Synthesis of Potassium Trioxalatoaluminate (III) Trihydrate

You have already concluded that the bubbles indicate that hydrogen gas is formed. Let's assume that two protons are reduced. Each receives an electron and they form a molecule of $\ce{H2}$:

$$\ce{2H+ + 2e- -> H2}$$

Where did the electrons come from? Did the aluminium powder dissolve?

$$\ce{Al -> Al^3+ + 3e-}$$

We are now having two equations, one for a reduction, one for an oxidation (of aluminium).

Let's add these two equations in a way that the electrons cancel out. For that, we multiply the first equation by 3, and the second equation by 2.

$$\ce{2Al +6H+-> 2Al^3+ + 3H2 ^}$$

Something's still wrong with the equation: You didn't throw the aluminium powder into a sea of protons, but suspended it in hot water. Let's add some $\ce{OH-}$ on both sides:

$$\ce{2Al +6H2O -> 2Al(OH)3 + 3H2 ^}$$

This is much better, but still not perfect: You did not describe a cloudy precipitate of a hydroxide being formed, and your first step took place in  the presence of potassium hydroxide:

$$\ce{2Al +6H2O +2KOH-> 2K[Al(OH)4] + 3H2 ^}$$

So far, we have explained the reaction of aluminium metal in aqueous alkaline solution: tetrahydroxoaluminate $\ce{[Al(OH)4]-}$ is formed.

Can you figure out the reaction with the oxalic acid on your own? Have a look at the structure of oxalate (Hint: bidentate ligand) and forget about the permanganate ;)

UPDATE

Let's add some oxalic acid, $\ce{HOOC-COOH}$ now. For the reaction, it doesn't make any difference whether it's a dihydrate, $\ce{H2C2O4*2H2O}$, or not. The reaction is performed in aqueous solution anyway. However, you might want to consider the difference in molecular weight between the anhydrous form and the dihydrate when you're planning to add a stoichiometric amount of oxalic acid.

$$\ce{[Al(OH)4]- + 3HOOC-COOH -> [Al(OOC-COO)3]^3- +2 H+ + 4H2O}$$


Can I make a mirror on PET plastic using Tollen's reaction?

I found a video on YouTube making a mirror on PET bottles.

Just that it consumes quite a lot of substance, and is not using Tollens reagent. My question is specifically about Tollens reagent.

What she uses is some "secret mixture" called RA-205.

It just proves that Ag can precipitate on PET bottles, which further supports the fact that will work with Tollens reagent. I'm will edit this answer once I execute my experiment, so this finding may only be somewhat helpful.


Will gamma radiation cause "thermal" decomposition of sodium bicarbonate?

There is an equilibrium between decomposition and recombination, i.e. as $\ce{CO2}$ is given off, it joins back to the  $\ce{Na2CO3}$ to form $\ce{NaHCO3}$.

  $\ce{2NaHCO3 <-->Na2CO3 + CO2}$

So for moderate doses, given at moderate rates, it's unlikely great pressure would develop, but you'd need to test to see if the drug is degraded by decomposition products.


Reduction of hemiaminal by triethylsilane / boron trifluoride

My organic chemistry is rusty by now, but the full mechanism for the reduction itself is likely something along these lines:

In a "typical" lactone / ester, the alkyl oxygen can't be lost so easily, but in this case there's a nitrogen lone pair that can assist in that. The resulting iminium ion is reduced by the silane. The acid at the end will remove the remaining boron trifluoride.


Concept of "non-superimposable mirror image" in chirality

Your hands are chiral, that is why you need two different leather gloves, one that only fits your right hand, and one that only fits your left hand.  If your hands were superimposable, then you would only need one kind of glove and it would fit both hands.


Boys function for Gaussian integrals in ab-initio calculations

I am not aware of any existing Fortran code for direct numerical quadrature of this problem, but it is worth pointing out that Mathematica can perform this integral symbolically:

Integrate[t^(2 n) Exp[-x t^2], {t, 0, 1}]

(* 1/2 x^(-(1/2) - n) (Gamma[1/2 + n] - Gamma[1/2 + n, x]) *)

where the $\Gamma$ function can be computed by exponentiating easy-to-find $\ln \Gamma$ functions, often called lngamma().  (One of them is the Euler function $\Gamma(z)$ and the other incomplete $\Gamma(a,z)$ function).

As a check:

With[
  {x = 1, n = 2},
  1/2 x^(-(1/2) - n) (Gamma[1/2 + n] - Gamma[1/2 + n, x])
  ] // N

(* 0.100269  *)

and numerical quadrature yields:

With[
 {x = 1, n = 2},
 NIntegrate[t^(2 n) Exp[-x t^2], {t, 0, 1}]
 ]

(* 0.100269  *)

One has to be a little bit careful, as the $\Gamma$ function is closely related to the factorial, and it can explode as the arguments get too large.  I would experiment and see whether you can get by with this analytic solution, which essentially takes you from numerical quadrature into evaluation of special functions.

The $\Gamma$ function approach is faster by over two orders of magnitude  than the quadrature approach.  This is critical when evaluating molecular integrals.  I have seen implementations where the Boys function is actually interpolated rather than evaluated, and there is discussion about the error in interpolation being almost insignificant if suitable interpolation basis functions are used.


Where is the probability of finding an electron in 1s orbital maximum?

You are right in saying that an s orbital is spherically symmetric, hence the probability density must not depend on the angles. Also, you are right in saying that $dV$ in polar coordinates contains $\sin\theta$. So what? You don't have to multiply or divide anything by that $dV$ (nor do you need to even know what it is). Old man Schrödinger took care of that for you. $|\psi^2|$ already gives you the probability density. Just as expected, it is a function of $r$ and not of the angles.


I am seeing Doublet of Doublets peaks when I should see just a doublet

Your question is how can the CH2 be split into a dd when the -OCHO shows no triplett, right?

Well, the second splitting at the $\ce{CH2}$ is very small, and the $\ce{CHO}$ looks like it is very nearly split. That´s not just a broad peak, it has a rather clear shape imo. If you do just a bit of zero filling and apodisation, the triplett might well (should) appear.

But generally, you are right. Every splitting must have its counterpart.

Zero filling: Adding blanks to the end of your FID. Artificially increases the resolution, i.e. point density of your spectrum.

Apodisation: Changing the gain of your FID towards the end, by multiplying it with a decreasing or increasing function. The former gives you a less noisy spectrum, the latter a better resolution (sharper peaks actually), each at the cost of the other. The mathematical term for that is "convolution".


Finding half-life of a nuclide based on gas release

For this we can first write the equation for the reaction again:
$$MH_2 => He_{(g)} + Noble gas + H_{2(g)}$$
Next we can find the moles of gas at 80 mins (I am assuming it is 80 minutes and not 82). You already did this and you got 0.045 moles. Since each mole of $MH_2$ forms 3 moles of gas, $\frac{0.045}{3}=0.015$ moles of $MH_2$ used in 80 minutes. Then we can use the first order integrated rate law now:
$$ln(N_f) = -kt + ln(N_i)$$
$$ln(0.025-0.015) = -k(80 mins) + ln(0.025)$$
$$k = 0.0115 min^{-1}$$
Now that we have k, we can solve for the half-life.
$$t_{1/2}=\frac{ln(2)}{k}$$
$$t_{1/2}=\frac{ln(2)}{0.0115min^{-1}} \approx 61 mins$$


Is it possible to generalize whether an organic molecule is chiral or not?

There is a general statement. If the molecule has any improper axis of rotation, it is achiral.


Molpro - Connection between atomic term symbols and system symmetry

You are trying to use the wf card to pass individual configurations. This is wrong in 2 ways:

The wf card does not specify which configurations to use, it specifies the symmetry of the wave function (or electronic state).
You don't need to tell Molpro which configurations to use. It will figure this out on its own based on the symmetry of the wave function you are requesting.
All you need is the IRREP of the $^1S_0$ state, which is totally symmetric.

{multi;
  occ,2,1,1,0,1,0,0,0;
  core,1,0,0,0,0,0,0,0;
  wf,6,1,0;
 }

However, this will give you different a state than expected. Add expec2,lxx,lyy,lzz; and check $\langle L^2\rangle$ in the output. It should be 6, telling us the $^1D_1$ state is lower in energy than $^1S_0$. So we need to request more states of that symmetry (use the state card). Also, we should make sure the $^1D_1$ is described appropriately, as it might negatively affect the targeted $^1S_0$ state. So you should add all of the 5 degenerate $^1D_1$ components. I leave that as an exercise for you (see also below).

Some further note about your configuration $(a_g)^2\otimes(a_g)^2\otimes(b_{3u})^1\otimes(b_{1u})^1=B_{2g}$ (Note that the symmetry of orbitals is commonly denoted by small letters, while capital letters are used for electronic states.)

Only IRREPs of same symmetry mix, so this one will not contribute to $^1S_0$.
It is breaking the spherical symmetry of the atom. This can only happen if there is some external perturbation, e.g. a magnetic or electric field.
It is however, one component of a degenerate (different) state. But you need to consider all the degenerate components to preserve symmetry. This is an issue of the symmetry reduction from $O_h$ to $D_{2h}$. For example you can calculate the $^3P$ state with

{multi;
  occ,2,1,1,0,1,0,0,0;
  core,1,0,0,0,0,0,0,0;
  wf,6,4,2;
  wf,6,6,2;
  wf,6,7,2;
 }

Check that you get 3 exactly degenerate $p$ orbitals, and 3 exactly degenerate electronic state energies.
If you would request each state separately, by running multi 3 times with only one wf card each, you will get 3 degenerate state energies, but the $p$ orbitals won't be degenerate.


Regarding the use of dilute sulphuric acid in this titration

A little bit of sulphuric acid is used when preparing iron(II) solution. This acid prevents the possible hydrolysis of $\ce{Fe^{2+}}$ ion according to : $$\ce{Fe^{2+} + 2 H_2O <=> [Fe(OH)_2(s) + 2 H^+}$$ In case this reaction occurs, the added sulphuric acid will react and destroy the precipitate $\ce{Fe(OH)_2}$ according to the inverse reaction : $$\ce{Fe(OH)_2 + 2 H^+ -> Fe^{2+} + H_2O}$$
Now, during the titration, the ion $\ce{H^+}$ is needed to carry out the reaction of permanganate on the ion $\ce{Fe^{2+}}$ . Hopefully you know this equation.
And concentrated sulphuric acid is not wanted, because it reacts strongly with water just during its dissolution. It produces a lot of heat.


Difference between dissociation energy and dissociation temperature for water?

One key is an assumption about various degrees of dissociation.

It states that it takes 493.4 kJ/mol + 424.4 kJ/mol to dissociate the bonds in a water molecule which gives a total of about 920 kJ/mol or about 51 MJ/kg.

That sounds like the energy required for $\ce{H2O <=> 2H + O}$.

But on this other Wikipedia page, I found: "at 2200 °C about three percent of all H2O molecules are dissociated into various combinations of hydrogen and oxygen atoms.

Here it sounds like they are considering less extreme dissociations than complete dispersal to individual atoms. For example, $\ce{H2O <=> H + OH}$ and perhaps even $\ce{H2O <=> H2 + O}$ or $\ce{H2O <=> 2H + \frac{1}{2}O2}$ and related equilibria may be happening.
In effect, if just one OH bond breaks in water, that only requires 493.4 kJ/mol and leads to the formation of species such as $\ce{OH}$ and $\ce{H}$ atoms.  Those species have a different heat capacity than $\ce{H2O}$ vapor.  The heat capacity page you linked to says:

The values above apply to undissociated states. At high temperatures above 1500 K dissociation becomes appreciable and pressure is a significant variable.

In the calculation you are doing you are not including a specific pressure variable either.  Heat capacity could also be a function of pressure.  Presumably the data you linked to is all for 1 atm of pressure, although it didn't really say.  Likewise, the claim that at 2200 °C, 3% of water is dissociated must also be pressure-dependent.  For which pressure is it true?
The final complexity is the entropic term.    The three dissociated atoms have 9 translational degrees of freedom but intact water only has three.
These pressure and entropy complexities matter because $\Delta G$ is what determines equilibrium, and $\Delta G = \Delta U + \Delta (PV) - T \Delta S$.  The bond dissociation energies you quote represent the $\Delta U$ term, so the calculation you are doing is implicitly assuming that $\Delta U \gg \Delta (PV) - T \Delta S$.
I think that your calculation is admirable in that it comes so close to the right answer while still neglecting all these complexities!


What does "stoichiometric amounts" mean?

Sometimes, practical chemistry does not work like it's supposed to on paper. A perfectly balanced reaction equation may dictate that two reactants should be in a 1:1 stoichiometric ratio, but in practice, you get much better yields or reaction times if you use an excess of one.

When that happens, you are no longer using a stoichiometric amount, as specified by the balanced equation, but a super- or substoichiometric amount.

Welcome to chemistry in the real world.


Can vicinal dihalide produce conjugated alkene in strong base?

Comprehensive Organic Transformations (2nd edition) p. 467 gives examples of converting 2,3-dibromobutane to buta-1,3-diene.

On the other hand, Reaxys has many examples of producing alkynes from vicinal dihalides, even when potassium t-butoxide is used.

According to Carey's Advanced Organic Chemistry (5th edition, volume A), p. 554, the direction of E2 elimination is determined by whether the mechanism is more similar to E1 or to E1cb. If it is similar to E1, then the thermodynamically more stable product dominates; if it is similar to E1cb, then the direction is determined by the more easily removed hydrogen atom.

On page 556, Carey attributes the Hoffman rule of t-butoxide to its stronger basicity, not to its bulky size, and states that stronger bases drive the mechanism toward the E1cb side.

Therefore I think the problem might be understood this way: If a strong base such as $\mathrm{NaNH_2}$ is used, then the elimination direction is determined by which hydrogen atom is the easiest to remove. Since bromine acidifies its geminal hydrogen, the alkyne product dominates. If a weak base is used, the diene product might predominate.


You can't do adiabatic reversible work at constant pressure.  The pressure has to change.  If you want to do adiabatic irreversible work at constant pressure, say by suddenly increasing or decreasing the pressure load to begin with, and then holding it constant during the process, this is sometimes referred to as adiabatic work at constant pressure, but it really isn't.  The final pressure will not be equal to the initial pressure that was present before you changed the applied load.


Before we can consider the reaction, we must consider the conversion of the reactants into the reacting species:

Now, we may consider the reaction occurring between these two species:

The product of this reaction is now easily converted into para-phenyl phenol by an acid workup.


What software is used to generate the PDB molecule of the month images?

These illustrations are by David Goodsell, and carry his famous look. Whilst I was not able to determine what specific software he uses, this page suggests that he probably creates his illustrations as a composite of renders and that the outlines and depth cueing are achieved by running a 'find edges' filter over the z-buffer of the scene, which can be output from many renderers.

VMD is a fairly powerful program (that F'x has contributed patches to!) capable of achieving this cel-shaded look with minimal postprocessing by using a GLSL Fresnel shader.

As F'x mentioned, QuteMol can do this style as well, however QuteMol has not been actively developed since about 2007 and is not terribly feature-rich, which is a shame as it's one of very few molecule viewers with competent ambient occlusion.

How to do it with VMD and Blender

In the first instance, I tried using the GLSL shader Goodsell style in VMD on the PDB file 1RON, a neuropeptide. I was not satisfied with the results, but this would probably look decent on a protein SES as illustrated on the aforementioned VMD site.

Note that I have used 2 VMD representations to separate out the style of the backbone and the sidechains, minus the hydrogens. I'm not going for a 1-to-1 clone of the referred style, just something thematically similar.

I have shrunk this VMD output down because as F'x pointed out in chat, GLSL shaded spheres are not antialiased. This said, if you are very tricky, you might be able to force antialiasing using a shader method like FXAA, which can be enabled in arbitrary programs in the very latest nVidia beta drivers (I know...). Alternatively, you can do what I did and manually oversample :D.

Next step - Blender

I output the VdW spheres as two separate VRML files, one for the backbone, one for the sidechains. These can be imported into Blender. I'm going to assume proficiency with Blender, however if you are not familiar, Blender is fairly easy to learn, especially versions 2.5 and later.

I've combined the VdW sphere meshes of each component into a backbone and sidechain object, assigned materials for each (just changing the colours of the default diffuse shader) and have applied one level of Catmull-Clarke subdivision to the meshes, to make the spheres more spherical than the default.

I've also removed the light source and turned on environmental lighting, which lights the scene isotropically and will generate some nice ambient occlusion while we're at it.

I now render this at high resolution, obtaining a standard render and the Z-buffer for the scene. I've scaled them down here. Note that if you want the spheres to be a flat colour, you should set the material to 'shadeless'. I like the subtly shaded look better.

Oh yeah, that looks nice. And here's the Z-buffer. Regrettably, the Z-buffer in blender has no antialiasing which is why I've rendered things huge and scaled down in Photoshop.

I now run the Glowing Edges filter on the Z-buffer and threshold to remove faint spurious edges inside the spheres.

Invert and level, set blendmode to multiply and you're done.

Not perfect, but not bad.

Bonus

Here's the shadeless version, which is closer to Goodsell's ideal.

The Journey Continues!

Poking around with the Toon shaders in Blender, I have found a super easy way to make Goodsell-esque figures!

If you configure your material like so:

You have a shader that will rapidly shade to black at angles far from normal with a lightsource. To make this work, you should disable environment lighting and position a bright point light behind the camera.

You will note that I have adjusted the Toon size to 0.98, which will create a shadow just shy of perpendicular to the light source normal. You will also note that I have set the specularity to zero, which gives an approximately flat finish to the sphere.

Render it up and you will get something like this:


Spontaneity and nature of attack of fluorine gas on aluminum

First and foremost! Do not do this reaction unless you are properly trained and have the appropriate safety equipment. Fluorine is one of the most dangerous substances out there, and one of the presumed products, $\ce{OF2}$, is also terrifying.
If you can calculate the free energy change for a reaction, you will know whether it is spontaneous. To do this, you need to have more information than you currently provide. We need to know the products of the reaction in order to write a balanced chemical equation. Then we can look up appropriate thermodynamic values for each substance in the reaction in order to calculate the free energy change. Below is what I presume to be the balanced equation for the reaction of fluorine with aluminum oxide:
$$\ce{Al2O3 + 6F2 -> 2AlF3 + 3OF2}$$
Using the NIST Chemistry Webbook, we can look up the enthalpies of formation and standard entropies of each substance so that we may calculate the enthalpy and entropy changes.
$$\begin{array}{l|c|c|c|c|c|c}
& \ce{Al2O3} & \ce{F2} & \ce{AlF3} & \ce{OF2}  \\ \hline
\Delta_f H^\circ\ \mathrm{(kJ/mol)} & -1675.7 & 0 & -1510.4 & 24.52 \\
S^\circ\ \mathrm{(J/K)} & 50.92 & 202.8 & 66.5 & 247.46 \\
\end{array}$$
We can use these data to calculate the free energy change of the reaction:
$$\Delta_r G^\circ = \Delta_r H^\circ - T\Delta_rS^\circ$$
If the free energy change is negative at that temperature, the reaction is spontaneous.


Zinc in the +1 oxidation state is $\text{[Ar]}3d^{10}4s^1$, and even in its highest, most common known oxidation state +2 (which the quoted values above seem to have forgotten) it's still $\text{[Ar]}3d^{10}$. No known zinc species in what we normally consider the realm of chemistry breaks that complete $3d^{10}$ subshell, and we would need a major revamp of our calculations and models if any ever does.  Moreover, the thermophysical properties of zinc also betray a loss of transition-metal character.  Zinc is just not a transition element.


What is the physical origin of conjugation?

I think you have posed an interesting question. Upon looking through some literature, I have chanced upon the answer to your question in Fleming (2010), p. 30-31. The diagram below was also taken from the discussion in the book.

What you have proposed are two ways of constructing the $\pi$ molecular orbitals of butadiene from the $\pi$ molecular orbitals of two ethene molecules. What is normally discussed is the one shown on the left, whereby we consider the interaction between the two filled MOs of exactly the same energy and a similar interaction between the two unfilled MOs of equivalent energy. Considering the bonding and antibonding combinations allows us to derive the four $\pi$ Mos of butadiene.

You also propose considering $\pi$ to $\pi^*$ conjugation as an alternative way of deriving the butadiene $\pi$ MOs. Diagramfmatically, what you are essentially referring to is shown in the MO diagram on the right. Firstly, this interaction cannot be the main interaction between the two ethene $\pi$ systems simply because it would result in the butadiene $\pi$ MOs comprising of two separate levels of two degenerate MOs each, which is quite different from the picture which we are aware of, which has 4 $\pi$ MOs of different energies. Secondly, note that this interaction is considerably weaker due to the large energy difference between the $\pi$ and $\pi^*$ MOs of an ethene molecule. Recall that for a strong interaction between any two orbitals, one of the important conditions is that the energy difference between them must not be too large. In fact, the strongest interaction is obtained when the energy difference between them is zero. This interaction would contribute to the energies of the $\pi$ MOs in butadiene but the contribution would surely be less than the one that is highlighted in the previous paragraph.

To construct the MOs of the butadiene, it is incomplete to just consider the stronger interactions shown on the left. For completeness, we would have to then mix the resultant MOs obtained from these two interactions (i.e. both left and right diagrams), factoring in the extent of contribution of each interaction. Naturally, we would be curious as to what is the effect of the weaker interaction on the MO energies derived from the stronger interaction. Fleming (2010) describes the effects rather detailedly in the text:

  Mixing these two sets together, and allowing for the greater contribution from the stronger interactions, we get the set of orbitals shown in the figure below. Thus, to take just the filled orbitals, we see that $\psi_1$ is derived by the interaction of $\pi$ with $\pi$ in a bonding sense ($\psi_a$), lowering the energy of $\psi_1$ below that of the $\pi$ orbital, and by the interaction of $\pi$ with $\pi^*$ in a bonding sense ($\psi_w$), also lowering the energy below that of the $\pi$ orbital. Since the former is a strong interaction and the latter weak, the net effect is to lower the energy of $\psi_1$ below the $\pi$ level, but by a little more than the amount ($\beta$ in simple Huckel theory) that a $\pi$ orbital is lowered below the p level in making the $\pi$ bond of ethylene. However, $\psi_2$ is derived from the interaction of $\pi$ with $\pi$ in an antibonding sense ($\psi_b$), raising the energy above that of the $\pi$ orbital, and by the interaction of $\pi^*$ with $\pi$ in a bonding sense ($\psi_x$), lowering it again. Since the former is a strong interaction and the latter weak, the net effect is to raise the energy of $\psi_2$ above the $\pi$ level, but not by as much as a $\pi^*$ orbital is raised above the p level in making the p bond of ethylene. Yet another way of looking at this system is to say that the orbitals $\psi_1$ and $\psi_2$ and the orbitals $\psi_3^*$ and $\psi_4^*$ mutually repel each other.

Does it really matter if we factor in the weaker interaction into our considerations? Yes, it does. If we were to not do that, we would observe that the stabilisation predicted would not be as great as the one which we would predict from just looking at the MO diagram shown on the left. In fact, as Martin writes in the comments, the interaction between orbitals would produce an antibonding MO that is more antibonding than the bonding MO produced is bonding. If we were to not consider the weaker interaction, then we would predict that $\psi_2$ is raised more in energy from the original $\pi$ level than $\psi_1$ is lowered in energy from the original $\pi$ level (i.e. $\ce {E_2}$ is larger than $\ce {E_1}$), and this is clearly not the case.

Of course, as you mentioned, we can also build the $\pi$ MOs of butadiene from first principles via the Huckel method. I am not very well-versed with the mathematics behind the method but I do believe that the method discussed above and the original Huckel method would converge in terms of the results obtained eventually. Hence, they are equally valid approaches to the problem.

Reference

Fleming, I. Molecular Orbitals and Organic Chemical Reactions (Reference ed.). United Kingdom : Wiley, 2010.


Using Hess's law to find molar enthalpy of formation without an initial reaction

Since you are asked to find the enthalpy of formation for ammonia, it's convenient to write the equation of ammonia synthesis normalized for $\pu{1 mol}$ of $\ce{NH3}$:

$$\ce{0.5 N2 + 1.5 H2 -> NH3}$$

To obtain this equation and apply Hess's law, the following linear combination of the provided equations should be used:

$$
\begin{align}
\ce{4 NH3 + 3 O2 &->6 H2O + 2N2} & ΔH_1 &= \pu{-1516 kJ} &&|\cdot(-0.25) \tag{1} \\
\ce{2 H2 + O2 &-> 2 H2O} & ΔH_2 &= \pu{-572 kJ} &&|\cdot 0.75 \tag{2} \\
\hline
\ce{0.5 N2 + 1.5 H2 &-> NH3} & ΔH_3 &= -0.25ΔH_1 + 0.75ΔH_2 \tag{3}
\end{align}
$$

So the final answer is

$$ΔH_3 = -0.25\cdot(\pu{-1516 kJ}) + 0.75\cdot(\pu{-572 kJ}) = \pu{-50 kJ}$$

The catch here is that you are asked to find molar enthalpy of formation, and calculating for the "convenient" equation with integer coefficients

$$\ce{N2 + 3 H2 -> 2 NH3}$$

you eventually would obtain the doubled value of $\pu{-100 kJ mol-1}$ referred to $\pu{2 mol}$ of ammonia (I almost fell for it, too).


How can isotope labelling be used to determine the mechanism of ether cleavage with hydrogen iodide?

If your products are simply an alcohol plus an alkyl iodide, then you really don't need the isotopic labelling. If you have an $\mathrm{S_N1}$ mechanism operating, then iodide ion attacks the tert-butyl cation, and you get the products

$$\ce{Me-O-^tBu + HI ->[S_N1] MeOH + ^tBuI}$$

If you have an $\mathrm{S_N2}$ mechanism, then iodide ion attacks the methyl group, displacing t-butanol:

$$\ce{Me-O-^tBu + HI ->[S_N2] MeI + ^tBuOH}$$

In both cases it should be trivial to take a proton NMR of the crude reaction mixture and determine which set of products are present.

The isotope label is more pertinent if you are forming two alcohols as the products. In that case, if the $\ce{^18O}$ remains attached to the methyl group (for example), it means that the $\ce{^tBu-O}$ bond was cleaved: this indicates an $\mathrm{S_N1}$ mechanism. In other words:

$$\ce{Me-^{18}O-^tBu ->[S_N1] Me^{18}OH + HO-^tBu}$$

Likewise

$$\ce{Me-^{18}O-^tBu ->[S_N2] MeOH + H^{18}O-^tBu}$$

However, you would have to be careful when running this experiment: if you leave it too long, especially under acidic conditions, the labelled oxygen will simply be exchanged with unlabelled oxygen from water by a degenerate nucleophilic substitution reaction, destroying any precious evidence that you had obtained:

$$\ce{R^{18}OH + H3^{16}O+ <=>> R^{16}OH + H3^{18}O+}; \quad \ce{R} = \ce{Me}\text{ or }\ce{^tBu}$$

where the equilibrium position lies to the right because of the larger concentration of $\ce{H3^{16}O+}$. Equilibrium tells us nothing about kinetics, though, so as stated earlier, you will have to stop the reaction (or monitor it using e.g. mass spectrometry, since NMR won't tell you the difference between oxygen isotopes) before this happens.


Why is oxygen oxidized with lead instead of being reduced?

Your confusion stems from that fact that you are trying to look at what reactions are favorable instead of the reaction as written.

Oxygen is readily reduced because it is a strong oxidant. You can check any reduction half reaction with oxygen on the left side. Chances are that the reduction potential will be fairly positive.

In the reaction you have written, oxide is being oxidized to oxygen. Nowhere (including your textbook) do we claim that this reaction is favorable or spontaneous in the forward direction. In fact, if you consult a table of free energies of formation (example), you would find an entry like:

$$\ce{PbO (s, red)}, \Delta G^{\varnothing}_{f}=-189.3\ \mathrm{kJ}\ \mathrm{mol}^{-1}$$
or
$$\ce{PbO (s, yellow)}, \Delta G^{\varnothing}_{f}=-188.5\ \mathrm{kJ}\ \mathrm{mol}^{-1}$$

These values indicate that indeed, the reverse reaction is the one that's favorable, the one where oxygen serves as an oxidizing agent and is reduced.

However, that is not the reaction as written. The notation of chemistry allows us to describe reactions that are hypothetical or unfavorable.


The molecular geometry argument is pretty common-sensical, and is true to a certain extent. But I don't think it's the cleanest argument. If you look at a typical ketone and its enol, you will notice that the enol has a more constrained geometry than the ketone:

In the keto tautomer, the highlighted carbon is free to go wherever it likes, since it's part of an sp3 chain. But in the enol tautomer, by virtue of the 120° bond angles in the alkene, the highlighted carbon must be in the same plane as the double bond.

Furthermore, it's not possible to claim that the ketone and the enol have the same geometries in the carbon chain. Many of the bond lengths in the two tautomers will be different: after all one has a C–C (and C=O) bond and the other has C=C (and C–O). Unlike resonance structures, both of these tautomers are things which exist independently, and they have different bond lengths.

The safest way is to go back to the IUPAC definition. That's a bit boring, but when you read these, you can see the effort that has gone into coming up with a definition that is (as far as possible) consistent with edge cases such as these. Tautomerism is defined as (https://doi.org/10.1351/goldbook.T06252):

  Isomerism of the general form $\ce{G-X-Y=Z <=> X=Y-Z-G}$ where the isomers (called tautomers) are readily interconvertible.

Acrolein and propadienol do match the bonding pattern required for tautomerism, but don't pass the second criterion of being "readily interconvertible".

The reason is fairly simple. The process of tautomerisation involves protonating the oxygen, and deprotonating an α C–H (which way round this happens doesn't matter). To deprotonate a proton that is α to C=O, the C–H bond must be oriented parallel to the C=O π system, because the new C=C π bond is going to be made from that C=O π bond and the C-H σ bond. This cannot be the case in acrolein, because of the extended conjugation between the C=C and C=O bonds. Indeed, the α C–H bond in acrolein is perpendicular to the C=O π system.


Phosphers have never been made with phosphorous just like lead pencils have never been made with lead. The words have a common origin but there is very little chemical relationship.

Phosphorus is a Latin name for the planet Venus and, figuratively, for any glowing object. Early chemists in the 17th century used the word to refer to any substance which glowed spontaneously or when heated. They characterised and named several different types of phosphoruses. When the element phosphorus was isolated from urine in the late 17th century it was called “English Phosphorus” or “Phosphorus of Urine” and later simply “Phosphorus”.

The older meaning of “phosphorus” and “phosphor” continued into the 19th century and, in the early 20th century, “phosphor” began to be used with more scientific rigor to refer to a material exhibiting phosphorescence.

Phosphorus (the element) is not phosphorescent; it emits light only insomuch as it spontaneously ignites in air.

Graphite (a.k.a. plumbago) was used as a substitute for black lead hence it was the "lead" in the pencil.


How to determine number of alpha hydrogens in a cyclic compound?

The relative stability of your two compounds is not determined by arguments based on hyperconjugation.  Rather the relative stabilities have to do with ring strain.  Placing a double bond in smaller and smaller rings increases the angle strain in the ring system.  A double bond, being roughly $\ce{sp^2}$ hybridized, would prefer to have bond angles around 120°.  This becomes difficult in 3- and 4-membered rings where the internuclear angle is 60° and 90° respectively.  Therefore molecules containing double bonds in 3- and 4-membered rings will be more strained than molecules with the double bond in a larger ring.

Edit: Response to OP's comment

No, the given solution is not the correct answer to the problem.  It is correct that structure B has more alpha (alpha to the double bond) hydrogens (4) than structure A (only 2 alpha hydrogens).  It is also true that hyperconjugation can be used to explain why alkyl groups attached to a double bond act to stabilize a double bond (see this earlier answer for an explanation), and that the double bond in B is more stabilized by hyperconjugation than the double bond in A.  However, hyperconjugation has a much smaller effect on the stability of these molecules than the ring strain effect.


Does caffeine isolated from coffee contain harmane?

Given that caffeine can be readily extracted from a variety of abundant and exceedingly cheap natural sources, I doubt it would be economical to synthesize. Various decaffeinated products are popular, and the caffeine removed during the decaffeination process is normally recovered, purified, and re-sold for use in various products, which makes the synthetic route even less economically attractive. As far as I'm aware, the vast majority of caffeine sold commercially is extracted from plant sources. You can do your own research to determine the exact statistics, but since your concern relates to caffeine extracted from natural sources that might also contain harmala alkaloids, I'll address that case specifically.

There are a number of different methods of caffeine extraction (see the Wikipedia page on decaffeination), but all use solvents and/or adsorbents at some stage in the separation and purification processes. From the perspective of chemistry, then, the question is how similar are the solubilities of caffeine and harmala alkaloids, and what percentage by mass do they constitute of the natural caffeine sources used? I was able to find one article that quantified the chemical constituents of coffee beans. If you examine the data table therein, you'll see that the main constituents of those beans are carbohydrates, lipids, amino acids, and chlorogenic acids, not one of which is a category that applies to the harmala alkaloids. Lipids, carbohydrates, and amino acids are generally the main constituents of virtually all plants, so I think it's probably safe to extend those statistics to other natural sources besides coffee beans. Purely as a matter of probability, it's reasonable to assume the impurities are likely to consist mostly of substances from those categories. With respect to solubility, the harmala alkaloids are mostly somewhat heavier nitrogenous bases which are less polar than the caffeine molecule (at least in their neutral, "free base" form). Given that, I'd expect that methods using water extraction, which appear to be popular, would separate caffeine from harmala alkaloids fairly efficiently (I found one paper indicating that harmala alkaloids are water insoluble, while caffeine is anywhere from moderately to highly soluble depending on temperature).

An impurity of 1% is already quite small, and that 1% comprises a variety of compounds, both organic and inorganic (the overwhelming bulk of which are likely not harmala alkaloids, given the natural composition of caffeine-containing plant sources I described above). Based on the above information, I'd suspect that only a tiny fraction of those impurities, if any, would be likely to consist of harmala alkaloids at a detectable level.


How to calculate the equilibrium constant for an esterification?

It's given that this whole process is taking place at constant volume, so you can concentrate instead on what the concentrations of the species are (no pun intended) at equilibrium. The information is there, you just need to work through it.

You have been given initial conditions:
$[{\rm CH}_{3}{\rm COOH}] = [{\rm C}_{2}{\rm H}_{5}{\rm OH}] = 1.2\;{\rm mol}; [{\rm CH}_{3}{\rm COOC}_{2}{\rm H}_{5}] = [{\rm H}_{2}{\rm O}] = 0\;{\rm mol}$

and final conditions, where you are told that $[{\rm CH}_{3}{\rm COOC}_{2}{\rm H}_{5}] = 0.8\;{\rm mol}$. This means that at equilibrium $[{\rm H}_{2}{\rm O}] = 0.8\;{\rm mol}$.

Your reactant equilibrium concentrations will be $1.2\;{\rm mol} - x$, where $x$ is given to you (I'm leaving that part to you. Hint: It's in the  sentence right above this one). Plug in your equilibrium values and you'll get ${\rm K}_{\rm eq}$ (also up to you).


How to esterify phenol with oxalic acid?

I don't know for sure, but it could work provided some conditions be fulfilled

use of oxalyl chloride
use low temperature, in order to try to suppress fragmentation of the intermediate (semi-ester) into the simple monoacyl chloride (by CO loss): in such cases $\ce{(COCl)2}$ behaves much like phosgene ($\ce{COCl2}$.
use the phenol as a dissociated salt (sodium, potassium) in polar solvents such as sulfolane or acetonitrile, this is optional and possibly risky, due to enhanced competitive "C-acylation" if ortho/para positions are free.
use Piridine-based solvents as reactivity enhancers (they convert acyl chloride to acyl immonium salts). Optionally, even stronger catalysts such as DMAP (p-dimethylamino pyridine) could work

Anyway, to sum up, the major risks stem from:

"C-acylation" (which is generally kinetically slower but irreversible and thermodynamically favored over "O-acylation"), if some suitable position is avail.
formation of diaryl carbonates (on extrusion of CO).
I guess this last drawback coul be at least reduced by working at low temperature.


On the plus side, you've clearly thought about this problem and have worked out a detailed synthetic route.  I think most of the steps would work as desired.  The two steps that concern me the most are 4 and 7.  I suspect your step 4 will give a mixture of products.  Perhaps with further experimentation you could find reaction conditions that would give you the desired product without  other products from undesired side reaction (elimination in the other direction and perhaps some rearrangement products too).  I see your step 7 as the most problematical.  Aluminum chloride, heat and the alcohol will produce a carbocation and it will undergo significant rearrangement creating a variety of elimination and addition (polymer formation) products.  So while your reaction scheme is plausible, in my opinion it is not efficient.


Coefficients and Parameters for contracted Gaussian basis sets

Take a look at the carbon STO-3G first. It means for each AO, you form it with three Gaussians. For carbon, you have a 1S, 2S and (three) 2P. 5 functions when finished.

The top line says you are defining the basis for carbon; no big deal.

Next line says you are creating an S shell, with three primitives, with a 1x scale factor.

Then you list your three rows of primitives.

The first column is the Gaussian exponent. The next column is the S expansion coefficient. These are the three Gaussians that make up your 1S "Slater" atomic orbital.

In the next section, you define SP. What this is, and I quote from the Gaussian website now: "Note that 2SP requests the best least-squares fit simultaneously to S and P slater orbitals and is not equivalent to separately specifying the best S and the best P expansions."

S and P (and all higher angular momenta, actually) can be related by a recursion relationship. It's an efficient way of computing integrals. I believe the Obara-Saika recursion takes advantage of this, though in Gaussian the integrals are formed using the PRISM algorithm.

The first column is the exponent for your Gaussian, the second column is the S expansion coefficient, and the third column is the P expansion coefficient. They just share the same exponent. Thus you have 4 + 1 (from before) = 5 AOs total formed with three primitives each.

To jump ahead to your last example, the pattern simply continues. The first section gives six primitives to form one 1S (the 6 in 6-311++G**). The second section gives three Gaussians for you valence S and P, the third section is one Gaussian for valence SP. This is the 31 in 6-31++G**. You have a split valence description, one with three primitives, one with a single primitive.

The next section is the diffuse function, hence the really small exponent in the first column. This is the ++ part of the 6-31++G**. The small exponent gives a really wide spreading Gaussian. Helpful for describing electronic activity far from the nucleus. (It's helpful to compare with the huge exponents in the first section, which gives really tight functions that describe the core electrons).

Finally, you have the D functions which are polarization functions. Carbon doesn't have a D valence shell, but we include them to describe polarization of electronic density in molecules.

Note that depending on your program, you either have 5D AOs or 6D AOs. This is the difference between spherical and cartesian representations, and is more of a computational trick than anything else. Gaussian uses 5D by default, except in some cases.


Does an onion completely remove "fresh" paint VOCs and other chemicals emitted?

I've never heard of onions, lemons or vinegar removing a paint smell, but I would guess that they'd just cover up the unpleasant odor with something nicer.

As to your later comment:

  I wouldn't want to remove the smell but leave the VOC or other bad elements there.

VOC stands for "volatile organic compound." (See the Wikipedia page on VOCs) Volatile compounds are ones that evaporate easily. Technically, the smell of that onion (or lemon, or vinegar) is a VOC, just not a particularly harmful one. (Not knowing the particular chemicals that make up a onion's scent, I can't say for sure that it's a volatile organic compound, but it's a pretty good bet.)

That fresh paint smell comes from VOCs (and potentially from other volatile compounds), so by the time the smell is gone, any VOCs you couldn't smell are likely gone too.


When you divide the mass of the solid by its density then you get the volume of the object which means not only the volume of the molecules also the volume of voids. Now the volume you get is = volume of the particles + volume of voids. So the radius you get is = radius of void + radius of molecule.  So in order to solve the problem you must which kind of packing is this. And then you should know it's packing efficiency. So now the volume you got should be multiplied by the packing efficiency to get the actual volume of molecules. If you say why do they have voids. Then I'll  say you can't fill a particular shaped object with small sphere's so packing efficiency can't be 100%. So all you need to know is the kind of lattice formed by the molecules of amelopectin. And now what you got in your question is the volume of single particle + the volume of void. Well if you wanted to get the radius from it then it would be error as I said before. But in that case also you had to put $V=\frac{4}{3}\pi(r^3)$= the volume of the single molecule. And in this way you get the radius as 6.02*10^-8cm. And then get the radius. But actually taking the molecules as cube gives you a better answer though this would not give the proper answer if amilopectin molecules aren't cubic. But certainly you can fill a shape with cubes more efficiently (without having much voids) so in that case $a^3=volume you mentioned which is 9.17*10^-22 cubic cm. That gives a as 9.7*10^-8cm as the side length.


Does hyperconjugation only arise from C–H bonds?

No.
Hyperconjugation is the donation of electrons from a one orbital to another, where at least one of those orbitals is a bonding orbital, and there is a net overlap between the orbitals.
A typical first example taught of hyperconjugation is that which occurs in the conformations of simple alkanes (e.g. ethane). When ethane is in the staggered conformation, electrons from the $\sigma_\ce{CH}$ bonding orbital are donated to the properly aligned $\sigma_\ce{CH}^*$ antibonding orbital of the adjacent carbon atom. The interaction of these two orbitals, the $\sigma_\ce{CH} \to \sigma_\ce{CH}^*$ lowers the energy of those electrons, and thus lowers the energy of the molecule.
Next up is to look at a longer alkane chain (e.g. butane), which will have $\sigma_\ce{CH} \to \sigma_\ce{CH}^*$ interactions like in ethane, but will also have additional interactions involving the $\sigma_\ce{CC}$ and $\sigma_\ce{CC}^*$ bonding & antibonding orbitals. Depending on which conformation of butane you are considering, it is possible to have $\sigma_\ce{CC} \to \sigma_\ce{CH}^*$, $\sigma_\ce{CC} \to \sigma_\ce{CC}^*$, and $\sigma_\ce{CH} \to \sigma_\ce{CC}^*$ interactions.
Continuing in a more general sense, these interactions are not limited to $\sigma$ bonding orbitals. In carbocations, which in principle have an empty p orbital ($\ce{C}_\mathrm{2p}$) on the electron-deficient carbon, there can be hyperconjugation in the form of $\sigma \to \ce{C}_\mathrm{2p}$. (Note that I didn't specify the atoms involved in the $\sigma$ bond; as long as the atoms form a stable $\sigma$ bond, this hyperconjugation interaction can occur, though how stabilizing it is will depend on the size of the orbitals involved.)
There are a variety of hyperconjugation interactions that can occur:

$\sigma \to \sigma^*$
$\sigma \to \pi^*$
$\sigma \to$ empty hybridized orbital
hybridized orbital $\to \sigma^*$
$\pi \to \sigma^*$

For example, in protonated formamide ($\ce{H-C(=O)-NH3+}$), there is hyperconjugation in the form of $\pi_\ce{CO} \to \sigma_\ce{NH}^*$ and $\sigma_\ce{NH} \to \pi_\ce{CO}^*$, as well as $\ce{O}_\mathrm{2p} \to \sigma_\ce{CN}^*$.


Where can I find ranges of isotope abundances?

I am aware of this range diagram about $\ce{^{81}Br}$ that appeared in the recurrent (open access) IUPAC recommendations of atomic weights in Pure and Applied Chemistry:

(source: doi 10.1351/PAC-REP-13-03-02)

which cites data of a publication in Environmental Chemistry extending a little to $\ce{^{37}Cl}$ (doi 10.1071/EN10090, open access), too.

Even if the elder (2000) IUPAC compilation in the same journal includes a graphical representation of the range in occurrence of $\ce{^2H}$ as known at this time:

(source: doi 10.1351/pac200375060683, open access)

it may be worth to read more in detail, as -- in contrast to the 2011 edition (32 pages) -- the 2000 edition (120 pages) includes a dedicated section "Isotope fractionation from natural processes" and "Element-by-element review of the standard atomic weights" for a detailed view.

Otherwise, if you were "just" interested in an overall overview of isotopes, their masses and their percentage of occurrence, NIST has a searchable dedicated database here.


Why are temperature and pressure represented as the number 2 in Gibbs' Phase Rule?

The underlying ah-ha moment is that the Gibbs phase rule works best in (constructing) a phase diagram. And those are usually 2-dimensional - with pressure along one axis and temperature the other. So the "2" simply reflects that 2 dimensionality. At 2 degrees of freedom, there are no constraints and the system can change either pressure or temperature and be stable without a phase transition. If you have 1 degree of freedom, then you either have a constraint on $P = k$ or $T = k$ or a "mixture" $f(P,T) = k$ (If P is changed, T must be a specific value) . At 0 degrees of freedom you are at a triple point $(P =k, T=k)$ or similar (there are a handful of named "points") - any movement in either direction means the universe is going to change the phases present or the components present.

So they are not really constants, but if the degrees of freedom are 0, then they are... "fixed" - or better yet "constrained".

The archetypical example is Ice-Water-Steam. There will be at least 3 lines in this diagram from 2 phases in equilibrium and the 1 component - and 1 point.

Out of the 3 components, lets take 2 and see what we get.

$$DoF = 1-2+2 = 1$$.

So we see that for 2 phases to be in equilibrium we have 1 constraint eating away at our DoF and we know then that there will be a fixed relationship between T and P here - we can change P and still have the equilibrium but then we have to change T as well with a specific amount. This creates a line in our T-P diagram. Since there are 3 ways to arrange this we get 3 lines (I+W) + (S+W) + (I+S)

So lets see what happens if we force all three phases to be at equilibrium:

$$DoF = 1-3+2 = 0$$.

So that is no wiggle room. There will be a point where this happens, a specific combination of P and T - where this equilibrium exists.

Now lets say I invent a new phase, and I call it Ice-9. Lets see if we can find a place where this coexists with all the others.

$$DoF = 1-4+2 = -1$$.

Nope. This is not stable. No equilibrium can exist which permits my Ice-9, regular ice, water and steam at the same time. So now we know that Kurt Vonnegut is just making stuff up, what a hack. The result can be seen here... 3 lines and a point. Don't mind the dotted stuff, that is because water is more complicated than what it is at first glance


Sigma bonds, pi bonds and the puzzling symmetry of methane

I am going to attempt to answer this, based on the link supplied by @Kartik and expanding on @Mitheron’s comment above.
As we know, covalent bonding is the process of “filling up” partially filled outer orbitals by sharing electrons between atoms. Carbon’s structure [He]2s22p2, or, more accurately, [He]2s22px12py1. When bonding occurs, however, two things happen:
1.  One of the 2s electrons gets “promoted” to a p orbital. This is an “excited” state in which the electrons have slightly more energy (though not enough to remove any of them). The configuration therefore becomes [He]2s12px12py12pz1, and voila, we now have 4 unpaired electrons available to bond with other atoms.
2.  The extra energy allows the orbitals to “hybridize”. In effect, the 2s and 2p orbitals morph into 4 identical orbitals of slightly higher average energy. These orbitals are called sp3 orbitals (one part s, three parts p). The electronic structure of the ready-to-bond atom is now [He] (sp3)4.
These four sp3 orbitals of course repel each other equally, and hence end up pointing towards the vertices of an imaginary tetrahedron. Hence the regular structure of CH4 and analogous compounds.
Returning to the question, the sigma bonds formed by carbon are neither from overlapping s or overlapping p orbitals, but from overlapping sp3 orbitals.
The thing that makes a bond a sigma bond is not the type of subshell from which the electrons “originated”. It is the fact that the orbital overlap directly, with the two atomic nuclei and the main concentration of the shared electron cloud in a straight line. Pi bonds, on the other hand, overlap “sideways”, with the main concentrations of the bonding electron cloud off to the sides of the bonded atoms.
Please chime in, people, if I have omitted or misunderstood anything.


What is the chemical equation for cooking rice?

Essentially when rice is cooked all you are doing is adding water to it and heating it. No true chemical reaction occurs.  A way you could think of this addition of water in an equation is:
$$\ce{(C6H12O6)2 ->[\Delta][water] (C6H12O6)2.xH2O}$$
Not the most accurate representation as water was not added to a crystal system but I hope this helps you understand that no real reaction is occuring.


There are a few wrong assumptions in your question that I would like to clear up.

Yes, d-orbital splitting is part of the reason for colour. However, $\ce{[FeF6]^3-}$ displays a beautiful d-orbital splitting which could display colour from the energy difference — but doesn’t. There is more to colour than d-orbital splitting; the most important terms are spin-forbidden and Laporte rules.
You somehow got the impression that ionic solids have nothing to do with coordination complexes. Please remove this idea from your brain; it is wrong. Ionic solids are equally coordination complexes as present in solution only that the environment and sometimes the composition of the complex is different. E.g, $\ce{CuSO4(aq)}$ is essentially a solvated sulphate ion plus $\ce{[Cu(H2O)6]^2+}$. When precipitated or crystallised as ‘$\ce{CuSO4 . 5H2O}$’, what you actually got was $\ce{[Cu(H2O)4(\mu{-}SO4)2].H2O}$ — clearly a coordination compound.

To answer your question, it depends on the nature of the compound you are precipitating — the colour can derive from cation, anion or their combination. For example, aluminium salts will be colourless unless the anion dictates otherwise. Likewise for zinc salts. For a very large number of salts where the cation is responsible for the colour, these will be the same as in solution. But sometimes ($\ce{PbI2, Fe4[Fe(CN)6]3}$ and others) you just have to know because only this specific combination will give this colour.


It depends a lot on the context.

For example if we are talking about synthetic chemistry the exact amount often isn't relevant, for example in case of solvents for the reaction. It doesn't really matter if you do it in 200 mL or 198 or 201 mL. Therefore no one cares to measure it exactly in the first place and no one really cares about those errors, especially since you would need to measure temperature as well.

In other cases, like many analytical methods, you need to know the exact amount which was used and in those cases you should definitely include errors.

In the end you need to include it if it's relevant information, it's common to omit it if it's not relevant.


Why do enthalpy changes of combustion and formation not take into account oxygen?

By convention elements in their most stable polymorph at standard conditions have at standard conditions (298.15 K) a standard enthalpies of formation $\Delta{}_fH^\circ$ of zero (one reference).  This is why thermodynamics discerns C(graphite) from C(diamond), and C(fullerene), for example.  Hence on first view it looks like zeros in your example (graphite, Hydrogen, Oxygen) got forgotten; but they are not.


Dihaloalkane reaction with sodamide/ammonia

You were correct up to this point, the strong base alcoholic KOH sets off elimination instead of substitution so you will not get an alcohol.
Accordingly your products A (presumably propene) and B are correct.
The amide/ammonia is an even stronger base than alcoholic KOH and will also produce elimination.  With a vicinal dihalide like your B, there is an added twist.  Read here to find out what happens.


According to one source, the major elemental constituents of cremation remains are calcium and phosphorus, presumably from the hydroxylapatite in bone (which consists primarily of calcium phosphate). The rest is mainly assorted trace metals, which exist in relatively minute concentrations in the human body and don't vaporize at the temperatures typical of a crematorium furnace.

Virtually all organic matter, comprising mainly oxygen, carbon, hydrogen, nitrogen, and sulfur, is oxidized into a variety of simple compounds and is removed as a mixture of gases in the furnace fumes (and perhaps some finely dispersed, light solid particulates).

Consequently, what remains is largely inorganic and non-calorific in nature. While the mineral and metal content of the ash can be dissolved and subsequently absorbed by various organisms in trace amounts, there's probably little or no organic material left that can be metabolically decomposed by bacteria, fungi, or plants as a direct energy source. Of course, there are chemical processes that can further break down the ash and isolate specific components via redox reactions (dissolution in strong acids to give salts, electrolysis, etc.), but obviously that's distinct from biodegradation.


The $\ce{C=N}$ bond has a dissociation energy of $\pu{615kJ/mol}$, while $\ce{N=O}$ has an energy of $\pu{607kJ/mol}$. The difference isn't much, so its difficult trying to comparing the energies without being given the exact values.

References:

Yoder, Claude. “Wired Chemist.” Common Bond Energies (D, www.wiredchemist.com/chemistry/data/bond_energies_lengths.html.


Dissolving collagen in water: Why does the weighing scale reading fall?

Because collagen does not react with water and you say, no (measurable) evaporation of water is evident, there are only two options as to what is happening:

Matter is converted into energy:
If virtually no evaporation or spillage took place, and your scale is working properly, this is what must have happened. Using the formula of mass-energy equivalence $E=mc^2$ (E = energy, m = mass, c = speed of light in a vacuum) we can calculate the amount of energy that should result from a "loss" of 2 grams:
$$ E = 0.002~kg\cdot(2.998\cdot10^8~m/s)^2 $$
$$ E = 0.002~kg\cdot8.988\cdot10^{16}~m^2/s^2 $$
$$ E = 1.798\cdot10^{14} (kg~m^2)/s^2 = 1.798\cdot10^{14}~J$$
so... that would be a LOT of energy.
Your scale is not working as intended.

I'd go with option 2.

Now on a serious note: We do have a scale (precision of 0.0001 g) in our lab that occasionally has the same problems. Sometimes I would measure an empty reaction flask and it would be fine but some other time I could watch the measured weight drop by the second, "losing" up to 100 mg. The scale was checked by a professional but he could not find anything wrong with it.

My advice is: try another scale, see if the problem persists.


Comparison between van der Waals radius and metallic radius

Metallic radius is half of the distance between the nuclei of two atoms when they are in crystalline structure and bond is formed between positively charged atoms, that means they are tightly packed in metallic lattice.

And van der Waals radius is half of the internuclear distance of two non-bonded atoms of the same element in the closest possible distance.

Don't compare both the situations using above diagrams.  The radii are calculated in two different situations.

As in case of metallic bond, the atoms are bonded tightly so their shells can overlap, but in case of van der wal  forces the bond is weak. Therefore, in case of van der waal force, the internuclear distance will be higher as compared to metallic bond.

Via:
https://clay6.com/qa/94625/covalent-radius-of-cl-is-more-than-its-van-der-waals-radius-true-or-false-

https://socratic.org/questions/why-is-van-der-waal-radius-greater-than-covalent-radius-for-covalent-compounds


Physical state of the carbon denoted with "gr"

The letters in brackets indicate state of aggregation (includes state of matter, crystal structure (solids) and atomicity (gases)), therefore $\ce{C_{(gr)}}$ most likely refers to graphite. Often one can also see $\ce{C(s, gr)}$ or $\ce{C(s, graphite)}$ There is a nice demonstration in [1, p. 248]:

  $$\ce{C(gr) + O2(g) <=> CO2(g)}$$
  Here $\ce{(g)}$ indicates that the compound is gas, [...], and $\ce{(gr)}$ that carbon is graphite (not many people will burn diamonds, even if they are chemists).

P. S. According to IUPAC recommendations, subscript notation is deprecated.

Metiu, H. Physical Chemistry: Thermodynamics; Garland Science, 2006. ISBN 978-1-315-31491-4.


CO2 + High humidity = transperent substance?

Does it stay like thay when you remove the pressure and take it to the normal air? if yes, then it is likely co2 is being absorbed into the plastic part, and that plastic part is swelling. You can avoid it by using a glass window for optical sensor. Or use H2O permeable but not CO2 permeable membrane if this sensor needs contact with the gas. Or use another type of sensor that does not contain this plastic. Or reduce CO2 pressure. I dont know any CO2+H2O compounds that would stay stable and solid at normal pressure. Methanol could be forming from those materials if energy is present, but it needs catalyst and is very unlikely to happen on its own.


Why does mercury decolourise a gold ring?

The grey colour is an amalgam of mercury and gold.

Mercury forms amalgams with many other metals. Some are used as chemical reagents in laboratory chemistry as they have different properties than the original metals involved. Gold amalgam is much greyer than gold. Silver amalgam has been used in dentistry.

Mercury has been used in the extraction of gold in mining as it can extract gold from low grade ore. But the recovery process involves distilling off the mercury which is dangerous and polluting.

Chemists used to work with mercury a lot and, when they did, they were recommended to remove their wedding rings or other gold jewellery as the mercury would damage the gold.

Once the amalgam is formed there is no easy way to reverse the process short of resmelting the ring. If the contact with mercury has been brief, you might be able to polish it to remove the amalgam layer, but this won't be easy.


Does changing the electrolyte concentration in the electrolysis of water affect the speed that hydrogen gas is produced at the cathode?

With decreasing electrolyte concentration, either the needed voltage raises at constant current, either the passing current decreases at the constant voltage.
Both happens because at lower concentration, a higher potential gradient across the electrolyte is needed to maintain the same current, as the smaller ion count requires  ions to move faster for the same current..
This potential gradient and the resulting potential drop acts against the external voltage. It pushes forced electrode potentials closer to their equilibrium potential. The net electrode reactions are then slower.


Does a rotary evaporator need a draft chamber and a vacuum pump?

I think you are taking too dogmatic a position overall.

1) Vacuum also increases the rate of evaporation, without boiling.  This is just practical.

2) Your claim that your system is "completely closed" is over confident. What if you are wrong and you have a leak?  Now you are free to dismiss this, but this is the safety issue your adviser is getting at.

In other words, it would be easy to incorporate your practice with industry best practice.


In what ways does the transfer of heat affect bond stability?

How does a certain added number of joules affect the orbital of a
  bonded electron

If you pump enough energy into an element\molecule you can promote an electron from one orbital to the next highest.  For example, the hydrogen atom has a single 1s electron, put enough energy in and you can promote that electron to the 2s orbital.  That means that, on average, the electron would spend more time further away from the nucleus.  In molecules, there are numerous "sub-levels" between each (molecular) orbital.  These "sub-levels" relate to rotational and vibrational energy.  So as you pump energy into the system the molecule will rotate and vibrate more violently.  If you put enough energy in, the bond will break and the atoms connected by that bond will fly apart.


Does the concentration of water accounts for determining the rate of a reaction?

Rate law vs. definition of rate

For this problem, you don't need the rate law. And, just for the record, you can't get the rate law from the stoichiometry - it depends on the mechanism and has to be determined empirically.

Definition of rate of reaction

The rate of reaction is based on the change of concentrations. For a homogeneous reaction (i.e. all species in the same gas phase, or all species part of the same solution), the definition is the following:

$$\mathrm{rate} = \frac{d[\ce{X}]}{\nu_X \cdot dt}$$

where $\nu_X$ is the stoichiometric coefficient of the species $\ce{X}$. This does not apply to heterogeneous reactions. For one, if a species is a pure solid or liquid, its concentration will not change. Also, the stoichiometry of a reaction refers to amount of substance - only if all species are in the same solution (or the same gas phase) does this translate into easy relationships of concentration changes (or partial pressure changes).

Definition of "rates of the chemical species"

This is not a technical term I am familiar with. In order to be able to answer the question with answer 3., you would have to define it as:

$$v_\ce{X} = \frac{\mid dn_\ce{X} \mid}{dt}$$

Once you define it such, it does not matter what the physical state of reactants and products is, it does not matter whether a species is a reactant or a product, the only thing you need to related the different rates are the stoichiometric coefficients.

So for example, the ratio of the amount of hydrogen and oxygen used is equal to the ratio of stoichiometric coefficients:

$$\frac{\Delta n_\ce{H2}}{\Delta n_\ce{O2}} = \frac{\nu_\ce{H2}}{\nu_\ce{O2}} = \frac{2}{1}$$


Electrolysis of Baking Soda separated by semi permeable membrane

The primary problem is too low permeability = too high resistence of the gloves, causing high voltage drop where it should be minimal one.

The gloves are not hydrophilic enough for electrolyte soaking to get low resistance.

The remaining voltage - after subtracting the glove voltage drop - is not high enough to cause electrolysis by significant current.

You may try to use a salt bridge, as a passthru vessel with porous glass ends, or a gel formed from an electrolyte solution.

You may also want to try a compartment created from a glued filtration paper, or other hydrophilic porous material, impregnated by an electrolyte, preferably sharing one of ions with both free solutions.

For the paper, be aware  of the same leveling on both compartments, otherwise. the paper will do its major purpose filtering. Eventually, more compact paper may be better, like the thick charts for paper painting.


Yes. Beer's law tells us that the absorption is the product of path length, extinction and concentration. Thus concentration is absorption divided by the product of path and extinction. The rate of change of concentration (first derivative with respect to time of concentration is the rate of change of absorption divided by the product of path and extinction. If the rate of change of absorption is negative the rate of change of concentration is negative. Concentration is decreasing.


Cells of fish contain proteins, and proteins are made of long chain of amino-acids, linked together by peptide bonds. In the presence of an acid (like phosphoric acid in your experiment or hydrochloric acid in our stomach), the peptide bonds are destroyed and the free aminoacids are released, so that they can be immediately assimilated by living organisms, like roots of plants if these acids are added to the ground as fertilizers.
But the phosphoric acid is not destroyed in the process. It still remains in the aminoacids mixture. As most acids, it is a corrosive substance that must be neutralized before being disposed in the ground. This neutralization is made with KOH, by one of the reactions $$\ce{H3PO4 + 2 KOH -> K2HPO4 + 2 H2O}$$ $$\ce{H3PO4 + 3 KOH -> K3PO4 + 3 H2O}$$ Whatever the reaction chosen, phosphoric acid is now transformed into a potassium phosphate, which can be assimilated. The advantage of this process is that now the fertilizer contains amino acids (bringing Nitrogen N to the soil), plus a salt bringing two important elements for fertilizing a soil : phosphorus P and potassium K. The three fundamental elements (N, P, K) for the agriculture are present in this fertilizer
To go back to the use of KOH, the equation shows that the amount of KOH to be added is twice or three times the amount of the initial phosphoric acid. That is why you have to add a huge amount of KOH to your mixture.


How many grams of acetic acid are present in 2.80 L of this vinegar?

Your math and stoichiometry are correct.
$2.80L*(\frac{1000 ml}{1 L})*(\frac{1.006 g}{ ml})*8.7$%$=245g \checkmark$


The scientist who coined the term polymer, Jöns Jacob Berzelius, used the word to refer to different substances which had the same empirical formula. In this sense, benzene is a polymer of ethyne (acetylene), because benzene $\ce{C6H6}$ and ethyne $\ce{C2H2}$ have the same empirical formula, namely $\ce{CH}$. In other words, all substances with molecular formula $\ce{(CH)_x}$ were polymers of the smallest unit in the family (in this case, $\ce{C2H2}$, because $\ce{CH}$ molecules are only known as extremely fleeting species on Earth, though they are abundant in space).
However, that was back in the 1830s, when we had no clue about atomic and molecular structure (in 1808, one of the eminent chemists of the time, John Dalton, defended $\ce{HO}$ as the formula for water). The meaning of the term has evolved over time to accommodate our increase in knowledge. The most current IUPAC recommendations on polymer nomenclature (Compendium of Polymer Terminology and Nomenclature - IUPAC Recommendations 2008) states the definition of polymer as:

A polymer is a substance composed of macromolecules.

And, fortunately, they also define a macromolecule as:

Molecule of high relative molecular mass, the structure of which essentially comprises the multiple  repetition  of  units  derived,  actually  or  conceptually,  from  molecules  of  low  relative molecular mass.

Note  1:  In  many  cases,  especially  for  synthetic  polymers,  a  molecule  can  be  regarded  as  having  a  high  relative  molecular  mass  if  the  addition  or  removal  of  one  or  a  few  of  the  units has a negligible effect on the molecular properties. This statement fails in the case of certain properties of macromolecules which may be critically dependent on fine details of the molecular structure, e.g., the enzymatic properties of polypeptides.

Note  2:  If  a  part  or  the  whole  of  the  molecule  has  a  high  relative  molecular  mass  and  essentially  comprises  the  multiple  repetition  of  units  derived,  actually  or  conceptually,  from   molecules   of   low   relative   molecular   mass,   it   may   be   described   as   either   macromolecular or polymeric, or by polymer used adjectivally.

Note  3:  In  most  cases,  the  polymer  can  actually  be  made  by  direct  polymerization of  its  parent  monomer  but  in  other  cases,  e.g.,  poly(vinyl  alcohol),  the  description  ‘conceptual’  denotes that an indirect route is used because the nominal monomer does not exist.

As you can see, there is quite a bit of detailing to try to keep things consistent. Observe that the definition speaks of high relative molecular mass, and this point is expanded upon in Note 1. With these considerations, benzene is not a polymer - it is certainly a very small molecule, and adding or removing a portion of it (presumably cyclooctatetraene $\ce{C8H8}$ and cyclobutadiene $\ce{C4H4}$, respectively) would result in a substance with drastically different properties.
In some sense, Berzelius wasn't really wrong to call benzene a polymer, because for all he knew back then, benzene could well have the formula $\ce{C20000H20000}$ (actually, it was not widely believed that "large" molecules could exist at all, with some resistance to the idea existing as late as the 1920s-1930s). He did the best with what knowledge was available to him. But now we can do better, and it just is more convenient and self-consistent to not label benzene as a polymer.
This isn't to say that, however, that benzene and ethyne are completely chemically unrelated. There is in fact a close relationship, in that ethyne can efficiently undergo a [2+2+2] cycloaddition reaction to form benzene. This is not exploited industrially, however, as there are much cheaper (and safer) ways of making it.


Can we do combustion caused by limited supply of air in a lab? How?

Hmm, I see a couple of problems in here:
Energy in the reaction is misunderstood:
In general, breaking bonds is endothermic, and making them is exothermic. In any combustion reaction we have bonds broken ($\ce{O2}$'s for example) and bonds created (forming $\ce{H2O}$, in this instance). The overall change in the energy defines a reaction to be endo- or exothermic; and in the case of combustion (any type) it's exothermic. If you want to write "heat" and "energy" in both sides of the reactions you face, it's going to look a little awkward. Note that in the biochemical reactions in which you see terms like "energy" and such in only the side of reactants or the products (e.g.: ATP energy releases) it's because they want to put emphasis on that particular intake or release of energy.
The equation you've provided for incomplete combustion is incomplete:
I don't see any balanced equations in there, do you? And, I can't get the reason for why you've written two $\ce{CO2}$s.
However, your second equation is about the complete combustion of ethyne, and is balanced and correct.
As for your answer, take a look at what wikipedia says:

The incomplete (partial) combustion of a hydrocarbon with oxygen produces a gas mixture containing mainly $\ce{CO2, CO, H2O}$, and $\ce{H2}$.

Depending on how much your burning sample will be out of reach of air, the mixture will vary in weight percent of the products of the reaction.
In a lab, you normally will get incomplete combustion reactions, for reasons:

Your sample is hardly pure hydrocarbon.
The air in the lab isn't in wild currents and overall, is "stale" in simple words.
The hydrocarbon doesn't get enough "requirements" for combustion and ignites incompletely.
etc.

Source

Source
A good sign (or alarm) of the incomplete combustion is the orange hue of fire, especially if it's accompanied by a visible "smoke". If the experiment requires complete combustion, we'll make use of burners such as Bunsen burner.
So, let's summarize: The incomplete combustion is the normal thing happening in the burning of ethyne. You usually do some modifications if you're to have a complete combustion reaction. When the examiner claims "with contact with adequate supplies of air", s\he only means that a complete combustion takes place, though this is rare in real life. Plus, ethyne is not a very good hydrocarbon to start with, because as CDC mentions, it has a potentially negative effect on the nervous system, causing headaches, asphyxia etc., as well as painful frostbites.


Naming alkenes with double bond branches

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the traditional name ‘vinyl’ for $\ce{CH2=CH-{}}$ is no longer used for preferred IUPAC names and is only retained for general nomenclature. The preferred name is the systematic name ‘ethenyl’.

Therefore, the name for the compound that is given in the question ‘3,7-dimethyl-5-vinylnona-1,8-diene’ could be found in the literature; however, the preferred IUPAC name in accordance with current recommendations is 5-ethenyl-3,7-dimethylnona-1,8-diene.


Is there an abbreviation for "something bulky"?

I've never come across one, but mostly those I come across will have "R" or "X", but somewhere in the reaction scheme specify what R is, like:

R = Me, Et, Pr, iPr...

So in your case, you could specify something like:

R = iPr or bulkier


Problems with precipitation of transition metal solution

Here's an idea.  Add the cations as their acetates or nitrates.  Control pH with acetate buffer. Fe(III) is a bad player.  "Ferric acetate," $\ce{[Fe3O(OAc)6(H2O)3]OAc}$, is water-soluble. You might sparge the solution with argon before sterilizing (sealed vessel) to prevent oxidation, and/or add ascorbate (or erythorbate, often biologically inactive) as anti-oxidant - if that is the problem.

Lactate chelates, and may give greater solubilities.  Fe(II) lactate is E585 via the European Food Safety Authority.

(Note that my singular contribution to biochemistry was telling a lab its phosphatase assay bombed because...they used phosphate buffer.  I can't say they were grateful for acquiring the knowledge.)


Iodine is the biggest halogen and by extension the most polarizable.

Diatomic iodine can be oxidized with nitric acid to facilitate electrophilic aromatic substitution.

Iodide ion, as found in sodium iodide, can also be oxidized by concentrated household bleach to facilitate EAS as well.

In the examples you give, there are strong electron withdrawing groups (such as oxygen, which is found in each of your examples). It is plausible that the oxygen has stripped the iodine of its electron(s) in each of your examples.


How do I calculate the ensemble-average net charge of an amino acid at given pH?

The Henderson-Hasselbalch relationship describing each ionizable group is:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{\ce{[A-]}}{\ce{[AH]}}$$

We can solve for the ratio:

$$10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})}=  \frac{\ce{[A-]}}{\ce{[AH]}}$$

However, we really want the fraction of protonated among the total (not the ratio of deprotonated to protonated).

$$10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})} =  \frac{[\mathrm{total}] - \ce{[AH]}}{\ce{[AH]}} = \frac{[\mathrm{total}]}{\ce{[AH]}} - 1$$

Add one to both sides:
$$10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})} + 1 = \frac{[\mathrm{total}]}{\ce{[AH]}}$$

Take the reciprocal:
$$\frac{\ce{[AH]}}{[\mathrm{total}]} = \frac{1}{10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a})} + 1}\tag{1}$$

This is still general for any acid/base group. For example, we could use it to calculate the charge of ammonia/ammonium ($\ce{NH3(aq) + H+(aq) <=> NH4+(aq)}$). At very basic pH, the charge would be zero, at very acidic pH, +1. To get the average charge at any pH, we take the charge at very basic pH and add the result of equation [1] using the $\mathrm{p}K_\mathrm{a}$ value of ammonium.

For any amino acid (or any other molecule with ionizable groups with $i$ different $\mathrm{p}K_\mathrm{a}$ values), you take the charge of the species at very basic pH (all groups deprotonated), plus the following:

$$\sum_i \frac{1}{10^{(\mathrm{pH} - \mathrm{p}K_\mathrm{a,i})} + 1}\tag{2}$$

This is only an approximation because there might be some cross talk between ionizable groups (i.e. if one group becomes negatively charged, it becomes more "difficult" for the neighboring group to become negatively charged). It also becomes more complicated for polyprotic groups, but all the groups in amino acids are monoprotic with water as a solvent.


Is NaCl in water considered "solution" despite of NaCl's having some precipitation?

When we are considering $\ce{HCl(aq)}$ and $\ce{NaOH(aq)}$. the system contain sufficient amount amount of solvent water molecules  in situ already present and this is more than sufficient to hydrate the produced $\ce{NaCl}$ in the reaction.
Hence the first equation
$$\ce{HCl(aq) + NaOH(aq) -> NaCl(aq) + H2O(l)}$$

will be the correct expression.


Can carbon subnitride be used as a fuel with similar properties to gasoline?

No
You wouldn't want to burn this thing anywhere, it burns like, 9010 degrees... Fahrenheit. It's nearly the temperature of the surface of the sun (around 900 degrees off). Not much can contain it. So unless you had some kind of super-metal, than you are out of luck trying to use it as fuel.


Why is a third body needed in the recombination of two hydrogen atoms?

Yes there is validity to the finding that two atoms of hydrogen form molecular hydrogen at very slow rates.  Review Gould, R.J., and Salpeter, Edwin E., Astrophysical Journal, Vol 138, pg. 393.  It is publicly available at: http://articles.adsabs.harvard.edu/pdf/1963ApJ...138..393G
The exact answer to your question is found on page 138 of that work, under the heading “II. Mechanisms for the Formation of Molecular Hydrogen.”
In space, molecular hydrogen is destroyed (by cosmic radiation) at rates that exceed its formation without some type of surface catalyst.  This conundrum motivated significant research to find the actual mechanisms of molecular hydrogen formation in natural settings.  More recent work can be found by Gianfranco Vidali, et.al., Journal of Physics: Conference Series:6, 2005:
https://iopscience.iop.org/article/10.1088/1742-6596/6/1/003/pdf
 This more recent work cites the Gould and Salpeter work. It appears to be still accepted as the explanation for the slow rate of formation of molecular hydrogen from atomic hydrogen.  According to this explanation, after ground state hydrogen atoms collide, the new hydrogen molecule will be in a triplet state, and it cannot decay into a singlet state because this is a forbidden transition.  A cursory search on the PES's of triplet states of diatomic hydrogen show that indeed these are shallow curves which would lead to rapid dissociation.


"4K-metal chelated" A chemical term used in Sagan's Cosmos

These guys are "cryogenic superconducting electrovores" who apparently have "no planetary communities" so my guess is that "4K" should probably be taken to mean "4 Kelvin" and "metal-chelated organic semiconductors" as that they are composed of various types of semiconductors whose molecules are organics chelated with a metallic ion. About the "Local Group polylogue" thing, it's probably fictional stuff like a sort of interstellar Commonwealth or something.


Observability of Orbitals and Orbital Energies

Excluding the trivial case of a one electron system, where the wavefunction for the system is an orbital (1 electron wavefunction) and so could be measured via the electron density, orbitals are not observables. There is no known Hermitian Operator that returns orbitals (or their energies) from its expectation value. So is that the end of the discussion? Not necessarily.

In a 2006 paper, Shabazian and Zahedi$^1$ argue that what constitutes an observable depends on the conventions we establish for describing a system. They give as an example the Quantum Theory of Atoms in Molecules (QTAIM) developed by Richard Bader$^2$. Strictly speaking, the normal formulation of Quantum Mechanics doesn't allow us to describe subsystems within a molecule. We form molecules from atoms, but once the molecule is formed, there are no longer atoms; rather there is just a single wavefunction which describes the whole system. This is horribly inconvenient from a classical chemists perspective, as much of our understanding of chemistry comes from subdividing molecules into important functional groups that give the major properties of the molecule. To resolve this, Bader proposed a definition of an atom in a molecule by defining boundary conditions for a atom based on flux of the electron density. This gives an unambiguous, experimentally measurable way of determining observables like atomic charge, which without this set of external conditions is not considered an observable at all. While this has as of yet not been done, Shabazian and Zahedi argue that orbitals could be made observables by establishing a similar set of external conditions.

So while orbitals are not currently observables, what is and isn't an observable can change as new theoretical models develop.

Shahbazian, S. & Zahedi, M. Found Chem (2006) 8: 37. https://doi.org/10.1007/s10698-005-8247-4
Bader, R. Chem. Rev., 1991, 91 (5), pp 893–928
DOI: 10.1021/cr00005a013


Why there are more examples of carbonylate anions than metal carbonyl cations?

Yes there are more carbonylate ions than metal carbonyl cations, but the latter do exist.  A salt $\ce{[Mn(CO)6^+][BF4^-]•SO2}$ has been synthesized in a superacid medium.

Metal carbonyl cations are often disfavored because they have to contend with counterions that are anions, which may compete with the carbon monoxide ligands.  Such a competition is tilted towards the anion winning out if the metal is positively charged.  Solvents with significant Lewis-base character are also apt to displace carbon monoxide from the coordination sphere, for a similar reason.  Carbonylate anions do not face this situation (or at least not as severely) because cations are more likely to bind with the solvent than with an an ionic metal center that is dispersing its charge through pi donation to the carbon monoxide ligands.

Going back to the referenced salt, we can see features that protect $\ce{[Mn(CO)6^+]}$ from such an ignominious fate.  First note the counterion, $\ce{[BF4^-]}$which is a very weak nucleophile and has its negative charge buried inside the relatively bulky fluorine "shield".  Then note the superacid medium; perforce the solvent is not competing as a Lewis base either.


What is the oxidation state of Mn in HMn(CO)5?

On negative oxidation states, in general

Although it's usually a topic that's covered relatively late in a chemistry education, negative oxidation states for transition metals[1] are actually quite alright. On the Wikipedia list of oxidation states, there are quite a number of negative oxidation states. Some textbooks have tables which only show positive oxidation states, but good authors will be careful to add a disclaimer along the lines of "only 'common' oxidation states are listed".

The key is in the ligand that the metal is paired with. In metal ions with positive oxidation states, the ligands that they pick up tend to be Lewis bases, like $\ce{H2O}$, $\ce{Cl-}$, $\ce{NH3}$, ... This should "make sense" intuitively, because the positively charged metal ion wants to gain electron density, and these ligands are happy to give electron density to it.

In $\ce{Mn(CO)5-}$, the ligand is carbon monoxide: a very poor Lewis base, but it does possess low-lying π* orbitals. In the parlance of coordination chemistry, CO is a poor σ-donor, but a great π-acceptor. This means it tends to seek out metal centres which are relatively electron-rich instead of electron-poor, and thus it makes sense that metals with a negative oxidation state are its preferred partners. [I'll defer a full explanation of the $\ce{M-CO}$ bonding to a proper inorganic textbook.]

It turns out that it's actually quite difficult to find examples of carbonyl complexes with positive oxidation states.[2] Nearly all carbonyl complexes have the metal in oxidation states of zero or lower: see, e.g. $\ce{[Cr(CO)6]}$, $\ce{[Fe(CO)4]^2-}$, $\ce{Ni(CO)4}$, and of course, $\ce{[Mn(CO)5]-}$. Another common feature is that all of these complexes obey the 18-electron rule.

On [HMn(CO)5]

So, back to the question of the oxidation state in $\ce{[HMn(CO)5]}$... IUPAC has the final say on oxidation states;[3,4] see also this writeup by Martin. Quoting from ref. 4,

  Oxidation state equals the charge of an atom after its homonuclear bonds have been divided equally and heteronuclear bonds assigned to the bond partners according to Allen electronegativity [...]

Following this, we have to assign the electrons in the $\ce{Mn-H}$ bond to the hydrogen, since hydrogen is more electronegative than manganese on the Allen scale (3.30 and 2.75 respectively). There is also evidence that in metal carbonyl hydrides like these, the hydrogen bears a partial negative charge, although this argument is not without its caveats.[5] Therefore, the oxidation state of $\ce{Mn}$ should be +1. However, IUPAC offers a way out of this:

  [...] except when the electronegative atom is bonded reversibly as a Lewis-acid ligand, in which case it does not obtain that bond’s electrons.

According to this, it is possible to have exceptions to the general algorithm for determining oxidation state in the previous quotation, and there is a case to be made for this exception being valid here: the electronegative atom ($\ce{H}$) is indeed bonded reversibly ($\mathrm pK_\mathrm a = 7.1$) as a Lewis acid ($\ce{H+}$). If we accept this line of argument, then $\ce{H}$ does not "obtain" the two electrons in the $\ce{Mn-H}$ bond, and the oxidation state of $\ce{Mn}$ returns to −1, which is perhaps more in line with our chemical intuition.

I leave it to the reader to draw a conclusion for themselves; but this is an excellent illustration of the most important lesson here: oxidation states have their limits, and should not always be taken as meaningful descriptors of the chemical bonding in a species, especially when ambiguity is present, since this makes their assignment essentially arbitrary. If we assign an oxidation state of −1 to hydrogen here, it doesn't magically make the complex any less acidic than it was before. And IUPAC are aware of it (ref. 4):

  The applications of OS in chemistry are wide and deal with a cornucopia of chemical compounds and materials. It is therefore not surprising that, for some compounds, one value does not fit all uses, or that dedicated measurements or computations are needed to ascertain the actual OS. In those rare cases when the most convenient OS becomes a matter of choice, this fact must be clearly stated.

Refs. 3 and 4 make for excellent (and fairly accessible) reading on the topic, for those wishing to find out more about these edge cases. Both can be accessed without a subscription.

Notes and references

Negative oxidation states for non-transition metals exist (e.g. alkalides), but these are probably more "exotic" than organometallic complexes.
They do exist, for example $\ce{[Ir(CO)6]^3+}$, although the bonding situation is markedly different (as evidenced by, e.g., IR spectroscopy).
Karen, P.; McArdle, P.; Takats, J. Toward a comprehensive definition of oxidation state (IUPAC Technical Report). Pure Appl. Chem. 2014, 86 (6), 1017–1081. DOI: 10.1515/pac-2013-0505.
Karen, P.; McArdle, P.; Takats, J. Comprehensive definition of oxidation state (IUPAC Recommendations 2016). Pure Appl. Chem. 2016, 88 (8), 831–839. DOI: 10.1515/pac-2015-1204.
Sweany, R. L.; Owens, J. W. Evidence for a negative charge on hydrogen for some metal carbonyl hydrides. J. Organomet. Chem. 1983, 255 (3), 327–334. DOI: 10.1016/S0022-328X(00)99326-4.


Finding the mole fraction of nitrogen in a mixture of nitrogen and oxygen

OP has given a good effort to solve the problem using correct path. Only loose point was not considering the diatomic nature of the gases as Safdar Faisal pointed out in a comment. Suppose the amount of $\ce{O2}$ in the gas mixture is $n_1 \ \pu{mol}$ and the amount of $\ce{N2}$ in the gas mixture is $n_2 \ \pu{mol}$ in volume of $\pu{1.0 L}$ container. Thus, two parallel equations can be setup as follows:
$$PV = nRT \ \Rightarrow \ n_1 + n_2 = \frac{PV}{RT} = \frac{4 \times 1}{0.082 \times 298} = 0.164 \tag1$$
$$\frac{32n_1 + 28n_2}{1} = 4.7 \ \Rightarrow \ 32n_1 + 28n_2 = 4.7 \tag2$$
To resolve two parallel equations, take $(1) \times 32 - (2)$, which gives you,
$$(32-28)n_2 = 0.164 \times 32 - 4.7 = 0.538$$
$$\therefore \ n_2 = \frac{0.538}{4} = 0.135$$
Thus mole fraction of $\ce{N2}$ is $\chi_\ce{N2}$, then
$$\chi_\ce{N2} = \frac{n_2}{n_1 + n_2} = \frac{0.135}{0.164} = 0.820$$
Therefore, the mole fraction of $\ce{N2}$ in the mixture is $0.82$ (hence, the mole fraction of $\ce{O2}$ in the mixture is $0.18$).


Can you steam clean your range hood, by adding dishwashing liquid to boiling distilled water?

Clogged? Failing to suck out fumes? Quantitatively, how bad is it? And in what manner did this situation arise? Slowly, over time? Or suddenly? Can you hear a fan running?
Grease has a high boiling point, and while it is theoretically possible to clog some fine filter, it is doubtful that 4 years of home cooking could stop a hood fan from operating. In commercial practice, hoods over grills are cleaned frequently from spattered grease, not so much for better air flow, but more for sanitation and prevention of possible fires from leaping flames.
If no filter is accessible from the stove end, perhaps the exit end is the cause for blockage. Exit openings frequently have flaps to prevent rain entry. Two possibilities: 1) the flap is stuck down and hinders airflow, or 2) the flap is stuck up and the tube is blocked by a bird nest or a pile of acorns. (Of course, this exit end will be so high that it is not accessible for cleaning, but that's another problem.)
There is a third situation. In my former house, the builders omitted an outdoor exit pipe for the microwave and fume hood. There was simply a fan which sucked the steam/vapor up off the stove, thru the filter and back at the cook. We generally turned the fan off. The idea of a fume hood was nice, but not really necessary.


Alcohol in the bloodstream modeled by differential equations

One such article that I found was published by the University of South Florida. The abstract for which reads:

  Given a set of differential equations describing blood alcohol content as a function time, we integrated the
  equations to obtain a general solution. The general solution equation depends on three free parameters: the initial
  concentration of alcohol in the stomach after ingestion, the rate of alcohol absorption into the blood stream and the rate
  at which the alcohol is metabolized by the liver. We fitted our solution to experimental data to determine the unknown
  parameters for a particular subject.

If this sounds useful to you, it can be found here.


Is the lattice point in ZnO crystal structure occupied by the molecule ZnO or the ion O^2-/Zn^2+?

You are falling in the rather common mistake of confusing the crystal structure with the lattice. The lattice is a mathematical object (it does not exist as a real object), whereas the crystal structure is the convolution of the lattice with a basis (an atom or a collection of atoms).
In some cases atoms are distributed exactly at the same positions of the lattice points; for example Cu has a face centered cubic structure where copper atoms occupy the same positions of the face centered cubic lattice. In this case Cu at 0,0,0 position is your basis.
In your case neither the Zn atoms, nor O atoms occupy lattice positions. Your basis is contituted of Zn at 1/3,2/3,0 plus O at 1/3,2/3,z; 'attaching' this basis to each lattice point (taking into account the correct distances of both Zn and O atoms from each lattice point), you obtain the final structure.
In any case, the picture that you posted does not represent the real hexagonal structure; rather, I would that is the trigonal polymorph.


Can five membered ring form through Friedel-Crafts alkylation?

Maybe your teacher means to say that because further reactions would happen due to the further activation of the ring due to the alkyl group as opposed to the deactivation due to the acyl group.
Also further catalyst induced polymerization reaction can occur apart from possible polyalkylation

In general alkylation on rings are not preferred because of the above reasons as well


The inert pair effect plays no particular role in the stability or instability of transition metals’ oxidation states. It is due to the large and rising energy difference between a shell’s s- and p-subshells which make the ionisation of the lower electrons more difficult. However, it is equally wrong to say that lower oxidation states be stabilised in lower periods: the most stable halide with the lowest oxidation state a halogen can have is fluoride ($\mathrm{-I}$).

All 3d-metals before iron can be oxidised up to their highest oxidation state while being strong oxidants in that state. This is not so true for the 4d-metals; an indication that the 3d electrons are better attracted to the nucleus. Indeed, the 3d orbitals do not contain radial nodes and their electrons are thus closer to the nucleus. The radial nodes in 4d and higher d-orbitals ‘add distance’ between the nucleus and the corresponding d-electrons (do not copy that wording: it is bad). It is this proximity to the nucleus and the consequential lesser extension of the orbitals into space that draws electrons into them, destabilising the higher oxidation states and stabilising slightly lower ones.

The same effect is true for f-elements; see Nicolau’s answer to a related question on f-element oxidation states’ stabilities.


Do metal ions of 4d and 5d series always form low spin complex?

While the vast majority of the compounds and complexes of the heavier transition metals are low spin (where that concept has any meaning) $\ce{PdF2}$ is one exception, which is unusual not only this way but also having $\ce{Pd^2+}$ in an octahedral environment (see e.g. Wikipedia). Thus your source is not quite correct.

As for the reason why 2nd and 3rd row transition metals are more likely to form low spin complexes than the lighter elements, the reason is given in the answer linked above in the comments. In fact, while the question may be different, the answer is almost a duplicate.


Why is the antibonding pi bond higher in energy than the normal pi bond?

Firstly, the + and – signs you see do not indicate charge. They mark the sign of the electron’s wave function. In an atom, the electrons always have a negative charge, and the nucleus always has a positive one.

When a bonding orbital is formed, the electron density is concentrated in the area between the nuclei. The electrostatic attraction between the electrons and two (instead of one) nuclei, lowers the energy of the molecule. Furthermore, the electron density between the nuclei shields the electrostatic repulsion between the two positive nuclei to a certain degree.

In an antibonding orbital, the electron density is moved away from the internuclear area: the electron-nucleus attractions are minimized and the internuclear repulsion maximized. Thus, the molecule is destabilized.