenheimer_approximation). Resonance structures can have different energy or be symmetric (=conformationally identical) to each other. If one resonance structure has higher energy than the others, that is just a hint that the actual structure might by more closely described by the other resonance structures.


Do almonds smell like they do because of cyanide?

Almonds smell like they do mostly due to the presence of benzaldehyde:

  This colorless liquid has a characteristic almond-like odor. Benzaldehyde is the primary component of bitter almond oil and can be extracted from a number of other natural sources.

Hydrogen cyanide also has an almond-like odor, but it is not as pronounced as that of benzaldehyde. From CDC|Facts About Cyanide:

  Cyanide sometimes is described as having a “bitter almond” smell, but it does not always give off an odor, and not everyone can detect this odor.

Benzaldehyde and hydrogen cyanide are both byproducts of the enzymatic catalysis of amygdalin, which is present in almonds (and other foods). Have a look at this abstract (Toomey VM, Nickum EA, Flurer CL. Cyanide and amygdalin as indicators of the presence of bitter almonds in imported raw almonds. J Forensic Sci. 2012 Sep 57 (5): 1313-7) for more about the cyanide-amygdalin-almond connection.

In terms of safety:

  For a 70-kg human, the lethal dose is estimated at 50 mL. Benzaldehyde has been classified as a hazardous substance by the United States Environmental Protection Agency.

and

  Benzaldehyde does not accumulate in any specific tissues. It is metabolized and then excreted in urine.

Safety issues from benzaldehyde consumption $\it{via}$ the almond consumption route appear to be minimal.

From our sister SE site Skeptics in re: "How many almonds does it take to get cyanide poisoning?":

  With bitter almonds, 8 - 32 almonds will give you the lethal dosage of cyanide. Bitter almonds yield about 6.2 mg of cyanide per almond and the LD50 for cyanide is 50 mg - 200 mg.

  This applies only to bitter wild almonds: considering that you're not dead, you most likely ate domesticated sweet almonds, which apparently do not have this problem. The sale of wild almonds may be illegal, in fact, though I can only find blogs and internet comments alluding to this.

  From wikipedia

  Wild almonds are bitter, the kernel produces deadly cyanide upon mechanical handling, and eating even a few dozen at one sitting can be fatal.
  ...

  While wild almond species are toxic, domesticated almonds are not

Given the evidence and the expectation that you won't ingest several kilograms of almonds in a day or so, I'm comfortable in endorsing the almond as safe for consumption.


Finding root-mean-square speed of neon using kinetic energy

As hinted in the comments by @AngusTheMan we use the equation
$$\mathrm{KE_{avg}} = \frac{3}{2}kT,$$
but ultimately we are looking for the $v_\mathrm{rms}$, so we'll also use
$$v_\mathrm{rms} = \sqrt{\frac{3RT}{M}}.$$
You've been given your $\mathrm{KE_{avg}},$ $R$ and $k$ are constants, and $M$ is the mass of 1 mole of $\ce{Ne}$. We can set all of these equal and the solution will come from
$$v_\mathrm{rms} = \sqrt{\frac{3 \cdot R\cdot\mathrm{KE_{avg}}\cdot\frac{2}{3k}}{M}}.$$


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


Can 32% hydrochloric acid be stored in a milk jug (HDPE)?

There is more than just material composition to container selection. Yes HDPE is compatible with hydrochloric acid, but you have to consider rigidity and failure probability. Think: would you want to store Concentrated Hydrochloric acid in a 2 mil bag? Probably not because the probability of puncture and resulting loss/cost from the spill would be sufficiently high. Though not as extreme your milk jug is along those lines as it is softer and less rigid as the original container.

There is also the issue of legal considerations. In the US I'm sure that milk jug would not meet the DOT requirements for transport, and OSHA prohibits using containers typically used for food as being used for chemical or waste storage.

That said for home use only, you will probably be okay to store it, certainly not recommended, but this is the real world. I would at least recommend that if you do use a milk jug, you at least use a basin that can hold all of the contents for secondary containment in case of a leak which milk jugs occasionally do  AND certainly label the contents.


Why does the first lithium stationary state have n = 9/4?

I think your calculation is correct. There is an error in this textbook.  All you have to do is to switch  between $\ce{Li^{2+}}$ and  $\ce{He^{+}}$. And 'll prove my point:

$\ce{Li^{2+}}$ and  $\ce{He^{+}}$ are hydrogen-like atoms.

For these species, the ionization energy is given by the equation:
$$E_\mathrm{i}=E_\mathrm{i}(\ce{H})\times Z^2$$

For $\ce{Li^{2+}}$:
\begin{align}
E_\mathrm{i}(\ce{Li^{2+}})&=E_\mathrm{i}(\ce{H})\times Z^2\\
19.6 \times 10^{-18}&=E_\mathrm{i}(\ce{H})\times 3^2\\
E_\mathrm{i}(\ce{H})&=2.18 \times 10^{-18}  \,\mathrm {J/atom}\\
E_\mathrm{i}(\ce{H})&=\frac{2.18 \times 10^{-18}}{1.6 \times 10^{-19}}\\
  &= 13.6 \,\mathrm{eV}
\end{align}

$13.6 \,\mathrm{eV}$ is well-known to be the first ionization energy of hydrogen. So, my assumption is true.
To solve the problem after correction:
 The ionization energy of $\ce{He^{+}}$ is given by the equation:
\begin{align}
E_\mathrm{i}(\ce{He^{+}})&=\frac{E_\mathrm{i}(\ce{Li^{2+})}\times Z^2(\ce{He^{+})}}{Z^2(\ce{Li^{2+})}}\\
E_\mathrm{i}(\ce{He^{+}})&=8.72\times 10^{-18}\,\mathrm {J/atom}
\end{align}
The energy of the first stationary state:
$$E_\mathrm{i,1}=E_\infty-E_1$$
But, $E_\infty=0$, we find:
$$E_1=-E_\mathrm{i,1}=-8.72\times 10^{-18}\,\mathrm {J/atom}$$


Sodium Sulfate or Sodium Carbonate for electrolysis in water

If you electrolyze a solution of sodium carbonate, it will be fine at the beginning. Later on, some problem may happen around the anode where $\ce{O2}$ is released. Indeed some acid ($\ce{H+}$) is produced in solution according to the half-equation : $$\ce{2H2O -> 4 H+ + O2 + 4 e-}$$ and these acidic ions $\ce{H+}$  react with the carbonate ions $\ce{CO3^{2-}}$according to $$\ce{CO3^{2-} + H+ -> HCO3^- + H2O}$$ In presence of the ion sodium $\ce{Na+}$, the ion hydrogenocarbonate $\ce{HCO3^-}$ reacts to produce sodium hydrogenocarbonate $\ce{NaHCO3}$ which is not very soluble in water, and may make a white deposit around the anode $$\ce{Na+ + HCO3^- -> NaHCO3(s)}$$ Furthermore if the electrolysis is continued a long time, a supplement of acid ions will react with these $\ce{HCO3^-}$ ions producing some $\ce{CO2}$ by the reaction  $$\ce{H+ + HCO3- -> CO2 + H2O}$$
$\ce{CO2}$ is a gas, and it is relatively soluble in water. In the beginning the $\ce{CO2}$ produced will stay dissolved in the liquid. But on the long run, $\ce{CO2}$ will get out of the solution and get mixed with the oxygen produced at this electrode. So instead of getting pure $\ce{O2}$ at the anode, you will obtain a mixture $\ce{O2 + CO2}$.
So you better use a solution of sodium sulfate $\ce{Na2SO4}$ instead of sodium carbonate. The above mentioned drawbacks do not occur with sodium sulfate.
In a $\ce{Na2SO4}$ solution, the ions $\ce{Na+}$ are moving towards the negative electrode (cathode), and the ions $\ce{SO4^{2-}}$ are moving toward the positive electrode (anode). They move but they don't touch their electrodes and they are not discharged. Instead that is water that either give or absorb electrons.
At the negative electrode (cathode) where $\ce{Na+}$ ions are approaching, the following reaction occurs :  $$\ce{2 H2O + 2 e- -> H2 + 2 OH-}$$ and the $\ce{OH-}$ ions neutralize the approaching $\ce{Na+}$ ions. Of course some $\ce{H2}$ bubbles are appearing on the cathode.
At the positive electrode (anode), the following reaction happens $$\ce{2 H2O ->  O2 + 4 H+ + 4 e-}$$ and the $\ce{H+}$ ions neutralize the approaching $\ce{SO4^{2-}}$ ions. Of course some $\ce{O2}$ bubbles are produced on the anode. Half as much as $\ce{H2}$, because $4$ e- produce $1$ $\ce{O2}$ and $2 \ce{H2}$
To summarize, electrolyzing a $\ce{Na2SO4}$ aqueous solution will produce $\ce{H2}$ and a solution of $\ce{NaOH}$ at the cathode. And it will produce $\ce{O2}$ plus a solution of $\ce{H2SO4}$ at the cathode.
Now if the anodic solution is stirred and mixed with the cathodic solution, the following reaction happens $$\ce{H2SO4 + 2 NaOH -> Na2SO4 + 2 H2O}$$ As a consequence, $\ce{Na2SO4}$ is regenerated, and the only chemical that is consumed in the whole process is water. The only product getting out of the whole process is the formation of $\ce{H2}$ at the cathode and $\ce{O2}$ at the anode.


Let us start with the following most important part of the IUPAC definition for the bond energy.

  bond energy (mean bond energy)

  The average value of the gas-phase bond dissociation energies (usually
  at a temperature of 298 K) for all bonds of the same type within the
  same chemical species.

I think it should be immediately clear now that the higher a bond energy is, the stronger is the bond, so that there is no any contradiction in a bond energy increase with the increase of the $\mathrm{s}$-character in hybrid $\mathrm{sp}^n$-orbitals.

Indeed, $\mathrm{2s}$-orbitals are lower in energy that the $\mathrm{2p}$ ones, and thus, the more $\mathrm{s}$-character a hybrid orbital has, the lower is its energy: $E_{\mathrm{sp}} < E_{\mathrm{sp^2}} < E_{\mathrm{sp^3}}$. The lower an atomic or hybrid orbital energy is, the lower is the energy of a molecular orbital in formation of which it participates, $E_{\sigma_\mathrm{sp}} < E_{\sigma_\mathrm{sp^2}} < E_{\sigma_\mathrm{sp^3}}$, and consequently, the higher is the corresponding bond dissociation energy, as well as, the average of them, i.e. the bond energy. So, the bond energy indeed increases as the s-character grows, say $E_{\ce{C(\mathrm{sp})-C(\mathrm{sp})}} > E_{\ce{C(\mathrm{sp^2})-C(\mathrm{sp^2})}} > E_{\ce{C(\mathrm{sp^3})-C(\mathrm{sp^3})}}$.

Below is the MO diagram for $\sigma$ $\ce{C-C}$ bond between two $\mathrm{sp}$-, two $\mathrm{sp^2}$-, and two $\mathrm{sp^3}$-hydridized $\ce{C}$ atoms that illustrates the logic above.


At equilibrium, the mass of reactants and products are equal. Does that mean the reaction stops?

At equilibrium, the reaction continues, but the rate of the forward reaction is equal to the rate of the backward reaction.

Let's consider the first order reaction $\ce{A <=>[k_f][k_b] B}$

So, the rate of the forward reaction is $k_f[A]$, and the rate of the backward reaction is $k_b[B]$. Initially, $[B]=0$, and $[A]$ is high, so the forward reaction is very fast, and the backwards reaction doesn't occur at all. As time passes, $[A]$ decreases and $[B]$ increases, so the forward reaction becomes slower and slower, and the backwards reaction speeds up. At one point, both reactions have equal rates ($k_f[A]=k_b[B]$). Here, both reactions take place, but for every time when an $A$ becomes $B$, another $B$ becomes $A$. The reaction is taking place, just that you don;t notice it since the net effect is zero.

Here's a diagram for how the rates of the reaction proceed for the formation of ammonia. Note that the rates never reach 0:


Can gold cause tin or other metals to corrode?

First off, read carefully:

  Different metals will increase the likelihood of corrosion.

It doesn't say:

  Different metals will increase the likelihood of corrosion of gold

With that in mind, it's called galvanic corrosion and occurs when contact between two (or more) dissimilar metals causes the more reactive one to corrode when it normally would not have or faster than it would on its own. This action is similar to a galvanic pile (though its only one cell) that has short circuited thus corroding the more reactive metal.

In your case it is definately the tin that will be corroded, not the gold as tin is more reactive.


Why can't the electrophile attack the chlorine atom in a halobenzene?

Given the charge distribution and the shape of the HOMO, I would be very surprised if an electrophile does not attack the chlorine. This is from a calculation on the DF-BP86/def2SVP level of theory.

If you account for the hydrogens effectively shielding the negative charges of the carbons, the only place left to go is the chlorine. (Natural Population Analysis, NBO 6.0)

The largest coefficient of the HOMO is at the chlorine (C1 ipso, C4 para):

Alpha occ 29 OE=-0.224 is Cl12-p=0.29 C4-p=0.24 C1-p=0.19 C2-p=0.09 C6-p=0.09

Generally I would expect it to be rather unreactive towards electrophiles.


Why are all the phenyl protons in benzyl alcohol equivalent in the ¹H-NMR spectrum?

You're right, those five hydrogens are not magnetically equivalent. You would expect to see three signals for the aromatic protons.

If you look closer you'll see that the aromatic signal is not a perfect singlet, but is rather more complicated with an additional small peak and a broader base, it is a multiplet. You have three distinct signals, but they are so close in chemical shift that they overlap and result in this multiplet.


Why is a ketone more nucleophilic than an ester?

The ester carbonyl carbon is a stronger nucleophile and less prone to nucleophilic attack than the carbonyl carbon in a ketone. I think you are trying to understand why the carbonyl in a ketone typically reacts faster with a nucleophile than the carbonyl in an ester.  Look at the resonance structures drawn below.  Both the ketone and ester have a resonance structure that places positive charge on the carbonyl carbon, however you can also draw a third resonance structure for the ester that removes some of the positive charge from the carbonyl carbon and places it on the ester oxygen. This makes the ester carbonyl carbon less positive (more negative, more nucleophilic, less prone to attack by a nucleophile) than the ketone carbonyl carbon (more positive, less nucleophilic, more prone to attack by a nucleophile).  A nucleophile has electrons it would like to share and will react preferentially with centers of low electron density (more positively charged).  Therefor a nucleophile will react faster with a ketone carbonyl than an ester carbonyl.

EDIT

I like Martin's posts, they make me think - and that is always a good thing. Here is where my thinking has taken me:

I believe Martin's approach (basically frontier MO) is fundamentally correct.  I would be more convinced that his actual numbers are meaningful if he were to calculate 1) the carbonyl carbon LUMO coefficient and 2) the HOMO-LUMO separation for the following series of compounds in order to see if his calculated values correlated with this order of reactivity of carbonyl compounds towards nucleophilic attack (we could use water as the nucleophile) - but Martin, I think such calculations might be a poor use of your time and resources.

acyl halide > acid anhydride > aldehyde > ketone > ester ~ carboxylic acid > amide > carboxylate ion

My argument based on how resonance affects the charge density at the carbonyl carbon is one approach used to explain ("teach") the above order of reactivity.  See for example page 4 on these UCLA notes.  I was just trying to keep it simple.
The issue I have with Martin's approach is, what should the reader do next week when he wants to know if an acid chloride reacts faster or slower than an ester.  While the resonance approach may not be rigorous, it does yield the correct order of reactivity and allows the user to make predictions while only using pencil and paper.
I am considering further editing of this post to use the resonance approach to explain differences in ground state stabilization of the carbonyl series.  Then, since they all go (more or less) to the same tetrahedral intermediate, one could explain the reactivity order based on differences in Eact.  This is another textbook method used to explain the carbonyl reactivity pattern.
Finally, what is the "correct" answer?  Is it the real truth as nature knows it, but in a form that is difficult for the common individual to use and extend; or is it a rough approximation of the truth that can be easily used?  I think the answer to that question depends upon who is asking the question, who is the audience.


Basicity of nitrogen heterocyclic compounds

Nitrogen inversion is not relevant to this discussion. A faster rate of inversion does not correspond to a greater delocalisation in the electronic structure, as lone pair inversion is not equivalent to resonance. The lone pair does not exist as a delocalised smear; it exists in two different localised forms.

The PDF on nitrogen inversion in the comments does not even bring up the topic of basicity, and quite rightly so.

Inductive effects explain why morpholine 2 is less basic, and resonance effects explain why pyrrolidone 4 is even less basic. However, since the pKaH's of piperidine 1 and pyrrolidine 3 are so similar (11.22 and 11.27 respectively), I think it is pointless to try and qualitatively rationalise a difference in their basicities. It is like trying to explain why the temperature in Spain one day is 11.22 °C and the temperature the next day is 11.27 °C. The relative basicity could well be different if you switch the solvent, which would invalidate any reasoning based on - for example - s-orbital character of the nitrogen lone pair, which I think is the only plausible qualitative argument here.

Simply looking at the data, though, the order you want is 4 < 2 < 1 < 3.


Separation of solvent from residue of material.

Dichlorethane has a boiling point of about 84°C - If I would have to separate EDC from a product / residue, I would use a rotovap with water bath configuration. This distillation is a very gentle way to separate solvents from product / residue.

Set the temperature of the water bath slightly higher than the boiling point of the EDC.

If you want to get fancy to make your solvent super pure again after the rotovap distillation, you could perform a fractional distillation on it - however in most cases I think this might not be necessary.


Why does Mg2+ have the ionic radii bigger than Ca2+

Source: http://upload.wikimedia.org/wikipedia/commons/thumb/9/9f/Atomic_%26_ionic_radii.svg/551px-Atomic_%26_ionic_radii.svg.png

Ca2+ is bigger than Mg2+


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

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