Relation in the sign (+/-) for gyromagnetic ratio and magnetic susceptibility

I should preface this with the caveat that I'm a chemist and not a physicist, so it's possible that I haven't understood your question correctly.

Magnetic susceptibility, $\chi$, is an electronic property, not a nuclear property, so one could simply write the question off as not making sense.  That being said, let's look a little closer as I believe we can demonstrate an answer to this question.

First, magnetic susceptibilities are typically divided into diamagnetic susceptibilities (which are negative and arise from paired electrons), paramagnetic susceptibilities (which are positive and arise from unpaired electrons), and then a few others (ferromagnetic, ferrimagnetic, etc. which arise from the cooperative behavior of unpaired electrons).

Second, the shell theory of the nucleus provides a nice picture when discussing the NMR properties of different isotopes as we can picture protons and neutrons existing in discrete nuclear energy levels and being paired or unpaired, much like chemists picture electrons in atomic or molecular orbitals with their discrete energy levels.  One of the neat results from this approach is that any isotope with an even number of protons and and even number of neutrons will have all of the protons paired up and all of the neutrons paired up, resulting in a total nuclear spin = 0.  Thus $\ce{^4He}$, $\ce{^12C}$, and $\ce{^16O}$ are all expected to be spin 0.  On the other hand, an isotope with an odd number of protons will have an unpaired proton; and an odd number of neutrons gives rise to an unpaired neutron.  Having a single unpaired nucleon gives rise to a half-integer nuclear spin (1/2, 3/2, 5/2, etc.) where the unpaired nucleon provides a unit of 1/2 to the angular momentum and any remaining integer spin comes from the nuclear equivalent of orbital angular momentum ($l=0,1,2,$ etc.)  Having both an unpaired proton and neutron gives you an integer-spin nucleus (1, 2, 3, 4, etc.) where 1 unit of angular momentum comes from the two unpaired nucleons and the rest comes from the shell theory equivalent of orbital angular momentum.

OK, so why I am boring you with all of this detail?  Well, nuclei with unpaired nucleons are our nuclear analogues to atoms or molecules with unpaired electrons.  Thus, all NMR active isotopes (spin > 0) can be considered to exhibit nuclear paramagnetism.  It's the presence of the unpaired nucleon that gives rise to a non-zero spin angular momentum.  On the other hand, all NMR inactive isotopes (spin = 0) would be expected to exhibit nuclear diamagnetism.

Now, at this point I'm venturing beyond what I actually know, as I have never seen any values reported for nuclear magnetic susceptibility.  However, one would expect that, analogous to the electronic case, nuclear paramagnetism should have a positive susceptibility and nuclear diamagnetism should have a negative susceptibility.  In other words, the only nuclei that should have negative nuclear magnetic susceptibilities have a spin of 0 and no gyromagnetic ratio.  Any nucleus that is NMR active should have a positive nuclear magnetic susceptibility; but, some NMR active isotopes have positive gyromagnetic ratios and others have negative gyromagnetic ratios.

So the answer to your question (if I've understood it correctly) would be NO, there is no formula relating the sign of the gyromagnetic ratio to the magnetic susceptibility.

In the case of normal (electronic) magnetic susceptibility, the nucleus plays no role (or at least such a small role that it's completely negligible) and in the case of the extremely weak nuclear magnetic susceptibility, all NMR active nuclei have positive susceptibilities regardless of the sign of their gyromagnetic ratios and all NMR inactive nuclei have negative susceptibilities and no gyromagnetic ratios.

As a footnote that you may or may not find interesting.  Picturing integer-spin nuclei as the nuclear equivalent of di-radicals gives a nice rationalization for why they are so rare (only ~8 stable isotopes with integer spin vs. ~112 stable isotopes with half-integer spin vs. I don't know how many isotopes with spin = 0).  Just like molecules, the di-radical isotopes are quite unstable, with some exceptions (e.g., $\ce{^2H}$ is spin=1 and has an unpaired proton and neutron, but is a stable isotope; similarly, molecular oxygen is probably the most abundant di-radical (note: you need M.O. theory to see why it's a di-radical, but you can easily show that liquid $\ce{O2}$ is paramagentic - go do a youtube search) and is relatively stable, at least compared to many other di-radicals).


What is the reaction between NaOH and urea?

Urea can be attacked by nucleophiles like any carbonyl. The immediate intermediate is a tetrahedral orthoacid which will break down to liberate ammonia once. The same process can be repeated a second time to liberate the second molecule of ammonia.

Technically, this is a hydrolysis catalysed by hydroxide. Hydroxide may be the reactive species but it is regenerated in the course of the reaction.

For urea to tautomerise to ammonium cyanate, a significant rearrangement and breaking a $\ce{C-N}$ bond would be required. Neither is likely in any way.


Estimate Partition Function from a simulation

What you're saying does make sense. You can, in fact directly estimate the partition function using nested sampling, with error bounds. See for example this paper http://journals.aps.org/prb/abstract/10.1103/PhysRevB.93.174108 . In your case without PBC it's actually even simpler. Also check out Livia's blog here https://liviabpartay.wordpress.com/publications/


How do I correctly typeset metastable radionuclide symbols?

According to the international standard ISO 80000 Quantities and units – Part 9: Physical chemistry and molecular physics, Amendment 1, a metastable nuclide is indicated by adding the letter m (in roman type) to the mass number of the nuclide, as in the following examples:

$$\mathrm{^{133m}Xe}$$
$$\mathrm{^{99m}Tc^+}$$


How to balance the reaction equation between copper (I) sulfide and nitric acid?

$$\ce{Cu2S + HNO3 -> S + NO2 + Cu(NO3)2 + H2O}$$
Let's balance this reaction using the half-reaction method.
The oxidation half reaction (O.H.R) would be:
$$\ce{\overset{+1}{Cu}_2\overset{-2}{S} -> 2Cu^2+ + S^0 + 4e-}\tag{O.H.R}\label{ohr}$$
The reduction half reaction (R.H.R) here is:
$$\ce{H\overset{+5}{N}O3 +e- -> \overset{+4}{N}O2} \tag{R.H.R}\label{rhr}$$
The final reaction would be $4\times\eqref{rhr} + 1\times\eqref{ohr}$.
Therefore the reaction would be partially balanced as:
$$\ce{Cu2S + 4HNO3 -> 2 Cu^2+ + S + 4NO2}$$
Now, we have $\ce{Cu(NO3)2}$ and not $\ce{Cu^2+}$, so we need to add the 4 more nitrate ions which implies 4 more molecules of $\ce{HNO3}$
So now, we have the equation as:
$$\ce{Cu2S + 8HNO3 -> 2 Cu(NO3)2 + S + 4NO2}$$
Now, we balance the hydrogens and oxygens, the left has 8 hydrogens and 24 oxygens, whereas the right has no hydrogens and 20 oxygens. Therefore, we need to add 4 water molecules to the right. Therefore, the final reaction becomes:
$$\ce{Cu2S + 8 HNO3 -> S + 4 NO2 + 2Cu(NO3)2 + 4 H2O}$$
Now, $\sum a = 1 + 8 + 1 + 4 + 2 + 4 = 20$ and not 23.


Explanation of steric hindrance and existence of PI5

The simplest explanation is that you can not fit five iodine atoms around a phosphorous atom.

Consider the pentachloride equivalent - this is already quite crowded. Since chlorine atoms have a smaller covalent radius than iodine, you can get five of them around the phosphorous. Of course, all this depends on the bond length of P-I and P-Cl and the size of the phosphorous atom.

Interestingly, there seems to be some suggestion of an Uranium Pentaiodide compound, for example this PubChem compound. However, there are no references on that page, so presumably this has never been synthesised either?

In general, steric hindrance means geometric conformations of a molecule that are impossible (or difficult to achieve) due to clashes between groups and atoms within the molecule.


Why is the total number of electrons in a neutral carbon atom different from the number of dots in a Lewis Dot structure for carbon?

Only the valence electrons are represented in a Lewis structure. The inner electrons are not important for the bonding, therefore not displayed.


Does >2e contribution of an atomic orbital (AO) to a hybrid AO of a molecule violate Pauli exclusion?

The answer to your question is yes, such a contribution violates the Pauli principle, and is hence not a meaningful wave function.

The demonstration of this calculation (paper) aimed at a different statement and is unfortunately very thin on the calculation details. Furthermore it is a bit difficult to distinguish between the different models they use.
However, I believe the key is actually given in the first sentence of the "Results and Discussion" section (see emphasis below). As far as I understand it, the formula presented in equations (2) are not normalised. Furthermore it neglects any ionic contribution and/or occupation numbers. This further manifests itself that the structure is about 600 kcal/mol higher in energy than the resonance hybrid.

  When the $\ce{3s}$, $\ce{3_{$px,y,z$}}$, and $\ce{3d_{$x^2−y^2$}}$ sulfur AOs are free to mix in any proportion to form six bonding orbitals on the central sulfur, the resulting energy is 1305.8 kcal mol1 (entry 1,
  Table 1) with sulfur orbital mixing as described by [..., equation (2)].
  John Morrison Galbraith, J. Chem. Educ. 2007, 84 (5), 783-787. (emphasis mine)

While the wave function constructed in equation (3) seems to obey the criteria of a physically meaningful wave function,  but it produces an energy even higher.

After that they suddenly jump into the MO picture of the argument, but not really explaining that. In this description one can clearly see in figure 3, that the s orbital only contributes to one molecular orbital.

While I completely agree with the basic premise of the paper, from my point of view it completely fails to present a compelling argument. For example: how did they come up with an energy of >300 kcal/mol of $\ce{SF6}$ relative to the $\ce{S + 6F}$ fragmentation? (And who uses Gaussian98 in 2007?)

I personally would rather recommend our own discussion:

Why do compounds like SF6 and SF4 exist but SH6 and SH4 don't?

Last but not least I would like to recommend some educational papers that I find more enlightening on the matter (no order):

Ronald J Gillespie, Bernard Silvi, Coord. Chem. Rev. 2002, 233-234, 53-62. The octet rule and hypervalence: two misunderstood concepts;
Ronald J. Gillespie, Edward A. Robinson, Inorg. Chem. 1995, 34 (4), 978–979. Hypervalence and the Octet Rule;
Thomas A. Halgren, Leo D. Brown, Daniel A. Kleier, William N. Lipscomb, J. Am. Chem. Soc. 1977, 99 (21), 6793–6806. Apparent octet rule violations, fractionality, and the interpretation of localized molecular orbital structures. Polarization and hybridization functions in chemical bonding.


What is the major product obtained on acidification of substituted epoxide?

The reaction of epoxyketone 1 affords diketone 2 under photochemical and thermal conditions ostensibly through the diradical.1 Boron trifluoride provides the diketone 3 through acyl bond migration and not via alkyl group migration as in your intermediate 3a. I have not located a proton-catalyzed reaction. Unfortunately, diketone 3 was not one of your choices.

1) John R. Williams, George M. Sarkisian, James Quigley, Aaron Hasiuk, and Ruth VanderVennen, J. Org. Chem., 1974, 39, 1028.


While I do not think that cocaine is a particularly interesting molecule, the crystal structure has been reported in Ronald J. Hrynchuk, Richard J. Barton, and Beverley E. Robertson, Can. J. Chem., 1983, 61, 481. There is also an open access paper of the XII IUCr Congress available, R. D. Hrynchuk, R. J. Barton and B. E. Robertson, Acta Cryst A, 1981, A37, C72. The crystal structure file has been deposited to ccdc.cam.ac.uk and can be downloaded free of charge. However, as far as I can see, it does only contain the information about the space group, etc, but no atomic co-ordinates.

However, the hydrochloride structure has also been reported in E. J. Gabe and W. H. Barnes, Acta Cryst., 1963, 16, 796-801. And the cif file can be also obtained at ccdc.cam.ac.uk.

I did a quick calculation on this geometry on the DF-BP86/def2-SVP level of theory, the plain xyz coordinates are attached at the end of this post. You can open them with molecular viewers such as Avogadro to create your own model.

The following, however, has been produced with the commercial ChemCraft program.

I do not know how far you want to go with your presentation, there are some pictures on wikipedia. I can also recommend checking the record of it on ChemSpider.

Geometry of cocaine hydrochloride:

symmetry c1
Cl      -2.051693000     -0.493009000      4.153090000
C       -2.451519000     -3.565091000      6.602382000
C       -6.214292000      1.164927000      8.772468000
C       -7.187531000      2.140573000      9.037661000
C       -8.160694000      1.917991000     10.029303000
C       -8.167271000      0.711328000     10.754343000
C       -7.200746000     -0.269358000     10.489848000
C       -4.535361000     -2.199809000      5.904903000
C       -4.965503000     -2.410764000      3.577976000
C       -0.472992000     -3.311748000      5.082266000
C       -3.681531000     -2.746076000      7.063993000
C       -3.208615000     -1.681350000      8.100725000
C       -1.959949000     -0.929634000      7.627840000
C       -0.886578000     -1.913758000      7.125382000
C       -0.580604000     -3.058883000      8.128726000
C       -1.633721000     -4.157723000      7.786625000
C       -5.209882000     -1.120147000      9.257751000
C       -6.215633000     -0.043947000      9.503197000
H       -2.776079000     -4.325870000      5.870251000
H       -1.141638000     -5.101415000      7.477122000
H       -5.458897000      1.318599000      7.989093000
H       -7.190344000      3.080452000      8.463274000
H       -8.921920000      2.687588000     10.235464000
H       -8.932096000      0.535630000     11.527409000
H       -7.186414000     -1.224606000     11.036739000
H       -4.915209000     -3.233215000      2.840991000
H       -4.408859000     -1.522439000      3.213266000
H       -6.015572000     -2.140911000      3.803492000
H        0.202195000     -3.979332000      5.655364000
H       -4.361351000     -3.437105000      7.610613000
H       -1.009855000     -3.900141000      4.313786000
H        0.111577000     -2.526413000      4.564425000
H       -2.986886000     -2.247300000      9.031392000
H       -2.216795000     -0.241161000      6.793930000
H       -1.551394000     -0.325210000      8.463897000
H        0.011514000     -1.355303000      6.795935000
H        0.453167000     -3.437387000      7.998115000
H       -0.655253000     -2.701279000      9.175171000
H       -2.290875000     -4.410651000      8.642544000
N       -1.471722000     -2.634200000      5.937688000
O       -4.202728000     -0.704253000      8.439562000
O       -5.256578000     -2.245544000      9.737495000
O       -5.375360000     -1.329144000      6.012186000
O       -4.299429000     -2.894414000      4.763398000
H       -1.903686000     -1.736871000      5.153339000


Would the density and radius of your metal appear to be higher or lower if you forgot to tap the flask to release air bubbles?

In this experiment, you were determining the average volume that a copper atom occupied in your sample of copper.  In order to do this, you had to know the number of copper atoms and the volume that number of copper atoms occupied.

The moles of copper was measured by gravimetrically determining the mass of the copper (e.g. by weighing it). From that you determined the number of moles of copper using it's atomic weight, and from there you calculated the number of copper atoms using Avogadro's number.

The volume occupied by your copper sample was determined by measuring how much water the copper displaced. So, if the volume of water increased by 10 $\mathrm{cm^3}$, then you knew your volume of copper was $\mathrm{cm^3}$. But air bubbles occupy space also.  If, for example, 1 $\mathrm{cm^3}$ of the total displaced volume were actually due to air bubbles, then you would actually only have 9 $\mathrm{cm^3}$ of copper, but you would mistakenly have used a value of 10 $\mathrm{cm^3}$.

So, the bottom line is that because you were calculating the molecular volume as the volume occupied by the copper per molar quantity of copper, your calculated molar volume would have been proportionally higher than the true value.


How to determine an oxidation number of an atom in the compound, that has at least two elements, which are not hydrogen or oxygen?

In this case, the potassium ions are +1 each, and the oxygens are -2 as normal.  That leaves uranium in a highly oxidized +6 state.  The piece of information I think is missing is that Group 1 ions are almost always in the +1 state, because they so readily lose their outer valence electron.  A molecule that is not an ion, like your example, must have an overall charge of zero.


What is the difference between "bromine" and "bromine water" in their reaction with an alkene?

From ChemGuide:

  Using bromine water as a test for alkenes

  If you shake an alkene with bromine water (or bubble a gaseous alkene
  through bromine water), the solution becomes colourless. Alkenes
  decolourise bromine water.

  The chemistry of the test

  This is complicated by the fact that the major product isn't
  1,2-dibromoethane. The water also gets involved in the reaction, and
  most of the product is 2-bromoethanol.

In essence, the electrophilic pi bond in the presence of bromine and water attacks diatomic bromine as usual, forming the usual bromonium ion intermediate.

After this step, however, we have two competing nucleophiles which can open the ring: bromide ion and water. The water is likely present in great excess, making it the more likely attacker (statistically).

Of course, the desired reaction below to form a dibromide also occurs. But it is in competition with other reactions such as the one listed above.

So usually electrophilic halogenation of alkenes takes place in an inert medium such as methylene chloride or carbon tetrachloride. These aren't inclined to act as nucleophiles, so they don't participate in ring opening, leaving the $\ce{Br-}$ ion to do its thing unhindered.


What kind of interacion is there between a nanoparticle and its functionalizing coating?

In small nanoclusters/nanocrystals/quantum clusters (my goodness, they have so many names nowadays) in the size range around 13, 55 and 144 metal atoms,1 the ligands enter a real orbital interaction (dative bond2). They are ligated to the metal core (Puls et. al., 2014).

In DNA-ligated gold clusters, the phosphate groups are able to bond to the metal core and substitute the phosphine ligands it had before (Yunping et. al., 2003).

In larger nanoparticles,3 the tendency of the particles to agglomerate is much smaller. Some of them can be measured with TEM (Transmission Electron Microscopy) without full agglomeration. This is not given for the small 13 metal atom clusters. The big particles enter interactions with their surfactants for which the main contribution is given either by van der Waals forces, or by electrostatic interactions.

1 Those are some of the most common numbers in gold cluster chemistry, like $\text{Au}_{13}(\text{PR}_3)_{12}$ for example. 55 and 144 can be found in thiolated gold clusters.

2 It has covalent portions as well as ionic. It is, however, better understood if you think of coordination complexes. Only in this case, there is not only a single metal atom in the centre

3 i.e. 5 nm, 12 nm. I stated this because I am not aware of any case in which a 13-metal-atom cluster was analysed by TEM (Transmission Electron Microscopy). The conditions are simply too harsh, the ligand shell is ripped off and the little clusters experience a very strong tendency to agglomerate. You need specialised TEM equipment to acquire meaningful data. Nanoparticles of a few nanometers are less prone to show such a fast agglomeration.


Is there a way to export detailed data about atom connectivity, dihedral angle, etc from GaussView?

To my knowledge, neither .xyz nor .com format preserve connectivity information by default. Gaussview (and many other programs) makes an educated guess based on atomic covalent radii and positions of atoms. Gauss .com may also store a set of internal coordinates used for optimization that gaussview derive from connectivity information. Gaussian .com may contain connectivity information if geom=connectivity option was used, but you cannot expect it. The format of connectivity section is documented in geom=connectivity suboption part of Gaussian input docs.

That said, some other formats like .pdb allow to store that data explicitly and are well documented. Also, Gaussian can be used to perform various forms of population analysis printing a lot of nice tables, some of them might be interpreted as connectivity information (then again, some cannot).


Your product is correct, but the mechanism isn't quite right. In the first step, one of the oxygens will be protonated (either is possible). The other oxygen "kicks out" the one activated in the first step. So far so good. However, in the next step, water will attack the "carbonyl" carbon to give a hemi-acetal and a hydroxyl group. Repeating a similar set of steps reveals the ketone and second hydroxyl.

The key difference here is that by adding to the pi system, water reacts with the C-O pi-antibonding orbital, which is the LUMO of the system


Is there any way to use syringe pumpe in pumping flow

This should be no problem when you use a syringe pump model with adjustable flow rate and sufficient volume (at least 30 ml).


The increasing order of reactivity of substituted benz0yl chlorides towards nucleophilic substitution

Electronic effects govern in the order
Resonance/Mesomeric > Hyperconjugation > inductive
+M or +R dominate over inductive. That's because +M involves π electrons which are more reactive than the sigma electrons. + I operates through sigma electrons. So therefore +M effect has priority over +I.
Therefore the order of reactivity towards nucleophilic substitution is 4>1>2>3


Blue flame has a hotter temperature. In physics there is a relationship called Wien's displacement law, which states
$$\lambda_{max}T=2.898$$
where $\lambda_{max}$ is the peak wavelength of light emitted (which is the wavelength you see) and T is temperature in kelvins.

This relationship can be rewritten $$T=\frac{2.898}{\lambda_{max}}$$, or just $T\propto \frac 1 \lambda$.

From a table we find that the wavelength of blue light is about 450-500 nm and yellow light is 570-590 nm. Since temperature is inversely proportional to wavelength, the light with the smaller wavelength has the higher temperature.

(Now you might quibble that Wien's law is for idealized blackbodies but it can be used to model most things which emit light because of their temperature regardless.)


Reaction of ethylacetoacetate with Iodine

Are you certain that the coupling itself isn't a radical reaction?

The bond energy of an ordinary $\ce{C-I}$ bond is in the range of $210\,\mathrm{kJ\cdot\,mol^{-1}}$, which isn't a hell of a lot. Moreover, the resulting radical of an 1,3-dicarbonyl compound is nicely stabilized.


Does Glycerol give silver mirror test with Tollen's Reagent?

Your guess of formation of Acrolein is a pretty good one, as indicated in this answer, but a boiling water bath will probably not give you the 160°C temperature you are looking for to convert glycerol to acrolein

This question is most similar to yours and has a pretty good answer by bon. The answer says:

  However, from personal experience, gentle heating of a primary alcohol with Tollens' will cause a small amount of oxidation and result in a fine black precipitate.

It is most likely that Glycerol is undergoing a small amount of oxidation to any of it's oxidation products (most probably glyceraldehyde, since that gives tollen's test) and this oxidized compound is giving a positive tollen's test.


Variation in atomic sizes in the transition elements

From Ni to Zn there is increase in size because of the completely filled 3d and 4s orbitals. Here the nuclear charge is increased with atomic number but due to more electron -electron repulsion in 3d and 4s orbitals  as compared to force of  attraction between nucleus and outer orbital electrons there is little expansion in size as compared to other elements in 3d series.


How does aluminium react with bases to form aluminates?

There's no positive charge on Al, yet it reacts with hydroxide groups floating around.

Your observation is correct. In the beginning, aluminium in its elemental state has the oxidation state 0. Apparently, that changes in the course of the reaction:
$$\ce{Al <=> Al^3+ + 3e-}$$

  Where will the electron from a non-ionic aluminium atom go?

Again, your observation is correct. If one species is oxidized, another one must get reduced. In this case, protons from the water are reduced to form hydrogen gas:

$$\ce{ 2H2O + 2e- <=> 2OH- + H2 ^}$$

If you combine both half reactions and balance the stoichiometry, you end up with the initial equation for the whole redox reaction.


Think of it this way: A saturated solution is the equilibrium. Say you had a supersaturated solution, formed a crystal out of it and somehow managed to get your solution below saturation.

But once that happened, your solution is no longer saturated and there is undissolved material on the ground (your crystal). So, always striving towards equilibrium, some of your crystal would get redissolved to re-reach the saturation.

In fact, crystallisation and equilibrium are not static but rather dynamic processes. So even if there is a crystal sitting at the bottom and it looks like nothing is happening, there is a constant flux of compounds redissolving from the crystal surface or recrystallising onto the surface out of the solution. Which is also why everything seems to strive towards the equilibrium: It is the state, where the speed of one direction matches that of the other direction.

So short answer: The solution above a crystal will be saturated, but no longer supersaturated.


Is s-p mixing referring to hybridization or is it the mixing of one atoms s orbital with the other's p orbital?

According to molecular orbital theory s and p orbitals can mix if they
  are close enough in energy to each other.

We can make that statement a bit more general so that we can apply it to all orbitals, atomic and molecular, by saying,

According to molecular orbital theory orbitals can mix if they are

close enough in energy to each other, and
have the same symmetry

  Does the mixing of the s and p orbitals occur between the orbitals of
  one atom (basically hybridization right?) or does the s orbital of one
  atom mix with the p orbital of the other (and vice versa)?

Good question, you are really thinking about this!  First off, yes, mixing of the s and p orbitals occurring between the orbitals of one atom is called hybridization.

When I think of these problems I

first, hybridize the atomic orbitals (s and p) on an atom to generate hybrid atomic orbitals
second, mix these hybrid atomic orbitals that exist on separate atoms to form a bond, a molecular orbital

But that's not how it really happens.  It is important to realize that hybridization only occurs in response to some type of interaction between two atoms.  In other words, only when atoms begin to approach and interact with one another does hybridization start to occur.  When hybridization starts to occur, bonding between the atoms starts to occur at the same time.  To first hybridize and then bond (like I think of it in my mind) is an artificial construct that allows me to sort things out.  Hybridization on an atom and bonding between two atoms both occur simultaneously.

  the s and p orbitals on one atom do not have the correct symmetry to
  mix effectively (the s orbital would simultaneously constructively
  interference with one lobe of the p orbital and destructively
  interfere with the other).

Yes they do.  As this picture shows we can add and subtract an s and a p atomic orbital to form two $\ce{sp}$ hybridized atomic orbitals.

picture source

  in $\ce{O2}$ there is no s-p mixing so why would oxygen mix its s and
  p orbitals when it's bonding with Carbon.

I think they are leaving oxygen unhybridized.  So, for example, they are mixing (adding and subtracting) a carbon $\ce{sp}$ orbital with an oxygen s orbital to form the $\ce{1\sigma}$ and $\ce{2\sigma}$ molecular orbitals in your MO diagram.

Edit: Response to OP's comment

  Do you know why s-p mixing raises the energy of the $\ce{2σ_{g}}$ orbital?
  (it lowers the $\ce{1σ_{g}}$)

Nice diagrams and a helpful explanation can be found here.  Basically, It has to do with the symmetry of the orbitals, like I mentioned above.  Look at your $\ce{1σ}$ orbital, when you add or subtract it with the $\ce{3σ}$ orbital, the symmetry is such that there is no constructive or destructive overlap. Therefore there is no interaction and there energies will not change.  On the other hand, the symmetries of the $\ce{2σ}$ and $\ce{3σ}$ orbitals are such that they can constructively and destructively overlap.  As a result they interact and their energies change, one is pushed up and one is pushed down.


Are all strong bases alkalis?

No, quaternary amine hydroxides such as tetramethylammonium hydroxide can also be strong bases.

  Can you not have a weak alkali?

According to "Acids and Bases" by Kristi Lew, beryllium hydroxide is the one and only exception to the "alkali bases are strong" rule.

I would add that many of the pKb values of the alkali hydroxides are very close to zero.  So you should only think of the them as strong in the sense of strong means pKb near or less than zero.  If you think of strong as full dissociated even in a concentrated solution then they (NaOH for example) would not be strong by that definition.

Also, for the alkaline earth metal hydroxides (which have two hydroxides per metal ion), it is only the first pKb that is strong.


What is the underlying chemistry behind the triboelectric effect?

You would've probably heard of the commonly given example, rubbing a glass rod with silk. The glass rod becomes positively charged, and the silk becomes negatively charged.
But why? That's your question I believe, and for that we need to zoom in a little further, until we can see the molecular structure.
Glass, is a polymer mainly composed of silicates. It is a network polymer, with a bunch of $\ce{Si-O}$ groups, as evident in it's structure:

(source: mechanicalengineeringblog.com)
You can observe that some of the oxygens are negatively charged, and are countered by sodium ions.
Now, lets observe silk,
Silk, like most natural fibers, are composed of proteins. These proteins exist as $\alpha$-helices. Sericin is the outermost protein in the silk structure, so I'll show a bit of it's structure,

Now, let's get to the rubbing part!
What happens when you rub them?
You're agitating the two molecules and bringing them in contact. Loose bonds can be broken off. Consider the $\ce{N-H}$ bond in the Sericin structure. It can be easily removed as a $\ce{H+}$ ion, since the resulting negative charge can be shared with the nearby oxygen in the adjacent carbonyl group. This leaves silk with a net negative charge.
This $\ce{H+}$ can now bind to the negatively charged oxygen atoms on glass, giving the glass rod, a postive charge.

I shall end my answer with a note:
Whenever in doubt, look deeper into the structure.


The question tells you
\begin{aligned}
&&v&=3\cdot10^7\:\mathrm{m/s}\pm1\%\\
&&\Delta v &= |(v+1\%v) - (v-1\%v)|\\
\implies&&\Delta v &= \frac{2}{100}v\\
\therefore&&\Delta v&= 2\cdot3\cdot10^5\:\mathrm{m/s}
\end{aligned}

Hence
\begin{aligned}
&&\Delta x\cdot\Delta p &\geq \frac{\hbar}{2}\\
\implies&&\Delta x\cdot\Delta (v\cdot m) &\geq \frac{\hbar}{2}\\
\implies&&\Delta x&\geq \frac{\hbar}{2\cdot\Delta v\cdot m }\\
\therefore&&\Delta x &\gtrapprox\frac{1.05\cdot10^{−34}\:\mathrm{J\cdot s}}{2\cdot6\cdot10^5\:\mathrm{m/s}\cdot1.6\cdot10^{−27}\:\mathrm{kg}}\\
&&\Delta x&\gtrapprox5.5\cdot10^{-14}\:\mathrm{m}\\
\end{aligned}


Why are complexes of type MA3B3 not optical active?

Orient the fac- isomer so one A atom is facing towards you and a space between two B atoms is below it.  Then the left half of this view is a reflection of the right half so you are actually looking down a mirror plane.
Because you could have chosen any of three A atoms in your view there are three mirror planes, which pass through a common rotation axis.  The axis passes through the face that has all A atoms at its corners and the opposite face which has all B atoms.  This symmetry is called $C_{\mathrm{3v}}$ in the language of point groups.


Mercapto- and sulfanyl- are both prefixes for use with the $\ce{-SH}$ group (known as a thiol), while thio- is used to denote the "sulfur" equivalent of an oxygen-containing functional group.

More details from the IUPAC Blue Book:

Mercapto- is mostly abandoned, and only used as a prefix for -SH
groups. This quote is from the older edition of the Blue Book. "When -SH is not the principal group, the prefix "mercapto-"
is placed before the name of the parent compound to denote an
unsubstituted -SH group."
Sulfanyl- "In these [1993] recommendations,
the prefix "sulfanyl-" is preferred to "mercapto-" which was used in
previous editions of the IUPAC Nomenclature of Organic Chemistry."
Thio- "In nearly all its applications "thio" denotes replacement
of oxygen by sulfur."


Which reaction to use for standard formation enthalpy?

The heat of formation refers to the formation of one mole of the compound from its elements in their natural states at $1\mathrm{~atm}$ and $298\mathrm{~K}$.  Your first example forms 2 moles and your second example uses ozone, which is not the "natural state" of oxygen at $298\mathrm{~K}$ and $1\mathrm{~atm}$.

Regarding your question about oxygen, it has nothing to do specifically with oxygen. The important thing is to have 1 mole of the compound on the right hand side of the equation.


When the heat is absorbed by the ice then ice will melt to form liquid water

Energy needed for melting and heating up the water

$$
Q = \Delta H_{\text{fus}} \cdot m + m \cdot c_p \cdot \Delta T
$$

solve for $m$:

$$
m = \frac{Q}{\Delta H_{\text{fus}} + c_p \cdot \Delta T }
$$

The molar heat capacity for water is around $75\,\frac{\text{J}}{\text{mol K}}$. You've been given the specific heat capacity which is around $4.18\,\frac{\text{kJ}}{\text{kg K}}$. All that's left to do is convert the enthalpy of fusion to $\frac{\text{kJ}}{\text{kg}}$ and you can solve the equation.

$$
\Delta H_{\text{fus}} = \frac{6.01\,\frac{\text{kJ}}{\text{mol}}}{18.02\,\frac{\text{g}}{\text{mol}}} = 0.3335\,\frac{\text{kJ}}{\text{g}} = 333.5\,\frac{\text{kJ}}{\text{kg}}
$$

Calculate $m$:

$$
m = \frac{1690\,\text{kJ}}{333.5\,\frac{\text{kJ}}{\text{kg}} + 4.18\,\frac{\text{kJ}}{\text{kg K}} \cdot 50\,\text{K}} = 3.12\,\text{kg}
$$

The combustion of 1 mol of propane supplies enough heat to melt 3.12$\,$kg of ice and and heat the resulting water up to 50$\,$°C


Which of the following have the same molecular formula as the molecule shown in the model?

Remember they are asking for the molecular formulas which are the same, not the structural formulas. The structural formulas are shown in the question, but for example the first option, if you count up the number of carbons and hydrogens you get the molecular formula $\ce{C7H16}$, so you'll need to check if the molecule shown has the same formula.

So for molecular formulas, it doesn't matter how the atoms are arranged; all isomers have the same molecular formula. Only in structural formulas do you get information on the arrangement of atoms in the molecule.

Note there is also condensed structural formulas which look similar to the molecular formula but is still a structural formula. For the first option the condensed structural formula would be $\ce{CH3CH(CH3)CH2CH2CH2CH3}$.


Why does lithium from the nitride whereas the other alkali metal forms the azide?

The formation of $\ce{Li3N}$ can be explained thermodynamically. The enthalpy of formation of the $\ce{N^3-}$ ion is extremely endothermic (it involves three successive additions of electrons) and with the larger counterions $\ce{Na+}$ or $\ce{K+}$ the lattice energy (which is proportional to $\frac{q_+ q_-}{r_+ + r_-}$) is not sufficiently large to compensate for the large $\Delta_\text{f}H$ of $\ce{N^3-}$. The same argument explains why magnesium also forms a nitride.

According to my very brief research on the Internet, lithium azide does exist but is unstable. It can be prepared via reaction of $\ce{NaN3}$ and $\ce{Li2SO4}$: see Acta Chem. Scand. 1957, 11, 581 for more details.


Nucleophiles and bases (elimination and substitution)

There are two possible definitions of base:

Substance which accepts a proton throughout the course of the reaction (catalyticly or stoichiometricly). This is basically the Brønsted definition. To accept a proton, the molecule in question needs a lone pair whose energy level and orientation is in a good range for a proton to attach (e.g: ammonia: good; phosphane: bad). We could also extend the definition Lewis-style.
Substance which generally may accept a proton in reactions. This is still basically the Brønsted definition only without the requirement of it actually happening. The molecule needs a lone pair whose energy level and orientation is in a good range for a proton to attach even if it doesn’t happen. We could also extend it to a Lewis-style definition.

Applying the first definition, nucleophilic substitution reactions usually do not require bases. However, the molecules that attack are generally bases: A nucleophile generally needs an accessable lone pair, as would a base.

Consider the case of an alcoholate. If it is methanolate, it is small and can attack nucleophilicly. However, we also know that it can act as a base to be protonated to generate methanol. Depending on the substrate, it would choose to do substitution or elimination reactions.

If the alcoholate is tert-butanolate, we have a large molecule. Technically, it still could attack as a nucleophile. However, it is usually too big to attack and rather abstracts protons to allow elimination reactions.

You cannot really draw a line. Both are bases, both could be nucleophiles.


Why do we need to assume that gas is made up of large number of small particles in kinetic molecular theory of gases?

To echo and expand on Porphyrin's comment, the idea of having a large number of particles shows up in KMT because this is a statistical theory. That is, everything KMT predicts is some average property, or, to be rigorous, the expectation value of a distribution.

Thus, if one only had a very small number of particles, you would never arrive at something like the Maxwell-Boltzmann distribution because you would have to treat each particle individually rather than averaging over a phase-space.

So, to answer your actual question, we need this assumption because it justifies us in doing things like the derivation in this link where they take the derivative of the unknown function (eq. 9.8). That assumes this function is continuous over all space because no constraints are put on the resultant function. The only way we can do that and expect to have our theory match reality is if reality has an approximately continuous distribution of particle speeds, i.e. a lot of particles present.

As it happens, $10^{23}$ particles really isn't that many in practice, so when you get down to very low pressures and temperatures (that is the number density $\frac{N}{V}$ is small) the theory still works because something like $10^6$ probably still makes a pretty continuous function.

As a totally unrelated aside, you might also wonder if the MB distribution is wrong because it allows particles to have speeds ranging from $0\rightarrow \infty > c$... I computed the special relativistic correction one time and it's so laughably small as to actually annoy you that you even took the time to do it.

I know this answer focused mostly on the MB distribution, but that is very central to KMT so I thought it would make for a representative though not exhaustive answer.

EDIT:

To be critical of my answer, however, I'm almost certain that much of KMT is derivable while only thinking about a single particle sampling its whole phase-space (perhaps all at once) and once could simply argue that the above function I mention taking the derivative of is a hypothetical distribution in which each of the infinite states need not be occupied at any given time. So basically taking a time average of the phase-space over infinite time gives this continuous distribution. That idea also lines up well with ideas to do with the partition function where every state could be equally likely but we get a non-even distribution because some states are more likely to be sampled.

Then perhaps it is just more physically instructive to think about a large number of particles.


It does not have much to do with entropy, rather with the way and place of ozone creation.

Stratospheric ozone is produced typically at altitude 20-30 km by UVC radiation with $\lambda \lt \pu{280 nm}$:

$$\ce{O2 + \nu -> 2 O}$$
$$\ce{O + O2 -> O3}$$

See ozone cycle as courtesy of @M. Farooq.

Ozone in lower troposphere troposphere is created by UVB (280-320 nm)  (+UVA >320 nm ??) mostly due catalytic effects of nitrogen oxides that come typically of the oxidative smog of the Los Angeles type:

$$\ce{NO2 + \nu -> NO2^{*}}$$
$$\ce{NO2^{*} + O2  -> NO + O3}$$
$$\ce{2 NO + O2 -> 2 NO2}$$


Use of cryolite and fluorspar in electrolytic reduction of pure alumina by Hall-Héroult process

According to the CRC Handbook of Chemistry and Physics, melting point of $\ce{CaF2}$ is $\pu{1418 ^\circ C}$, of $\ce{Al2O3}$ (in its corundum form) is $\pu{2053 ^\circ C}$, and of $\ce{Na3AlF6}$ is $\pu{1009 ^\circ C}$ (lowest of the all three). Thus the cryolite should melt first.
I have read mentions of the problem you asked in the book Nonaqueous Electrochemistry (you can find information about it here. For the best of my summarising ability, I can conclude that the role of the cryolite and other salts added to the reaction mixture is that they decrease the melting point of alumina. As cryolite's melting point is much lower than that of alumina as cited above, when the cryolite melts it acts as a solvent to "dissolve" the $\ce{Al2O3}$, by the process of complexation. Recall how silver chloride dissolves in aqueous ammonia or lead iodide dissolves in excess iodide. The book states that:

Cryolite is used as solvent for alumina because it has high chemical stability and a high solubility for alumina.

In detail:
When pure cryolite ($\ce{Na3AlF6}$) melts, it dissociates into fluoride ions and various lower fluoroaluminates:
$$\ce{[AlF6]^3- -> [AlF4]- + 2 F-}$$
$$\ce{[AlF6]^3- -> [AlF5]^2- + F-}$$
(etc.)
However, in the presence of alumina, the dissociation process is complicated by the fact that the products arise from $\ce{Na3AlF6}$ dissociation can react with $\ce{Al2O3}$ creating oxofluoroaluminates, and it is those complexes that directly get electrolysed (cathodically reduced) rather than the $\ce{Al2O3}$ itself. By various means of calculation and experiments (Raman spectroscopy etc.) they proposed some following reactions (or equilibria):
$$\ce{Al2O3 + 4 [AlF5]^2- -> 3 [Al2OF6]^2- +  2 F-}$$
$$\ce{Al2O3 + [AlF5]^2- + F- -> 3/2 [Al2O2F4]^2-}$$
$$\ce{2 Al2O3 + [Al2O2F4]^2- + 4 F- -> 2 [Al3O4F4]^3-}$$
...
When you add calcium fluoride (or magnesium fluoride) into the mixture, the following reactions also occur:
$$\ce{MgF2 + n F- + (2-n) [AlF5]^2- -> [MgF_{(2+n)}(AlF5)_{(2-n)}]^{(4-n)-} }$$
$$\ce{CaF2 + (3-n) [AlF5]^2- -> [CaF_{n}(AlF5)_{(3-n)}]^{(4-n)-} + (2-n)F-}$$
The electrolytes in the electrolysis process is obviously "$\ce{Na3AlF6}$", but in fact they always add an excess of $\ce{AlF3}$ alongside other salts such as $\ce{CaF2}$ ($\pu{3-5 \%}$), $\ce{LiF}$ ($\pu{0-4 \%}$) and $\ce{MgF2}$ (also $\pu{0-4 \%}$). In fact, those salts decrease the solubility of $\ce{Al2O3}$ rather than increase it; that's why they only add a little of those salts.
Then why adding those salts?
There are a couple of reasons for this. Firstly, addition of those salts increase the density of the electrolyte. Because aluminium is already produced in the liquid state (melting point $\pu{660 ^\circ C}$), if the density of the two components are too similar, it is impossible to achieve a good separation-by-weight. Kvande and Rorvik had derived an equation of the melt density as a function of temperature and salt composition:
$\pu{d = 2.64 - 0.0008T + 0.18(wt. ratio NaF/AlF_{3}) - 0.008(wt. \% Al_{2}O_{3}) + 0.005(wt. \% CaF_{2}) + 0.008(wt. \% MgF_{2}) - 0.004(wt. \% LiF)}$
As you can see from the above equation density of the mixture increases with CaF2 content.
The other reason is that those salts increase the interfacial tension at the liquid Al-electrolyte boundary. The higher interfacial tension is, the less does aluminium re-dissolve in the electrolyte (yes it can!), thus better current efficiency. For example: at the boundary this reaction occurs: $\ce{Al + 3 NaF -> AlF3 + 3 Na}$. Adding more $\ce{AlF3}$ (or other fluorides) helps surpressing this reaction, and also may increase the $\ce{Na+}$ concentration in the electrolyte - the researchers found that $\ce{Na+}$ ions are the primary charge carriers.


Is it possible to crystallize a solid ingot of metal from an (acidic) aquaous solution?

I think you've just described the general concept of Electrowinning?

There are a large number of ways to recover metals from solutions or molten salts.

Producing monocrystalline metals is a different matter - crystal grains in metals tend to be microscopic, but there are techniques that permit growing large crystals of metals such as the Czochralski process or zone melting. Monocrystalline metals sometimes have attractive mechanical properties not achievable in polycrystalline metals - for example, Callister Materials Science and Engineering 6e p. 225 has a nice photo of a single-crystal jet turbine blade that is highly resistant to creep failure with respect to conventional blades.


Does a first order reaction really get completed only at infinity?

To add to the other answers that only address the mathematical behavior of the first order rate equation close to $t=\infty$, let me address what actually happens physically.

Before you reach $[A]_t=0$, the first order rate equation actually breaks down. It isn't valid anymore, because one of the assumptions on which it is based, namely that the reaction rate is deterministic, breaks down.

When you get to really low quantities of (one of) your reactants, reaction events, i.e. two molecules meeting, are blind luck. Actually, for high quantities this is also the case, but then the behavior is a physical version of the law of large numbers: a probabilistic system behaves deterministic on average.

You can simulate this basically by flipping coins whether a reaction occurs or not. If you do this for a simple $A$ to $B$ reaction you get graphs like the following:

Here the chance of reaction per timestep for each molecule $A$ is 10%. You can see that the curves at low number of molecules $N$, are jumpy (and actually would look different everytime I rerun my calculation), while for $N=1000$ you already see the fairly smooth and familiar first order behavior. Just note that close to t=10 also the $N=1000$ case is behaving jumpy again as you can see from the zoom below.


Why are the majority of cleaning solutions basic?

Dishwashing agents are usually anionic detergents—i.e. they are alkylbenzenesulfonates.
They consist of two parts:

Sulfonate part which is hydrophilic(water loving i.e. dissolves in water)
Alkylbenzene part which is lipophilic(fat loving i.e dissolves in fats and oils)

Thus, when in water, they dissociate to give $\ce{RSO3-}$ and $\ce{Na+}$ .  $\ce{Na+}$ is from a strong base (making it a weak conjugate acid), while $\ce{RSO3-}$ is from a weak acid (making it a weak conjugate base), thus the resulting solution is basic.

The sulphonate ion bonds to water, while alkylbenzene bonds to oil, protein, or fat . Thus fats, oils, proteins, and grease dissolve in water

The following is the mechnism of a dishwashing agent:

These are the ingreadients mentioned on a dishwashing bottle in my house:

As you can see, sodium dodecylbenzenesulphonate (an alkylbenzenesulfonate) is one of the main ingredients.

  Do acid make good cleaning solutions as well?

Yes they can make a good cleaning solution, but not as a dishwashing agent.

According to Wikipedia ,

  Acidic washing agents are mainly used for removal of inorganic deposits like scaling. The active ingredients are normally strong mineral acids and chelants. Often, there are added surfactants and corrosion inhibitors.


What can we say about the Gibbs free energy if it decreases or not?

I think the confusion may be because you are conflating two ideas about Gibbs free energy.

The first is that Gibbs energy is a thermodynamic potential - this means that on a given surface in phase-space (the NPT surface), it is the "driving" force for, well, everything. Including chemical reactions.

The second idea is that the net change in Gibbs energy when going from reactants to products indicates the direction that the reaction will (eventually) go, and that its magnitude tells you the maximum amount of work that could be done by that reaction.

The second idea follows from the first - this becomes easy to understand if you just picture the thermodynamic potential as a gravitational potential, since we all have a good intuitive understanding of gravitational potentials.

Imagine a hill representing the thermodynamic potential (in this case Gibbs energy):

The rock represents a reaction coordinate - say, the concentration of a reactant. What will happen to the rock? It will go downhill. What will its maximum energy be when it reaches the bottom? Whatever the difference in potential energies is. What is the minimum energy it would need to get back the top? Again, whatever the net difference is in potential energies.

When you ask about $dG$ being negative, you are asking about the differential change - in other words, what is the slope? Obviously, if the reaction proceeds, the slope must be negative - "things roll downhill"

When you ask about $\Delta G$ being negative, you are asking whether the net change is positive or negative. The system can never do more work than the net change in $\Delta G$, because that is the maximum change in thermodynamic potential energy.

For your problem, the reaction involves moving an electron against a potential - the cell potential. The reaction can provide $\Delta G$ amount of energy with which to move that electron.


How large is the energy difference between keto- and enol-form of guanine and uracil?

I am not an expert in the field, but a quick literature check turns up a good amount of papers on the topic. In particular, I found this paper [1] which nicely answers the question for guanine:

So, in the gas phase, the most stable enol tautomer has a 0.9 kcal/mol free energy difference with the most stable keto tautomer. That free energy difference is very exacerbated in aqueous solution, with now a ∆G° = 8.7 kcal/mol. We also notice that the most stable keto tautomer is not the same in the gas phase and in solution, and that both keto and enol have many tautomers close in free energy, showing the limits of the simple keto vs. enol line of thought.

Regarding uracil, the first reference that comes up in a bibliographic search is this paper [2]. I only quote the abstract:

  The effects of the solvent on the tautomeric equilibria of cytosine and uracil are studied using Onsager's reaction field model in the framework of density functional theory. […] Our results are in good agreement with available experimental results and confirm that the polarization of the solute by the continuum has important effects on the absolute and relative solvation energies.

pKa Values of Guanine in Water: Density Functional Theory Calculations Combined with Poisson-Boltzmann Continuum-Solvation Model, Y. H. Jang et al., J. Phys. Chem. B 2003, DOI: 10.1021/jp020774x
Solvent effects in density functional calculations of uracil and cytosine tautomerism, L. Paglieri et al, Quantum Chem., 2004, DOI: 10.1002/qua.560560517


How does acetone help to make a liquid cleaner stronger?

Well, acetone is at the same time polar (so it likes water) and lipophilic (so it likes fat). As such, it is a very good solvent.


Do Slater's rules apply to neutral lithium (at least approximately)?

You need to use the factor of 0.85 for each of the $\ce{1s}$ electrons:
$$Z_{eff} = 3 - (2\times 0.85) = 1.3$$
as per rule #2 in the Wikipedia article:

  If the group is of the [s p] type, an amount of 0.85 from each electron with principal quantum number n one less than that of the group, and an amount of 1.00 for each electron with principal quantum number two or more less.

The $\ce{1s}$ electrons both qualify as having "a principal quantum number $n$ one less than that of the group" (here, the principal quantum number of the group of the $\ce{2s}$ electron is 2). Here is a nicely worked example for further edification.


Way to confirm dichromate ion presence where other ions are messing with the color

The best reagent for detecting Cr(VI) in aqueous solution is the diphenylcarbazide, which produces a red color in presence of Cr(VI). The sensibility limit is as low as 0.2 ppm Cr(VI). This method has been developed one hundred years ago. See
1) M. A. Moulin, Dosage du chrome, Bull. Soc. chip. 31, 295 (1904).
2) H. Crossley, Diphenylcarbazide. An indicator for the titration of iron with dichromate. Analyst 61, 164 (1936).

The reaction with alcohols to produce aldehyde and acids does work. But it is not very sensitive. It needs at least some millimoles of chromate to be detected. The reaction with diphenylcarbazice is about 1'000'000 times more sensitive.


Why is the speed of ideal gases mass dependent while the kinetic energy is not?

As the mass increases, the velocity decreases, yes that is true. But how does the kinetic energy remain the same?

The increment in the mass is balanced by the decrement in the velocity. In other words, the velocity changes as the mass changes in such a manner that the kinetic energy of the molecules of the gas will remain to be same no matter what. It's something like the multiplication of two variables always giving the same constant value. Keeping in mind that this is true in an ideal environment.


Proton NMR (1D) acquisition time can vary by orders of magnitude depending on concentration (sample amount), instrument sensitivity and relaxation times. For a standard experiment it is on the order of minutes.


How to convert ethanal to 2-hydroxy-3 butenoic acid?

Once you have got your acrolein from the aldol condensation of ethanal and formaldehyde you can react it with $\ce{NaCN, HCN}$ under kinetic control to get direct addition of the cyanide to the enone, forming a cyanohydrin[1]. The nitrile can then be hydrolysed in aqueous acid to yield the carboxylic acid.


Why is the ratio between Silicon and Oxygen 1:3 in single chain silicates?

Silicon has 2 complete oxygen atoms ($2\times1=2$) and 2 equally shared oxygen atoms ($2\times1/2=1$), total 3.


What would chemistry seniors like to hear about industrial hygiene?

To keep this answer as far from opinionated as possible, I'd suggest some of the W's:

What is industrial hygiene? (I've never heard of the term before this post and actually had to look it up myself.)
How do you do industrial hygiene? Procedures, risk assessment...
Why is industrial hygiene important? You might give examples where disaster was averted thanks to IH, and an example where it would have helped if done right.
Where is industrial hygiene applied? (I guess the office building will have less rigorous guidelines than the production facility.)
When did industrial hygiene start to be a thing? A small historical overview could be given, you could even give a who to this one if it was some distinct person who came up with the idea.

Always good is a personal anecdote, something you witnessed or someone (whom you trust as a source) told you, highlighting the bullet points above.


What is the difference between chemical equilibrium and dynamic equilibrium?

Chemical equilibrium is a type of dynamic equilibrium, but not every dynamic equilibrium is a chemical equilibrium.

In a chemical equilibrium there is no change on the macroscopic scale. That means that if you look at the system it seems like nothing is happening, but at molecular scale there are reactions going on and the rate of forward reaction = rate of backward reaction.

There is another example of dynamic equilibrium (although not the best one, as it is not a closed system):
Imagine a bath into which water is flowing in at constant rate. Water is also flowing out of the bath at the bottom with the same rate.
This is an example of dynamic equilibrium, which has nothing to do with chemistry. If you look at the bath, the water level is constant, so on the macro level nothing is happening, but there is flow of water, so on micro level water from the tap is replacing water flowing out of the bath.


How to determine which aqueous solution has the largest freezing point depression

Freezing point depression is a function of the amount of dissolved substance — but substance must be treated as distinct particles rather than compound. This is an important distinction because ionic compounds, when they dissolve in water, typically separate into hydrated anions and hydrated cations — two distinct species as far as freezing point depression is concerned.

Thus, for ionic compounds you first need to determine how many individual ions are generated and then multiply that by the overall concentration. In your examples:

$1~\mathrm{M}\ \ce{KCl}$. One cation, one anion. Thus, there are two moles of dissolved particles per mole of compound; giving $2~\mathrm{M}$ effective particle concentration.
$0.75~\mathrm{M}\ \ce{C6H21O6}$. This compound is not ionic. Thus, it does not dissociate and one mole dissolved gives one mole of particles. $0.75~\mathrm{M}$ effective particle concentration.
$0.5~\mathrm{M}\ \ce{MgCl2}$. This again is ionic but we get two moles of anions per mol of compound in addition to one mole of cations. Thus three moles of ions per mole of compound. Effective particle concentration $1.5~\mathrm{M}$.
$0.25~\mathrm{M}\ \ce{Ga2(SO4)3}$. Here, we have a total of two moles of cations and three of anions per mole of compound. Thus, the effective particle concentration is five times the original or $1.25~\mathrm{M}$.

The actual formula for freezing point depression is based on a constant factor of degrees per dissolved mole; and this analysis was used to confirm that ionic compounds separate into ions and to determine molecular weight in the past.


Is LiF a weak electrolyte in acetic acid?

According to Acid-Base Equilibria in Glacial Acetic Acid. III. Acidity Scale. Potentiometric Determination of Dissociation Constants of Acids, Bases and Salts J. Am. Chem. Soc., 1956, 78 (13), pp 2974–2979:

  It is generally agreed [references 3 and 4] that acids, bases and
  salts are only slightly dissociated in glacial acetic acid.

While the article does not specifically address lithium fluoride, it has quantitative data for lithium chloride.

Lithium Chloride dissociates with a pK of 7.08 +/- 0.02 in acetic acid.

So considering the high concentration of LiF in the problem (5 in 100 grams), if LiF is similar to LiCl and the other salts studied, 1 rather than 2 is correct.

This knowledge is not part of a standard general chemistry curriculum, so unless a specific book or lecture provided relevant information, I wouldn't expect students to know this.

Furthermore, the solubility of LiF in acetic acid is only 0.84 grams per kilogram according to Glacial Acetic Acid as a Non-aqueous Solvent for Metal Fluorides, so the problem is not realistic. (Only 0.084g of the 5.00 g would dissolve).  Also, HF is a strong acid in acetic acid according to that paper, so LiF might even be an exception to the general observation that salts do not dissociate in acetic acid.

Also, keep in mind that in the liquid phase, acetic acid exists as mostly cyclic dimers, which gives the solvent some non-polar characteristics.


Chromatography: Why does the insoluble compound get pulled along?

Chromatography only works when the affinity for the paper vs. affinity for the water is in a dynamic equilibrium.  Suppose there are three compounds in the ink, called $\ce{I}$, $\ce{N}$, and $\ce{K}$.  Lets call the  spots on the paper fiber that bind ink molecules as $\ce{P}$.  Then let's write the dyanmic equilibria this way:

$$\ce{I(aq) + P <=>> PI} $$
$$\ce{N(aq) + P <=> PN} $$
$$\ce{K(aq) + P <<=> PK} $$

This means that any given $\ce{I}$ molecule will be absorbing to paper, then becoming unstuck from the paper, then absorbing to a (probably different) site on the paper, then desorbing again and returning to aqueous solution.  The same is true of $\ce{N}$ and $\ce{K}$.

There are two further facts that we need to realize to understand the chromatography.  The first is that the equilibria are balanced at different points.  This is why the size of the arrows in the reactions I drew above are different.  As drawn, component $\ce{I}$ has stronger affinity for the paper and a lesser affinity for water, as indicated by the longer rightward top arrow and the shorter bottom left arrow.  The reverse is true for component $\ce{K}$, and $\ce{N}$ is somewhere in between.

The second fact is that the water is moving due to the capillarity you mentioned. So when the compounds are in aqueous solution, they will flow with the water.  When they are stuck to the paper, they will not move.  Since all of the molecules are very rapidly transitioning between paper-bound and water-"bound" states, they will all move some of the time, and be still another part of the time.  The key is that the different components spend, on average, different fractions of time moving vs. being still.  This makes their average speed different.

As for the question of why they eventually come to rest, this is because the capillarity-driven flow eventually runs out of steam (i.e. gravitational forces compete with the capillary forces after the fluid rises far enough), and the water itself stops flowing very much.  Thus the bands of separated molecules eventually settle into a relatively fixed position.

So to answer your title question, truly insoluble compounds would not be separated by chromatography.  Compounds need to be soluble to at least a degree in order to be separated.


CO : CO₂ ratio when given amounts of carbon and oxygen react with each other

You should have written:

$$\ce{C + O2 -> CO2}$$

and

$$\ce{C + \frac{1}{2}O2 -> CO}$$

Then you should have determined what fraction of the carbon reacts by by the first reaction and what fraction of the carbon reacts by the second reaction.


The unit of $k$ is $\frac{\mathrm{concentration}}{\mathrm{unit~time}}$. It is not necessary that you use any specific unit for concentration or time, but most often $t$ is measured in seconds, and the unit of $k$ is $\mathrm{mol~L^{-1}~s^{-1}}$.


The $\ce{Li^+}$ ion substitutes the $\ce{OH}$ in $\ce{COOH}$ first. The $\ce{C=O}$ is then reduced by $\ce{AlH_3}$ to give an aldehyde. See below:

Another $\ce{LiAlH_4}$ then reduces the aldehyde to alcohol:

I found the pictures here.


Origin of broad CO stretch in infrared spectrum of tricobalt cluster

I found an reliable article (http://www.ilpi.com/organomet/carbonyl.html) describing wide spectrum of metal carbonyls. Since Chemistry SE won't open that website, I have attached scheme with a range of carbonyl stretching frequencies in the article for your benefits:

The IR stretching of carbonyl can be varied based on the structure. According to the website:

  In the IR, typical stretching frequencies are: (i) Uncoordinated or "free" $\ce{CO}$: $\pu{2143 cm^{-1}}$; (ii) Terminal $\ce{M-CO}$: $\pu{2125}$ to $\pu{1850cm^{-1}}$; (iii) Doubly bridging ($\mu_2$): $\pu{1850}$ to $\pu{1750 cm^{-1}}$; (iv) Triply bridging ($\mu_3$): $\pu{1675}$ to $\pu{1600 cm^{-1}}$; and (v) Semibridging: somewhere between terminal and $\mu_2$.

In literature, most tricobalt clusters were discussed included with other metal cations. For example: (a) Rational synthesis of tricobalt-molybdenum and -tungsten butterfly clusters with alkyne ligands in Journal of Organometallic Chemistry, 2002, 659(1-2), 142-150 (https://doi.org/10.1016/S0022-328X(02)01720-5) and (b) Clusters as Ligands. 5. Tricobalt Cluster Alkoxycarboxylates of Titanium and Zirconium Exhibiting Novel Structures and Properties in Organometallics, 1997, 16(24), 5289–5301 (https://pubs.acs.org/doi/abs/10.1021/om970837b).

You can also read: A. S. Goldman, K. Krogh-Jespersen, Why Do Cationic Carbon Monoxide Complexes Have High $\ce{C-O}$ Stretching Force Constants and Short $\ce{C-O}$ Bonds? Electrostatic Effects, Not $\sigma$-Bonding, J. Am. Chem. Soc., 1996, 118, 12159-12166 (https://cdn-pubs.acs.org/doi/10.1021/ja960876z).

Since you have completely described preparation procedure of your tricobalt cluster, I assume the broasness of carbonyl absorptions are due to different structural aspects of carbonyl on $\ce{Co}$ and other metals in the cluster. It is also possibility that you have a mixture of clusters. Note that range of $\nu_\mathrm{CO}$ could be $\pu{2100}$ to $\pu{1300 cm^{-1}}$, based on stereo arrangement (see above references).


Molar conductivity of transition metal complexes solutions

Charge transport depends on the presence of charge carriers, and an increase of the concentration of charge carriers may increase the electric conductivity.  This is the basis of the answer given to you which (implicitly) assumes each of these complexes dissociate, e.g.

$\ce{K4[Fe(CN)6] <=> 4 K+ + [Fe(CN)6]^{4-}}$

in your solvent completely.  In this view, charge transport happenz without moving these carriers on a macroscopic level (like  the salt bridge used to connect electrochemical cells). You may imagine a bucket brigade, instead people exchanging buckets of water, ions shuttle electrons.

On the other hand, there are examples where charge transport is coupled to the transport of matter (e.g., ions) on a macroscopic scale.  Proteins and nucleic acids, for example, may be separated by this (e.g., gel electrophoresis); their different size (and shape, and charge) quite noticeably influence how quickly they travel in an electric field.


How can subtilisin still function without its catalytic triad?

If the substrate still binds to the protein, other parts of the mechanism are still in place. Specifically, the main chain amide groups that help to stabilize the tetrahedral intermediate are still present. In the absence of serine, water will attack the carbonyl. In the presence of serine and the absence of the histidine, serine will still be the nucleophile, but a less efficient one.

Look up oxyanion hole if this part of the mechanism is unfamiliar to you.


How to calculate the enthalpy of formation of hydrazine from tabulated values?

In order to "make" hydrazine, $\ce{H2N-NH2}$ from hydrogen and nitrogen atoms, you have to

break $\ce{N2}$ once, giving 2 nitrogen atoms
break $\ce{H2}$ twice, giving 4 hydrogen atoms
form one $\ce{N-N}$ bond
form four $\ce{N-H}$ bonds

You have all the enthalpies that you need. Just pay attention to the coefficients and the signs.


One could think of the Fermi surface and conduction bands as an expression of the degree of metallic bonding, where metals such as aluminum or silver have overlapping empty and filled bands, allowing for electrical conduction, while semiconductors have a small gap between filled and conduction bands, offering more resistance.

As @Mithoron states, metallic bonding is a bulk property. Metals behave differently as nanoparticles -- for example, bulk silver is reflective, but nanometer particles of it are black, as in photographs.


How to determine which compound will get oxidised and which will be reduced in a cross cannizzaro reaction?

It correlates very strongly with ability to form a hydrate as @Jan's answer says, which is why formaldehyde is usually the sacrificial oxidised component. In general electron withdrawn aldehydes are more easily oxidised, electron rich aldehydes reduced.

This paper from J. Org. Chem. here compared several pairs of aldehydes and the conclusions were:

  m-nitrobenzaldehyde is (relatively) extensively oxidised as contrasted to p-anisaldehyde, which is much less extensively oxidised - i.e., much more extensively reduced.  On the bases of this and prior investigations, the five aldehydes studied can be arranged (in order of diminishing susceptibility to oxidation): m-nitrobenzaldehyde, furfural, p-bromobenzaldehyde, benzaldehyde, and p-anisaldehyde. With reference to benzaldehyde and the substituted benzaldehydes, this order correlates with expected combined resonance and inductive influences of the substituent groups on the formyl group, as these should contribute to ease of a) hydroxide ion attack and b) hydranion transfer


How come when it rusts iron can undergo oxidation without the prescence of a another metal to undergo reduction?

Rusting is a redox reaction.
$\ce{4 Fe + 3 O2 -> 2 Fe2O3}$

Redox reactions mean one chemical is being reduced and one is being oxidized. Reduction = electron gain, oxidation = electron loss. The reason it is called oxidation is because oxygen is so good at reducing other chemicals (it wants electrons). How do you tell what gets electrons in a compound? Oxidation numbers.

Oxidation numbers signify how many electrons each atom has. -n means it has n more than neutral state, +n means it has n less than neutral state. In molecules which have only one element, oxidation state is 0. So Fe/O2 both have oxidation state 0. There are some rules on oxidation states in compounds which are complicated, basically usually group 1 is +1, group 2 is +2, O is -2, and group 17 is -1. In this case each O in the end is -2, and Fe is +3. So each Fe lost 3 electrons.


Why doesn't 1,3-dichloro-2,4-dimethylcyclobutane have a plane of symmetry?

Ivan already stated quite briefly in the comments, that the planes you drew would transform an upward facing chlorine to a downward facing chlorine and vice versa, and that the same holds for the methyl groups. This is of course true and the first and most obvious point, why there is no symmetry plane.

A more close investigation will, however, point out another important feature of the compound itself. The cyclobutane ring itself is not planar. In the (electronically) lowest conformation of the molecule it has no symmetry whatsoever (technically $C_1$ symmetry), not even inversion symmetry like your book states. This is due to the fact, that the four-membered ring is puckered. This is best visualised by looking at the 3D structure of the compound. There are multiple programs that will allow you to view such a structure and I will xyz coordinates at the end of this posting.

The following image will hopefully already show you what I am talking about. It was optimised on the DF-BP86/def2-SVP level of theory.

In order for the the molecule to have inversion symmetry, the shown angle needs to be exactly $0^\circ$. The following animation shows the molecule rotating hopefully further clarifying this point.

You can clearly see the puckering of the cyclobutane ring.

One final comment: For a molecule like this it really is necessary to specify the stereochemistry with the name. There are other stereoisomers, which do indeed have a symmetry plane. I leave it to you to figure out which of them do.

   Except the last one, all of them have at least one plane of symmetry.

Addendum: Transition state structure

The symmetric ($C_\mathrm{i}$) transition state is on the DF-BP86/def2-SVP level of theory only about $\Delta G (298.15~\mathrm{K}) = 8~\mathrm{kJ/mol}$ higher in energy than the puckered asymmetric forms. It is therefore worthwhile to have a quick look at its structure. In the following graphics the red ball indicates the inversion centre.
Below that is an animation of the transitional mode. It briefly pauses at the puckered structures (these are extrapolated, not optimised structures) and the transition state to make it easier to see the movement in between them.
I would expect this process to be readily available at room temperature and therefore almost impossible to distinguish the different lower lying structures.

Coordinates of the lowest structure:

18
G09D.01; E(RB-P86/def2-SVP/W06) = -1154.66760674 A.U.
C        0.051334000      1.246599000      0.089432000
C       -0.768087000      0.117382000      0.789824000
C        0.002884000     -0.950933000     -0.020424000
C        1.086640000      0.142749000     -0.293350000
H        0.449081000      2.008047000      0.790972000
H       -0.535742000      0.060683000      1.871003000
H       -0.532992000     -1.105843000     -0.982729000
H        1.517474000      0.183811000     -1.312785000
C        0.343625000     -2.291820000      0.610089000
H        0.939420000     -2.156389000      1.535300000
H       -0.586821000     -2.839917000      0.866159000
H        0.932752000     -2.926935000     -0.083624000
C       -0.606103000      1.926898000     -1.110977000
H       -1.475939000      2.532182000     -0.784546000
H        0.116584000      2.600709000     -1.616245000
H       -0.977177000      1.195914000     -1.858831000
Cl       2.502295000      0.083512000      0.845211000
Cl      -2.570729000      0.126790000      0.662589000

Coordinates of the transition state (X is a dummy atom indicating the inversion centre)

19
G09D.01; E(RB-P86/def2-SVP/W06) = -1154.6659061600 A.U.
C       -0.188954000      1.025076554      0.415795307
C       -0.873066584     -0.383572154      0.501227861
C        0.188954000     -0.991627446     -0.457606693
C        0.898153416      0.404477846     -0.509590139
H        0.242239584      1.289691385      1.404629307
H       -0.864241753     -0.820445015      1.518694431
H       -0.267326416     -1.206068615     -1.446440693
H        0.939502247      0.849712985     -1.522875569
C        0.953473154     -2.212350139      0.042720768
H        1.426842139     -2.018382400      1.026791199
H        0.267219307     -3.079692431      0.145048491
H        1.760843738     -2.494108569     -0.663022216
C       -0.974378846      2.203987861     -0.147249232
H       -1.806836262      2.489927431      0.529225784
H       -0.309030693      3.083873569     -0.266301509
H       -1.418479861      1.964027600     -1.135500801
Cl       2.591747000      0.489781723      0.126797539
Cl      -2.591747000     -0.506506277     -0.064080461
X        0.006271708      0.013588700     -0.012543416


Assigning Cahn-Ingold-Prelog priority numbering rules for 2 groups

When you've reached the end of your first group (after resolving the double bond), you're next comparing no atom (on the ghost C connected to C2) to a hydrogen atom (on the 4-methyl group), and any atom has a higher priority than no atom, so your second group gets the higher priority.


Is there any way of representing a dative/coordinate covalent bond?

Generally, you wouldn't describe the bonds in an ammonium ion as dative bonds. They are usually considered to be proper covalent bonds. The term "dative bond" is typically used for bonds in (transition) metal complexes between ligands and a metal center. But for those complexes Lewis structures have lots of weaknesses and are rarely a good representation of the actual bonding. Yet they are used for the sake of convenience. Often an arrow from the ligand to the metal center is used to indicate a dative bond.

(source: uni-erlangen.de)
But just as often you see normal "covalent bonds" (straight lines) like in traditional Lewis structures being used.
Often both kinds of representations are even intermixed.
For a more detailed view on the subject of bonding in coordination complexes and its representation have a look at this answer of mine.


If you add sodium salt to carbonated water (sparkling mineral water), could you produce sodium bicarbonate?

Assuming $\ce{Na}$ is used to denote sodium metal and $\ce{NaCl}$ is used to denote table salt, the following reactions in water occur:

\begin{align}
\ce{Na + 2H2O &-> 2Na+ + 2OH- + H2(g)} \tag {1}\\
\ce{NaCl(s) &-> Na+ + Cl-} \tag{2}\\
\ce{CO2_(g) <=> CO2(aq) &<=> H2CO3(aq) <=> H+ + HCO3-} \tag{3}\\
\ce{H+ + OH- &-> H2O} \tag{4}
\end{align}

So (1), (3), and (4) would give you sodium bicarboante if you dried the solution.

(2) and (3) would give you just sodium chloride back if you dried the solution.


Unfortunately, your confidence in what a chemist can, is quite exaggerated. Though flattering.

Analyzing the entire range of parameters that you list, requires several analytical techniques, including sample preparations like homogenization, extraction etc.

A technique like Near-infrared spectroscopy can give an easy and fairly good estimate of some major components (protein, fat etc) in selected types of food. FOSS Analytical (and probably several others) purchases equipment for this.

Routine methods exist for analyzing components like vitamins, pigments or metals, but they require a laboratory setup and often very expensive equipment.


What is the maximum pressure that JB Weld can withstand?

This is only a partial answer, addressing the maximum pressure a JB weld can withstand. I am still thinking about the reason for the strength so I will leave it to the 'real' chemists among you for now.

According to Repair Products the JB weld in fully hardened state has the following properties (in psi):

Tensile Strength:   3960
Adhesion:           1800
Flex Strength:      7320
Tensile Lap Shear:  1040

Tensile strength is simply the pressure needed to break the material by pulling it apart. Flex strength is the pressure needed to bend the material which, in case of a perfectly homogeneous sample would equal the tensile strength.

The adhesion is the pressure needed to pull the bond from the surface when pulling perpendicular to the surface (parallel to the bond). The tensile lap shear is also an 'adhesion', but measured when pulling parallel to the surface (perpendicular to the bond).

Since the latter is the lowest it means that this would be the maximum pressure the JB weld could withstand, but only if it is applied exactly perpendicular to the bond. If the direction is in a different direction than you will get some mixture between the adhesion and the tensile lap shear pressures.

What is clear from these numbers is that the bond itself will not break but rather it will detach from the materials it is keeping together.


Most metal nitrates decompose on heating to give metal oxides. Decomposition of silver nitrate is different as it yields elemental silver.
$$\ce{2AgNO3 -> 2Ag + O2 + 2NO2}$$
Usually the metals which lie very low in the activity series decompose to give their elemental form.
I don't think it's possible to extract metals from just any nitrate by strong heating. They will yield their respective oxide which won't give the element rather they'll just fuse (melt) just like stable metal oxides like $\ce{Al2O3}$ does.


Why is the first ionisation energy of a sodium cation bigger than that of a neon atom?

A sodium cation $\ce{Na+}$ and a neutral neon atom $\ce{Ne}$ are isoelectronic species; meaning they have the same number of electrons and also the same electronic configurations.

Yet, the two have different nuclei: most prominently, sodium always has an additional proton when compared to neon. Therefore, only one of the isoelectronic species (neon) is neutral, the other is positively charged.

It may now become obvious why the ionisation enthalpy of $\ce{Na+}$ is much larger than that of neon. Different ways to express this all boil down to the fact that it is harder to remove a negatively charged electron from a positively charged species than it is to remove one from a neutral species.


How does concentration affect The Position Of Equilibrium?

You statement of Le Chatelier's Principle is a perversion that isn't true. The equilibrium does not depend on the sum of the concentrations of the reactants relative to the sum of the concentration of the products.

Your analysis is correct.


Explaining stereochemistry in a sigmatropic 1,3-alkyl migration

Problem #1 is that if you interchange the methyl and hydrogen groups, you get diastereomers, not enantiomers. But this is a relatively minor point. I assume your question is about the diastereoselectivity, not enantioselectivity (as ron rightly pointed out there is no enantioselectivity in this reaction).

The analysis you have drawn is somewhat similar to a Fukui frontier molecular orbital analysis. In that picture, for a reaction to be thermally allowed, you need to check for constructive bonding overlap between the HOMO and the LUMO, within geometrical reason. This is achieved by connecting lobes of "like" shading: i.e. shaded lobe with shaded lobe, and unshaded with unshaded.

You don't necessarily have to draw it stepwise (although I gather it's your own way of doing it). You can simply draw both interactions in the starting material, like this. The bottom interaction is not an issue. The top interaction, however, is an issue.

Problem #2 is that you tried to connect the large unshaded lobe of the σ* orbital with the white lobe of the π orbital. That is not geometrically reasonable: if you connect these two lobes, then you need to somehow end up with the new σ bond being formed underneath the ring. In other words, you would have to have an antarafacial migration of the alkyl group.

Problem #3 is that you have concluded that the reaction must be suprafacial based on electron counting. Note that in this context, the reaction being suprafacial means that the migrating alkyl group stays on the same face of the ring. In fact, the reaction must occur in a suprafacial manner simply because of geometrical constraints: there's simply no way that the alkyl group is going to migrate to the opposite side. Technically it could, but you would have to break so many bonds in the transition state that it would not be possible.

[Note that this has a different meaning from the suprafacial or antarafacial components in a Woodward–Hoffmann analysis.]

So what is the solution? Well, because the reaction is suprafacial, you need to use the shaded lobe of the π HOMO, which is on the top face of the ring – the same face as the migrating alkyl group. So then for constructive bonding you need to also use the shaded lobe of the σ* LUMO.

Now one might argue that just like how the unshaded lobe of the π HOMO was geometrically inaccessible, so should the shaded lobe of the σ* LUMO. But we already know that the C–C σ bond is going to break in the course of the reaction, so you're going to go through a pathway whereby the bond behind it (with an arrow pointing towards it) can rotate such that the shaded lobe of the σ* can overlap with the shaded lobe of the π* orbital. In fact, that's not too different from what you drew. The transition state is pretty similar to the intermediate in your stepwise reaction – it just has a few more dotted lines.

If you now "join" the shaded lobes then you can see how the correct stereochemistry arises.

In general my preferred way of analysing pericyclic reactions is to use Woodward and Hoffmann's original rules. Here, there is no shading of lobes: you connect the lobes within geometrical reason, then count the number of $4q+2$ suprafacial components, and the number of $4r$ antarafacial components. If the total is odd, the process is thermally allowed; if it is even, the process is thermally forbidden.

I am not going to explain this approach in detail - it is too long for here and it is adequately covered in textbooks which you should have access to - but I will upload the two diagrams I drew:


Sodium Silicide (NaSi) Synthesis, Reversability

The reaction of $\ce{NaSi}$ with water is both highly enthalpically and entropically favored—the reaction produces 5 molecules of $\ce{H2}$ gas for every 2 $\ce{NaSi}$ molecules that react, and the enthalpy of reaction is –175 $\mathrm{kJ\ mol^{-1}}$. The reverse reaction would require ludicrous amounts of energy and is likely not even kinetically feasible.

It would be especially pointless to attempt the reverse reaction when $\ce{NaSi}$ is generated easily enough from $\ce{Na}$ and silica gel. Liquid sodium-potassium alloy absorbed onto silica gel results in a fine black powder that spontaneously ignites in humid air. Heating $\ce{Na}$ and silica gel to 165ºC with continual agitation produces a shelf stable, yet still highly reactive $\ce{NaSi}$. When slowly heated to 400ºC, air stable $\ce{NaSi}$ may be produced, though with less reducing power than those produced via the other two methods. It will, however, still react with water to produce $\ce{H2}$ and $\ce{ Na2Si2O5}$.$^{[1]}$

$^{[1]}$ Alkali Metals Plus Silica Gel:  Powerful Reducing Agents and Convenient Hydrogen Sources


Difference between intermediates and transition states

An intermediate is a short-lived unstable molecule in a reaction which is formed inbetween the reaction when reactants change into products.

Whereas, transition state is just the state before formation of new molecule(involves breaking of bonds of reactants and formation of new ones)

An intermediate differs from a transition state in that the intermediate has a discrete lifetime (be it a few nanoseconds or many days), whereas a transition state lasts for just one bond vibration cycle. Intermediates may be unstable molecules (in which case they are called reactive intermediates) or highly stable molecules.
The difference between them can be better described through the energy profile diagram.

Transition states are local energy maximums and have partial bonds. This might be one of the reasons why they cant be isolated as intermediates.


Why do higher fluorine oxyanions not exist?

There are two factors at play here:

Fluorine is more electronegative than oxygen. In oxyanions, the central atom carries a substantial positive charge due to the oxygen atom pulling away electron density. In oxyfluorides, oxygen actually donates electron density to fluorine and oxygen would logically be the central atom. There are theoretical studies into $\ce{OF3+}$ but no successful syntheses yet.
The fluorine atom is tiny, even by period 2 standards. There's simply not enough space around the proximity of the atom to fit three or more other atoms.


Absolutely not. It means that for the mixture to be boiling at 110°C, it must not contain any benzene.


Potentials only tell you about thermodynamic stabilities. They do not tell you anything about the kinetics of the system, which would correspond to how quickly the system reacts.


Why does the ionization energy decrease anytime the atom size increases?

The ionization energy is the lowest amount of energy required to remove one electron(the most loosely attached electron) from each atom in one mole of an element in gaseous state to form gaseous (positive) ion.

There are certain factors that have an impact on the ionization energy value.

To understand the answer to your question, you must understand the effect of all three

Atomic radius (Inversely proportional to I.E. Value)
Shielding effect of electrons (Inversely Proportional to I.E. Value)
Nuclear Charge (Proportional to I.E. Value)

The Atomic radius is the distance between the to-be-removed electron and the nucleus. The greater the distance, lesser the force of attraction. Hence as this distance increases the electrons are easier to remove; lesser force is required to remove it from that attractive field of the nucleus. (This answers your question) Even the smallest change in this distance will cause a change in ionization energy for example The first ionization energy of Magnesium and Aluminum. The ionization energy should increase as the nuclear charge increases across the period but the I.E. of Aluminum is lesser than that of magnesium because there is a change in orbital which increases atomic radius, decreasing I.E.

Now going for a bit more information.
The Shielding effect of the other electrons decreases the I.E. because it doesn't let the outer most electron feel the full force of attraction.

IF both the first factors are CONSTANT (atomic radius and shielding effect of electron) only then the last factor comes in play.
As the atomic charge/Nuclear charge increases, there is a general increase in ionization energy because of the increased force of attraction by the extra proton. for example, again across the period, the general trend is increase in ionization energy because there is no (significant) increase in atomic radius or shielding effect so the I.E. increases proportional to the Nuclear charge.
(the decrease in sulfur is because of the pair of electron in the p orbital repelling each other. Both electrons apply a force on each other. The looser of the two, is removed when energy is applied.)


What is the absolute configuration of carbon 4 in glucopyranose?

therefore it will retain that chirality as it forms the hemiacetal.

This statement is incorrect. Many chiral centres retain their CIP-descriptor upon reactions that don’t influence them directly, many others do not.

How do you assign CIP priorities? You basically had the same problem in the linear form of D-glucose. You move along the chain in order of priorities, finding the first priority difference and using that to assign the higher-priority side. On C4, you have (read this from left to right, i.e. everything with the same indention at the same time):

an oxygen (a)
the carbon C3

an oxygen (a)

a hydrogen.

the carbon C2 (b)

an oxygen (a)
a carbon C1 (b) <-- this one is higher priority than on the other side!
a hydrogen (c)

a hydrogen (c)

the carbon C5

an oxygen (a)

a hydrogen

the carbon C6 (b)

an oxygen (a)
a hydrogen <-- less priority than carbon
another hydrogen.

a hydrogen (c)

a hydrogen (d)

Following the same method in the pyranose form we get:

an oxygen (a)
the carbon C5

an oxygen (a)

a carbon <-- this one is higher priority than the other side!

the carbon C6 (b)
a hydrogen (c)

the carbon C3

an oxygen (a)

a hydrogen <-- this is lesser than carbon!

the carbon C2 (b)
a hydrogen (c)

a hydrogen (d)

Therefore, the absolute configuration of said atom changes from (R) to (S).

This is actually a pretty common thing that can happen in highly functionalised molecules with very similarly modified carbons. Because we cannot distinguish at any proximate position we need to go far down the chain and a single distal modification can change a lot.

Other examples that would have changed the absolute configuration of carbon C4 without changing its geometry would be:

methylation at C5
silyl protection at C5
oxidation at C5

All of these would make C5’s priority higher than C3’s.


Percent of silver by mass in an unknown sample

Noah--

You're good to go. I was taught to use unit analysis to help me make sure I was working through these problems correctly:

82.46 mL KSCN solution * (0.25 M KSCN/1 L) * (1 M SCN- / 1 M KSCN) * (1 L / 1000 mL) = 0.020615 M SCN- (mL cancels, M KSCN cancels, L cancels)

0.020615 M SCN- * (1 M AgSCN / 1 M SCN-) * (1 M Ag+ / 1 M AgSCN) * (107.8682 g Ag / 1 M Ag) = 2.2237 g Ag ( M SCN- cancels, M AgSCN cancels, M Ag+ cancels)

2.2237 g Ag / 7.563 g sample = 0.294 g Ag per g sample * 100% = 29.4% Ag by mass

I found that it helps if you read your equations out loud you can work through the logic more easily. If you set the equation up as fractions, you can see the units that cancel each other. For the most past, if you can get the right units for your answer, you're doing it right.

And you get to ignore the nitric acid (as well as the potassium) because they don't factor into your analysis.


What chemical elements does a nuclear explosion release into its mushroom cloud?

According to The Preparatory Manual of Explosives, the mushroom cloud is produced by the partial vacuum left after initial expansive shock wave. That implies that it is primarily the condensation of water that makes it visible.  These clouds form in high explosive as well as in nuclear detonations near the ground. See https://books.google.com/books?id=J55D3HcgPuoC&pg=PA98#v=onepage&q&f=false.


Bond angles mainly depend on the following three factors:

Hybridization: Bond angle depends on the state of hybridization of the central atom
Hybridization: $\ce{sp^3}$, Bond angle: $109^\circ$, Example: $\ce{CH4}$
Hybridization: $\ce{sp^2}$, Bond angle: $120^\circ$, Example: $\ce{BCl3}$
Hybridization: $\ce{sp}$,   Bond angle: $180^\circ$, Example: $\ce{BeCl2}$
Generally s- character increase in the hybrid bond, the bond angle increases.
Lone pair repulsion: Bond angle is affected by the presence of lone pair of electrons at the central atom. A lone pair of electrons at the central atom always tries to repel the shared pair (bonded pair) of electrons. Due to this, the bonds are displaced slightly inside resulting in a decrease of bond angle.
Electronegativity: If the electronegativity of the central atom decreases, bond angle decreases.

Good Side note:

Triple bonds repel other bonding-electrons more strongly than double bonds.
Double bonds repel other bonding-electrons more strongly than single bonds.

Heres a nice read into the different components (extra)