Acidity/basicity of oxides across the periodic table

It is one of the characteristics of the metals as for electrical conductivity, heat conductivity, mechanical properties. All these basically originate from the outer electrons to be "loosely" bound to the nucleus, when compared to non metals. Bonds in metal oxides are ionic, with positive charge on the metal. The same is for their hydroxides, i.e. their compounds containing an OH group. This leads to their basic dissociation in water  giving the stable hydroxide ion.

Contrary non metals have tightly bound electrons and are prone to fill their outer shell (Octet rule). On reaction with a non metal as oxygen the situation is therefore a bit more complicated depending on what specific element the non metal is. Non metals oxides are essentially molecular, although the covalent bond is polar to various extents. However, their compounds containing an OH group will retain electrons on the fragment consisting of the two non metals (of which, one is oxygen of course) leading to acid reaction, i.e. H+

(Note that HClO would be better wrote as ClOH).

It should be clear that things are going as ionisation potential - electron affinity - atom size (distance of the outer electrons from the nucleus). For a better explanation I would need more time, it is basically a big chapter of introductory general chemistry, which goes under the name of periodic law and properties of the elements, or something like that.


The Joule-Thomson experiments occurs with no change in enthalpy.

Suppose that at the left of a porous plug there is a pressure $p_1$ and temperature $T_1$ and $p_2,T_2$ to the right of the plug, as $p_1>p_2$ the gas moves left to right. The experimental configuration must ensure that pressures remain constant and that the experiment is performed under adiabatic conditions when $q=0$.

If a volume $V_1$ of gas moves from the left to right the work done/mole is $W=p_1V_1-p_2V_2$. This is the difference between the work of compression on the left of the plug and work recovered on expansion on the right. If the gas were ideal then $w=0$, but real gases are not. The gas expansion is also adiabatic so that no heat leaves or enters then $q=0$ and the change in internal energy $\Delta U$  is equal to the net work $$\Delta U =U_2-U_1 = p_1V_1-p_2V_2$$ therefore $$U_2+p_2V_2=U_1 + p_1V_1$$
As $H=U+pV$, then  $$\Delta H = H_2-H_1=U_2 +p_2V_2-U_1 -p_1V_1 =0$$

The Joule-Thompson  coefficient $\mu$ is defined, as you write, $\left ( \partial T/\partial P \right)_H$ and this measures how much the intermolecular interactions make the gas differ from an perfect gas. Most gases cool when passing from high to low pressures at room temperature.

Notes:

The coefficient can be rewritten in other forms using
$$ \left ( \frac{\partial T}{\partial p} \right)_H \left ( \frac{\partial H}{\partial T} \right)_p \left ( \frac{\partial p}{\partial H} \right)_T =-1$$
then
$$ \mu C_p= -\left (\frac{\partial H}{\partial p}  \right)_T $$

as
$$ \frac{dH}{dP} = C_p\frac{dT}{dp} +V-T\left (\frac{\partial V}{\partial T}  \right)_p$$
then if $\alpha=(1/V)(\partial V/\partial T)_p$ is the coefficient of expansion then at constant T, $(\partial H/\partial p)_T= V(1-\alpha T)$ which means that the Joule-Thompson coefficient can be written as $\mu=(V/C_p)(\alpha T-1)$ (This last equation has been used as a way of measuring absolute temperature because $V,\mu$ and $ \alpha $ are all measurable quantities.


Closed shell molecules can have degenerate states?

Degeneracy is usually based on symmetry. You can map the binding orbitals for each bond onto the other bond. So you can't distinguish between the molecular orbitals and its energies.

Jahn-Teller doesn't apply for linear molecules because there is no distortion which can reduce the symmetry and reduce the system energy as well.


How to convert Sodium Carbonate to Sodium Bicarbonate at home (CO2 scrubber)

A process commonly used to teach $\ce{CO2}$ scrubbing to chemical engineering students involves not sodium bicarbonate, but sodium hydroxide ($\ce{NaOH}$).

The reaction is exothermic and is as follows:

$$\ce{CO2 (g) + 2 NaOH (aq) -> Na2CO3 (aq) + H2O (l)}$$

The aqueous sodium carbonate is subsequently treated with calcium hydroxide to precipitate out calcium carbonate:

$$\ce{Na2CO3 (aq) + Ca(OH)2 -> 2NaOH (aq) + CaCO3 (s)}$$

Disclaimer:

This is usually done with much higher concentrations of $\ce{CO2}$ (e.g. >10%) than you would find in the atmosphere (~400 ppm). The point of scrubbing in a chemical plant is to reduce emissions to a level that is still much higher than atmospheric $\ce{CO2}$ levels but lower than governmental standards. I suspect that you would not see a noticeable change in ambient $\ce{CO2}$ levels from putting something like this in a place with such (relatively) low $\ce{CO2}$ to begin with. There is a very low driving force to push gaseous $\ce{CO2}$ into the liquid and solvate it when the partial pressure of $\ce{CO2}$ is low.

There exists a unique relationship between the partial pressure of $\ce{CO2}$ above a solution, the dissolved carbon in the solution, and the pH of the solution, shown here (own work):

Note that the equations on the right-hand side are in solution.

So if you wanted to see what was going on in a controlled environment with no other species in the solution / that could enter the solution, you could just use a pH probe to see how much carbon was pushed into the solution!


Paramagnetism is a form of magnetism whereby certain materials are attracted by an externally applied magnetic field. Valence bond theory (VBT) and hybridisation doesn't really do a good job at predicting whether a molecule is paramagnetic or diamagnetic (isn't attracted by an external magnetic field). This is why molecular orbital theory (MOT) is so useful as it is successful at predicting whether a molecule is paramagnetic.

For a molecule to be paramagnetic, it needs to have an overall magnetic moment meaning that it needs an unpaired electron. If all the electrons are paired, then the molecule is diamagnetic. So by seeing whether a molecule has an unpaired electron, we can predict if it is paramagnetic or not.

Now lets consider the example of $\ce{O2}$. Experimentally, $\ce{O2}$ is known to paramagnetic. According to VBT, $\ce{O2}$ should look like this:

]1

As you can see, all the electrons are paired. Therefore VBT predicts that $\ce{O2}$ should be diamagnetic.

Now lets examine how the electrons are arranged according to MOT. In MOT, unlike VBT, it involves the creation of bonding and anti-bonding MOs. MOs are basically the supposition of the wavefunctions of atomic orbitals. In $\ce{O2}$ the 2s AOs of each oxygen atom constructively and destructively overlap with each other while their 2p AOs also constructively and destructively overlap with each other The resulting MOs for $\ce{O2}$ looks like this:

Now that we got the MOs, all we have to do is fill them with electrons using the same method that we use for AOs. By doing that we get:

Note that we have 2 unpaired electrons. Therefore MOT correctly predicts that $\ce{O2}$ should be paramagnetic, unlike VBT which predicts that $\ce{O2}$ is diamagnetic.


As, your question isn't much clear and it needs to be focused little more.
From current state of this question, I can provide you the answers for specific questions that I get to understand from its current state.

$\ce{H_2 \ \ C=C=O}$
What is the group of the  molecule?

Well this is known as ketene.

how to characterize it and how this is dissimilar to ketone?

Ketenes are generally very reactive, and participate in various cycloadditions. They will also undergo [2+2] cycloaddition reactions with electron-rich alkynes to form cyclobutenones, or carbonyl groups to form beta-lactones.

Ketones engage in many organic reactions. The most important reactions follow from the susceptibility of the carbonyl carbon toward nucleophilic addition and the tendency for the enolates to add to electrophiles.

See this and this

If we add $\ce{O2}$ with it then will we get $\ce{CO2}$?

No! I hope structure of ketene and wikipedia articles make it clear to you why. :)
Also see this.


Phenomenon where fumes can relight an extinguished flame

It looks like you accidentally created a combustion mixture.  I checked the Material Safety Data Sheet (MSDS) for several mosquito coils, and in addition to the insecticide, one of them also contained the oxidizer potassium nitrate $\ce{(KNO3)}$.  The burning candle likely had some vaporized paraffin (fuel) in the flame, so when you directed the $\ce{KNO3}$ vapor into the flame, you had heat + fuel + oxidizer = combustion!  You literally set that vapor on fire.  I'd be careful with that: although the MSDS indicated a small amount of $\ce{KNO3}$ in the coil, you could still be injured if you were not prepared.


Below is the preparation of the Wittig reagent you are given. Yes, the intermediate is a phosphonium salt, with a formal positive charge on phosphorus. Treatment with base (BuLi) deprotonates at the adjacent carbon giving a phosphorus ylide.

As for the Wittig reaction itself, one postulated mechanism (depicted in Organic Chemistry by Maitland Jones) is shown below. Nucleophilic attack of the ylide onto the aldehyde gives an open betaine intermediate, which closes to form an oxaphosphatane. The strained 4-membered ring decomposes readily to give an alkene and triphenylphosphine oxide.

The stereochemistry of the Wittig is related to the nature of the ylide used. Unstabilized ylides, such as the one here, predominantly give the Z-configured alkene, while stabilized ylides (usually through conjugation to a carbonyl) give an E-configured alkene as the major product. With substrates similar to the ones here, I obtained a 4:1 ratio of Z:E products, so the selectivity is not perfect, as K_P points out.

The stereochemical outcome is intimately related to the mechanism, and according to my copy of Jerry March's Advanced Organic Chemistry (4th ed), there is no evidence for the betaine intermediate. Instead, the oxaphosphatane is formed directly through a [2+2] cycloaddition mechanism. This mechanism is attractive because the stereochemical demands of the cycloaddition account for the preference of the (less stable) Z-product. However, this does not explain why different ylides give different selectivities.


Why is atomic orbital one electron wave function? Why distance from centre is proportional to angular wave function?

OK, let's address these one by one.

Yes, it should be a two-electron wave function, i.e., the one that depends on 6 spatial coordinates. Just imagine the amount of trouble required to simply write it down, let alone produce a meaningful plot or do anything else with it. So we go with two one-electron functions instead (see Hartree-Fock method). That's an approximation, of course, but it would suffice for now.
Yes, in angular wave function plots, it is the numerical value of $\Theta\Phi$ that is actually plotted by means of the distance from the origin. There is no physical reasons behind this. It is not a physics question at all. It is a question about drawings made by humans on pieces of paper. We may plot whatever we like against whatever else we like. If you find the type of plot which gives you better intuition of what the spherical harmonics look like, you are welcome to use those. (Hint: people have been using these for quite a while now.)
The "not the distance" part of the quote is cryptic to me. I can only guess it is a botched attempt to point out that the plot is not related at all to the radial part of the wave function, and hence to the distance from the electron to the origin.
Yes, that would require a four-dimensional graph, which we can't do on paper. Wait, we can't even do a three-dimensional graph on paper. That's why we resort to cross-sections and other weird plots in the first place.

So it goes.


Is the subtrahend in the Nernst equation related to ohmic drop?

In short: No.

The Nernst equation describes the thermodynamic equilibrium state. By definition, there is no current flowing through the cell. Therefore, there is no ohmic drop in any term within the Nernst equation.

The Q term relates to the activities of the reactants and products at equilibrium.


Solubility of solid in a liquid depends on the nature of the solid.

It depends on whether the dissolution process is exothermic or endothermic. Using Le Chatelier's principle, if the temperature is increased for an exothermic process then dissolution decreases and if the temperature is increased for an endothermic process then dissolution increases.

I am going to stick to the case of $NaCl$, if you want to know more about Le Chatelier's principle, click on the link below and look for the 2nd answer https://physics.stackexchange.com/questions/311409/why-do-most-gases-dissolve-easier-at-colder-temperatures/311429#311429

There are some solids for whom change in temperature doesn't affect the degree of dissolution.

There are some substance which on heating, change on a particular temperature from one polymorphic form to another (like $\alpha$ to $\beta$ as in the case of ammonium nitrate) or from one hydrated form to another ($CaCl_2.6H_2 O -> CaCl_2.4H_2 O$) or from hydrated to anhydrous form ($Na_2SO_4.10H_2O -> Na_2SO_4$).

Generally, dissolution process in a solvent is a disintegration process. This process needs energy. In such cases, energy is absorbed. But in some cases, besides the process of breaking or ionization, there is hydrate formation. During hydration heat is evolved. The net result is that heat is either evolved or absorbed.

There are also cases in which heat of separation of ions in just equal to the heat of hydration and there is very little heat effect as in the case of $NaCl$. The heat of solution of $NaCl$ is very small as the heat of ionization is nearly equal to the heat of hydration. Therefore, temperature doesn't affect the dissolution of $NaCl$ as the heat of hydration almost equalizes heat of dissolution.


What is hybridisation of XeF6 in solid state?

Forget about applying hybridization outside the second row, especially in 'hypervalent' compounds. I know, that it is common to use and sometimes works, but it is incorrect.
The $\ce{XeF6}$ molecule is a hard spot. While, indeed, experimental data suggest that it adopts distorted octahedral geometry in the gas phase, there is evidence that the minimum is very shallow.
The octahedral structure of $\ce{XeF6}$ (which is probably a local minimum, or at least found to be one in more than one calculation) can be pretty easily described in terms of three-center four-electron bonds.
There is no good agreement on the nature of the stereo-active electron pair in $\ce{XeF6}$. It seems, that repulsion with core d-shells is important, but the recent articles I could get my hands on has little to no rationalization on the fact. Still, it must be noted that the minimum is very shallow, meaning that rough qualitative theories like VSEPR and MO LCAO are too rough for a good rationalization anyway.
The solid state typically involves a lot more interaction, and, as result, compounds may adopt very different structures. The simplest one example I can think of, I could point to $\ce{SbF5}$ for example. A monomer adopts trigonal-bipyramidal structure. However, in the solid state, a tetramer with octahedrally coordinated antimony and four bridging fluorines is formed. In terms of VSEPR, it is a move from $\mathrm{sp^3d}$ to $\mathrm{sp^3d^2}$. Similarly, $\ce{XeF6}$ has a crystal phase (one of six) that involve bridging atoms.


Why is copper a better conductor than iron?

This is, ultimately, a question on solid state physics rather than chemistry. Further, the OP indicates that they are in high school, which kind of limits the depth of the answer that might be useful to them. However, I will try to make a simple, yet detailed answer.

As atoms are brought closer and closer together, their electron clouds overlap and interact. When a crystalline solid is formed, one abandons consideration of the atomic energy levels. Instead, one seeks out the Bloch functions to describe the electonic states - the Bloch functions explicitly consider the symmetries of the crystal. Further, they represent extended electron states, those that extend across the entire crystal. They represent the allowed (energy and momentum) combinations of electrons.

For a metal, the highest Bloch states occur as a band of states which is only partially filled. Unoccupied states lie just above states with electrons. To move an electron about, it needs just a little bit of energy, and off it can go. (Fully filled bands do not allow conduction - for any electron going one way there is one going the other way. This needs a good grounding in solid state physics to fully grasp).

OK, in general then the conductivity should just be the number of electrons near the line between occupied and un-occupied. Kind of. This line is called the Fermi surface, and things get more complicated quickly. The shape and connectivity of the Fermi surface impact conduction dramatically. Lets take a deeper look at copper and iron.

Copper is a simple noble metal with a beautiful near-ideal Fermi surface (see Ashcroft and Mermin's Solid State Physics book, or C.Y. Fong et al., Comparison of band structures and charge distributions of copper and silver). It turns out that the Fermi surface of copper is almost exactly what one would expect from the free electron sphere. Electrons can move easily in any direction, the surface is fully connected, life is good.

Iron is much more complicated. Following along from J. Callaway and C.S. Wang, Energy bands in ferromagnetic iron one finds two major differences. First, because iron atoms indeed have individual magnetic moments, the overall band structure is split into two different spin states for the electrons. So, each spin state is separated from the others, reducing how each can move (relative to copper, were such splitting does not exist). Second, and more importantly, the Fermi surface looks nothing like a free electron sphere. Instead, for iron the Fermi surface is not fully connected, comprising instead of multiple 'pockets' of electrons that do not communicate directly with the other pockets (and each spin state has a different set of pockets). To move electrons around, they have to 'jump' from one pocket to another through scattering. This makes it much harder to move the 'free' electrons in Fe around to get net conductivity.

An analogy would be the difference of crossing a stream using either (1) a nice smooth bridge (copper) vs (2) jumping from rock to rock (iron). The band structure for copper just makes it so much easier, so the resistivity is much lower.


It is indeed nitrobenzene.

You really should look at the compounds, literally, because chemistry is ultimately an experimental science.

That being so, you can still note that the conjugated nitro group with its electronegative atoms gives relatively low-lying vacant orbitals into which electrons can be excited.  These orbitals can be accessed by absorbing short wavelength (violet/blue) visible light leading to a yellow color.


How do you determine the amount of acrylamide in foods such as baked fries?

Check out the methodology page for the FDA section on acrylamide quantitation, which goes into considerable detail.

In short, they start by mashing up some fries, and then extract and purify a fraction containing the acrylamide through a number of steps leading up to liquid chromatography. They then use quantitative mass spectrometry to determine the amount of acrylamide present.

They add a small known concentration of isotopically labelled acrylamide (i.e. acrylamide with all the carbons as carbon-13, which has very low natural abundance and results in a molecule that is approximately 3 g/mol heavier). This shows up in the mass spectrum as an offset peak. Then they can integrate the normal acrylamide mass spectrum peak and the labelled peak and obtain the concentration from the relative peak areas.


How do I convert benzoic acid to m-nitrobenzyl alcohol using 2 steps?

Use the meta-directing effect of $\ce{-COOH}$: nitrate benzoic acid first.
Reduce m-nitrobenzoic acid with $\ce{LiAlH4}$. The nitro group is not attacked, but when you calculate the equivalents of  $\ce{LiAlH4}$ needed, don't forget that you are reacting an acid with a hydride.

Edit 1:
Nitroalkanes or nitroalkyl-substituted arenes can be converted to the corresponding amines by $\ce{LiAlH4}$. This is usually not the case for nitroarenes. Here, catalytic hydrogenation, e.g. over $\ce{Pd/C}$ is the method of choice.

Edit 2:
Let's play, change the order of events and first reduce benzoic acid to benzyl alcohol. The hydroxymethyl group isn't m-directing, but what will happen to it under nitration conditions ($\ce{H2SO4}$ and $\ce{HNO3}$)?

Under these pretty acidic conditions, elimination of water from the starting material furnishes benzyl cations, which might react with benzyl alcohol in different ways, such as (multiple) Friedel-Crafts alkylations or etherification. I'd dare to say that the result is a mess.


What order would these compounds elute in a 3:1 hexanes:ethyl acetate TLC run?

Making predictions about elution order is always guesswork but based on my experience I would expect acetic acid to streak off the baseline, The benzyl alcohol to be of low Rf but above the acetic acid, the benzyl acetate Y to run high and I really don't know about the acetic anhydride, I suspect it will hydrolyse on the plate to give you more acetic acid so you will not see a separate spot. How are you planning on visualising these?


cis and trans are older notations that express roughly same concept of E and Z.

Explicitly, E and Z have absolute meaning and are assigned based on precedent, where the Z is assigned to the case where the highest priority substituents are on the same side of the alkene/double bond and E being opposite.  These notations explicitly refer to a pair wise relationship.  It is not normally said '... the group E to the ...'

cis and trans refer to substituents of context, with cis being substituents on the same side of the alkene/double bond, and trans being opposite.  Frequently these refer to the most important nomenclaturally, but not always.  Additionally cis and trans are used in discussions such as '...the group trans to the ...' describing a relation of one object relative to the other.  cis and trans are no longer considered IUPAC-correct nomenclature, but still have some usage for descriptive discussion.


why is the pressure exerted by ideal gas on the walls of the container more than that of real gas

For neutral gases, the intermolecular forces that we observe tend to be attractive. Significant repulsive forces don't really show up until your particles have charge.

The other difference between ideal gases and real gases are that real gases take up space, but the volume deviation from the ideal gas law is usually swamped out by the attractive intermolecular forces. This causes a real gas to exert less pressure than an ideal gas, and also is the reason matter undergoes phase changes.


Ammonia decomposition on platinum surface

Thanks to Poutnik's comment, I arrived to the answer and I would like to share it.

$$\ce{2NH_3 ->N_2 + 3H_2}$$
$$\theta_{NH_3}=\frac{K_{NH_3}p(NH_3)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$
$$\theta_{N_2}=\frac{K_{N_2}p(N_2)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$
$$\theta_{H_2}=\frac{K_{H_2}p(H_2)}{1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)}$$

Since the hydrogen binds strongly on the surface, we can safely determine that $K_{H_2}>>1$, wich means the denominator $1+K_{NH_3}p(NH_3)+K_{N_2}p(N_2)+K_{H_2}p(H_2)\approx K_{H_2}p(H_2)$. So the surface coverage of ammonia can be written as $\theta_{NH_3}=\frac{K_{NH_3}p(NH_3)}{K_{H_2}p(H_2)}$. By definition $v=-\frac{dp(NH_3)}{dt}=k_0\theta_{NH_3}=k_0\frac{K_{NH_3}p(NH_3)}{K_{H_2}p(H_2)}$, defining $k=k_0\frac{K_{NH_3}}{K_{H_2}}$ we arrive at the desired equation.

Feel free to edit.


Is NMR possible using nitrogen as the nucleus?

Of course yes -- because of the spin $\ce{^{15}N}$ and the much lesser used $\ce{^{14}N}$ nuclei posses:

The image is from here, the site provides some useful additional  overview to the source mentioned by @orthocresol.

P.S. If there were no NMR active nitrogen nuclei, 3D NMR would not this widespread in protein research:

(source)


SN2 reactions and inversion of configuration in compounds with more than one chiral centre

Here I have a compound with 2 chiral centers. There is only inversion at Carbon with iodine undergoing $\mathrm{S_N2}$ reaction. The second carbon does not undergo $\mathrm{S_N}$ reaction. There is no change in its configuration.

Example 1

In the example below ,

[Ref : https://en.chem-station.com/reactions-2/2016/05/neighboring-group-participation.html]

before a nucleophile attacks externally, OH of carboxylic group attacks internally via $\mathrm{S_N2}$. The 3 membred ring opens via another $\mathrm{S_N2}$ attack by hydroxide ion. The net result is retention of configuration .

Example 2

In this example , two chilral carbons are at neighboring carbons. As explained earlier , Sulphur attacks internally via $\mathrm{S_N2}$ attack. Again bromide ion attacks via another $\mathrm{S_N2}$ leading to retention of configuration ( Anchimeric assistance ).

[Ref : http://www.chem.ucla.edu/~harding/IGOC/A/anchimeric_assistance.htmll]

Example 3

This example also behaves similarly to one discissed above.

[Ref: Molecules 11(4):212-8 • April 2006   DOI: 10.3390/11040212 ]

In conclusion,

If there exists no lone pair donating group at neighboring carbon to leaving group, then $\mathrm{S_N2}$ attack takes place with inversion of configuration (Example 1).
If there exists a lone pair donating group (Example 2 and Example 3)at neighboring carbon to leaving group, then retention of configuration is seen.


difference between hexane and hexane fraction

The difference is simple. Hexane, if that is what the label says it is, should mostly be n-hexane. But hexane fraction is any hydrocarbon that distills at close to the same temperature as n-hexane. This mostly means a mixture of other six-carbon hydrocarbons (e.g. 2-methylpentane) most of which boil between 55 and 70 °C. You might find that the mixture contains some C-7 hydrocarbons as well.

For many reactions what you need is some volatile hydrocarbon and it isn't that important that it is a mixture. Occasionally, though, this will affect a crystallisation (especially if the hydrocarbon is bound in the crystal lattice). Pure hexane will work better in that case.

The reason why they are sold as different things is that pure hexane is a lot more expensive to produce as it will require careful distillation to separate it from the other, similar boiling point, hydrocarbons. So you don't want to use it if the cheaper mixture works. The cheaper mixture will just be a fraction obtained from crudely distilling petroleum at around 60–70 °C.


Is sugar necessary for the sweetening effect of food?

There are plenty of sweet things that are not sugar(s)

People have a sweet tooth (or, to put it another way humans like sweet things). This is an evolutionary adaptation because fresh fruits are both good food (both nutritionally and in providing lots of energy) and rare in pre-modern food gathering cultures. So we are adapted to like things that are sweet as eating them is good for us (in moderation).

In the modern world we can have as much sweet stuff as we want but this is no longer a survival advantage as too much sugar provides more energy than we need to survive or thrive and eating it just makes us fat.

So the modern food industry has adapted to give us sweet-tasting things that don't overload us with unneeded energy. Many of those food ingredients mimic the sweetness of sugar without containing many calories, so allow us to eat sweet food without the same risk of consuming too many empty calories and getting fat.

There are many chemical that taste sweet but are not sugars. The successful ones tend to tase a great deal sweeter than the typical sugars found in fruits and related plants (glucose, sucrose, fructose and others). For example, aspartame is hundreds of time sweeter than sucrose but is a peptide not a sugar; Saccharin is an aromatic suphimide and is also hundreds of times sweeter than sucrose; acesulfame K is similar to saccharin.

There are also sugar-like sweetness that the body cannot digest but which taste sweet. Steria is a complex sugar with sweetness but few calories; Sucralose is a modified sugar the body can't digest so tastes sweet without containing the calories of sucrose.

The point of the claims on "sugar-free" foods is that they are sweet but non-calorific. Plenty of compounds can mimic the sweetness of real sugars like sucrose but are calorie free either because they are far sweeter (so much less stuff is required for the same level of sweetness) or because the body can't digest the compound and turn it into unneeded calories.


irreversible adiabatic process against constant external pressure?

In this irreversible adiabetic process, only the work done on the system will be =$-P_{ext}dV$. Because in a reversible process we actually consider $P_{ext} = P_{int} + dP$, and hence we neglect the $dPdV $ term as it is negligible.  Considering this, if we go by the definition of adiabetic process and first law of thermodynamics, we will have $$0 = nC_vdT  + P_{ext}dV$$ as the heat changed in an adiabetic process is $0$. Now simply taking one term to other side and integrating we have $$nC_v \int_{T_1}^{T_2} dT = -P_{ext} \int_{V_1}^{V_2} dV$$  where $P_1,V_1,T_1$ are initial conditions, and $P_2, V_2,T_2$ are final conditions of the system.Now we put $V_1 = \frac{nRT_1}{P_1}$ and $V_2 = \frac{nRT_2}{P_2}$, so we will have $$C_v(T_2 -T_1) = -P_{ext}R(\frac{T_2}{P_2}-\frac{T_1}{P_1})$$ now taking $T_2$ associated terms to one side and $T_1$ related terms to other side we have $$T_2(C_v + \frac{P_{ext}R}{P_2}) = T_1( C_v + \frac{P_{ext}R}{P_1})$$ and thus we have our desired result as, $$T_2=[\frac{C_v + (\frac{P_{ext}}{P_1})R}{C_v + (\frac{P_{ext}}{P_2})R}]T_1$$


How are the parameters of a force field obtained for the bond angle HCH?

You don't necessarily need a potential energy curve to fit a force field. I answered a related question about fitting a force field from quantum calculations.

Let's assume you have a harmonic angle term. You could either do this in terms of the angle bending or the distance between the end atoms, but let's take the angle for our example:

$E_{angle} = k (\theta - \theta_0)^2$

The first parameter you need is the optimal angle $\theta_0$ which you can get from a geometry optimization, experiment, etc.

Your question stems from the second parameter, $k$ - the force constant for whatever this angle type happens to be (e.g. H-C-H in methane for your example).

There are a few ways to get the force constant, but it comes from the second derivative, right? So you can get that from a Hessian or vibrational calculation. My favorite discussions of this center around the "Badger's rule for angle force constants, e.g."

"Maximally diagonal force constants in dependent angle-bending coordinates. II. Implications for the design of empirical force fields" J. Am. Chem. Soc. 1990, 112, 12, 4710-4723

  Unfortunately, most such force fields are defined in well-determined
  sets of internal coordinates, whereas empirical potentials use larger
  sets of dependent coordinates. This paper illustrates a unique
  “localized” representation of the angle-deformation potential in
  dependent coordinates which is exactly diagonal for in-plane bending
  at trigonal-planar centers and is nearly diagonal for angle bending at
  tetra coordinate centers.

Modern force field fitting methods typically use scripts that minimize the differences between an in-development parameter set and a set of Hessians and/or experimental data.

As I mentioned in the other question, there's a huge pile of such papers, many with code to derive force fields from quantum chemical data.

QubeKit J. Chem. Inf. Model. 2019, 59, 4, 1366-1381 - code at GitHub
QuickFF J. Comput. Chem. 2015, 36, 1015– 1027 - code at GitHub
ForceBalance J. Chem. Theory Compute. 2013, 9, 1, 452-460 and J. Phys. Chem. Lett. 2014, 5, 11, 1885-1891 - code at GitHub
ForceFit J. Comput. Chem., 2010 31: 2307-2316
Parfit J. Chem. Inf. Model. 2017, 57, 3, 391-396 - code at GitHub


I was wondering if there was a difference between salt formation equation and production.

Absolutely! In this case, the latter means that you're trying to turn a side product (waste) of a technical product into another marketable product.

In order to understand, why the oxidation of $\ce{FeSO4}$ is interesting, you have to realize that it is a side product in the synthesis of titanium dioxide ($\ce{TiO2}$). Here, ilmenite ($\ce{FeTiO3}$) is processed with concentrated sulfuric acid to yield $\ce{TiOSO4}$ as the valuable intermediate and $\ce{FeSO4}$ as "waste".


How DFT-D3 incorporates coordination number (CN) into C6

I will quote the paper quite a bit, but I'll try and summarize a bit after the quotes and equations. You might want to start at the bottom and work backward, a lot of this is just for later reference.

First, some background. Equation 3:

$$
E^{(2)} = \sum_{AB} \sum_{n=6,8,10,\dots} s_{n} \frac{C_{n}^{AB}}{r_{AB}^{n}} f_{d,n}(r_{AB})
$$

  Here, the first sum is over all atom pairs in the system, $C_{n}^{AB}$ denotes the averaged (isotropic) $n$th order dispersion coefficient (orders $n=6,8,10,\dots$) for atom pair $AB$ (...)

So $A$ and $B$ are nuclear indices, and together specify an arbitrary pair.

  Although the $C_{6}^{AB}$ values can be compute easily for any pair of free atoms by using Eq. (5) [the Casimir-Polder integral] in principle, this would lead to a rather inconsistent treatment of dispersion in and between molecules.

For reference, here is equation 5:

$$
C_{6}^{AB} = \frac{3}{\pi} \int_{0}^{\infty} \alpha^{A}(i\omega) \alpha^{B}(i\omega) \, d\omega
$$

Instead they calculate the integrals over pairs of hydrides, subtracting the effect of molecular hydrogen (equation 10), in order to capture the effect of being bonded:

$$
C_{6}^{AB} = \frac{3}{\pi} \int_{0}^{\infty} d\omega \frac{1}{m} \left[ \alpha^{A_{m}H_{n}}(i\omega) - \frac{n}{2} \alpha^{H_{2}}(i\omega) \right] \times \frac{1}{k} \left[ \alpha^{A_{k}H_{l}}(i\omega) - \frac{l}{2} \alpha^{H_{2}}(i\omega) \right]
$$

  At first sight this new approach seems to be a disadvantage because it leads to reference molecule dependent (ambiguous) coefficients. However, as will be discussed in detail below (...), it opens a route to system [coordination number (CN)] dependent "atomic" $C_6$(CN) coefficients. (...) We propose here a radically new "geometric" approach that is based on the concept of a fractional coordination number. (...) We base our ansatz on a fractional CN for an atom $A$, that is a sum of a "counting" function over all atoms in the system,

Equation 15:

$$
\text{CN}^{A} = \sum_{B\neq A}^{N_{at}} \frac{1}{1 + e^{ -k_1 ( k_2 (R_{A,\text{cov}} + R_{B,\text{cov}}) / r_{AB} - 1) }}
$$

  where $R_{A,\text{cov}}$ is a scaled covalent (single-bond) radius of atom $A$ (and analogously for $R_{B,\text{cov}}$).

So the CN$^A$ is some bond distance-dependent scaling factor for the specific atom $A$ under the influence of all other atoms $\{B\}$. Here is how they incorporate their coordination number scaling into the dispersion coefficients. Following emphasis mine:

  Turning back to the dispersion coefficients we propose the following general approach. For each element in the Periodic Table, at least one reference molecule is used to compute $\alpha(i\omega)$ values in Eq. (10). (...) If an element is commonly investigated, and/or often found in different chemical environments, we suggest to use several representatives for which values are computed. In the case of carbon for example, ethyne, ethene, and ethane can be used for CNs between 2 and 4. Values computed for $\ce{C-H}$ and the carbon atom complete the set. For the atom of interest in any of the reference systems also the CN is calculated and stored. Using Eq. (10) the $C_{6,\text{ref}}^{AB}$(CN$^A$,CN$^B$) values are computed for this pair with the two atoms in their specific environments. These serve as supporting points in an interpolation procedure to derive the coefficient for any combination of fractional CN values. Because we base our approach on pair-specific coefficients $C_{6}^{AB}$, a two-dimensional (2D) interpolation scheme must be used. After extensive testing we propose a simple Gaussian-distance ($L$) weighted average,

Equation 16:

$$
C_{6}^{AB}(\text{CN}^{A},\text{CN}^{B}) = \frac{ \sum_{i}^{N_A} \sum_{j}^{N_B} C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B}) \times L_{ij} }{ \sum_{i}^{N_A} \sum_{j}^{N_B} L_{ij} }, \\
L_{ij} = e^{ -k_3 [ (\text{CN}^A - \text{CN}_i^A)^2 + (\text{CN}^B - \text{CN}_j^B)^2 ] },
$$

  where $N_A$ and $N_B$ are the number of supporting points (= number of reference molecules) for atoms $A$ and $B$, respectively. The CN$^A$ and CN$^B$ are coordination numbers for the atom pair $AB$ in the system of interest, and the CN$_i^A$ and CN$_j^B$ are those for the two reference systems $i$ and $j$, for which $C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B})$ is the precomputed value. (...) For most of the elements only two reference values are necessary to cover the typical bonding situations. For "important" elements such as carbon, however, more values (5 in this case) seem appropriate. Currently we have computed in total 227 reference systems for the elements up to $Z=94$ which leads to about $2.6\times10^4$ difference values for $C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B})$. They are computed once and read at the beginning of any computation.

Conceptually, here is how a calculation would work:

Your SCF has converged and it's time to calculate the -D3 correction. This table is read in from a file into some data structure so that all the $C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B})$ can be looked up easily.
Loop over the atom pairs $AB$ of your system as part of equation 3. For atom $A$, calculate its coordination number CN$^A$ using equation 15. Do the same thing for atom $B$. In the equation, $B\neq A$ means all atoms other than $A$, not atom $B$ in the pair!
For this pair, the $C_{n}^{AB}$ in equation 3 is really $C_{6}^{AB}(\text{CN}^{A},\text{CN}^{B})$, which requires evaluating equation 16. What equation 16 does is interpolate between those precomputed $C_{6,\text{ref}}^{AB}(\text{CN}_{i}^{A},\text{CN}_{j}^{B})$ for a series of coordination numbers. Because there are two CNs, the interpolation is on a 2D grid, rather than a line or some higher dimension. The interpolation function is chosen to be the $L_{ij}$. From the above quote, each $i,j$ sum is only about 2-5 points, so this is very fast; even the $\ce{C-C}$ grid will only total $5\times5=25$ points.

So, to answer your question,

  According to Fig. 5 in the paper, it seems the AB pair is put in some reference system and then based on the precomputed C6 values in the reference system we could get the C6 in the environment.

This is exactly correct. It is done by interpolation on a grid of reference values, where the coordinates/points on the grid are the coordination numbers, and there is a grid for every possible $AB$ pair.

$\tiny{\text{Phew.}}$


I can't speak about the etymology of the word, since I don't speak any of the languages in question (plus, it it is purely concerned with that, I believe this question is off topic here).

However, if you are interested in the contents of gasoline, I can tell you a little something about them.

Gasoline (or petrol if you're not from North America) is typically a mixture of hydrocarbons with 4 to 12 carbon atoms per molecule. It is a comprises of alkanes, cycloalkanes, alkenes, or in industry parlance parrafin, naphthenes and olefins.

Moreover, as you suspected, it does contain aromatics in general and benzene in particular, and most countries set limits on their content in the fuel. For example, in the EU the benzene content should be under $1\%$ (v/v) for automative grade gasoline$^{[1]}$ Benzene, Toluene, and Xylene (BTX) are important petrochemicals, and valuable chemical feedstocks, and most are often recovered during catalytic reforming.

EU: Fuels: Diesel and Gasoline


Is it safe to look at a mercury gas discharge tube?

Could it have caused irritation or injury? Possibly, but not likely, depending on a few factors:

The intensity of the emitted ultraviolet.
Proximity to the light source.
Length of exposure.

You'd need to calculate the dosage integrated over time.
That said, it is very unlikely any harm was done, because of the following:

The power supplies are limited to to only a few mA current. The total power is the arc voltage drop, ~30 to 100 VAC, times the current. For the supply abvove, that would be 100 VAC * 0.01 A = 1 W. One watt is low power for a mercury lamp.
Demonstration spectral lamps, such as this capillary mercury vapor lamp, usually have thick borosilicate glass capillaries. Borosilicate glass does not transmit shorter wavelengths well. Most of the UV would have been absorbed by the thick capillary wall.
The length of time was given as "several minutes", so the integrated intensity over time (i.e., total energy absorbed) was low.
That you state you were, "within one foot of the discharge tube," ~30 cm, is of concern, because some older discharge tube power supplies had exposed high voltage terminals. Though limited to 10 mA, considered below a dangerous current for electric shock, in most cases, one could still get burnt by the ~10,000 VAC arc, should one come in contact with exposed terminals, or could be startled and hurt by a muscular reflex action. Likely, though, the ends of the tube and associated wiring were safely enclosed.

So, all things considered, there might have been fumes in the lab or other cause of irritation, but I believe the UV light was not likely an issue.


R,S Configuration of asymmetric carbon atom when group priority 3 and 4 are not on wedge bond?

To do this, Put the lowest priority group as far away from you as possible.

In the 1 example, if I look at 4 from between 1 and 2 it would be farthest away from me. And i will see that the 3 group is on top, 2 on my left and 1 on my right.

And now if i move from highest priority group to lowest (1-2-3) it would be in clockwise sense. So the configuration is R.

2nd and 3rd example has S and R configuration respectively.


Why can't we rotate a carbon while checking for chirality?

It simply depends on what is the subject. The book asks for the displayed conformation, which we agree is chiral.

The question would have been more tricky to answer if it were about optical activity of a sample.

In this case we know the conformer can not be isolated at room conditions but this is not a general rule.

And one can always cool down, at least in principle.


In terms of how they got the relation for the diagonal elements, I believe it is relatively straightforward. Given a list of atomic energies (energy of atom relative to separated nucleus and electrons), one can try to fit these values to a function of nuclear charge. If you plot a few values, it suggests trying an exponential fit (I don't know why all the papers say polynomial fit, as this could not lead to a noninteger exponent). So, you can try to fit the values to a function of the form $aZ^b$ using, for example, least-squares regression. You should wind up with something pretty close to the equation given in your question, with some variation because I don't know what atomic energies they used to do the fit. (I'm also not sure why the function isn't negative since the atomic energies are negative. My best guess is the sign winds up not being important as a descriptor in the network).

As to why they choose to use these values along the diagonal, they describe the system in terms of the energies of the individual atoms, along with the energy of interactions between atoms given by the off-diagonal terms. The original Rupp et al. paper compared two Coulomb matrices by looking at the difference between their eigenvalues, with small differences implying similar molecules. These eigenvalues can be thought of as effective atomic energies within the molecule, that is, energies of the individual atoms with the influence of the surrounding atoms included. Similarly, the eigenvectors would describe fragments of the molecule or an effective atom which account for the influence of neighboring atoms.


Chlorinating specifically one carbon on a symmetrical alcohol

It might be overkill for your purposes, but there is a nice reaction that coverts secondary alcohols to $\alpha$-chloroketones in a one-pot synthesis, using trichloroisocyanuric acid (for instance, they got 75% yield of 2-chlorocyclohexanone from cyclohexanol).  The paper is 'Direct conversion of alcohols to $\alpha$-chloro aldehydes and $\alpha$-chloro ketones' (Jing, Daniliuc, Studer 2014, Org. Lett. 16 (18) pp. 4932-4935).

  Direct conversion of primary and secondary alcohols into the
  corresponding α-chloro aldehydes and α-chloro ketones using
  trichloroisocyanuric acid, serving both as stoichiometric oxidant and
  α-halogenating reagent, is reported. For primary alcohols, TEMPO has
  to be added as an oxidation catalyst, and for the transformation of
  secondary alcohols (TEMPO-free protocol), MeOH as an additive is
  essential to promote chlorination of the intermediary ketones.

Trichloroisocyanuric acid has the advantages that it's inexpensive, widely available and easy to handle.


Will the sulfur-iodine cycle and Haber Method work at lower pressures and temperatures?

Yes this is possible but you must remember it is a reversible reaction so if more pressure creates ammonia, applying less will cause the reaction to reverse creating some of the reactants again.


Acidic and alkaline salt solutions: why do some salt form acid/bases while others don't

Without trying to go into too much detail this has a two part answer.

First a salt by definition has an anion and a cation. So ordinary table salt, $\ce{NaCl}$ is a chemical salt, but table sugar is not a salt. So lets think of the cation as $\ce{C^+}$ and the anion as $\ce{A^-}$.

So we start out with solid salt which in chemistry is written as $\ce{CA_{(s)}}$. When the salt dissolves the two ions are hydrated which means that water molecules surround each ion and separate the pair. the chemical reaction is written like this: $$\ce{CA_{(s)} -> C^+_{(aq)} + A^-_{(aq)}}$$

Typically the (aq) subscripts are left off, but implied.

The second part of the answer involves the nature of the cation or anion. To understand if the pH of the solution will change depends on whether the cation is able to "donate" a hydrogen ion, or the anion is able to accept one.

$\ce{NH_4^+}$ is a cation and is able to donate a hydrogen ion to the solution. So the pH will drop and the solution becomes more acidic. $$\ce{NH_4^+ + H_2O -> NH_3 + H_3O^+}$$

$\ce{CO_3^{2-}}$ is a anion and is able to accept a hydrogen ion from the solution. So the pH will increase and the solution becomes more basic. $$\ce{CO_3^{2-} +H_2O-> HCO_3^- + OH^-}$$

When water, $\ce{H_2O}$, makes or breaks bonds the process is called hydrolysis.


Sodium carbonate ($\ce{NaCO3}$) is basic and gives solutions with a pH of around 10.  Sodium bicarbonate ($\ce{NaHCO3}$) creates solutions with a pH of approximately 7.3.  This is in fact what the human body uses to regulate pH.  You want to used the sodium bicarbonate since it will yield a nearly neutral solution.


What gas are the bubbles that evolve in the redox reaction of copper(II) chloride and aluminum?

The gas bubbles are very likely just air that was dissolved in the water previously and now gets released. I would expect most of the bubbles to actually be nitrogen, followed by oxygen, argon and then some carbon dioxide — that is assuming you used deionised water which should not have significant amounts of carbonate and hydrogencarbonate dissolved.

The copper particles merely supply suitable nucleation sites for the gas.

The theory on aluminium oxide being oxidised to pure oxygen is not correct. Oxygen is a very oxidative element meaning that you require strong oxidants to oxidise oxide anions back to oxygen. With ‘strong’, we’re talking the oxidative strength of potassium permanganate. Aluminium’s oxidative powers are nowhere near that strong.

Copper(II) chloride and aluminium oxide do not react with each other; only metallic aluminium beneath the passivisation layer is able to react with copper(II) ions.


What gives rise to the color of sapphires?

The blue of sapphires is produced by the elements titanium and iron that are both present. There is a process called charge transfer causing the transfer of an electron between Ti$^{4+}$ and Fe$^{2+}$ to Ti$^{3+}$ and Fe$^{3+}$ to fit together with the Al$^{3+}$ of the corundum lattice. Those ions give the blue sapphire its colour.

Note that the ions in a crystal are different from isolated ions in the gas phase, they tend to have larger absorption bands instead of sharp lines. Also the distances in the lattice are important and may modify the colour, anisotropy can lead to colour dichroism, the sapphire may be blue viewed from one side and green from another.


Is there a metal for which its gas is necessarily a plasma?

In principle, all that is required is for a substance to evaporate from a liquid into a plasma directly is for its ionisation energy to be significantly smaller than the average kinetic energy of its particles at its "boiling point". Or in other words, as a very rough estimate this is true when $\mathrm{k_BT_{boiling}≳EI}$.

Going further down in the periodic table, elements tend to have higher boiling points and lower ionisation energies. After reaching the superheavy element region (Z>100) things get much more complicated due to strong relativistic effects, but the elements from Rf to Hs (Z=104 to 108) are expected to have the highest boiling points at ambient pressure of any known chemical substance, possibly well past $\mathrm{6000\ K}$. However, their calculated ionisation energies of $\mathrm{6-8\ eV}$ suggest they would only ionise considerably at temperatures 10 times higher still ($\mathrm{1\ eV/k_B= 11605\ K}$), meaning there is still a long way to go.

As others have mentioned, increasing the pressure is a reliable way to push boiling/sublimation points higher, up to where they can match and surpass the energy regime required for ionisation. However, this would require absurdly high pressures; for example, the boiling point of tungsten at $\mathrm{1000\ bar}$ is still a mere $\mathrm{9000-10000\ K}$ (ref). Thus, a very crude extrapolation would suggest the direct conversion of liquid (or more likely solid) tungsten to a plasma would happen around $\mathrm{90000\ K}$ under a pressure of $\mathrm{20\ Mbar}$ (20 million atmospheres!).


Some pointers:

  the hexagon structure signifies an aromatic

Not quite. The hexagon structure depicted with alternating carbon-carbon double and single bonds is aromatic. It is called benzene and is very commonly encountered in organic chemistry. It can be represented by two resonance structures as shown below and is sometimes also represented with a circle inside a ring of singly bonded carbons to show the delocalised $\pi$ system.

However, the are other molecules depicted as hexagons which are not aromatic such as cyclohexane.

Additionally, there are other molecules which are aromatic which are not benzene, and many which are not hexagonal but you probably won't come across these at an introductory level.

  My first guess was carboxylic acid, but with further googling, I'm now thinking it's an amide, because it contains nitrogen and oxygen

You are correct that it is an amide. The amide functional group contains a nitrogen adjacent to a carbon-oxygen double bond. The carboxylic acid group is similar but the nitrogen is replaced by an $\ce{-OH}$ group. Amides can be primary, secondary or tertiary depending on the number of carbons that are bonded to the nitrogen.

  alkane is $\ce{CH4}$, alkene is $\ce{H2C=CH2}$, and an alkyne is $\ce{HC#CH}$

These are the simplest example of alkanes, alkenes and alkynes. They are called methane, ethene and ethyne respectively. However, there are more complicated molecules involving these functional groups. In general, alkanes have only carbons-carbon single bonds and carbon-hydrogen single bonds. Alkenes are distinguished by having a carbon-carbon double bond denoted by two lines between the two carbons which look like an equals sign. Alkynes have a carbon-carbon triple bond which is denoted by three lines between the carbons. Therefore, your example is an alkene.


How do we determine molecule chirality in these examples?

Although you have arrived at the correct answers and seem to have followed a reasonable process for getting there, it's important to be precise with the use of terms like symmetry and chirality. For a molecule to be chiral, it must be non-superposable with its mirror image. It's important to really embed this definition before talking about symmetry and things like chiral centres. For example, it is possible for molecules to be chiral without any chiral centres, and it's possible for molecules with multiple chiral centres to be achiral.

the guaranteed way to determine if a molecule is chiral: draw (or build a model) of the mirror image and determine whether they are superposable

Now, symmetry is a different, but related, topic. A molecule with a point or plane of symmetry will be achiral. However, molecules with a rotational axis of symmetry can still be chiral, eg. many often-used chiral ligands are C2 symmetric.

So, in answer to your question, "What am I missing? - I would just say: the rigor in use of the terminology. Your approach of drawing different conformers of the molecules with two chiral centres and looking for a plane of symmetry (that shows the molecule is a meso compound, and therefore achiral) is sound and led you to the correct answers.


Unsymmetrical ketones can produce two different enolates, as shown in your reaction scheme.  The enolate pictured in the top row of your drawing is referred to as the thermodynamic enolate, while the enolate shown in the bottom row of your drawing is the kinetic enolate.

More highly substituted double bonds are thermodynamically more stable than less substituted double bonds. Therefore, the thermodynamically favored enolate is the one that gives rise to the most highly substituted double bond in the enolate intermediate. The thermodynamic enolate will be formed when reaction conditions are chosen (see below) to allow equilibration of all enolate intermediates; product formation is then thermodynamically controlled.

The kinetically favored enolate is the fastest formed enolate, the one formed by abstracting the most acidic proton (proton acidity follows primary > secondary > tertiary).  In your reaction, removal of the proton on the secondary carbon would be kinetically favored (more acidic) over removal of the proton on the tertiary carbon atom (less acidic).

Reaction conditions can be selected to favor kinetic (aprotic solvents, strong bases, low temperature) or thermodynamic (protic solvents, weaker bases, higher temperature) enolate formation, giving the chemist control over selecting which product will be formed.

If your reaction were run under conditions favoring kinetic enolate formation, then we would expect to observe product formed from following the bottom reaction line.  If the reaction were run under conditions favoring thermodynamic enolate formation, then we would expect to obtain products from following the top line of your reaction scheme.

So if your professor said that pathway "A", the top line in your drawing, is favored, then the reaction is being run under conditions favoring formation of the thermodynamic enolate.


Introductory chemistry book with calculus prerequisite

Most introductory physical chemistry courses and books don't rely very heavily on prior chemistry knowledge, and there is quite some math to do there. I think you can start most subjects there independently from scratch (QM, thermodynamics, kinetics).

To better prepare us for quantum mechanics, my (chemistry) faculty offered a two-semester lecture for "freshmen" where the professor derived the Schroedinger equation with us from first principles. The lecture was basically self-contained, both mathematically (introducing analysis) and chemically (first-years do basic chemistry and qualitative inorganic analytics, there is hardly any overlap with QM).

With students that are better in math than we were, this could work very well, and faster. :-)

I'm afraid I have no english book for this. The regular physical chemistry books don't dive into the actual mathematical derivation so much. We used "Wedler, Lehrbuch der physikalische Chemie".

Btw.: Students that are good in math AND chemistry are very valuable. Be prepared for your colleagues from the chemistry department to be after them. ;-)


This can be done by first reating the phenol with $\ce{HNO2}$ Which gives us this

Then reacting with dil $\ce{HNO3}$ we get 90% yield of para nitro phenol

This happens because $\ce{N=O}$ has a very strong $- I$  effect thus it concentrates the electrons more towards the para position so on reacting with dil $\ce{HNO3}$ we get max yield of para

P.S - Another way of getting only para is using a protecting agent but  i cant recall the reaction now. I will post it later


When making a calibration curve, do we have to use the wavelength at maximum absorbance?

No it doesn’t. Measurements can in principle be made at any wavelength.

If there are two colored specimens in the same solution, it is actually better to measure at the wavelength where the difference between the spectra is as high as possible.

Measuring at a peak has some advantages: the absorbance is high, giving a good signal/noise ratio, and if the wavelength varies a little it has less effect on the absorbance when measured at a peak than at a slope of the curve.


Yes, that second resonance structure, the one on the right, is a significant contributor to the overall description of the molecule.  That resonance looks a little different because the p orbitals on the alpha and beta carbons are on $\ce{sp}$ hybridized carbons, rather than the $\ce{sp^2}$ hybridized carbon more commonly seen in enones.   As this second resonance structure explains, alpha-beta unsaturated enones and ynones are both subject to attack by nucleophiles at the electrophilic beta carbon.

Resonance structures, like the one pictured below, explain the reactivity seen in enamines and yneamines (the triple-bond analogue); however, now the positive charge is on the nitrogen and a negative charge on the beta carbon.  Therefore these nitrogen compounds are subject to attack by an electrophile at the nucleophilic beta carbon.


How do I calculate the pressure of a known liquid in a sealed container heated above boiling point?

Note: for convenience, "gas" refers to any gas at a temperature beyond its boiling point; "vapor" refers to any gas evaporated from its liquid state, also implying that the liquid itself is below its boiling point, such as the 150 degrees water in your example.

Dalton's law of partial pressure would come in handy here. The law states that the partial pressure generated by each type of gas particles sum up to the total pressure in a sealed container. The partial pressure of "gasses" can directly be calculated using the ideal gas law. Of course the partial pressure of "gasses" is 0Pa if there is no gas but only vapor in your container.

Now the partial pressure of the vapor of liquids. In a sealed container, if left to attain dynamic equilibrium (which is the steady state in this case) the partial pressure of the vapor of any liquid would be equal to its equilibrium vapor pressure, which is a function of temperature only. Though not excessively accurate, the Antoine equation can be used to estimate this:
$\log P = A- {B \over C+T}$
where P is the equilibrium vapor pressure, A, B and C are substance-specific constants and T is the thermodynamic temperature of the substance. The three substance-specific constants can be checked up on the internet, or provided by other sources, whereas conducting experiments can be a last resort (as it is a very tedious experiment that you need the whole curve of the function to approximate the constants).

If the liquid in question is a mixture of more than one liquid substance, then you'd need a third law - Raoult's law to determine the partial vapor pressure of each component of the mixture:
$p_i = p_i^\star x_i$
where $p_i$ is the partial vapor pressure of component $i$ of the mixture, $p_i^\star$ is the equilibrium vapor pressure of a pure sample of component $i$(in itself, not in a mixture), and $x_i$ is the mole fraction of $i$ in the liquid.
At dynamic equilibrium, combining Dalton's law of partial pressure and Raoult's law yields the expression for total vapor pressure:
$p_{total} = \sum p_i^\star x_i$


Electrovalent bonds are easiest to identify. If a compound is made up of a metal and non-metal/non-metallic radical (like carbonate), then, 99.99% times, it contains electovalent bond. If a compound is made up of 2 or more non-metals/non-metallic radicals, then it contains covalent bond. Coordinate covalent bonds appear mostly with compounds containing Hydrogen element. To identify the coordinate covalent bonds, you can draw the branched structural formula of the compound and see if the shared pair of electrons are coming from the same molecule.


Why do tealights burn violently when nearly burnt out

There are two factors that probably matter here. One is that, for most of the life of the tea light, only a small amount of wax is liquid and actually burning. Most of the flame is liquid wax burning in the wick with a small pool of liquid wax below the wick. But later, when most of the solid wax has gone all the wax will be liquid. The second is that the matchstick can act as a second wick.

When the wick is still vertical (and held there by the presence of solid wax) the area of melting is kept small as the flame is vertical and the area of melting is constrained. Later, the wick may fall over as there isn't enough solid wax left. Also the heat can spread wider as the volume of wax decreases as it is also conducted more effectively through the aluminium container (wax isn't a good conductor of heat). The net result is more liquid wax near the end of the candles life.

Then two other factors may intervene. If the wick falls over, there is now a larger area of wick where burning can happen and the whole wick (not just the tip) can burn. Sometimes the matchstick will now also act as a wick, increasing the area of flame. Both these effects make more flame and more heat in a sort of runaway reaction which might also cause enough heat to reach the flash point of the wax leading to an even bigger flame. Even if the flash point isn't reached, the flame from a bigger wick area and a match acting as a wick can be a lot bigger than the flame form a normal vertical wick.


How to get a "rechargable" chemical reaction for a glow stick

Chemoluminescent processes based on the reaction of diaryloxalates with hydrogen peroxide in the presence of a fluorescent dye are not reversible.

While the dye remains unchanged, both the oxalates and the hydrogen peroxide are consumed. "Recharging" a device (glow stick) isn't possible. When performing the reaction in a flask, only could probably prolong the light emission by subsequently adding portions of the starting materials over time.

Photoluminescent glow-in-the-dark devices based on the emission of long-lived excited triplet states can indeed be "recharged" by holding the device in the light again. But this has nothing to do with the reactions happening in glow sticks.


Decarboxylation under acid, water, and extreme heat

Yes, your analysis is correct. This is a common sequence in the malonate synthesis of carboxylic acids and acetoacetate synthesis of methyl ketones. After taking advantage of the acidic, doubly alpha-proton and using the resulting enolate as a nucleophile, the ester(s) are hydrolyzed with acid. A carboxylic acid with a carbonyl in the beta-position can be thermally decarboxylated to give an enol, which typically tautomerizes to the keto form.


Does glycerin promote bacterial growth the same as water?

If I understand correctly if the concentration of water is greater on the outside of a bacterial cell the water will go inside the bacteria allowing it to multiply or grow faster. Does glycerin have the same effect?

Usually the point of adding salt to water is to lower its activity, in order to make the water concentration outside the cell lower than on the inside. This leads to water wanting to escape the cell, causing osmotic pressure to swell the cell.  With strong enough salt, the osmotic pressure causes water to flow out of the cell, so much that the cells dry up and aren't able to function.

Water is miscible with glycerol, which makes glycerol an osmolyte.  That means that glycerol would have the same effect as salt.  Water mixed with glycerol is lower in concentration than pure water.  Putting a happy aqueous bacterium into 100% glycerol will cause immense osmotic shock, probably resulting in death for most bacteria, due to too much of their intracellular water trying to escape to the extracellular, water-free (but glycerol rich) environment.

The downside of using glycerol is that at low concentrations (say 10 g/L or even 100 g/L) it is a great food for many types of microorganisms.  Sure at >1000 g/L it is a deadly osmolyte, but at low concentrations, its a yummy food.  Salt would merely lose its effectiveness as it gets diluted, but glycerol on dilution becomes a potent agent that would enhance microbial growth.  For this reasons, and others (costs, viscosity) it is not often used (outside of laboratory settings)


Can a substituted cyclohexane progress through SN1?

I think it's difficult to say for sure without actually making experiments. It should depend a lot on the conformation of the reactant, the substituents, the leaving group and also on the solvent. Nevertheless, I would imagine that it should be possible in some cases if you have a good leaving group.

In the case of a substituted cis-decalin, for example (I suppose you actually meant to refer to cis because in the trans isomer both hydrogens are stuck in axial positions), the elimination of the leaving group must indeed induce some strain, but maybe not that much. The structure could adapt with small changes in angles all over the molecule and both cycles would keep chair-like conformations. A tertiary carbocation would also be quite favorable.

So yes, in my opinion it should be possible in some cases, specially if heating is employed or you have a good leaving group. However, I should reinforce the fact that it really depends on many factors. This type of system can get quite complex if you consider the influence of the solvent, the influence of the substituents and the possibility of side reactions, including eliminations reactions, which commonly compete with substitution (on the presence of bases).


Kinetic isotope effect in nitration of benzene

Page 513 of Organic Reaction Mechanisms by Bansal references the work of Melander in Arkiv för Kemi, 2, 211 (1950). This states that the rates of nitration for benzene vs benzene-d6 are about the same, implying that the addition of the nitronium ion is rate controlling and not the weakening of the C-H / C-D bond. However in case of sulphonation H containing benzene reacts faster than D containing benzene.


Why does Pb have a higher electronegativity than Sn?

The explanation is actually well-known.

Elements in Period 6 get f orbitals for the first time. 4f subshell has relatively poor shielding effects, thus it's harder for the atoms to lose the outer electrons, increasing electronegativity. (similar reasons also account for some chemical properties of Period 4 elements, like $\ce{H2SeO4}$'s oxidizability)
Relativistic effects also contribute. For a very informal (and actually incorrect) understanding, just imagine that the electrons in these heavy atoms move fast and their masses increase and their energy decreases.


Most people find alkaline taste bitter and "bad tasting". There are some exceptions; shellfish, particularly conch, may have a pH of more than 8.0. This is also the approximate pH of bile, which tastes particularly foul when retching.
Crackers made with baking powder or baking soda also may be somewhat alkaline, as is egg yolk.
However, your question is valid... why are the vast majority of foods neutral or somewhat acidic? Why do most enjoy the taste of grapes, plums and citrus fruits, which may have pH below 3.0? Does it have to do with metabolism? With the preservative effects of low pH?


Find second acid dissociation constant at a given pH

For starters, I’m going to define your molecule as being $\ce{HA}$ and its two ionised forms to be $\ce{A-}$ and $\ce{H2A+}$. This means, we are observing the following equilibrium:

$$\ce{H2A+ <=>[H+] HA <=>[H+] A-}$$

As you correctly calculated, you have $10~\mathrm{mmol}$ of $\ce{H_nA^m+}$ and $12~\mathrm{mmol}\ \ce{HCl}$ that you add to your solution.

$12~\mathrm{mmol} > 10~\mathrm{mmol}$. Thus, there are two possibilities:

we started with $10~\mathrm{mmol}\ \ce{HA}$. We ended up with $10~\mathrm{mmol}\ \ce{H2A+}$ and $2~\mathrm{mmol}\ \ce{HCl}$
we started with $10~\mathrm{mmol}\ \ce{A-}$. We ended up with $8~\mathrm{mmol}\ \ce{HA}$ and $2~\mathrm{mmol}\ \ce{H2A+}$

$\mathrm{pH}~7.4$ is only consistent with possibility 2; in case of 1 it would mean $2~\mathrm{mmol}\ \ce{HCl}$ controlling the $\mathrm{pH}$ which would give us $\mathrm{pH} < 3$. Thus, we are well within the operating scheme of $\mathrm{p}K_\mathrm{a,1}$ and we can easily use the Henderson–Hasselbalch equation:

$$\begin{align}\mathrm{pH} &= \mathrm{p}K_\mathrm{a} + \log \frac{[\ce{HA}]}{[\ce{H2A+}]}\\
7.4 &= \mathrm{p}K_\mathrm{a} + \log \frac{\frac{8~\mathrm{mmol}}{V}}{\frac{2~\mathrm{mmol}}{V}}\\
7.4 &= \mathrm{p}K_\mathrm{a} + \log 4\\
7.4 &= \mathrm{p}K_\mathrm{a} + 0.60\\
6.8 &= \mathrm{p}K_\mathrm{a}\end{align}$$

You did indeed identify the spot where you went wrong. Just because a compound is dibasic does not mean you should double its concentration. As I showed above, $10~\mathrm{mmol}$ of original molecule, whether it was entered as $\ce{H2A+}, \ce{HA}$ or $\ce{A-}$ will remain $10~\mathrm{mmol}$; only the number of attached protons will differ.


Are the total number of electrons present in the central atom of triiodide equal to 10?

Looking at the wikipedia article here, your explanation seems to be exactly right:

  The ion is linear and symmetrical. According to VSEPR theory, the central iodine atom has three equatorial lone pairs, and the terminal iodines are bonded axially in a linear fashion, due to the three lone pairs bonding to the central iodine-atom. In the molecular orbital model, a common explanation for the hypervalent bonding on the central iodine involves a three-center four-electron bond.

Thus, the central iodine atom has $3\cdot 2$ valence electrons from the lone pairs and 3 three-center four-electron bonds, so that I speculate you end up with $3\cdot 2$ VE + $4/3 \cdot 3$ VE = 6 VE + 4 VE = 10 VE. Which also fits to the reference given by you.


First off, in neither triiodide nor periodate have the structures you drew. They all assume octet expansion on iodine which has been disproven. It was traditionally explained by the ‘participation’ of iodine’s empty 5d orbitals but these are energetically too far removed and do not take part in bonding to any noticable extent.

Second off, it is wrong to give a formal charge to the central iodine of triiodide. In fact, it is the two outer iodines that carry the negative charge. The structure should be given as follows with a four-electron-three-centre bond between the two iodines:

$$\ce{I-I\bond{...}\overset{-}{I} <-> \overset{-}{I}\bond{...}I-I}\tag{1}$$

Because iodine adheres to the octet rule, there is no contribution by any double-bonded structure. That would create a superoctet structure.

In periodate, there is no question that oxygen is far more electronegative than iodine and that iodine’s oxidation state of $\mathrm{+VII}$ does not go well with a formal or partial negative charge. Thus, it is only logical for the outer oxygen atoms to bear the negative charge. A better way to draw it would be in adherence to the octet rule, i.e. four $\ce{O-}$ substituents and a central iodine with a $3+$ formal charge.

There is nothing inherently wrong with a formal charge on the central atom, though. Take for example tetrahydridoborate $\ce{BH4-}$, the addition product of borane and hydride. Here, boron should only have three valence electrons but if bonded to four hydrogens it must have four. Thus, it carries a formal negative charge. (In reality, the hydrogens carry the negative charge, though, since they are again more electronegative than boron.)


To get one thing out of the way first: toxic does not always mean the same thing. Some things are acutely toxic (e.g. carbon monoxide), some things are toxic upon accumulation (e.g. lead). Some things can get removed from the organism given enough time (e.g. methanol), some things will stay in your body ‘forever’ (e.g. lead). Some things kill you quickly (e.g. cyanide) while others will cause cancer that will kill you in maybe decades (e.g. benzene). And the organism can cope with some things given small enough concentrations (e.g. nitrite) while others are harmful at even the smallest concentrations (e.g. dimethyl mercury). Paracelsus is typically quoted in German with the saying:

Alle Dinge sind Gift, und nichts ist ohne Gift; allein die dosis machts, daß ein Ding kein Gift sei.
(All things are poison and nothing is without poison; only the dosis determines that a thing may not be poison.)

Thus, the question is only which category oxygen falls into: both in terms of acute/chronic/instant/accumulative/… and in terms of $\mathrm{LD_{50}}$ values.

Oxygen is, in principle, a very reactive gas. It is very electronegative and often reacts from its neutral $\pm 0$ oxidation state to compounds in which it features a $\mathrm{-II}$ oxidation state.
Life and living matter builds on being able to selectively use this type of oxidation enthalpy for its own uses, i.e. to build up complex structures (proteins, DNA) or to overcome entropic problems. From its beginning, ‘life’ was basically a method of using inspecific chemical energy for its own uses. Spontaneous reactions out of life’s control are typically suppressed as well as possible, workarounds made and so on. For example, aldehydes are useful, reactive chemical species that can be derivatised in may different reactions. Living organisms tend to make use of similar chemistry (as least partially) but it also tends to avoid the generation of free aldehydes as much as possible. Indeed, one way life keeps aldehydes ‘safe’ is by masking them as Schiff bases or as acetals (e.g. sugars). Life relies a lot on in situ generation of reactive species.
Indeed, imagining an evolution of the first living being in an oxygen-rich surrounding is very hard. When the first life forms came into being, the atmosphere and the conditions on Earth were generally much more reductive, much less oxidative than today. The first living organisms probably used chemosynthetic energy sources that did not rely on oxygen. It took a while for organisms to develop photosynthesis. This had quite a number of beneficial effects for those organisms: the light energy source is much more easily accessible, carbon dioxide to build up organic molecules was plentiful and these organic molecules could then be synthesised according to the organism’s needs either as building blocks or as an energy source. Photosynthesis only has one disadvantage: It has a strongly toxic by-product which can ruin the carefully-tuned enzymes your cell has by oxidising their metal cores, their sulfur atoms and more; and it can react with other molecules in the cell in unwanted ways. It had to be removed from the cell and the first photosynthetic life forms indeed proceeded to dump it into the environment.
At first, the dumped oxygen would react with reactive species and be chemically bound e.g. in rocks at the ocean floor. Partially, it was also absorbed and diluted in the oceans. After a certain oceanic concentration was reached, oxygen was released into the atmosphere and started to accumulate in land-side oxygen sinks by similar mechanisms. After all these oxygen sinks (‘dumps’) were filled, atmospheric oxygen levels began to increase dramatically within a few million years.
The effect that this had on existing life forms — even during the early evolution of oxygen-generating organisms — was catastrophic (hence the term Oxygen Catastrophy). Most life forms — obligate anaerobic ones — would have become extinct and paved the way for an entirely new generation of species: oxygen-tolerant and finally aerobic ones. The obligate anaerobes had no method to control cellular oxygen while later forms had these mechanisms in place.
The great advantage of having molecular, atmospheric or ocean-dissolved oxygen as an energy source is that it can supply a lot more energy than non-oxygenic processes: think of the amount of heat generated by burning and compare that to most other chemical processes to understand the difference. However, any organism relying on the use of oxygen must also cope with the by-products that can occur during oxidation — both desired oxidation procedures and undesired, spontaneous ones. Aerobic organisms tend to have a large battery of peroxidases and superoxide-dismutases for the controlled degradation of peroxides $\ce{R-O-O-H}$ and superoxides $\ce{H-O-O^.}$ — even though these species autodegrade almost at diffusion-controlled rates. The reason is that these reactive species, generated by the use of oxygen, would create too much havoc to allow them to degrade themselves; organisms must degrade them specifically and with minimal danger to the cell as quickly as possible.
And this is also the question why high oxygen levels can be considered toxic. Oxygen reacts with organic molecules present in the cell in an undesired fashion to generate the more reactive peroxides and superoxides. If they cannot be captured and destroyed quickly enough, they create more havoc by randomly oxidising further parts of the cell in undesired fashions potentially causing protein malfunction, DNA damage, cell wall damage and more. The higher the exterior oxygen concentration, the higher the lung oxygen concentration and the higher the blood oxygen level and cellular oxygen level. The higher the cellular oxygen concentration, the more damage it can perform to a cell and the harder it is for the peroxidases and superoxide-dismutases to keep the reactive species under control. Thus, more damage occurs leading, finally, to that bit of damage that can no longer be controlled.
A catch-it-all term for this is oxidative stress. All the different reactions are too numerous and too diverse to be presented here (or even on a medium as Wikipedia), but they all involve the reactive peroxides, superoxides and hydroxyl radicals $\ce{HO^.}$ (which I didn’t mention yet, but which are also terribly reactive).

Tl;dr: oxygen is great in principle, but too reactive. Cells can cope well with ambient oxygen levels and keep the reactive species under control. A rise in oxygen levels will generate too many harmful, reactive species that the cell can no longer control.


What is the molecular shape of Sulfur Pentafluoride Anion?

Yes, $\ce{SF5^-}$ does infact have an octahedral shape. It's structure would be

Yes, it is similar to the structure of $\ce{BrF5}$


I think your question relates to the field of combinatorial chemistry. This is an important area especially for pharmaceutical companies that want to try out lots of different structures to see if they have therapeutic potential.

Lets talk specifics. In the case of known chemistry - e.g. formation of esters from alcohols and carboxylic acids, you can identify the possible reactions. Start with a system where A, B, C, and D are different alcohols and E, F, G, and H are different carboxylic acids. Under the right conditions (acid catalyzed dehydration), the alcohols can combine with the acids all possible ways to give 16 different esters (AE, AF, AG, ... etc.). But A (an alcohol) will not combine with the other alcohols B, C, or D -- and acid E will not combine with acids F, G, or H. Under more harsh conditions, however, they could be induced to react within each group (to form ethers or acid anhydrides).

If you are talking about a random selection of compounds - especially ones with more complex structures and more functional groups, and a variety of reaction conditions, the outcome would be much less predictable.


Would equal masses of gold and copper result in the same volume change if added to water?

"Added to water" is ambiguous.

If the metal objects are shaped and placed such that they float (eg: as a metal boat does, or a light hollow sphere), then they will each displace the same volume of water (the volume of water equivalent to their equal common masses).  So long as they float, they will displace the same amount of water no matter their shape.

If the masses objects are shaped and placed such that they sink (eg: consider a solid sphere or cube), then they will in general displace different volumes.

Bonus.  The second part is slightly more complex than it seems when it comes to calculating how much water they would displace.  If they have no dry internal voids, their respective displaced volumes are simply the mass of each divided by its density.  Others have given the simple calculation.

If they were shaped such that they had dry voids yet were small enough to sink (think of a heavy hollow sphere with overall density greater than water so it sinks, or an overturned metal boat with small air pockets), then the displaced water will be somewhere between the above two calculations (fully floating calc and no-voids calc).

Note that it would be possible to construct the two objects such that they sank and yet displaced the same amount of water, eg: two hollow spheres with the same outer diameter but different internal diameters, both heavier than the same volume of water.  But that would be a very trick question indeed!


How does the process of protein denaturatation work?

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

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


Can you lower the alcohol concentration of a drink using vegetable oils?

Liquid-liquid extraction is a widely used method, but it's normally used the other way round: You add e.g. ether to a solution of your valueable substance X in water, separate the phases, and evaporate the ether to get pure X. Add some more ether to the aqueous phase, etc, two-three times.

Now you can be quite sure that you have recovered all of your substance X, but also your aqueous phase will be saturated with ether. No problem if you'd discard it anyway, not so good for drinking. :-/

It's also bad if X, like ethanol, is well soluble in both phases. For hard liquor, the whole mixture is likely totally miscible, and for beer, you'll need a lot of oil to get any significant effect. And then what do you do with the ethanolic oil afterwards?


Thorium dioxide can be dissolved in the mixture of hydrofluoric acid and hydrochloric acid, as described by Kluger and Lieser in 1981.1

Proceed with caution since hot condensed acid are known to be highly corrosive and hydrofluoric acid is also highly toxic and penetrating.

Kluge, E.; Lieser, K. H. Dissolution of ThO2 in HCl/HF-mixtures. Radiochim. Acta 1981, 29 (2–3), 167–168. DOI: 10.1524/ract.1981.29.23.167.


What makes metals malleable and ductile?

The metals are malleable because they don't make molecules : they exist as individual atoms. In a metallic piece, the atoms are identical, and they are piled up like layers of beads in the bottom of a flat box, each bead touching its neighbors. The first layer of beads covers the entire surface of the surface. You may imagine a second layer identical to the first one covering it, then a third layer, and so on. Hopefully you see that is not very difficult to push a whole layer, or a whole pile of such layers from the side. A whole layer can slide easily on a lower layer. This would not break the whole piece, which gets distorted but not broken. It would be much more difficult to do it if suddenly one or some of these beads are replaced in a given layer by bigger beads or by non-spherical objects. ${}{}{}{}{}$


Based on the structure, this drug is weakly basic. It has an amine moiety, which if protonated, would likely have a $\mathrm{p}K_{\mathrm{a}}$ value of around 9.5.


Detailed description of hydroxy(phenyl)methanesulfonic acid?

Hydroxy(phenyl)methanesulfonate is a compound of a class commonly called "bisulfite addition products" or "bisulfite adducts", since bisulfites are commonly used for the reaction you have mentioned. But there's an equilibrium in aqueous environment between metabisulfites and bisulfites
$$\ce{S2O5^{2-} + H2O \rightleftharpoons 2HSO3^{-} }$$
in favor of the bisulfite, this is why metabisulfites can be used as well, probably.

The bisulfite anion is proposed to have two tautomers:

The form at the left has a lone electron pair at the sulfur atom, and can act as a nucleophile.

The carbonyl/aldehyde group $\ce{>C=O}$ has a partial positive charge at the carbon atom.

The reaction in now no mystery, it's a nucleophilic addition to the carbonyl group.

If I may cite a crude reaction mechanism from Clayden/Greeves/Warren's Organic Chemistry, short chapter Bisulfite addition compounds:

And from the March's Advanced Organic Chemistry:

  The reaction is reversible (by treatment of the addition product
  with either acid or base) and is useful for the purification of the starting compounds, since the addition products are soluble in water
  and many of the impurities are not.

If you want more details (kinetics, etc.), there is freely available couple of scientific articles, Sulphur isotope effects in the bisulphite addition reaction of aldehydes and ketones

I. Equilibrium effect and the structure of the addition product
II. Bond-formation effect


As you have already mentioned, the key to the answer of this question is the seniority order for classes that are expressed by suffixes.

Esters are senior to hydroxy compounds (including alcohols). Therefore, the preferred IUPAC name for the corresponding glycolic acid ethyl ester with an intact hydroxy group is ethyl hydroxyacetate. (Note that according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the locant ‘2’ in ‘ethyl 2-hydroxyacetate’ is not cited.)

However, anions are senior to esters. Therefore, the anion formed by subtracting a $\ce{H+}$ from the hydroxy group of ethyl hydroxyacetate is named as ethanolate. The ethanolate is substituted using the usual principles of substitutive nomenclature. The prefix ‘oxo’, denoting $\ce{=O}$, is used to indicate a carbonyl group when the group cannot be cited as suffix. The group $\ce{-O-CH2-CH3}$ is indicated by the prefix ‘ethoxy’. This leads to the systematic name 2-ethoxy-2-oxoethan-1-olate for the anion and potassium 2-ethoxy-2-oxoethan-1-olate for the salt.


If your anode/cathode are from copper, then the blue chemical in the solution is copper chloride (CuCl2).  Copper chloride can be toxic. So you should have safety equipment.

As for rust removal in general, I'd recommend you to use Citric acid (Squeezing lemons for the lemon juice isn't efficient since it will contain 5-6% Citric acid - so its better to buy powdered Citric acid from a store).


Why does the pH of this solution rise instead of fall when exposed to the atmosphere?

At that pH, you're well outside the buffering capacity of both phosphate and boric acid, so a $\mathrm{pH}$ rise from $5.5$ to $5.9$ corresponds to a net increase in base of about $2\ \mathrm{\mu M}$.  That implies to me that it's likely something going on with one of the first five ingredients in your recipe, or the $\ce{HCl}$ added to adjust the $\mathrm{pH}$.

I can think of two scenarios:

Similar to Brinn's answer, perhaps a small amount of $\ce{HCl}$ has evaporated out of solution.
If your $\ce{Fe-EDTA}$ was added as $\ce{Fe(II)-EDTA}$, oxidation to $\ce{Fe(III)-EDTA}$ could explain the pH rise, by reducing water to hydrogen gas and hydroxide ion:

  $2\ \ce{Fe(II)-EDTA} \rightarrow 2\ \ce{Fe(III)-EDTA} + 2\ e^-$

  $2\ \ce{H2O} + 2\ e^- \rightarrow \ce{H2} + 2\ \ce{OH-}$

I don't know offhand how thermodynamically or kinetically feasible #2 is.  With all of the other metals present in solution, though, it's not unreasonable to imagine that some sort of catalysis might be possible.


Does lowering temperature of a gas make it easier to compress?

Compressing a gas by applying an external force means to work against the force the gas exerts on the walls of its container, i.e. its pressure.

The definition of compressibility

$$\beta = -\frac{1}{V}\frac{\partial V}{\partial p} \tag{1}$$

for an ideal gas with the well-known equation of state

$$p V = n R T \tag{2}$$

gives

$$\beta = -\frac{1}{V}\frac{\partial V}{\partial p} = \frac{1}{V} \frac{n R T}{p^2} = \frac{1}{p} \tag{3}$$

which is independent of the temperature $T$.

For non-ideal gases there will be higher order terms that also depend on temperature.

It is also worth noting that at the critical point the compressibility of a real gas is infinite. So if you are below the critical temperature you will be able to compress the gas more easily at a state (the critical point) with higher temperature.


Must a good leaving group be a weak base, or a weak nucleophile?

It is a common misconception that good leaving groups should be weak nucleophiles. However, there is a simple counterexample: iodide is known to be both a good nucleophile and a good leaving group in $\mathrm{S_N2}$ reactions.

However, it does make sense for the leaving group to be defined in thermodynamic properties. Consider the reaction equation for a nucleophilic substitution following an $\mathrm{S_N2}$ mechanism:

$$\ce{Nu- + R3C-Lg -> [Nu\bond{...}CR3\bond{...}Lg]^\ddagger -> Nu-CR3 + Lg-}\tag{1}$$

Nucleophilicity, i.e. the ability of $\ce{Nu-}$ to attack the reaction centre is, as you noticed, a kinetic property: the nucleophile has to approach the electrophilic carbon atom from a certain direction in a certain angle with appropriately constructed orbitals. The leaving group itself does play a slight role here in determining the energy of the $\unicode{x3c3}^*$ orbital and thus whether it is energetically favourable for the nucleophile to attack.

However, the leaving group’s principal characteristic is to diffuse away after the reaction. If it is diffusing away, its kinetic properties no longer influence how the reaction proceeds as it is not attacking anything anymore. Rather, a favourable overall reaction enthalpy is achieved if $\ce{Lg-}$ is able to stabilise the negative charge well (or lone pair, if e.g. $\ce{Lg} = \ce{NR3+}$). This is indeed a thermodynamic property, not a kinetic one. And stabilising a negative charge directly correlates with being a weak base.


Comparing acidities of substituted and aromatic carboxylic acids

In the case of 3-chloropropanoic acid vs. formic acid, I suspect the disparity in expected acidity can mostly be explained by solvation effects. The 3-chloropropanoic acid is much bulkier than formic acid, and hence interacts with solvent molecules differently than formic acid does. There are two essential components to the thermodynamics of solvation:

Due to steric hindrance, it may be the case that the negative charge is less accessible to solvent molecules, and thus more weakly stabilized. I suspect the difference in enthalpy here is actually negligible, but I'm not able to find a reference at this time.
Perhaps more importantly, there is an entropic effect, in that larger molecules tend to require a larger number of solvent molecules to solvate them. These solvent molecules form highly ordered, cage-like structures, which is entropically unfavorable. I believe similar underlying mechanisms are at work in, e.g., the hydrophobic effect, micellization, protein folding, etc.

As Klaus Warzecha points out, if the chlorine were not as far-removed from the carboxylate moiety (since induction rapidly diminishes with growing distance), the strength of the inductive effect would trump the solvent effects.

In the case of benzoic acid, there are probably similar solvent effects occurring. Additionally, the benzene ring is actually electron-donating within the $\pi$-electron system of the molecule, which can only be destabilizing to the carboxylate conjugate base. Unlike in, e.g., phenol or aniline, the excess negative charge cannot actually be delocalized into the ring. Experimentally, the electron-donating effect of the ring (equivalently, the electron-withdrawing effect of the carboxylate) is confirmed by the fact that benzoic acid is severely deactivated in electrophilic aromatic substitution reactions by comparison to ordinary benzene. The benzene ring's electron-withdrawing effect within the $\sigma$-framework (rationalized, according to valence bond theory, by the greater s-character of $\mathrm{sp^2}$ carbons) of the molecule is probably quite weak. There may also be Coulombic destabilization between the strong $\delta^+$ of the carboxylate carbon atom and the $\mathrm{sp^2}$ carbons of the aromatic ring, again due to greater s-character, although that's speculative.


About oxidation states, I find it easier to follow the extended definition, as it is explained in this question: Electronegativity Considerations in Assigning Oxidation States

Look at the first compound $\ce{Cr(OH)3- }$. What is the most electronegative element? Assume it gets all the electrons to fill the octet.

   Oxygen, it shall have the oxidation state $-2$

What about the other atoms?

   Hydrogen can only receive or lose an electron, in this case it is bonded to the much more electronegative element oxygen, it shall have the oxidation state of $+1$
Chromium gets the oxidation state of $+2$, since $\ce{O}(3\cdot -2) + \ce{H}(3\cdot +1) -\ce{e-}(1\cdot-1)=-Ox(\ce{Cr}) = +2$

Look at the hydrogen peroxide ion, apply what you know about Hydrogen to calculate the oxidation state of oxygen.

   Hydrogen gets $+1$ (see above)
 Charge has to remain, i.e. $-1$
 Oxygen gets $-1$, since $\ce{H}(1\cdot +1) -\ce{e-}(1\cdot-1)=-\frac12Ox(\ce{O})=-1$

Now the hydroxyl ion is straightforward.

   Hydrogen gets $+1$ (see above)
 Oxygen gets $-2$

Look at dichromate and apply the above.

   Oxygen gets $-2$, most electronegative
 Chromium gets $+6$, since $\ce{O}(7\cdot -2) -\ce{e-}(2\cdot-1)=-\frac12Ox(\ce{Cr}) = +6$

In total that will give you:

   $\ce{\overset{{+II}}{Cr}~(~\overset{{-II}}{O}~\overset{{+I}}{H}~)3{}^{-} + \overset{{+I}}{H}~\overset{{-I}}{O}_2{}^{-} -> \overset{{+IV}}{Cr}_2~\overset{{-II}}{O}_7{}^{2-} + {}^{-}\overset{{-II}}{O}~\overset{{+I}}{H} }$


Intrinsic Reaction Coordinate - Converting To Reaction Coordinate with Units of Distance

Intrinsic reaction coordinate (IRC) is not some internal coordinate of a molecular system (such as, bond length, bond angle, etc.), rather it is a curvilinear coordinate that describes the intrinsic reaction path (IRP), which is the reaction path along the direction of the gradient. Only for some simple reactions one can visually (and approximately) identify IRC as one of the internal coordinates, but in general that is not possible.


Is it possible to prevent methane combining with oxygen?

I doubt that the conditions in an Anaerobic digester (organic acids, ammonia (!), etc) allow algae to grow. If they do, the oxygen they set free will create oxic conditions and stop the AD - Methanogens are obligate anaerobe bacteria.
In nature, Methanogens and Algae inhabit different habitats - methanogens in the mud at the bottom of a swamp, algae (maybe) in the water above.

Processes for separating $\ce{CO2}$ from $\ce{CH4}$ usually work on the $\ce{CO2}$ (absorption in amines or water under high pressure), so it would be hard to pull the $\ce{CH4}$ from the headspace of your vessel - but given the biological points, this is a non-issue.


Benzyl Carbocation Stability: Would electron-withdrawing group on the ring destabilize the carbocation?

The reaction shown here is a Finkelstein reaction and SN2 reaction mechanism is followed.is an organic reaction that uses an alkyl halide exchange into another alkyl halide through a reaction wherein the metal halide salt is used. This reaction takes place at an equilibrium process by taking the advantage of poor acetone solubility in metal halide salt that is newly formed. The mechanism of the Finkelstein reaction is single-step SN2 reaction with stereochemistry inversion.

The success of this reaction depends on the below conditions.

Nucleophilicity of your halogen
Nature of leaving group
Carbon-halogen bond
Alkyl Halide reactivity

Holtz and Stock showed that electron-withdrawing groups increase the rate of SN2 reactions. This can be ascribed to stabilization of the transition state by withdrawal of some of the electron density. So NO2 will make reaction faster.

Ref:ADVANCED ORGANIC CHEMISTRY: REACTIONS, MECHANISMS AND STRUCTURE, 4TH ED


Unknown alcohol sample. Chromic acid negative but iodoform positive?

I would recommend to repeat all the experiments one more time. The reasons for such results may be:

You had dirty test tube for ceric nitrate test, and it was really false positive. You could have a methyl ketone, which gives negative chromic acid test and positive iodoform test.
You added too much chromic acid, and had low amount of your "alcohol". That caused all alcohol to be oxidized, but that blue-green color was too faint, and you didn't notice it because of chromic acid excess.

Moreover, if your "alcohol" is immiscible with water, that means it is one of the higher alcohols. You could also try to identify it (if it's really an alcohol) by smell: those alcohols have pretty distinctive smells. x.x