Can other substances be dissolved in a saturated solution? If I have a liter of water fully saturated with sucrose would it be possible to dissolve something like salt or any other substance in the water? Or when the solution is saturated, is it impossible to dissolve another solute in it?

Saturating a liquid with one solute does not mean that the liquid will no longer dissolve another solute. However, you can expect the solubility of the second solute to be different, generally lower, than in the neat solvent.

One relevant concept here (though not specifically applicable to sucrose), in the case of ionic solutes, is the common-ion effect. According to this Wikipedia article:

  The common ion effect is responsible for the reduction in the solubility of an ionic precipitate when a soluble compound containing one of the ions of the precipitate is added to the solution in equilibrium with the precipitate. It states that if the concentration of any one of the ions is increased, then, according to Le Chatelier's principle, some of the ions in excess should be removed from solution, by combining with the oppositely charged ions.

Regardless of whether the solute is ionic or not, when you are given a solubility value, for example, the solubility of some compound in water as $\pu{g solute/100mL water}$, this value is only relevant to the solubility in pure water. Once you have saturated water with sucrose, you then have a very different solvent system. There are no simple means of determining what the solubility of a particular solute will be in a solution saturated with another solute. However, there will, in general, be some solubility of additional solvents except as in the case described for the common ion effect.

Please don't hesitate to ask for clarifications in the comments below if I have misunderstood your question or left anything out.

One of the 'reasons' something dissolves is because the solution has a thermodynamically higher energy (more negative free energy) than the two separate compounds. Ionic compounds dissolve in water into two (or more) separate ions, so things are a bit more complicated than molecular dissolution like sugar in water. The simplest case is two chemicals which can be considered "non-polar". If you are familiar with the various van der Waals forces, then we can describe these as those chemicals which have only London dispersion interactions. These can be thought of as quantum mechanically induced transient (instantaneous) polarization of the electron clouds of two adjacent atoms. This would be true for two atoms with the same electronegativity, for example. Saturated hydrocarbons are in this category (methane, ethane, propane, etc.) since the H and the C atoms making them up have nearly the same electronegativity (but not quite), which suggest there aren't any permanent polarizations. So given two hydrocarbons, A and B, there won't be much stronger interaction between a molecule of A and another A molecule compared with a molecule of B (or the interaction between two Bs). This is only true up to a point. High molecular weight hydrocarbons aren't infinitely soluble in methane, for example, because of steric (geometric) influences. Big molecules (generally, polymers) are better able to align themselves with similar big molecules (compared to smaller ones which "bounce around" in random orientations. So you see (perhaps) that even the simplest case isn't subject to complexities. The next step towards complete charge separation is induced dipoles. It is possible for a ion or even a molecule with a polar group to change the charge distribution around a neighboring group or molecule or atom. For a given electric field, the induced dipole depends on how "loose" or "tight" the electrons are being held by that molecule (or group (molecular fragment) or atom). Explaining this in further detail is too complicated for this, I think. The next step (of course there aren't really steps, there's all sorts of "in-between" cases) is permanent dipoles. Now, depending on the geometry two dipoles can "cancel" and a molecule may have no net dipole. CO2 is an example. Although both C=O bonds are polar, the molecule is linear ← → so while the end oxygen atoms are more electronegative than the middle carbon both positive and negative center of charge is in the middle (centered at the C atom) so the dipole effects are much diminished compared to say H2O where the molecule is not linear, more of a V shape so that the partially negative charge on the central O atom is separated from the partial positive charge center (which is midway between the two H atoms, meaning on the geometric bisector of the central angle but away from the O atom. (the fact that water has a permanent dipole makes it a great material to use in a microwave oven, the microwaves make those little puppies dance like crazy (heats the heck out of them) and so makes microwave ovens effective at heating foods (the ones that contain water, anyway).) HCl, NH3 also have permanent dipoles (but ammonia does some funky stuff with its geometry). Materials with permanent dipoles tend to be soluble in water because the electric field orients the polar H2O molecule (negative electric field attracts the H atoms more than the O atom, and positive electric field the opposite). Water molecules surround and solvate polar and ionic chemicals. So, I've already mentioned the "last" step, which is full charge separation - ions. Of course there actually are ions which have non-integer charge, remember covalent bonds are basically shared electron pairs. There tend to be lots of water molecules oriented around each ion, which means there are fewer available to surround other ions, and other polar groups. So, that means more of one ionic solute will tend to reduce the solubility of another ionic solute. The important caveat here is the assumption that the ions (both cations and anions) aren't increasing the free energy (in a negative direction). If for some reason Na+ interacted with the -OH (hydroxy group) of sugar (and sugars have lots of hydroxy groups) and improved the thermodynamics then salt would increase the solubility of sugar and vice versa. I don't know of any cases where this happens. In fact, salts are frequently used to "salt out" organic compounds (meaning they reduce the solubility), sometimes very dramatically. So I've mentioned everything except what happens when you dissolve two polar chemicals in water. I've pretty strongly implied that they'll reduce each other's solubility. (BTW, you understand that when you've saturated a solution with one ionic species, and add another, that some of the first will preciptiate, as discussed above, right?) Well, that gets a little murky if one or both of the solutes are present in large enough concentrations to affect what the solvent is. Sucrose is soluble in water at 2100 grams per liter at 25°C. So, given a saturated solution, is the solvent the water or the sucrose. Which is solvent, which solute? I could argue it either way and in fact there's no one answer - it depends on the context. So, if you've got such a solution, what would happen to the solubility of some other polar compound? It turns out the answer is "it depends" there are examples of it decreasing solubility and examples where it increases solubility. No simple answer for highly concentrated solutions. (compounds where the solute approaches solvent concentration).


Isotopes and electronegativity Is there any difference between electronegativities of isotopes of the same element. For example consider the above compound. Will all the resonance structures be equivalent?

Also if I place the above compound in water (which will act as a Bronsted acid), which carbon will get the H atom?

(Star represents $\ce{^14C}$ atom)
This figure represents three out of the five possible resonance structures. Of these five structures, four have a negative charge on a $\ce{^12C}$ atom whereas one has negative charge on $\ce{^14C}$ atom.
As the first premise, let's consider stability of the two kinds of resonating structures. The more stable the resonating compound, the more likely it is to be protonated.

We can compare stability using the premise of electronegativity (or rather charge density). According to this property, the more heavy an atom, the larger is its volume. If volume increases, the charge density decreases making it less likely to be able to stabilize the negative charge (becomes less electronegative). Due to this we can say that the resonance structure with $\ce{^14C}$ having the negative charge is less stable and so is less likely to be protonated.
Now, moving onto the second premise, the number of resonating structures containing the negative charge on $\ce{^14C}$ and $\ce{^12C}$ respectively.
If you notice only one resonating structure has a $\ce{^14C}$ with the negative charge, whereas there are $4$ resonating structures containing $\ce{^12C}$ atom with the negative charge. This would mean that probabilistically (considering stability to be equal) The resonance structure having a $\ce{^12C}$ with the negative charge would be more likely to be attacked.
Now, if we combine the two premises, we see that both stability and frequency of the resonance structure containing $\ce{^12C}$ with negative charge is greater its $\ce{^14C}$ counterpart. Concluding, we can say that the resonating structure with $\ce{^12C}$ stabilizing the negative charge would be more likely to be protonated.
Further Reading on how stability of isotopic compounds can be determined :  Inductive effect of hydrogen isotopes

Yes, it is there due to the increased size of nucleus. But, the difference in the electronegativity of the two isotopes would be so small that you can easily neglect it.
The resonating structure will be different in $\ce{^14C}$ and $\ce{^12C}$. The reason is $\ce{^14C}$ is radioactive (a very small effect) so it may cause  a difference between the chemical properties of the two resonating structures.


What is SPDF configuration? Recently in my chemistry classes, the teacher spoke about SPDF configuration and then said that we'll be taught about it in higher classes.

But I'm sorta curious to know that what is SPDF configuration and is there something like - obtaining electronic configuration in SPDF format?

s, p, d, f and so on are the names given to the orbitals that hold the electrons in atoms.  These orbitals have different shapes (e.g. electron density distributions in space) and energies (e.g. 1s is lower energy than 2s which is lower energy than 3s; 2s is lower energy than 2p).

(image source)

So for example,

a hydrogen atom with one electron would be denoted as $\ce{1s^1}$ - it has one electron in its 1s orbital
a lithium atom with 3 electrons would be $\ce{1s^2~2s^1}$
fluorine has 9 electrons which would be $\ce{1s^2~2s^2~2p^5}$

Again, as we read from left to right the energy of the orbital increases and the superscript shows the number of electrons in the orbital. You can read more here.

They refer to the secondary quantum numbers (L) :

The S holds one orbit which contains 2 electrons
The P holds three orbits which means 2×3=6 electrons
D holds five orbits 2×5=10 electrons
F holds seven orbits 2×7=14 electrons


Hazard and precautionary statements Is there a scientific database where one can find all the H and P statements for all chemicals? I would like to avoid using wikipedia but I can't find a database from a trusted source.

The most comprehensive one (so not "all" chemicals) that I've seen is the Aldrich website. By law they have to supply MSDS for everything they sell. However, they may ask for a lot number or the function may not work unless you're logged in as a customer.


What is meant by a 'hydrophilic' sphere? So, I know that a hydrophilic material is defined as those with a special affinity for water, opposite to
hydrophobic, those that naturally repel water.

I am investigating the impact of spheres with water surfaces, and the splashes caused by them. However, one research paper focuses on the difference between the fluid dynamics of hydrophilic spheres and hydrophobic spheres, and the difference in the cavities they make when they sink underwater.

I understand what a hydrophobic sphere is, but what is a hydrophilic sphere? Like a rolled up ball of tissue paper? Or the water absorbing waterbeads/orbeez that get bigger(the ones children play with)?

The answer to your question depends on the size level (microscopic or not). But it sounds like you are referring to spheres at a size that you can hold, or larger.

A hydrophilic sphere could simply be a sphere of some solid material (like plastic), coated with a layer of hydrophilic material/substance.

This page on hydrophilic coatings has some information on what this material may be.

Here is a educational reference: '19.2 COORDINATION CHEMISTRY OF TRANSITION METALS, to quote:

  The coordination sphere consists of the central metal ion or atom plus its attached ligands. Brackets in a formula enclose the coordination sphere; species outside the brackets are not part of the coordination sphere. The coordination number of the central metal ion or atom is the number of donor atoms bonded to it. The coordination number for the silver ion in [Ag(NH3)2]+ is two (Figure 3). For the copper(II) ion in [CuCl4]2−, the coordination number is four, whereas for the cobalt(II) ion in [Co(H2O)6]2+ the coordination number is six. Each of these ligands is monodentate, from the Greek for “one toothed,” meaning that they connect with the central metal through only one atom. In this case, the number of ligands and the coordination number are equal.

So, in the presence of a water ligand, the adjective 'hydrophilic' is appropriate. The sphere, itself, is the central metal ion and associated ligand enclosed in brackets, where per my readings, the brackets are sometimes missing.


How can meta-cresol be made, given that the hydroxyl group is ortho,para-directing? How is it possible for m-cresol (3-methylphenol) to be synthesized?

The methylation of phenol would lead to the ortho- and para-isomers instead.

Here is a synthesis of m-cresol starting from benzene.  It achieves the desired meta orientation of the final substituents by introducing a meta-directing substituent (the aldehyde group) early-on in the synthetic scheme. In the nitration step, the meta isomer is by far the major product (o:m:p 19:72:9), but a distillation or recrystallization would be needed (or you could carry all the isomers along and separate later, but you'd probably have the widest range in boiling points at this stage with the nitro group present). The only novel step is the double Wolff-Kishner used to reduce both the aldehyde and nitro groups in a single reaction.

From the comments above, its clear you are asking how m-cresol is prepared.
I know one method, which involves diazotization of m-toluidine.

In this way we should make m-toluidine first but it making m-toluidine is more difficult than m-cresol.

m-toluidine is prepared by m-nitrotoluene but unfortunately m-nitro toulene is minor isomer in nitration of toluene and this is really so difficult to separate it.
But you can prepare M-cresol thorugh this process, though the yield of the reaction will be small.

Though I will be looking for more answers on this question if someone knows another way of preparing this particular compound.
Hope this helps.


What is the correct way to verify a structure's geometry, for example for benzene? My goal is to compare a calculated bond length with experimental data. Benzene is obviously common enough to expect sufficient experimental data to be available for it on the internet.

I have found two sources for the C-C bond length in benzene, and the numbers differ very slightly. This source says 140 pm for benzene, while this source says 139 pm for benzene. Both claim their number comes from x-ray diffraction.

Is the discrepancy possibly attributable to a difference in solvent? Is a difference of 1 pm even significant? What is a good way to know for sure which quantity is correct?

A deviation within the low picometres is nothing to worry about, there are many reasons for this.

Primary literature, like peer-reviewed journals, will always publish an analysis of the obtained values along with any error bars, i.e. how accurate the measurement is. When you have disagreeing values it is worthwhile to have a look at these, too.
Secondary literature, like encyclopaedias, will usually only quote values and might not include all details due to their general reference nature.

Quite often if you have different values, the conditions of the measurement simply were not the same. Also keep in mind, that within the recent years a lot of progress has been made for spectroscopic measurements. Hence you can find values from older publications that are significantly different from more modern values.

For example Tamagawa et al. reported $r_g(\ce{C-C}) = 139.9 \pm 0.1~\mathrm{pm}$.[1] They used a method, which combines diffraction and spectroscopic data. This article was published in 1976 and it was introduced with the following paragraph. I think that pretty much summarises the predicament of modern structural analysis.

  It is often found that the structural data of the most basic molecules are not very up-to-date because such molecules were studied in the early stages of the development of experimental methods for structure determination. Benzene is one such molecule.

And since then many more studies on benzene have probably been published. Chemspider.com lists (currently) seven data sets[2] and I am confident this is just the tip of the iceberg.

In a quite recent publication by Budzianowski and Katrusiak the crystal structure of benzene has been studied at various pressures. They pressure froze the molecule(s) in situ in phase I.[3]
We learn from this, that there is more than one phase in which benzene crystallises. And a little more digging leads us to a publication by Piermarini et al. dealing with the crystal structure of benzene II at 25 kbar.[4] But I will not go into that and rather focus on the former publication.
This paper illustrates nicely that there cannot be only one true value for the compound. I will reproduce a part of table 2 from the publication here.
$$
\text{Bond lengths (}\mathring{\mathrm{A}}\text{) and angles (}^\circ\text{) for benzene in phase I.}\\
\begin{array}{llll}
             & 0.3~\mathrm{GPa} & 0.7~\mathrm{GPa} & 1.1~\mathrm{GPa} \\\hline
\mathrm{C1—C2}     & 1.40 (4)   & 1.397 (8)        & 1.420 (16)       \\
\mathrm{C2—C3}     & 1.38 (8)   & 1.384 (10)       & 1.41 (3)         \\
\mathrm{C3—C1^i}   & 1.38 (5)   & 1.354 (10)       & 1.32 (2)         \\\hline
\end{array}\\
\text{Symmetry code: (i)} 1-x; -y; -z.
$$

As you can see even in a single crystal structures, the bond lengths are not equal. This is obviously due to packing effects and intermolecular interactions. When we talk about a single value for the carbon carbon bond length in benzene, then we assume an idealised, even rigid, molecule.

And there are more ways to obtain the geometry of a molecule: IR, microwave, Raman to name a few.[5] Nowadays we can employ computational chemistry and try to predict bond lengths, but there is a catch to that, too. These calculations take place at absolute zero and no pressure and we have a couple of other approximations in the mix. You can have a look at the Computational Chemistry Comparison and Benchmark DataBase, which lists values from 138.3 to 144.3 pm for the carbon carbon bond.[6] For example, my own calculation on the MP2/def2-SVP level of theory gives me a carbon carbon bond length of 140.0 pm. But that is a value that corresponds to a stationary point on the potential energy surface. We all know that this molecule has thirty vibrational modes, which are not yet accounted for. Below is just one, one that I find particularly funny.

TL;DR Don't worry about deviations of about 1 pm. When there are larger differences, look at the original publications about the conditions and trustworthiness of their measurements. Always consider the error margins on your own experiments when you compare values. It also depends on how accurate you need your value to be before you go and dig deep into publications from 50 years ago. (For what it is worth, you could round 139 pm to 140 pm.)

Notes and References

Tamagawa, K.; Iijima, T.; Kimura, M. Molecular structure of benzene. J. Mol. Struct. 1976, 30 (2), 243–253. DOI: 10.1016/0022-2860(76)87003-2.
CSID:236, http://www.chemspider.com/Chemical-Structure.236.html (accessed November 18, 2018)
Budzianowski, A.; Katrusiak, A. Pressure-frozen benzene I revisited. Acta Crystallogr Sect B 2006, 62 (1), 94–101. DOI: 10.1107/S010876810503747X.
Piermarini, G. J.; Mighell, A. D.; Weir, C. E.; Block, S. Crystal Structure of Benzene II at 25 Kilobars. Science 1969, 165 (3899), 1250–1255. DOI: 10.1126/science.165.3899.1250.
Wikipedia has a longer list: Molecular Geometry.
Unfortunately I am unable to generate a permalink from the database: http://cccbdb.nist.gov/ I welcome you to play around a little by yourself.
However, I'll include the summary from the geometry comparison between the C1–C2 bond length:

  distance is atom 1 (C) to atom 2 (C)
  Experimental bond length is 1.397 Å     Calculated bond lengths range from 1.383 to 1.443 Å
  The bar charts show differences (theory - experiment) and the range is −0.046 to 0.046 Å

This is just a logical explanation but this might or might not be wrong.
Due do resonance in benzene if we look at canonical forms, at a single moment there are alternate single and double carbon carbon bonds..due to rapid interchange between these positions the average(or experimentally calculated) value of carbon-carbon bond in benzene  will oscillate between $\mbox{139}$–${\mbox{140 pm}}$.


Can potassium nitrate and sulfuric acid be used to nitrate organics? If we are doing a classic nitration to produce glyceryl trinitrate, one uses a mix of concentrated sulfuric and nitric acids. Additionally, one can produce concentrated nitric acid by distilling it from a mixture of potassium nitrate and sulfuric acid.
So, is it possible to do a nitration by mixing the glycerol with sulfuric acid and then adding potassium nitrate?
Disclaimer
In case anyone is wondering, I am not going to try this, and I do know how to make many better and safer explosives since I used to be a member of the alt.engr.explosives group way back.

A simple, but dangerous path to nitrating a metal oxide (say MgO) is to treat with NO2 gas (caution: toxic and corrosive), as to quote Wikipedia on Nitrogen dioxide:

NO2 is used to generate anhydrous metal nitrates from the oxides:[10]
$\ce{MO + 3 NO2 -> M(NO3)2 + NO}$

where heating mix of Cu, KNO3 and H2SO4 will generate NO2.
A safer path to nitrating, starts with the introduction of the powerful  radicals, like the hydroxyl radical (•OH). The latter can transform commonly available nitrate salts  (like KNO3) to the nitrate radical (•NO3) especially in the presence of an acid catalyst per the reaction:
$\ce{•OH + NO3- <=> OH- + •NO3}$
A supporting source reference, Nitrate radicals and biogenic volatile organic compounds: oxidation, mechanisms, and organic aerosol. Per the paper's abstract:

Oxidation of biogenic volatile organic compounds (BVOC) by the nitrate radical (NO3) represents one of the important interactions between anthropogenic emissions related to combustion and natural emissions from the biosphere.

Also, see The Reactivity of the Nitrate Radical (•NO3) in Aqueous and Organic Solutions, by
Stephen P. Mezyk, Thomas D. Cullen, Kimberly A. Rickman and Bruce J. Mincher, per the abstract to quote:

ABSTRACT
Rate constants for the nitrate (•NO3) radical reaction with alcohols, alkanes, alkenes, and several aromatic compounds were measured in aqueous and tert‐butanol solution for comparison to aqueous and acetonitrile values from the literature. The measured trends provide insight into the reactions of the •NO3 radical in various media. The reaction with alcohols primarily consists of hydrogen‐atom abstraction from the alpha‐hydroxy position and is faster in solvents of lower polarity where the diffusivity of the radical is greater. Alkenes react faster than alkanes, and their rate constants are also faster in nonpolar solution. The situation is reversed for the nitrate radical reaction with the aromatic compounds, where the rate constants in tert‐butanol are slower. This is attributed to the need to solvate the NO3− anion and corresponding tropylium cation produced by the •NO3 radical electron transfer reaction. A linear correlation was found between measured rate constants in water and acetonitrile, which can be used to estimate aqueous nitrate radical rate constants for compounds having low water solubility.

With respect to the hydroxyl radical, the UV photolysis of a photocatalyst like TiO2 in aqueous KNO3, could be a path to the powerful, but very transient, hydroxyl and nitrate radicals.

Nitration reactions rely on the absence of water to proceed efficiently, in fact part of the purpose of the sulfuric acid in the mixture is as a drying agent to remove the water produced by the nitration reaction.  Because potassium nitrate (and potassium sulfate for that matter) is not meaningfully soluble in concentrated sulfuric acid, it is impossible to create the dry solution required by this reaction in this way.


How will the periodic table change due to redefinition of the kilogram (May 2019)? The definition of a kilogram has recently been changed. What effect, if any, will this change have on data on the periodic table?

Molar masses

The molar mass of the carbon-12 isotope used to be 12 g/mol exactly (by definition). Now, it is experimentally determined, with the current value $\pu{11.9999999958(36) g mol-1}$ (CODATA 2018). So the molar mass of carbon-12 changed. The standard atomic weights (relative atomic masses) shown in the periodic table are not affected by this. However, atomic weights (dimension one) are no longer exactly equal to molar mass divided by 1 g/mol. Instead, they differ by a tiny bit (you would need 9 or 10 significant digits to see the difference).

No visible changes

Here is an animated gif of a portion of the periodic table at ptable.com over the past four years (Jan 23rd 2016, Jan 10th 2017,Jan 6th 2018, May 2nd 2019). Nothing much has changed near carbon:

The standard atomic weights change a bit each year, as better estimates of the isotope compositions of elements with multiple stable isotopes are made. Also, the publishers of this specific periodic table decided to truncate atomic weights to 5 significant figures in 2018. The biggest change occurred when elements in period 7 were discovered and named.

The last image is from May 2nd 2019, before the change of the kilogram (and mole) definition. On May 29th 2019, the data shown at ptable.com is identical.

I have not checked for every element, but I don't expect any difference to be significant as the goal of any international convention on that subject was to keep "backward compatibility" (as we would say for computers) with existing measures.

Well, of course, the 7th or 8th decimal can change, but it won't change the face of the world. At least, we won't suddenly grow up 4cm :-)


Why is fluorine deactivating while OH is not? Both have 2p-2p overlap and oxygen and fluorine have similar electronegativities but OH is strongly activating while fluorine is deactivating

F is the most electronegative element of the periodic table.Due to its -I effect, it shifts the electron density of any alkyl group towards itself, resulting in less density of electron cloud over the alkyl or phenyl group.This deactivates the ring for electrophilic substitution reactions.However , oxygen also being an electronegative element , has capacity to donate its lone pair electrons forming a back bond in case their is a deficiency in alkyl group.In case of cyclic rings that is in aromatic structures , the lone of oxygen is in  resonance with the ring during attack of the reagent , that is during formation of sigma complex.However this is not shown by fluorine due to its high electronegativity.Do not get confused with other halogens.Cl ,Br and I are ring activating groups as they also can form the same back bond as oxygen.


Is it possible to compress an atom to infinite density? I know, by nature, normally atoms are stable. But I wonder, if it is possible to compress an atom, say the Hydrogen atom, to infinite density by applying external force? After all, electrons and protons would attract each other by electrostatic forces and nuclear forces are also there at least up-to a certain distance...

Has it been done?

The reason I thought of it, was because I was trying to find the ratio of the amount vacuum to the amount of matter in the whole universe...

We do not know. Physicists THINK that there ought to be a fundamental limit in scale for space-time that occurs near $10^{-33}$ centimeters and $10^{-43}$ seconds; often called the Planck Scale. There is also a unit of mass associated with this scale which is about $10^{-5}$ GRAMS or $10^{19}$ Billion Electron Volts (BeV or GeV). It is a simple matter to construct a 'Planck density' by computing $10^{-5}$ grams / $\text{(} 10^{-33} \text{centimeters)}^{3} = 10^{94}$ grams/cc. Many physicists think that this represents the highest density for any natural compression mechanism, because at higher densities, space-time itself transmutes into some type of quantum mechanical system for which 'density' is not a meaningful concept. So, if fundamental particles are truly without any structure larger than the Planck scale, then they can be packed together into a 'system' with a density of $10^{94}$ grams; the highest achievable density for energy/mass in this physical universe. Even within black holes, the famous 'Singularity' may not have infinite density, but merely a finite density near the Planck value. This is how 'quantum gravity' is thought to regulate, or eliminate real physical singularities; something which ordinary general relativity is not able to do.

I read it in a book Elegant Universe, by Brian Greene. This book has a whole seperate chapter for this.

But we know Nuetron stars can be compressed down to extreme Densities,and extreme spherical  radius diameters, caused by the extreme compression,via Gravitational collapse,we know that any Mass therefore can be compressed to these extremes, because these very compressed masses already exist in the partial vacum of space,don't they, so some force has compressed there mass,into an extreme density. I suggest (E=MC2) has a new meaning, and that is, ("That Energy is equal to Mass Compressed Squared,or Spherically compressed"). You must Spherically compress, as you do when Bubble Collapse accurs,during implosion, with it's extreme tempatures, and extreme pressures created,one would expect that extreme density ocurred momentarially,before explosion or rebound ocurred then the transfer of energy,takes place via Photon emission, which ocurrs during implosion.So one would expect a chandresaker limit exists for a mass compressed or a Proton's Nuclei, to be Compressed,the BEC of the meson behavior is a Super Critical Fluid, at these extreme Pressures and Tempatures which is BOTH a Gas and a Liquid behavior, so it is Compressable,one only needs to read the work of the scientist/inventor who has discovered Hydrinos,which are below the normal energy level's,and are entirely new what i call Negative energy levels, only they are an absorbent only, or energy sink energy level's, all 9 of them,which at the last Jump from  the lowest energy level, one would have to assume the Chandresaker limit would take over. So Implosion and the Centripedal compression of the Vortice,or mini Black Hole compressession, compresses mass,and compresse's the Nuclie. Just as it does in a super massive black hole,only this on the micrpscopic level,the mini black hole on the Atoms Shell, allows for the Phonon resonance,to cavitate the Nucleons into Resonace,and if one Cavitates with Coherent optical metals like Tantalum, to resupply back onto itself the resonace in a loppback circuit,a closed loop,totally reflective/Optical properties of reflection, one creates a Laser Waveguide tube, with an end Mirror which momentarily PIN'S charge, zero time, zero mass time/Zero spacetime.Zero energy amplification, no loss of time, no loss of energy,Pure Potential only,this is why Resonance amplify's,is because of NO Time Loss Zero Time Loss,the evanescent wave, is transductive only, in one direction, down into,and the Phonon resonance does exacltly this,then couples to the evanescant field,resonates with the evanescent field, it Pins, and collimates, or lenses optically, to be re- supplied back onto itself,to create a coherent beam, of extreme energy which is coherently tuned, to spacetime/masstime,and also Compressed Nuclei,which is now a new energy level in the case of a Hydrino, it's tuned,and Amplified, or coherent  resonant energy,Created by (MASS COMPRESSION) via cavitation, and Implosion,the Vortice centripedal compression. So (E=MC2) Really always Meant this, and that was what Einstein Always meant, by this equation. That (Energy is equal to Mass Compressed Squared,or Spherically compressed Mass) just like the Atomic shape charge used to compress a Nucler Bomb first,before the other reactions take place, the Direct Mass to energy conversion,or (the Missing Mass) is Simply the Compressed Mass, the new Radius,and the new Density, the Loss was because it was compressed,then converted to Photon energy, and sent away and Photon emitted the excess,  as Highly energized,extreme energy of those extreme emitted Photon energies, as electron Volts Ev, extreme energy Photon emission, created by the MassTime Spacetime conversion compression/Pinning of, which then gets converted to, the the new compressed energy Level.


Product of an intramolecular aldol condensation When nonane-2,8-dione is treated with base, the product 1-(2-methylcyclopent-1-en-1-yl)ethan-1-one forms according to the following scheme: The arrow denotes the hydrogen that is first attacked by the base. But why does the reaction not take place like this? —
Why is the arrow-marked hydrogen less acidic than the previous one?

The two protons are of similar pKa and you can get deprotonation at both sites with equal ease, however it is important to understand that most intramolecular aldol reactions are reversible and thus give rise to the thermodynamic product i.e. the most stable product. [6]-ring formation is always preferred over [8]-ring formation 1, 2. If you prepare the [8] product by other means, and subject it to the aldol conditions you will get the [6] product. The only exception to this is when you set up the reaction to be under kinetic control e.g. by using LDA, then you can get a mixture.

Maybe the hydrogen in the second case is the allylic carbon containing hydrogen, so it is more stable. More stable means less reactive, which indicates less acidic. Allylic carbon means sp2 carbon beside another carbon.


Inductive effect and bond strength What is the relation between the inductive effect on bond length and strength.

how would one rank the bond strength of

$\ce{CH3-H}$, $\ce{CH3CH2-H}$, and $\ce{(CH3)2CH-H}$

I know that alkyl groups donate electron density by induction, but how does electron density in a bond affect its length or strength?

Just as hyperconjugation stabilizes carbocations and radicals by sharing the electron density of adjacent $\ce{C-H}$ bonds with the unoccupied/partially occupied orbital, hyperconjugation destabilizes carbanions. So more substitution means less acidic and more ionizable. I had a hard time finding a single resource listing bond lengths, but they seem to correlate with bond-dissociation energy. The wikipedia page for methane states that its $\ce{C-H}$ bond length is 108.7 pm and an answer by Jan says ethane’s is 109.4 pm.

$$ \small
\begin{array}{lcc}
\hline
\text{Hydrocarbon} & \mathrm{p}K_\mathrm{a} ^{[1]} & \text{BDE (kcal)}^{[2]}\\
\hline
\ce{CH3-H} & 48 & 104\\
\ce{CH3CH2-H} & - & 98\\
\ce{(CH3)2CH-H} & 51 &  95\\
\ce{(CH3)3C-H} & 53 & 93\\
\hline
\end{array}
$$

 $[1]$  Evan’s pKa Table
$[2]$ Wikipedia: Carbon-Hydrogen Bond

since the inductive effect leads to ionic character in the bond,the increase in -I effect decreases the bond length.


How many valid resonance forms can be constructed for aniline? Based on the above picture, it seems that there are four. However, I have two questions:

Aren't the top right and bottom left forms simply mirror images of each other, and thus considered the same?
If not, would an alternate version of the top left form be a fifth structure, if the double and single bonds were flipped?

Your four structures are correct and you could also include your suggested fifth structure (your first and fifth structures are comparable to the resonance structures of benzene). While some of the structures are mirror images, we generally think of particular atoms as being fixed when writing out resonance structures. You can think of the carbons as having a number (which they technically would if you named the resonance structures IUPAC style), which leads to the "mirrored" structures not being identical since the double bonds are between different carbons.

I believe there are only three resonance structures for aniline. The upper right and lower left structures are the same. Flipping the single and double bonds within the ring (your suggested fifth structure) would not make another resonance structure because the ring is a benzene ring. The pi bonds forming the double bonds are actually delocalized. They are "equally shared" throughout the ring.


SN1 reaction in a polar aprotic solvent I'm seeking a $\ce{S_N1}$ reaction that happens in a polar aprotic solvent.
Do you know any good simple $\ce{S_N1}$ reactions which satisfy the condition?

I think something like potassium 2-methylpropan-2-olate and 3-bromopentane in $\ce{SO(CH3)2}$ should do the trick.


What makes the cyclic structure of glucose more stable than the open chain structure? Is it tautomerism? Or some other factors involved?

You get to form a C-O $\mathrm{\sigma}$ bond at the expense of a C-O $\mathrm{\pi}$ bond. The single bond has a higher bond energy, even though it is somewhat de-stabilized by the anomeric effect. It's still an overall win, despite taking the entropic hit to confine your molecule into a ring.

The carbon atoms in the gulucose ring each have four covalent bonds ......it follows,therefore,that the gulucose molecule will be at its most stable when all carbon atoms can arrange themselves , so that their bond angles are all close to 109.5 degree .


Why is the units of rate constant dependent on the order of the reaction? In many textbooks, it is written that

For order (m + n), the rate constant, k has units of mol1−(m+n)·L(m+n)−1·s−1

However, by the Arrhenius equation, k can be expressed as

k = PZexp(-Ea/RT)

and its unit doesn't seem to change with the order.
I think if reaction velocity, v is defined as

v = k([A]/cref)m([B]/cref)n

like the definition of equilibrium, the unit of k becomes mol·L−1·s−1 and it looks more reasonable. Am I wrong?

The rate has units $\displaystyle \frac{concentration}{time}$ and is $\displaystyle \frac{d[C]}{dt}$ and the right hand side of the equation has always to have the same units as the left hand side, so for first order reaction consuming C;
$\displaystyle \frac{d[C]}{dt} = -k_1[C]$
and the rate constant $k_1$ therefore has units $\displaystyle \frac{1}{time} $ usually $\mathrm{s^{-1}}$ and for second order
$\displaystyle \frac{d[C]}{dt} = -k_2[C][C]$
the units of rate constant $ k_2$ are $\displaystyle\frac{1}{time \times\,concentration}$ which is $\mathrm{dm^3\,mol^{-1}\,s^{-1}}$.
The units for other examples follows in the same way.
In the Arrhenius equation as the exponential is dimensionless the other terms must have the units of the rate constant, which means units of PZ in your notation. These units must always depend on the order of reaction as described above and if they don't then there is an error somewhere.

By the Arrhenius equation, $k$ (rate constant) can be expressed as,

$$ k = \rho Z \times \exp\bigg(-\frac{E_a}{RT}\bigg) $$

where $\rho$ (aka. steric factor, or probability factor) is defined as the ratio of $A$ (Arrhenius contant or pre-exponential factor) and $Z$ (collision frequency) (Ref. 1).
And, units of $A$ are same as that of units of $k$ (Ref. 2), hence the units of $\rho Z$ is same as the units of $A$, or $k$.
Reference

Steric factor - Wikipedia: para 1
Pre-exponential factor -Wikipedia: para 2


Effect of drinking ultra-pure water What would be the effect if someone were to drink ultra-pure water with an electrical resistivity of $18 \, \mathrm{M} \Omega \! \cdot \! \text{cm}$?

Would they immediately die?  Would they just need to pee more?  Would $\ce{CO2}$ from the air (after the bottle is opened) and whatever's in saliva dissolve into the water making it much less pure before it gets into the important parts of their digestive tract?

What would be the threshold for bad things to happen — a teaspoon, a liter, 12 gallons in 10 minutes (at which point I assume regular water would kill you)?

I'm not sure what the science here would be.

What would be the effect if someone were to drink ultra-pure, 18 M-ohm water?

Not much, although if they drank many gallons of the water it could be a problem.

  Would they immediately die?

No.

  Would they just need to pee more?

Probably.

  Would $\ce{CO2}$ from the air (after the bottle is opened) and whatever's in saliva dissolve into the water making it much less pure before it gets into the important parts of their digestive tract?

It would become orders of magnitude less pure as soon as it mixed with their saliva and then their gastric juices.  Same as regular water.

  What would be the threshold for bad things to happen -- a teaspoon, a liter, 12 gallons in 10 minutes (at which point I assume regular water would kill you)?

I don't think the effect would be that different from drinking regular water.  All over the internet there is this idea that super pure water is somehow too pure or dangerous, no good evidence or mechanism is ever presented.

Water is toxic primarily because it dilutes the sodium and potassium ions in your body, leading to Hyponatremia or possibly the far more dangerous Hypokalemia.  The normal level of $\ce{Na+}$ in blood plasma is about 135 millimolar (mM), and the normal level of $\ce{K+}$ is about 5 mM.

A very rough approximation to understanding how toxic water can be would then be to compare the concentration difference of sodium (or potassium) between the ingested water and the blood.  A 25 micromolar -- i.e. 25 μM or 0.000025 molar solution of sodium hydroxide has a conductivity of about 6.7 microsiemens per cm, more than 100-fold higher than "pure" 18.2 megohm water (0.055 microsiemens per cm).  According to this rough difference-based model, the ability of the very dilute sodium ion solution to cause problems is proportional to (135 - 0.025) mM, and the ability of the "pure" water to cause problems is (135 - 0) mM.  For nearly biological or medical conditions of interest, 135 mM is not really different from 134.975 mM, and so ultra-pure water is not really more dangerous than everyday drinking water.

Would they immediately die?

No.

I agree, no. But if long time drink pure water, such as 18 M-ohm water, it could cause shortage of u-elements, such as Zn, Se, Fe,..., even all those very,very small quantity, but important for the life, the health...
Normally speaking, drink normal water is the best.

the question is how to get the pure water, 18 M-ohm? Could be expensive.
It can be said using two methods: Physical or Chemical. Steam method is physical. Ion change is chemical method. Both methods changed property of natural/original water. not good. especially, chemical method. More or less, chemical reaction take place, the products changed. pH changed. some thing changed,...    natural/original is the best, need the water, parents, grand parents, grand and grand parents' drinking.


Order of reactivity of carbonyl compounds towards nucleophilic addition Arrange the following according to reactivity towards nucleophilic addition reaction:

I have already seen this post regarding the reactivity. It helped me to some extent.

$\ce{Cl}$ increases the positive charge on the carbonyl carbon increasing its reactivity.

So (B) > (A) > (the remaining 3)

Further, aldehydes are more reactive than ketones. But I should also consider steric factors. In compounds (C), (D), and (E), compound (D) has highest carbocation stability but also has highest steric hinderance.

Compound (C) has lowest steric hinderance but low carbocation stability as well. How can I arrance (C), (D), and (E)?

To all those who say steric effects dont play a role: This is taken from page 147 of organic chemistry book by Clayden:

  The same structural features that favour or disfavour hydrate formation are
  important in determining the reactivity of carbonyl compounds with other
  nucleophiles, whether the reactions are reversible or not. Steric hindrance and
  more alkyl substituents make carbonyl compounds less reactive towards any
  nucleophile; electron-withdrawing groups and small rings make them more reactive

Carbocation doesn't come into the picture at all, when we are discussing about nucleophilic addition reactions. Steric hindrance will also not play a role because the compounds are planar (attack is going to take place on $sp^2$ hybridised carbon). The nucleophile is going to attack only from the top and bottom* as shown:

The only factors that come into picture in the options are the electron withdrawing or releasing nature of the groups attached to the carbonyl carbon. If the group is $e^-$ withdrawing in nature, the deficiency on the carbon atom increases, making it susceptible to nucleophilic attack.

So, the correct order is : B > A > C > E > D

P.S.:* Since the attack is from top and bottom, both of which are equally likely, the resulting product is a racemic mixture (if the carbon on which attack happens is chiral after the attack).


Can hydrofluoric acid etch glass at a pH of 7.0? Will it still etch glass when very dilute? Will the reaction still proceed slowly? At what pH will the reaction stop completely?

From Wikipedia on hydrogen fluoride: "Although hydrofluoric acid is regarded as a weak acid, it is very corrosive, even attacking glass when hydrated."
On hydrofuoric acid: It etches glass "by reacting with silicon dioxide to form gaseous or water-soluble silicon fluorides. "

It seems fishtank glass can be etched even when the water is kept at a pH of 7.0.
"You might try a razor scraper, to see if that will remove any of it. If not, there is a very high chance they are due to etching. Etching is caused by the action of water, especially acidic water, over a long period of time. "

In $\ce{SiO2}$ etching, there are two active fluorine-containing species, $\ce{HF}$ and $\ce{HF2-}$.

The rate of etching at 25 degrees C in Angstroms per second is:

$$9.66[\ce{HF2-}] + 2.50 [\ce{HF}] -0.14$$

according to A Study of the Dissolution of SiO2 in Acidic Fluoride Solutions J. Electrochem. Soc. vol. 118, pp. 1772-1775.

The relavent equilibrium constants at 25 degrees C and ionic strength of 1M are

$$\frac{[\ce{H+}][\ce{F-}]}{[\ce{HF}]} = 0.0013M $$

$$\frac{[\ce{HF}][\ce{F-}]}{[\ce{HF2-}]} = 0.104M $$

Because of the -0.14 term in the rate equation, only at enormously high F- concentration would there be etching at pH 7 (such as in this situation).

This study was concerning high (~1M) concentrations of fluorine species, so it might not be completely relevant for low concentrations or very long time periods.

Certain acids are corrosive at a pH of 7.0. These compounds do not involve H or OH, so are not neutralised when diluted in water to a pH of 7.0.

"SiFW (silicofluoride treated water) is associated with serious corrosion of lead-bearing brass plumbing, producing elevated water lead (PbW) at the faucet."

"Silicofluorides (SiFs), fluosilicic acid (FSA) and sodium fluosilicate (NaFSA), are used to fluoridate over 90% of US fluoridated municipal water supplies. ...New data refute the long-prevailing belief that PbW contributes little to children's blood lead (PbB), it is likely to contribute 50% or more."[1]

[1]: Journal of Neurotoxicology(2007): Confirmation of and explanations for elevated blood lead and other disorders in children exposed to water disinfection and fluoridation chemicals http://www.ncbi.nlm.nih.gov/pubmed/17420053


chemical possibility in oil - 8th graders have questions for a project I have a team of kids (8th graders) that are doing the First Lego League competition this year. As part of that they have to come up with a possible solution to a problem that relates to the human water cycle.  They need to contact experts and share their thoughts and get feedback of their solution.  We would greatly appreciate any feedback.  I've listed what the problem is they are trying to solve and their suggested solution below, and then some questions.

Problem:  Oil spills in water (caused by various things - accidents, hurricanes, etc)

Their proposed solution:  Create an environmentally safe chemical that can be put in the oil, that if there is a leak, as soon as the oil comes in contact with water, the chemical causes the oil to solidify thus sealing the leak or at least slowing it down.  The solidified oil would be buoyant so that it could be easily skimmed out of the water.  The solidified oil could also be reprocessed back into usable oil.

Questions the kids have:
    Does such a chemical currently exist that can do this?
    If so, why is it not being used - any downside to using it?
    If not, is it possible to invent such a chemical?  Now or in the future?
    What kind of components could be used to do create such a chemical?

We really appreciate any response you can give.
Thank you so much!

A chemical that can be added to crude oil so that the oil solidifies upon contact with water to either seal leaks or to make the oil recoverable sounds like a very good idea. Unfortunately, such a chemical is exceedingly unlikely to exist.

When processing crude oil, you typically want the oil itself and not any additives. That means, the chemical in question would have to be added in concentrations as low as possible. Therefore, it’s reaction with water must be very strong. But this in turn means that the traces of water present everywhere (yes, even inside your sample of crude oil) would almost be enough to cause the reaction. Therefore, outside of water-free atmospheres (used in laboratories under inert gas to perform water and air-sensitive reactions) this chemical would hardly be usable.

A second problem is getting the chemical reasonably dissolved. Crude oil is a very viscous mixture unless heated. To use the chemical effectively, it would need to be added almost immediately and distributed well in the sample of crude oil — which would mean immediately heating the crude oil to such temperatures that efficient mixing is possible (likely under pressure to not lose the more valuable but also more volatile petrol components) right after getting it out of the ground. Especially on oil rigs, I see major difficulties with this process; and it is basically on oil rigs that you would want this chemical the most.

Finally, there is the ‘how would it work’ problem. I have extreme difficulty imagining a reaction with water that would lead to solidification. The only idea I can come up with is a compound that would be liquid at standard conditions but adds water to become a solid which crystallises out of the crude oil phase. However, this solid would likely not do you the favour of crystallising where you need it and you would need so much of it that my first argument becomes invalidated. Therefore, I conclude that such a compound cannot exist.

If such a compound exists, I’m sure we would have heard of it or it would be in use (or have been in use if there are significant and severe environmental issues with it that are worse than the issues with crude oil itself.

The skimming equipment was designed and used many years ago; except it did not involve any chemical additions. Oil floats , A sponge-like continuous belt runs through the water preferentially absorbing oil phase . It then goes through rollers that squeeze out mostly oil by adjusting pressure. Then the belt passes second rollers that squeeze out water.  And the belt passes into the oil slick again. I don't know if it has been a commercial success. I will bet a lot of money the EPA would outlaw the chemical you want as they have for those chemicals used to break-up oil .


Confusion on electrode potentials Standard electrode potential values can be used to judge the reducing ability of a metal. When I was introduced to this topic a simple picture was painted: you put a metal into a solution of its ions. An equilibrium is set up. You do so for another metal. If you now hook up the two electrodes and measure the voltage, you can tell which one is 'more negative' hence more electron donating.
I was taught like it is covered here :
  https://www.chemguide.co.uk/physical/redoxeqia/introduction.html

My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'( As in this plate would have more negative charge). How has the voltage helped us determine the difference in the tendencies of these two metals?

My question is that suppose we have two metals, A and B. A produces +1 ions. B produces +2 ions. Now even if the two have the same tendency to lose or gain electrons, the B plate would be 'more negative'.

That is not a problem because the tendency of the metal to lose or gain electrons has no direct bearing on the reactivity of the metal or the reduction potential. You are thinking of this on an "per atom of metal" basis, but the reduction potential is on an "per electron" basis. One way to see that is that the cell potential does not change when you double the coeffients in an equation:

$$\ce{Ag+ + 1/2Mg -> Ag + 1/2 Mg^2+}$$

has the same cell potential as

$$\ce{2Ag+ + Mg -> 2Ag +  Mg^2+}.$$

(This is saying something like the potential of a D cell battery is the same as that of a AAA cell battery, even though the D cell battery is much bigger). On the other hand, the Gibbs energy of reaction would be twice as large for the second reaction compared to the first. One way to think about this is that the work the second reaction would do (if this is set up as a voltaic cell) is twice that of the first reaction because two electrons are transferred instead of one (across the same cell potential).

  How has the voltage helped us determine the difference in the tendencies of these two metals?

What we really want to know is which of the two metals would be reduced and which would be oxidized. For that, we need to know the potential (or in your picture, the excess or deficiency of electrons in the respective electrodes). This will tell you the direction of the reaction, and the cell potential when connecting the two electrodes and the two electrolytes to allow the reaction to proceed.

You can measure the potential at next to no current, so the stoichiometry of the reaction does not matter at this point - you just have to consider the charge separation that already has happened at the electrodes before you connect them.

Which plate would be negative depends on concentration of ions, as the plate with 1 electron transfer would have double rate of the potential change with the ion concentration at the electrode.

But if at particular concentrations both electrodes have the same potential, the cell voltage will be zero and neither of plates would be negative wrt the other.


Do hybrid orbitals exist in unbonded molecules? What would they look like? For example, the ground state of a neutral carbon atom could be notated as:

$$ [\ce{He}] \underset{\ce{2s}}{[\uparrow \downarrow]} \underset{\ce{2p}}{[\uparrow \vert \uparrow \vert \; \;]} $$

I know $\ce{s}$ orbitals are shaped like spheres so the two electrons in the $\ce{s}$ orbital should probably form a spherical shape.

But how should the electrons in the $\ce{p}$ orbitals be distributed?

I think in the end result a neutral carbon atom is either shaped in a trigonal-planar arrangement or a linear arrangement.

So do hybrid orbitals exist in unbonded molecules? What would they look like?

I think this would have interesting results for collisions between atoms and how probable certain reactions are.

Hybrid orbitals do not exist. Individual atoms have electronic configurations which can be explained by considering atomic orbitals. Molecules have electronic configurations which can be explained by considering molecular orbitals. Hybrid orbitals are just one mathematical way of arriving at molecular orbitals from combining atomic orbitals. They are a pure mathematical fiction and have no physical meaning (this does not mean they aren't useful).

Some schools or textbooks teach that in order for a reaction to occur energy is required to 'promote' electrons into hybrid orbitals to create some sort of excited atom prior to bonding. This does not happen in real life and is not a useful way of looking at bonding. What does happen is that as atoms approach each other, their atomic orbitals start to overlap significantly, giving rise to a new set of molecular orbitals (in quantum mechanics you are changing the potential in which the electrons reside and so the allowed energy states change). If the electrons in the new molecular orbitals are lower in energy than in the atomic orbitals then the reaction will proceed.

As for the electron distribution of atoms - it is spherically symmetric. The two $\ce{2p}$ electrons will reside in different orbitals because this does not require pairing of their spins (which is a higher energy state). However, it is meaningless to ask which $\ce{p}$ orbital they are in because all three orbitals are degenerate and so are indistinguishable.

... the ground state of a neutral carbon atom could be notated as:
$$ [\ce{He}] \underset{\ce{2s}}{[\uparrow \downarrow]} \underset{\ce{2p}}{[\uparrow \vert \uparrow \vert \; \;]} $$
I know $\ce{s}$ orbitals ... form a spherical shape. But how should the electrons in the $\ce{p}$ orbitals be distributed?

The answer from a physically point of view is unambiguous. With the help of the mathematics for spherical harmonics and in the limitation of the Cartesian coordinate system the shapes of the probability to find the electron in some volume around the nucleus (with some percentage of say 90%] are computated for the p-subshells of the Neon- and respectively Argon-elements as follows:

Despite their name, spherical harmonics take their simplest form in Cartesian coordinates, where they can be defined as homogeneous polynomials of degree l in (x,y,z) that obey Laplace's equation. (Wikipedia)

So according the quote from Wikipedia some other solutions in noncarthesian coordinates should exist. In the German version of WP some forms are outlined, of which the lower left might be of interest:

There are 8 volumes computed, all of the same symmetricity to the center of the sphere, 4 with a sign “+” and 4 with a sign “-“. Isn’t that a closer solution to the $sp^{x}$ hybridizations? The sign shows the direction of both the electrons magnetic dipole moments and the electrons spin. The electrons of the noble gases Ne and Ar are in a perfect equilibration of their 8 electrons of the “$sp^{3}$” shells. It is obvious that such a solution doesn’t correspond with the founded solutions in the scientific literature and we have to reject this solution.
Your question, how the electrons of single atoms are distributed around the nucleus is only of theoretical interest. Has anyone ever measured the shape of single atoms respectively the indisturbed outer electrons from these atoms? A the end you are interested in the electron distribution of molecules. From the form of the lattice one could conclude about the electron distribution for molecules and from this knowledge one can predict the existence and compound structure of new molecules.
With this in mind the above working assumption (Arbeitshypothese) about noble gas configuration with 4 (spin up) + 4 (spin down) perfectly balanced electrons on the edges of a cube has some advantages. A good example is the crystal structures of cubic lattices. Diamond as a form of carbon has this lattice too:

Simply take the $sp^{x}$- hybridization as the standard AND imagine the molecules you are interested in as compounds between their atoms with electron configurations of filled cubes for each atom. Take for example the Noble gas compounds. For some the imagination about cubic - or octahedral (which is the same, see Platonic solides) - distribution works perfect and for others such model doesn’t fit the founded geometries:

Images from Wikipedia

I think in the end result a neutral carbon atom is either shaped in a trigonal-planar arrangement or a linear arrangement.

From the point of physics the the answer for a carbon atom is clear: two electrons are in the s-subshell and two electrons are in atoms he p-subshell.

So do hybrid orbitals exist in unbonded molecules? What would they look like?
I think this would have interesting results for collisions between atoms and how probable certain reactions are.

If you want to go in the direction of a scientist this could be your area of work. The question is a astonishing good.


How to neutralize the stink of decomposed fish? Recently I discovered for myself an interesting way to easily produce home made plant fertilizer that is quite effective (tested myself on the plants). It's fish emulsion (so can be used also in a hydroponic system) - 1 part fish wastes(skin/bowels) to 2 parts water placed in an airtight container in a sunny spot for ca. 1-2 weeks. The problem - it stinks unbearably. Both - the gas that is produced during the decomposition and the liquid(/fats).

Is it possible to neutralize the stench somehow without loosing the fertilizing qualities of the emulsion? Especially of phosphorus.
1.1 Ideally of both - the gas and the liquid. But at least of the liquid - as that is what is used actually and then continues to spread the smell after being added to the plants.
1.2 Ideally it should be done using some usual household substances/wastes - maybe adding of lemon/orange peels along with the fish wastes? Or maybe ash/charcoals?

Ideally the resulting emulsion should be acidic as many plants like the pH range of 5 - 6.5 .

Plants like fully decomposed (composted) organic materials just like humans prefer cooked food. The solution which you are preparing, is going to hurt the plants rather badly. For example, cow dung is a good organic fertilizer but fresh manure will burn the plants. Why don't you bury this fish waste in the ground or soil, mix it with old leaves and let it compost for 1-2 months? No smell should arise because soil bacteria and perhaps worms/insects will take care of it. Above all everything will be well covered with soil.
Do not add vinegar or lemon juice to your liquid (which should go to waste anyway). Vinegar has salt and plants don't grow well with salty water.

Actually, I discovered a way of eliminating fish odor that was apparently generated from an old can of salmon, which was in the process of bursting and leaking. The can was hidden on a lower shelf in my kitchen.
I originally believed there was an issue with something on the floor and kept on treating the floor until I produced results.
The exact process I employed, however, remains the subject of a future patent.
But, I will disclose that the apparent active agent is Singlet Oxygen (or, O(1D)). The latter could interact with oxygen in air further creating reactive oxygen species (see, for example, Equation (4) presented here). Apparently, O(3P) may be formed from severe collision quenching of O(1D) atom with oxygen and acts as the major oxidant, for example, in the work by Huan Yue and colleagues ‘Exploring the working mechanism of graphene patterning by magnetic-assisted UV ozonation’. Note, O(3P) is also known as highly reactive ground-state 3P oxygen and a form of atomic oxygen.
Of practical importance, singlet oxygen has a half-life of around 45 minutes and should be considered toxic, so treated areas should be evacuated prior to treatment. On contact with water, it forms hydrogen peroxide.
Note, unlike other de-odorizing agents that disguise smells, this path actually results in a chemical breakdown with no toxic or carcinogenic residual products.
Also, an associated 2008 patent: "Use of photosensitizer for cleaning air, water or contaminated surfaces and in filters, where the photosensitizer is covalently linked and/or linked over physical interaction to surface and activated by visible light to form singlet oxygen" discussed here.


Identifying if reaction is reduction or oxidation. Is there an error in this question? I am looking at this reaction:

I am trying to determine if it is reduction or oxidation.

In the initial molecule I calculate the overall oxidation numbers as:

ox(Cl)*6 + ox(H)*6 + ox(C)*6 = 0
-6 + 6 +ox(c)*6 = 0
hence, ox(c) = 0

In the subsequent molecule, I calculated the oxidation numbers as:

ox(Cl)*4 + ox(H)*6 + ox(C)*6 = 0
-4 + 6 +ox(c)*6 = 0
hence, ox(c) = 6/2 = 1/3

However, my textbook says the oxidation number of ALL carbon in the second molecule is 2. I am very confused by this, since if all the carbons have an oxidation number of 2 then the molecule would be charged overall? Then I am unsure if my oxidation numbers can be fractional. Is my textbook wrong, and I wrong or are we both wrong?

Overall, what I am really trying to get at is if the molecule is being oxidized or reduced, and how many electrons are being transferred. Is there a better way of doing this?

Thank you kindly

Before the reaction, every carbon is attached to to other carbon atoms, one hydrogen atom and one chlorine atom. So oxidation state of each carbon = 0+0+-1(due to C-H bond) +1(due to C-Cl bond)= 0; average oxidation state of carbon is also zero.
In the product only two carbon atoms suffer a change in bondings; now each of these two carbon atoms is attached to one carbon through double bond, another carbon through single bond and a hydrogen atom; so O.S. of this carbon = 0+0+-1 = -1; so to conclude four carbon atoms remain in zero oxidation states, where as the doubly bonded carbon atoms are in -1 oxidation state. so average oxidation state of carbon = -2/6 = -1/3

P.S.:Fractional oxidation state results only when we take average, and that too in some cases


Residual hydroxyl groups in silicone resins Silicone resins are solutions of branched-chain siloxanes containing residual hydroxyl groups. What is meant by "residual hydroxyl groups"?

Silicones are produced by hydrolyzing chlorosilanes( in which chlorine atoms are replaced the hydroxyl groups). These hydroxyl groups undergo condensation polymerization with removal of water molecules to form polysiloxanes or silicones.

Now, residual hydroxyl groups in your question means that some hydroxyl groups has not undergone condensation yet. Solvent is evaporated and resin is heating resulting condensation of residual hydroxyl groups to form cured resinous film.

It means there are hydroxyl groups that have not bound to (think: intermolecular forces) or reacted with something else that are still present on the siloxane chains.


Dissolution of iron(III) hydroxide Problem
From the theoretical tour of the Mendeleev Olympiad in Chemistry:

a) Calculate the minimal amount of sulfuric acid that should be diluted with water to yield $\pu{100 mL}$ of solution that could completely dissolve $\pu{0.4 g}$ of rust.

Answer

 a) After the dissolution of a $\pu{0.4 g}$ portion of rust, the total concentration of iron(III) in a $\pu{100 mL}$ portion of solution is $\displaystyle\frac{\pu{0.4 g}}{\pu{107 g mol-1}\times\pu{0.1 L}} = \pu{0.037 M}.$ As there are no competing reactions (sulfate ion does not form stable complexes with iron), it is easy to find the minimal solution acidity after the dissolution using the solubility product $K_\mathrm{s} = [\ce{Fe^3+}][\ce{OH-}]^3$:

 $$[\ce{H+}] = K_\mathrm{w}\left(\frac{[\ce{Fe^3+}]}{K_\mathrm{s}}\right)^{1/3}  = \pu{0.0097 M},$$

 that is $\pu{0.49 mmol}$ of sulfuric acid in a $\pu{100 mL}$ solution. In addition, $\pu{0.037 M}\times 3\times\pu{0.1 L}\times 1000/2 = \pu{5.6 mmol}$ of sulfuric acid has reacted with iron hydroxide, therefore, the total amount of sulfuric acid required for the dissolution is $\pu{6.1 mol}$ (2 points).

My thoughts
This task seemed incredibly easy, but the solution overcomplicated some aspects, from my point of view.
The concept of "rust" was simplified in this problem since it was stated from the beginning that here "rust" means, in fact, $\ce{Fe(OH)3}.$ So, I wrote the chemical equation (which, because it is a neutralization reaction, should have an astonishingly large constant and is an equilibrium practically shifted fully towards the products):
$$\ce{2 Fe(OH)3 + 3 H2SO4 -> Fe2(SO4)3 + 3 H2O}$$
$\ce{Fe(OH)3}$ has a molar mass of $\pu{106.87 g/mol}$. Finally, the amount of sulfuric acid is
$$n(\ce{H2SO4}) = \frac{\pu{0.4 g}}{\pu{106.87 g mol-1}}\times\frac{3}{2} ≈ \pu{5.61 mmol}.$$
After checking the solutions, not only I realized I were wrong (even though my final answer was part of theirs), but I also were confused by the explanations given. For some reason, the solubility product was included, even though its constant is so insignificantly small in comparison to the constant associated with the reaction I wrote above.
What I thought was happening in this system was that $\ce{H2SO4}$ would neutralize the rust and that was basically it. While I am familiar with the concept of equilibrium and I understand that $\ce{H2SO4}$ doesn't nullify the concentration of $\ce{Fe(OH)3},$ in the light of the huge constant the neutralization is expected to have, the solubility product, $K_\mathrm{sp},$ should be negligible.
However, it seems that not only it is not negligible, but it also seems to be able to influence the final result significantly. But why is that? Why do we bother calculating the additional acidity from the hydroxide ions resulted in the solubility reaction? Why can't we solve the problem stoichiometrically, the way an inorganic chemist would approach it? Please correct my reasoning wherever you consider it wrong.

The $K_\mathrm{sp}$ of $\ce{Fe(OH)3}$ is varied from source to source, but a reliable source gives the value of $2.79 \times 10^{-39}$ at $\pu{25 ^\circ C}$. We'll use this value throughout the calculations.
Suppose, $s$ amount of $\ce{Fe(OH)3}$ dissolves some in water according to its $K_\mathrm{sp}$, but assume water is not ionized:
$$\ce{Fe(OH)3(s) <=>[H2O] Fe^3+(aq) + 3OH-(aq)}$$
$$K_\mathrm{sp} = s \times (3s)^3 = 27s^4 \ \Rightarrow \  \therefore \ s = \left(\frac{2.79 \times 10^{-39}}{27}\right)^{\frac 14} = 1.008 \times 10^{-10}$$
This is $[\ce{Fe^3+}]$ in water. Thus, $[\ce{OH-}]$ would be $3s = 3.024 \times 10^{-10}$.  This value is much smaller than autoionization values of water. Therefore we have to consider $[\ce{OH-}] = 1.00 \times 10^{-7}$ for this calculation. Hence, if revise the calculation accordingly:
$$K_\mathrm{sp} = s \times [\ce{OH-}]^3 = (1.00 \times 10^{-7})^3s \ \Rightarrow \  \therefore \ s = \left(\frac{2.79 \times 10^{-39}}{1.00 \times 10^{-21}}\right) = 2.79 \times 10^{-18}$$
Thus, when autoionization of water is considered, $[\ce{Fe^3+}]$ in water at $\mathrm{pH} \ 7$ is much smaller (the actual value). I did these calculations to show the effect of autoionization of water. Similarly, when real $[\ce{Fe^3+}]$ is high in the solution like the case (a) here, $K_\mathrm{sp}$ plays a role.
Let's re do the calculation again considering $[\ce{Fe^3+}] = 0.037 M$. If all $\ce{Fe(OH)3}$ stay dissolved, them we can find the $[\ce{OH-}]$ in the solution using $K_\mathrm{sp}$ calculations:
$$K_\mathrm{sp} = [\ce{Fe^3+}][\ce{OH-}]^3 =  0.037 \times [\ce{OH-}]^3 \ \Rightarrow \  \therefore \ [\ce{OH-}] = \left(\frac{2.79 \times 10^{-39}}{0.037}\right)^{\frac 13} \\ = 4.22 \times 10^{-13}$$
Thus, you need to keep $[\ce{OH-}] = 4.22 \times 10^{-13}$ to keep the solution away from precipitating back to $\ce{Fe(OH)3}$. Thus, $[\ce{H+}]$ should be at:
$$ [\ce{H+}] = \frac{K_\mathrm{w}}{[\ce{OH-}]} = \frac{1.00 \times 10^{-14}}{4.22 \times 10^{-13}} = 0.024$$
Thus, after completely dissolving rust, you need to add additional $\ce{H2SO4}$ to keep the $[\ce{H+}]$ of the $\pu{100 mL}$ solution at $\pu{0.024 M}$. I think you can calculate that amount in $\pu{mmol}$ easily now (Keep in mind that you already added $\pu{5.6 mmol}$ of $\ce{H2SO4}$ to dissolve $\pu{0.4 g}$ of rust).
Note: Any deviation from the given answer should be due to the $K_\mathrm{sp}$ value used here (The given answer didn't give that numerical value).

The solubility product of $\ce{Fe(OH)3}$ is about $\ce{10^{-39}}$. So its influence on the real solubility of $\ce{Fe(OH)3}$ is not significant. The following equilibrium constants are much more important : $$\ce{[Fe(H_2O)_6]^{3+} <=> [Fe(H_2O)5(OH)]^{2+} + H+ ; \ ...\ K = 10^{-3.05}}$$ $$\ce{[Fe(H_2O)_5(OH)]^{2+} <=> [Fe(H_2O)4(OH)_2]^{+} + H+ ; \ ...\ K = 10^{-3.26}}$$  Another difficulty is the fact that these ions has a tendency to dimerize, according to : $$\ce{2[Fe(H_2O)_6]^{3+}<=> [Fe(H2O)_4(OH))]_2^{4+} + 2 H+ + 2 H_2O \ ...\ K = 10^{-2.91}}$$ So it is not easy to calculate the pH of hydrolyzed Fe(III) solutions.
Ref. R.N. Greenwood, A. Earnshaw, Chemistry of the Elements, Pergamon Press, Oxford $1984$, § $25.3.4$, page $1265$.


Do non-redox reactions exist? Redox reaction is a type of chemical reaction, and is the result of electron transfer between chemical species. But, all chemical reactions somehow involve electron transfer!

So, are there chemical reactions without electron transfer?

There are many, here are some examples

Ion interchange reactions of various types, such as precipitation

$\ce{NaCl + AgNO3 -> AgCl \downarrow + NaNO3 }$
Polar molecule insert/ejection. The most common case is hydrolysis and reverse reactions.

$\ce{PCl5 + 4 H2O ->  H3PO4 + 5 HCl}$
Reorganization of bonds between atoms of same type, such as catalytic benzene synthesis from acetylene

$\ce{3 C2H2 -> C6H6}$
Ligand exchange

$\ce{ [Fe(H2O)6]^{3+} + 6CN- -> [Fe(CN)6]^{3-} + 6 H2O}$
polar oligomerization

$\ce{ H2O + n CH2O -> HO-(-CH2-O-)_{n}-H}$

Since redox is defined as gain or loss of electron or electron density it is impossible to have a non-redox reaction. If a reacting species must bond with a different species post-reaction, the electron density must shift due to differences in electronegativity.


Mechanism of the Gatterman Reaction Aryl diazonium salt on treatment with copper and the corresponding hydrohalic acid, gives the aryl halide. This reaction is called the Gatterman reaction.

What is the mechanism of this reaction? Is it similar to Sandmeyer reaction?

The Gattermann reaction you mentioned in the question uses Copper powder and Hydrohalous acid to substitute diazonium group with a halide in an aromatic diazonium salt.
It uses $ArSN_1$  pathway, similar to Sandmeyer Reaction which uses Copper Halide and corresponding Hydrohalous acid instead of copper powder.
As diazonium group is a very good leaving group, it departs leading to a phenyl carbocation which is very unstable, this is in fact the slow step of the reaction.
The Nucleophile (Halide in this case) attacks the phenyl carbocation leading to the formation of Halobenzene(fast step).
Nitrogen gas and Hydrogen chloride are also formed.


How are SN2 transition states stabilised by adjacent double bonds and carbonyl groups? It is my understanding that transition states in an SN2 reaction have some filled and some empty orbital character, as we have the backside of the C-X σ* beginning to fill with electrons from the HOMO of the nucleophile, forming a C-Nu σ bond in the process.

If we have an adjacent carbonyl group to the α carbon, I have learnt that the TS is then only stabilised by the  C-O π*, i.e. by electron withdrawal and not by the C-O π.

However, with adjacent double bonds, I have learnt that the TS is stabilised by both the C-C π and the C-C π*.

I was wondering why this would be the case, as looking at the shape of the orbitals, there seems no reason that the C-O π could not align itself so that the partly-empty p orbital in the TS is parallel to the lobes of this MO, therefore stabilising the TS as new lower energy MOs form.

Also, with regard to the best representation of these structures, I always see partial negative charges on the Leaving group and the Nucleophile in the transition state, but never the positive charge located on the α carbon. Surely if with adjacent carbonyl groups we only have stabilisation by electron withdrawal, it is valid to put the partial positive charge there?

Unsaturated groups conjugated to the site of attack (aromatic rings, alkenes, carbonyl groups) all tend to have the effect of increasing the rate of an SN2 reaction at that centre.

As an example, displacement by nucleophilic iodide in acetone as a solvent is ca. 30x faster for allyl chloride than ethyl chloride, with benzyl chloride being faster still.[1]

The effect was initially thought to be steric, given that unsaturated groups are sp2 hybridised (and therefore planar) and allow the nucleophile a less hindered approach, but many empirical and computational studies now dispute that, with a stereoelectronic effect[*] widely accepted (though sterics may play a small role).

Empirical Evidence for a Stereoelectronic Effect

One of the earliest indications that the effect was stereoelectronic was via a series of experiments conducted by Payne (shown below) [2].The open (acyclic) compound reacts ca. 8000x faster than the closed (cyclic) compound. To understand why, we have to consider the orbitals involved...

The Orbital Explanation

If we accept a stereoelectronic effect (coming back to Payne’s findings later),  we then have to decide whether the effect is a ground state effect (weakening of the C-X bond, raising the energy) or a transition state effect (stabilisation of the TS, lowering the energy). To quote Fleming (emphasis my own):[3]

  [This conjugation] may partly operate by raising the energy of the starting material, but most probably [the conjugating groups] are primarily operating to lower the energy of the transition structures.

In practice, computational studies [4] prove this to be the case, with the pi bond interacting with the principal axis of the SN2 transition state (sigma-pi conjugation):

Calculation of bond lengths also agrees with this mode of conjugation, with the C-C bond length decreasing upon stabilisation of the TS:

Hopefully, by this point, Payne's experimental results are obvious. In the cyclic structure, there is no way of allowing the pi-orbitals on benzene to correctly align with the SN2 transition state (given that we know the TS is linear, not bent).

One final thing to notice is that in both the alkene and carbonyl case (as in your question) the dominant effect is thought to be as above (sigma-pi conjugation).

—

[1]: March, J.; Smith, M. B. Advanced Organic Chemistry Wiley: Hoboken, New Jersey, 2007

[2]: King, J. F.; Tsang, G. T. Y.; Abdel-Malik, M. M.; Payne, N.C. J. Am. Chem. Soc. 1985, 107, 3224

[3]: Fleming, I. Molecular Orbitals & Organic Chemical Reactions:Reference Edition Wiley: Hoboken, New Jersey, 2010

[4]: Bach, R. D.; Soddens, B. A.; Wolber, G. J. J. Org. Chem. 1986, 51, 1030

[*]: Sorry Jan, couldn’t help myself with the bold, I’m not shouting, I promise…

In case of Beta Carbonyl groups, Carbon  attached to Oxygen through a double bond has a partial positive charge due to inductive effect and the nucleophile in most cases have negative charge. Therefore due to the electrostatic force between this partial positive charge and negative charge of nucleophile (electrostatic forces are very strong forces in chemistry), the nucleophile is attracted towards reaction site i.e. near the carbon bonded with leaving group. This increases the rate of reaction.

The double bond has a high electron density. Therefore, it helps the nucleophile pull the small carbon loop in the backside attack.
Same applies for the ajdacent lone pair.


How to dissolve chalk (CaCO3)? I've read that vinegar can dissolve $\ce{CaCO3}$, I know that extreme heat can decompose it, but how to dissolve $\ce{CaCO3}$ at home? I need to dissolve chalk in my home because of repairing something. Can someone help me do it without heating $\ce{CaCO3}$? Any replacement for that?

$\ce{CaCO3}$ is sedimenting when mixed with water, and therefore is not easy to remove with hot water. Nevertheless several methods exist that actually dissolve it that are better than vinegar:

You could add hydrochloric acid again to convert it into soluble Calcium chloride, according to equilibrium:
$$\ce{CaCO3 (s) + 2 HCl (aq) <=> CaCl2 (aq) + H2CO3(aq)}$$
You could use carbonic acid instead
$$\ce{CaCO3 (s) + H2CO3 <=> Ca(HCO3)2}$$
this is the reason shells are getting dissolved by an increase of the amount of carbon dioxyde in solution:
$$\ce{CaCO3 (s) + CO2 (aq) + H2O <=> Ca(HCO3)2}$$
But I recommend Sulfamic acid ($\ce{H3NSO3}$), is frequently used for removing rust and limescale, replacing the more volatile and irritating hydrochloric acid, it may be a little more expensive but is certainly not toxic.

Basically $\ce{CaCO3}$ is  not  soluble  in  water but it  can  disperse in  water by  using a dispersing  agent. The  solution looks like  milk .  The  viscosity of  solution  up  to  solid content or  quantity of $\ce{CaCO3}$ in  solution.  That is my  opinion.


What are the products of the dissociation of sodium bicarbonate in water? What is the relative pH of the solution? I had a recent question on a test that asked what the products would be if sodium hydrogen carbonate were dissolved in water. I had a few candidate answers

$\displaystyle\ce{NaHCO3 -> Na+ + HCO3-}$, but that one doesn't involve water at all
$\displaystyle\ce{NaHCO3 + H2O -> Na+ + OH- + H2CO3}$, but doesn't $\ce{H2CO3}$ decompose into $\ce{H2O + CO2}$?
$\displaystyle\ce{NaHCO3 + H2O -> Na+ + OH- + H2O + CO2}$, but that has water on both sides of the equation.

Part two of the question asked whether the solution would be acidic, basic, or neutral. What is the proper reaction for this?

It's not as intuitive as it seems and your questions are all legitimate questions, but sometimes even good arguments can't be used as evidence in chemistry.

  $\ce{NaHCO3 -> Na+ + HCO3-}$, but that one doesn't involve water at all

Look at this reaction:
$$\ce{NaCl <=>Na+ +Cl-}$$

Even this reaction doesn't "involve" water in the schematics but is right, we assume that is a dissociation of a salt in water.

  $\ce{NaHCO3 + H2O -> Na+ + OH- + H2CO3}$, but doesn't $\ce{H2CO3}$ decompose into $\ce{H2O + CO2}$?

Yes, but in fact this is an equilibrium:
$$\ce{H2CO3 <=> H2O + CO2}$$

And then you have to take into account another important equilibrium many time ignored:
$$\ce{CO2 (aq) <=> CO2 (g)}$$
So the amount of $\ce{CO2}$ present in the solution depends on by the type of system and the pressure exerted. Like a Coca-Cola bottle: if the system is closed, the $\ce{CO2}$ remains dissolved otherwise the equilibrium shift to the right and the $\ce{CO2}$ "escapes" until the rate between the $\ce{CO2}$ molecule dissolved and the $\ce{CO2}$ molecule escaped forms the solution is the same.

  $\ce{NaHCO3 + H2O -> Na+ + OH- + H2O + CO2}$, but that has water on both sides of the equation.

Okay, it is not a net ionic equation but the stoichiometry is still right. Maybe water is used to indicate that we are talking about a reaction that takes place in water.

Which reaction is not right?

All the reactions here are right, I think maybe they put some formal inaccuracies to help you, but in fact they are all part of series of equilibria that normally occur and can be described as:
$$\begin{multline}
\ce{NaHCO3 + H2O <=>[A] Na+ + HCO3- + H2O <=>[B] Na+ + H2CO3 + OH-} \\
\ce{ <=>[C] Na+ + OH- + H2O + (CO2(aq)<=>[C_2]CO2 ^ (g))}\end{multline}$$

The question should be more specific asking for the ultimate products of the reaction because all these are products of the dissociation of sodium hydrogen carbonate and because are equilibria is probable that all the species are present in different amounts when the reaction reach the equilibrium. However, the ultimate products are described by the third answer. And this equation can give you a hint about the acidity of the solution. When $\ce{CO2}$ escape from the solution (equilibrium $\mathrm{C_2}$ goes to the right) the concentration of the manly acid species goes down but the hydroxides from the equilibrium $\mathrm{B}$ are still present so at the end the solution in an open system with an atmosphere with a low partial pressure of $\ce{CO2}$ will have a higher pH, will become mildly alkaline.

First, ionize the salt as you have shown. NaOH is a very strong base so Na+ must be a very weak acid. It is solvated. Being a singly-charged, rather large cation, it stops there. If it were $\ce{Al^{3+}}$, coordinated water would be acidic by charge withdrawal from that brutally small trication.

Bicarbonate is the salt of the first ionization of weak carbonic acid. It will then be a modestly weak base by hydrolysis, as you have shown (reversible). Bicarbonate solutions are weakly basic. If you add acid (even weak acetic acid) they fizz. Boiling the solution will drive out carbon dioxide.

Carbonate is the salt of the second ionization of weak carbonic acid. It is a strong base.


Which has stronger hydrogen bonds: water or ice? As ice is the solid form of water and it has more hydrogen bonds than water, because it's oxygen atoms are precisely tetrahedrally positioned and each oxygen is hydrogen bonded by four neighbouring oxygen atoms.

This changes the density of ice by expanding it, but does it change the strength of hydrogen bonds of ice? Or more precisely, which one has strong hydrogen bonds, water or ice?

The quick answer is that ice has stronger hydrogen bonds than liquid water on average.$\require{mediawiki-texvc}%for angstrom unit$

As has already been brought up, this question is not easy to answer comprehensively as it may very well depend on temperature and pressure in ways that are not immediately obvious. That being said, I will answer this question for the most obvious case which is looking at hydrogen bond strength when water is just barely a liquid (just above $\pu{0~^\circ C}$) and when water is just barely frozen (just below $\pu{0~^\circ C}$). Also, as you may or may not know, ice freezes into many different forms which all have different lattice arrangements. I cannot speak to unusual forms of ice such as the one with a cubic lattice, but what I say is immediately applicable to forms like proton-ordered hexagonal ice (I think this is ice IX). Regular ice is called ice Ih.

It is worth looking at the very comprehensive reference [1], which discusses details of water from the dimer all the way to the many forms of ice which can form. The important points for us are only the following structural values, however:

In ice Ih, the average $\ce{O-H}$ bond length is $\pu{1.01 \AA}$ and the $\ce{HOH}$ bond angle is $109.5^\circ$. The trend which has been observed experimentally is that in going from the gas-phase water monomer, through the clusters, to the liquid, and finally to ice, the bond angle slowly increases to this tetrahedral value. Additionally, the average bond length in the liquid is shorter than in the solid, but you will have a hard time finding reliable reported values of this number from experiment because it is very difficult to determine the position of hydrogen atoms from scattering experiments in non-solid phases (this is mostly because of an ambiguity as to which oxygen atoms a hydrogen should be associated with in the scattering data). Nonetheless, there are reliable average $\ce{O-O}$ distances from scattering experiments.

In the liquid, the average $\ce{O-O}$ distance is $\pu{2.85 \AA}$ while in ice Ih it shrinks to $\pu{2.74 \AA}$. Intuitively, it is very hard to understand how this would happen if it is not simply due to increased hydrogen bond strength.

We can look at an even simpler case though and see that increased hydrogen bond strength causes the $\ce{O-O}$ distance to decrease in ice. This example is a comparison of the water dimer and water trimer. In the water dimer, the $\ce{O-O}$ distance is $\pu{2.976 \AA}$. In the water trimer, the $\ce{O-O}$ distance is about $\pu{2.82 \AA}$. This decrease in distance and the corresponding red-shift in vibrational frequency are given physical meaning in the Ref. [2].

Ref. [2] gets at a very important point for us, one which Ref. [1] also uses to explain the shrinking in $\ce{O-O}$ distances in ice. This is that hydrogen-bonding in water highly cooperative. What do I mean by this? This is exactly the topic of Ref. [2]. The key idea is this: when you allow three water molecules to mutually hydrogen bond, as in the ring shape of the water trimer, the increase in attractive interaction strength is more than simply three times the interaction strength of a water dimer. By attractive interaction strength, I mean that if you subtract out the initial decrease in binding energy which comes from distorting the water monomers. Again, all of this is laid out in detail in Ref. [2], but we learn the general rule that many-body interactions in water are very important.

So, finally, I feel we have said enough to explain why the knowledge that a shrinking $\ce{O-O}$ distance and widening $\ce{HOH}$ angle tells us that ice has stronger hydrogen bonds than liquid water. In the first case, the decreasing $\ce{O-O}$ distance means that every pair of monomers experiences greater nuclear repulsion. This must be at least offset by a greater hydrogen bond strength in order for this geometry to be favorable. Part of the way this repulsion is offset is that as the system cools and forms a solid, the hydrogen-bonds are allowed to become more ideal. This is why the bond angle widens. The tetrahedral arrangement of atoms means that the $\ce{OH--H}$ hydrogen-bond angle is much closer to 180 degrees on average than in the liquid where it frequently deviates fairly significantly.

There is one last point to make. It can be estimated both from accurate simulations and from experiment what is the average dipole moment of a water monomer in bulk liquid and in ice. It is found that the dipole increases going from the liquid to ice. This additionally tells us that the average charge on each hydrogen atom must have increased from in the liquid. We are guaranteed this because we know the bond angle increased, which should generally decrease the dipole moment (a linear configuration has no dipole). Instead, the bond angle increases, so we get more ideal hydrogen bonds, and the charge increases, so the hydrogen bonds would be stronger even if they were no more ideal than in the liquid!

Hopefully this rather long-winded answer helps you understand how we can take the evidence at hand to arrive at the conclusion that ice has stronger hydrogen bonds than liquid water.

Now to the question of different forms of ice and different temperatures and pressures. I would think that pretty much across the board, every form of ice will follow the same trends we have seen here and hence have stronger hydrogen bonds. One strange case is cubic ice, which has $\ce{OHO}$ angles somewhere close to 90 degrees. This is very strange, but should lead to a large increase in the dipole of each molecule and probably increase in positive charge on each hydrogen atom, so even then the hydrogen bonds should be stronger than in the liquid.

There may be temperatures and pressures where some of these arguments break down, but the $\ce{O-O}$ distance should smoothly decrease to that in the liquid and the angle should close to that in the liquid as the pressure increases, so one would expect the hydrogen bond strength to decrease to that in the liquid and never overshoot it.

References:

Ludwig, R. Water: From Clusters to the Bulk. Angew. Chem. Int. Ed. 2001, 40 (10), 1808-1827. DOI: 10.1002/1521-3773(20010518)40:10<1808::AID-ANIE1808>3.0.CO;2-1.
Xantheas, S. S. Cooperativity and hydrogen bonding network in water clusters. Chemical Physics 2000, 258 (2-3), 225-231. DOI: 10.1016/S0301-0104(00)00189-0.

Water has stronger hydrogen bonds than ice does.

Liquid water is denser than ice. Since water and ice are both made of H2O molecules, the fact that water is denser means the H2O molecules are closer together in water than they are in ice. This means the intermolecular forces attracting one H2O molecule to another must be stronger in water than they are in ice. Since hydrogen bonds are the primary intermolecular forces in H2O, the hydrogen bonds in liquid water are stronger than those in ice.

Remember that hydrogen bonds are an electrostatic attraction (an attraction between a positive charge and a negative charge). The strength of an electrostatic attraction depends on the magnitude of the charges and on the distance between the charges. Since the distance between molecules is greater in ice but the magnitude of the charge is the same, the strength of the attraction is lower.


The most acidic hydrogen among ethane, ethene, ethyne and allene Among
$$\ce{CH3-CH3}\qquad\ce{CH2=CH2}\qquad\ce{HC≡CH}\qquad\ce{CH2=C=CH2}$$
I need to find out the one having highest acidic hydrogen.
As per my observations, there are two equivalent resonating structures possible in case of conjugate base of $\ce{CH2=C=CH2}.$ However, the correct answer is given as $\ce{HC≡CH}$ due to inductive effect of sp hybridization of carbon.
But the priority order in determining stability of a conjugate base is:

Aromaticity
Number of equivalent resonating structures
Resonance
Hyperconjugation
Inductive effect

As such, the correct answer should have been the allene. Where did I go wrong?

The inductive effect will always make the stronger acidic compound, especially with an sp hybridization as it will ultimately stabilize the electrophile best in solution. Though resonance will also contribute to the stability, a structure similar to allene will have more resonance as it will have more bonds focused in one area so close to the electrophilic site.
I think you over resonance, its an easy mistake, but my notes from Orgo I show that the difference in pKa from an sp2 to an sp hybridization are roughly a factor of 2.
TL:DR, Induction beats resonance.


Methods of splitting carbon dioxide into elements Is it possible to split $\ce{CO2}$ into C and O? If so, what are some methods and their efficacies?

This question originated because I was wondering if it was possible, and if so, how, to convert ice from the ice caps on Mars (made up of frozen $\ce{H2O}$ and $\ce{CO2}$) into oxygen.

I assume you meant convert $\ce{CO2}$ into $\ce{O2}$.

Lithium peroxide ($\ce{Li2O2}$) is a substance that does this. It absorbs $\ce{CO2}$ and converts itself into $\ce{Li2CO3}$ and $\ce{O2}$:

$$\ce{2Li2O2 + 2CO2 -> 2Li2CO3 + O2}$$

It was used as an air purifier in the spacecraft to scrub carbon dioxide. It offers it's bonus of releasing oxygen as well as being non-hygroscopic, so it won't remove the moisture content in the air.

But at any day, you would need a lot of lithium peroxide if you want to make Mars habitable.

Edit: The partial pressure of carbon dioxide in the Martian atmosphere is $\pu{575.8Pa}$. Using the ideal gas law, and temperature of Mars is $218K$, we get amount of substance of $\ce{CO2}$ as $\pu{0.32mol/m^3}$.

Such an amount would require $\pu{14.65g}$ of lithium peroxide per cubic meter of martian atmosphere. According to SigmaAldrich, that costs about ₹21.22 (33 US cents) per cubic meter.

Plants do a pretty good job of using solar energy to convert CO2 to O2. But the intensity of sunlight on Mars is weak and you would need a very well-insulated greenhouse to do this sustainably. I would start with Antarctic algae and see where that takes us.


Why does sodium hydroxide react with aluminum oxide? I have seen all the chemical formulas for the reaction between sodium hydroxide, water, and aluminium. I understand that the hydrogen gas comes from the aluminum reacting with the water and the sodium hydroxide just breaks down the aluminum oxide. But why does sodium hydroxide break it down? Also since aluminum oxide is amphoteric can you do this reaction with a strong acid instead of a strong base?

If you use a "strong acid" to break down the passive oxide you need to be careful which acid you choose.  Sulfuric or nitric acid is oxidizing and thus could maintain the oxide.  Hydrochloric acid would be best, among common mineral acids, for actually getting to the metal while being relatively easy to handle.

C2H6O2 is a chemical formula.
HOCH2-CH2OH is a chemical formula.
CH4 + 2O2 → CO2 + 2H2O is a chemical equation. You should learn the difference.
Reaction products are usually lower in Free Energy than the reactants. If it's helpful, then the answer to your question of why does NaOH react with Al2O3 is: because the products are lower in free energy than the reactants. In my more experienced view, this is a circular answer: of course the products have more negative delta G, but that doesn't answer "why?". Science is much better answering "how?" than "why?". But at your level of understanding, you should be aware that speaking about the oxide coating on aluminum and its reactions with acids and bases is (arguably) categorically different than speaking about Al° and its reactions. You've asked about the oxide. Al+3 exists in aqueous solution as [Al+3].6H2O with its counter ion (in HCl(aq) is would be the Cl- anion) and as [Al(OH)4]- with the counter ion (in your case it would be Na+ ). BOTH of these are stable, one in acid, one in base (it's obvious which is which, right?). So, the question is: is the dissolution of Al2O3 spontaneous in acid (in base)? The answer is:"in sufficiently low (sufficiently high) pH it is." Strong base attacks skin, so does strong acid. The reactions are different, they are NOT opposites. I don't see anything unusual in the fact that a material reacts with strong acids and strong bases.  My guess is that one reason you might see a dichotomy between acids and bases is because in water solutions you don't generally have both simultaneously. (at least, since the equilibrium constant of water is 1E-14, in most solutions of chemical interest (contrasted with biologically interesting situations), either acid or base is present in overwhelming concentrations...). While dealing with Bronsted acid/bases, it's sometimes convenient to consider acids as opposites to bases, that approach has limits beyond which other definitions of acidity or basicity are more useful. HTH