Can a strong electrolyte be sparingly soluble?

It means that by just adding it to water wont make it dissociate into ions(sparingly soluble) but if current is passed through its solution, it will easily dissociate and give ions. These cases look rare.

But Curt.F's reasoning looks better :

  In the low concentration regime where the salt is "soluble", it completely dissociates, but that the solubility limit is just low.


Relative stability of keto- enol form of acetyl chloride

The stability of keto-enol compounds depends on the solution they are present in.

Keto compounds are stabilized due to inter-molecular hydrogen bonding when present in a polar solvent (for example water).

Enol compounds are stable when present in non-polar solvents, in which intra-molecular hydrogen bonding stabilizes them.


Free software for calculating the net charge of an amino acid as a function of pH?

I like MarvinSketch. You can run it in your browser or on your computer. Under Insert > Template, you can find all the amino acids. Once you've inserted one, select Calculations > Protonation > Isoelectric point and click OK. You'll get a window that looks like this showing the net charge versus pH (I picked a pretty boring one, but other amino acids produce more complex curves):


Can R-3-methyl-4-heptanone be enantioselectively synthesised from 4-heptanone?

Enders SAMP/RAMP hydrazone chemistry will do this(1).

The Wikipedia article on this technique is a good introduction:

(1): Corey, E. J.; Enders, D. (1976). "Applications of N,N-dimethylhydrazones to synthesis. Use in efficient, positionally and stereochemically selective C-C bond formation; oxidative hydrolysis to carbonyl compounds". Tetrahedron Letters. 17 (1): 3–6.


Is entropy of surroundings a state function?

In reality, when we talk about the surroundings, what we are referring to isn't really the entire universe, outside what we call "the system."  What we are really referring to is a subregion of the entire universe, wherein it is possible to carry out changes reversibly without encompassing the whole rest of the universe.  So, when we talk about the entropy change of the surroundings, we are not including spontaneous processes occurring light years away in other galaxies.  And, when we say that the entropy change for system plus the surroundings is zero, we are really only including the selected subregion of surroundings chosen to represent our sub-universe.


What kind of molecules do I smell when draining cooked potatoes or rice?

What a good question. But unanswerable by me and my prior knowledge and skill in google-fu. I have some general answers. This is a complicated area, since smell is both a physical issue and an individual and unmeasurable issue. Olfactory science is at a surprisingly basic level, given how far we are generally. I tried investigating what compounds in rice provide odour, the answer seems to be; way more than I could possibly have thought. Check out this article - while it is rough to digest a scientific articl unless you have basis knowledge just look at that list at page 505-506. There are many aromas in rice!

As for what functions they perform - looking at the list it is probably impossible to generalize down to what functions such a wide array of compounds do. Certainly, some compounds that are odorous has a primary function of being odorous, attracting pollinators and attracting mates. For rice, which has been around humans since time immemorial (at least 9k BC) they have probably been through a selection process, the most benevolent smelling rice plants being selectively bred, but this is speculation on my part. Other compounds have other functions and just happen to have a strong odour. Generally, to be odorous, the compound must have some volatility. What can also be said generally and very broadly is that alcohols and aldehydes / ketones are more often odorous than other compounds. There are also certain inorganic compounds that are (and often strongly so), Ammonia, Sulfide and the "garlic" series (S-Se-Te) for instance.

Good question. If you learn more, let us know.


I haven't done that in decades (or so), but I'd bet that you did right!

The result from google only lists the final outcome, not the order in which the orbitals are filled. Except for some examples $\ce{La}$, $\ce{Ac}$, etc., the Aufbau principle is reliable. So: Follow the yellow brick road :)


Will KOH work in place of NaOH as the electolyte in the electrolysis of water?

Using $\ce{KOH}$ rather than $\ce{NaOH}$ will work just fine.

Solutions of $\ce{KOH}$ are actually about $\pu{5\%}$ more conductive than solutions of $\ce{NaOH}$ at the same molar concentration. In terms of mass concentration, a given mass of $\ce{KOH}$ will have about $\pu{30\%}$ fewer equivalents of reactant than the same mass of $\ce{NaOH}$ because of the difference in molecular weights.

The combination of the difference in molar conductivity and molecular weights between $\ce{NaOH}$ and $\ce{KOH}$ just means that a given concentration of $\ce{KOH}$ should be about $\pu{35\%}$ less conductive than the same mass of $\ce{NaOH}$. Of course you can increase the concentration of $\ce{KOH}$ relative to the amount of $\ce{NaOH}$ you had planned to use if you like.

Please don't hesitate to ask for any clarifications in the comments below, and best of luck!


Is there any way to carry out voltammetry without a potentiostat?

This is a valid question. This project is entirely possible, but I would not advise it for a high school research project based on my experience building these. Much better projects are available.

A nonreactive electrode, for example gold or platinum, is required for good results.

Knowledge of op-amp design, or some other feedback control system, which is taught in advanced electronics courses is needed.  The control system fixes the potential between the the reference and the working electrode by drawing current from the third electrode.  The current drawn is measured to give some information about the system.

The Wikipedia article shows the circuit need, https://en.wikipedia.org/wiki/Potentiostat.

The open source cheapstat is available: http://web.chem.ucsb.edu/~kwp/cheapstat/

Integrated versions of the circuit are available. http://www.ti.com/product/LMP91000.  Evaluation kit ~$100 http://www.ti.com/tool/LMP91000EVM.

Information on making electrodes: https://publiclab.org/notes/JSummers/01-09-2014/potentiostat-notes-5-how-to-make-low-cost-electrodes

It is possible to build a potentiostat without advanced knowledge of electronics; however, this runs into costs. Getting experiments done with the equipment, resources, and funding available is something every scientists faces.


Which has the higher chemical shift E/Z alkene?

Are you talking about 1H NMR? In the general case, it is not so easy to say - it depends on all substituents, with geminal substitution usually contributing most to change in chemical shift. trans and cis substitution will affect chemical shifts differently, but will depend on the electronegativities of all groups concerned.

For your example, there will be very little difference between chemical shift for these two methyl groups, but the methyl trans to the alkyl chain is probably slightly downfield shifted. Coupling constants for the two methyl groups would be slightly different. In acyclic systems, 4J couplings from a methyl to a cis or trans proton can range from about -3 to +2.5 (but usually 0.8 to 2), and usually |Jtrans|>|Jcis|. Of course, nOe information would also be useful.

If you are talking about 13C NMR, a methyl peak trans to a carbon chain will be downfield shifted from a methyl group trans to a proton.


If "Gas" belongs to "State of Matter" then what does "Metal" belong to?

"Metal" is a classification used by chemists and usually means something along the lines of "this electrically conductive, usually solid element that is deformable".

The other way around: Chemists use "gas" to refer to a state of matter, not some defined gas (for example the atmosphere).

Putting these two together: "Metal" belongs to the periodic table of elements, while "solid", "liquid", "gas" and "plasma" are merely words to describe what state the element or compound is in. It is possible to have liquid, solid or even gaseous metals.

Edit: If you want a name for a variable that is either in the state Metal, Half-Metal or Metalloid and Non-Metal, I'm not sure such a name even exists. Feel free to make one up that suits your needs, comment your code sufficiently and it will not be a problem. "Metallicness" or something along those lines would be my choice.


Why does barium sulphate not dissolve in HCl while barium sulphite does?

$\ce{BaSO_3}$ has the same property as $\ce{BaCO_3}$ :
When reacting with an acidic solution, these two substances produce a gas which can get out of the solution, and displaces the equilibrium to the right  : $$\ce{BaCO_3 + 2 H^+ ->  Ba^{2+} + CO_2(g) + H_2O}$$
$$\ce{BaSO_3 + 2 H^+ -> Ba^{2+} + H_2O + SO_2(g)}$$ This allows these two substances to be soluble in $\ce{HCl}$ solutions. As a contrary $\ce{BaSO_4}$ has not this possibility to produce a gas in an acidic solution : it is and remains insoluble in $\ce{HCl}$ solutions.


What is the displayed formula for methylpropene?

"Propene" is what defines the main chain of three atoms so that the maximum number of carbon atoms in any other chain cannot exceed this number. Hence your proposition that

  there would be a one-carbon side chain on the first carbon

is wrong as in this case there would be a 4-membered chain of either 1-butene or 2‐butene:

In order to comply with the methylpropene name, one can only attach methyl group to the second carbon, and not to the terminal one:

To sum it up, your textbook is correct.


Enthalpy of the reaction between hydrogen and oxygen

The confusion is partly caused by careless use of quantities and units. The value $H = -484\ \mathrm{kJ}$ denotes an enthalpy. However, what is actually meant is the enthalpy per one mole; i.e. the molar enthalpy $H_{\mathrm m} = -484\ \mathrm{kJ/mol}$.

In the definition of the molar reaction enthalpy, the ‘per mole’ does not refer to any particular substance in the equation. Instead it refers to the entire reaction as a whole. Therefore, the reaction must be specified for which this quantity applies. In this case, the enthalpy of $484\ \mathrm{kJ}$ is released when $2\ \mathrm{mol}$ of hydrogen gas react with $1\ \mathrm{mol}$ of oxygen gas to form $2\ \mathrm{mol}$ of gaseous water:
$$\ce{2H2(g) + O2(g) -> 2H2O(g)}\qquad\Delta H^\circ = -484\ \mathrm{kJ}$$
(By way of comparison, the corresponding value for liquid water is about $-572\ \mathrm{kJ}$.)

Thus, the molar enthalpy relating to the amount of hydrogen is
$$\frac{\Delta H^\circ}{n(\ce{H2})}=\frac{-484\ \mathrm{kJ}}{2\ \mathrm{mol}}=\frac{-242\ \mathrm{kJ}}{1\ \mathrm{mol}}=-242\ \mathrm{kJ/mol}$$
i.e. the released enthalpy per $1\ \mathrm{mol}$ of hydrogen gas is $242\ \mathrm{kJ}$.

However, the given question is asking about $n(\ce{H2}) = 0.50\ \mathrm{mol}$ of hydrogen (and accordingly $n(\ce{O2}) = 0.25\ \mathrm{mol}$ of oxygen). The corresponding enthalpy is
$$\begin{align}
0.5\ \mathrm{mol}\times\frac{\Delta H^\circ}{n(\ce{H2})}&= 0.5\ \mathrm{mol}\times\frac{-484\ \mathrm{kJ}}{2\ \mathrm{mol}}\\[6pt]
&= 0.5\ \mathrm{mol}\times\frac{-242\ \mathrm{kJ}}{1\ \mathrm{mol}}\\[6pt]
&= 0.5\ \mathrm{mol}\times -242\ \mathrm{kJ/mol}\\[6pt]
&= -121\ \mathrm{kJ}
\end{align}$$


How do I separate graphite powder from light oil?

"Any ideas on how I can seperate and re-use both items?"
The devil is in the details:

what is particle size and concentration?
what is the light oil?
what do you mean by reusing them?

Depending on particle size (pumped) filtration (paperfilter, colloid filter, celite ) should work as well as centrifugation as was recommended by @Jon Custer. Is it that urgent? Also, cannot you just try it instead of asking it?

What is the oil?

Whatever you do, your graphite will be oily. It does matter if this oil has high-boiling point/waxy components... On the other hand, you may able to use something with low boiling point and just remove it under vacuum / heating, maybe can skipp the whole filtration part.

What do you mean by reusing them?

As I said, there will be remains of oil on the graphite. Also, most probably, you will have fine powder of graphite remains in the oil whatever you do. How do you clean those up it depends on what do you need it for. Do you need to wash the graphite and re-filter the oil? You should know/test that. Also, cost-wise does all this fuss make any sense? Neither graphite nor the oil sound something expensive/special. Also, the more unique, expensive they are, most probably their cleanup will be more demanding (as you need to shoot for higher purity).


What exactly does κ stand for in this equation connecting solubility and molar conductivity?

Nomenclature changes and authors don't always follow the rules.  The more typical use now is $\Lambda_m$ for the molar conductivity and $\Lambda_m^0$ for the limiting molar conductivity.

The molar conductivity, $\Lambda_m$, is given by the equation

$\Lambda_m = \dfrac{\kappa}{c}\tag{1}$

where:

$\kappa$ is the specific conductance (which is what you measure...)
$c$  is the concentration (which you know because of how the solution was made)

Now typically the specific conductance, $\kappa$, is plotted against concentration. For dilute solutions, there will be a linear relationship where the slope of the line is the limiting molar conductivity, $\Lambda_m^0$.  However at high concentrations the relationship becomes non-linear and the actual conductance is less than what the linear relationship would predict. That is because of the formation of cation and anion clusters. The local effect is thus to make the cluster uncharged (or charged less is the charges don't balance to zero).

Now the author of your book specified a sparingly soluble salt, so at low concentrations $\Lambda_m \approx \Lambda_m^0$, and we can substitute for $\Lambda_m$ to get:

$\Lambda_m^0 = \dfrac{\kappa}{c}\tag{2}$

which can be rearranged to give:

$c = \dfrac{\kappa}{\Lambda_m^0}\tag{3}$

If the conductivity of the solution is much greater than that of pure water then this is fine. The measured specific conductance will be directly proportional to the concentration since $1/\Lambda_m^0$ is a constant.

However the author is talking about a salt of limited solubility, so the conductivity of pure water is not negligible. Going back to the plot of specific conductance, $\kappa$, against concentration, $c$, we can determine that at 0 concentration there is some conductance which I'll call $\kappa_w$ which is the conductance of pure water. So for a salt with limited solubility the equation is:

$c = \dfrac{\kappa - \kappa_w}{\Lambda_m^0}\tag{4}$

If you look back at equation (2) you can see why equation (3) doesn't work for solutions of very low conductivity. When $c=0$ equation (2) blows up since you can't divide by $0$.

So the author used $\kappa$ in two different ways which is very poor writing. We could define $\kappa'= \kappa - \kappa_w\tag{5}$ then $c = \dfrac{\kappa'}{\Lambda_m^0}\tag{6}$


Confusions about oxidation and reduction

The formal definition of an oxidation/reduction is linked to the loss/gain of electrons. Considering the addition or loss of hydrogen and oxygen is not a global rule, but rather a trick generally used in organic chemistry. If you consider organic molecules, thus mainly composed of carbon, "adding" oxygen atoms suggests an oxidation of the carbon, as oxygen is more electronegative than carbon, while "adding" hydrogen atoms suggests a reduction, as hydrogen is less electronegative than carbon.

An example would be the successive oxidation of ethanol to ethanal, then acetic acid : $$\ce{CH3CH2OH ->[-2 e^-][``-2\text{ H}"] CH3CHO ->[-2e^-][``+1\text{ O}"] CH3COOH}$$


Why is the lone pair of pyridine's nitrogen atom not part of the aromatic ring?

As follows from our discussion in the comments to the question, pyridine uses only a single electron in the p-orbital in order to comply with Huckel's rule. Kudos to Orthocresol:

  Huckel's rule demands that there be 4n+2 pi electrons in the ring. Each carbon contributes one, making for a total of 5 from the carbons. If your nitrogen lone pair is in the ring, then you will have seven pi electrons. $7≠4n+27≠4n+2$ for $n ∈Z$

Of course, electrons know nothing of Huckel and his rule, but apparently they position themselves into the most energetically suitable configuration. I am not yet qualified to explain why exactly it is the most suitable one.


When titrating to find the concentration of a dissolved acid, does the volume of solvent matter?

At very great dilutions, impurities could have some effect on the end point. For example, even previously-distilled water will absorb some small amount of $\ce{CO2}$ from air, making it slightly more acid. However, if the aspirin is present in reasonable amount, that would have negligible effect. So, as stated by @MaxW, consistency, with a moderate theoretical excess of solvent, is important, but don't use greatly excessive amounts of solvent.


How do you calculate the heat of a reaction given a table of heat of formation values?

I can obtain the correct answer using your values. Maybe double check for mistakes and typos?

Here are my quick calculations:
The overall reaction is:

$\ce{NaHCO3 + NaOH -> Na2CO3 + H2O}$

The equation used is:
\begin{equation}
\Delta H_\text{reaction}^\circ = \sum {\Delta H_{f}^\circ(\text{Products})} - \sum {\Delta H_{f}^\circ(\text{Reactants})}
\end{equation}

$\Delta H_\text{products}^\circ= (\pu{0.008 mol}\times\pu{-286 kJ/mol} +\pu{0.008 mol}\times\pu{-677 kJ/mol}) = \pu{-7.704 kJ}$

$\Delta H_\text{reactants}^\circ= (\pu{0.008 mol}\times\pu{-692 kJ/mol} +\pu{0.008 mol}\times\pu{-230 kJ/mol}) = \pu{-7.376 kJ}$

$\Delta H_\text{reaction}^\circ = (-7.074 - (-7.376))~\pu{kJ/mol} = \pu{-0.328 kJ} = \pu{-328 J}$


Effect of sodium hydroxide solution on copper plate

Regarding the copper metal itself, the only down side I see of using a $\ce{NaOH}$ is that it is unlikely to remove some of the oxides that would be covering any exposed copper.  The initial etching process would have mainly produced water-soluble $\ce{CuCl2}$, likely with a bit of insoluble $\ce{CuO}$.  $\ce{CuO}$, $\ce{CuCO3}$ and some $\ce{Cu(OH)2}$ would have likely formed on any unprotected copper surfaces over time.  $\ce{NaOH}$ could remove $\ce{Cu(OH)2}$ and $\ce{CuO3}$, but not $\ce{CuO}$.  It also should be unreactive toward the copper plate itself.

A common method of putting a good shine on tarnished copper is with weak organic acids like citric acid.  Lemon juice and table salt make a favorite home copper cleaning solution.  Again, this is only with regards to the cleaning oxides from the copper metal itself, not the ink residues coating it.

My bottom line advise is to do what I’ve seen in your last comment that you plan to do; test a portion of the bottom side of the plate with the $\ce{NaOH}$ solution to verify that this solution cleans the ink residues without harming the copper.  I think you can do this with confidence that the copper metal will be unharmed.  If the ink residues are removed and you want to remove any remaining oxidation to create a fresh, shiney copper plate, you could dissolve just a bit of salt into lemmon juice and wipe the plate with this solution until shiny!


What is a stereochemically active or inactive s orbital?

Generally, these terms are used to refer to lone pairs rather than s-orbitals.

In the context of VSEPR theory, lone pairs usually affect molecular geometry. For example $\ce{NH3}$ adopts a trigonal pyramidal structure, because of the presence of the lone pair on nitrogen; if there was no lone pair, then it would be trigonal planar, cf. $\ce{BF3}$. These are so-called stereochemically active lone pairs, since they affect the stereochemistry at nitrogen.

In a few rare cases lone pairs do not affect the molecular geometry, and these are termed stereochemically inactive. The only example I can think of, off the top of my head, is in $\ce{[XeF8]^2-}$, which should have 8 bond pairs and 1 lone pair around xenon. Despite the presence of the lone pair, the dianion adopts a square antiprismatic structure, without any obvious position in space for the lone pair to occupy. Looking in Shriver and Atkins' Inorganic Chemistry (6th ed.) there are a couple more example of $\ce{[SeF6]^2-}$ and $\ce{[TeCl6]^2-}$, which both have octahedral geometries despite having six bond pairs and one lone pair around the central atom.

In VSEPR theory, these stereochemically inactive lone pairs are rationalised by placing them in an s-orbital, which is non-directional. An electron pair in an s-orbital has a spherically symmetric electron density, and hence cannot influence the shape of a molecule. However, it is not exactly the orbital that is stereochemically (in)active, it is the electrons, and hence the more frequently seen term is "stereochemically (in)active lone pair".

There is one more context which should be mentioned, because it occurs frequently, and that is in tin(II) and lead(II) compounds. The electronic configurations of these two ions end with $n\mathrm{s^2}$, and this s-orbital lone pair tends to generate some distortions in stereochemistry.

A great example is that of the oxides. Solid $\ce{MgO}$ has a structure in which $\ce{Mg^2+}$ is octahedrally coordinated by six $\ce{O^2-}$ ions (rock salt/halite structure), but $\ce{SnO}$ and $\ce{PbO}$ have an unevenly distributed, square pyramidal coordination of $\ce{M^2+}$ by four $\ce{O^2-}$ ions. Traditionally this is attributed to the presence of a stereochemically active lone pair on Sn(II) or Pb(II).

This description cannot actually be complete, though, because if the geometry is solely determined by the lone pair on the metal, then one would expect PbS to have the same structure - and it doesn't. In fact, PbS has the same structure as MgO, with regular octahedral coordination, so clearly the anion has a role to play. A full explanation is beyond the scope of this answer, but one way of rationalising this is by invoking a second-order Jahn–Teller distortion. The interested reader may consult Chem. Soc. Rev. 2011, 40 (9), 4455–4463.

However, the "stereochemically active lone pair" is a recurrent theme in Sn(II) and Pb(II) structural chemistry (sometimes Tl(I) and Bi(III) also). For example, the metallocenes $\ce{SnCp2}$ and $\ce{PbCp2}$ adopt a bent structure, rather than the linear structure of $\ce{MgCp2}$ (and the more famous $\ce{FeCp2}$). Again, though, VSEPR theory is not the best way of explaining any of this; a proper explanation would involve MO theory.


XRD-Shorter distance between planes than actual radius of atoms/or cell parameter

Does this help?

The distance between planes will be less than the distance between atoms in many cases because the atoms don't have to stack directly on top of one another in each plane.


Is the Lowdin Orthogonalization used in diagonalizing the atomic orbitals really a basis change?

Consider a nonsingular, linear transformation $\mathbf A$ of a set of vectors arranged into a matrix $\mathbf V$ $$\mathbf V'=\mathbf {VA}$$

We can say the matrix $\mathbf V'$ is orthogonal (and thus that we have linearly transformed the vectors to form an orthogonal basis) if $$\mathbf {V'^{\dagger}V'}=\mathbf {A^{\dagger}V^{\dagger}VA}=\mathbf {A^{\dagger}SA}=\mathbf{1}$$

In general, this occurs when $\mathbf{A}=\mathbf{S}^{-1/2}\mathbf{B}$, where $\mathbf{B}$ can be any unitary matrix. The Lowdin orthogonalization commonly used is just the case where $\mathbf{B=1}$, so we can see that we are properly changing the basis.

But your issue is specifically with transformations of the density matrix, where the transformation matrix is not unitary in general. The standard change of basis formalism would suggest that to go to AO basis from MO, we would do a transformation something like this:
$$\mathbf{C}\mathbf{D}^{MO}\mathbf{C}^{-1}=\mathbf{D}^{AO}\mathbf{S}$$
Ensuring orthogonality of the MOs requires that $\mathbf{C^{\dagger}\mathbf{S}\mathbf{C}}=\mathbf{1}$, which we can sub in to the equation above to give: $$\mathbf{C}\mathbf{D}^{MO}\mathbf{C^{\dagger}\mathbf{S}\mathbf{C}}\mathbf{C}^{-1}=\mathbf{C}\mathbf{D}^{MO}\mathbf{C}^{\dagger}\mathbf{S} =\mathbf{D}^{AO}\mathbf{S} \to \mathbf{C}\mathbf{D}^{MO}\mathbf{C}^{\dagger}=\mathbf{D}^{AO}$$

So the form in terms $\mathbf C$ and its adjoint is a direct result of the conventional similarity transform combined with the orthonormality condition on the MOs.

This latter derivation was adapted from$^1$ which can be found here.

As to why we avoid using transformations involving the inverse of the transformation matrix, I suspect Zhe is correct that it is a matter of cost, since obtaining the transpose/adjoint of a matrix requires almost no work at all, while an inverse is relatively difficult to obtain.

T. Helgaker et. al Chemical Physics Letters 327 (2000). 397–403


Firstly I would like to expand a bit on what a colligative property is, in order to bring more understanding and also be able to apply this to a named physical property.
Colligative properties depend mainly on the number of particles in a solution.
The values of the colligative properties are approximately the same for equal
concentrations of different constituents in solution regardless of the species or chemical nature of the constituents. Examples are osmotic pressure, vapor pressure lowering, freezing point depression, and boiling point elevation (as you have noted).
Furthermore, physical properties of substances are not just limited to the colligative nature. In fact a number of properties have been defined:

additive (depend on the total contribution of the atoms in the molecule or on the sum of the properties of the constituents in a solution e.g molecular weight).

constitutive (depend on the arrangement and partly on the number and kind of atoms within a molecule e.g interfacial characteristics).

other thermodynamic physical properties are known (extensive, intensive)

Having gained this background, it is now evident that surface tension (a force that pulls the molecules of the interface together) is dependant on the arrangement and also on the nature of atoms or molecules , therefore is more of a constitutive property rather than a colligative property.
I have included a screenshot to better visualise the phenomenon:

Hope this helps.


Regarding the ionization extent of sulfuric acid solutions:

$\text{1)}$ The first ionization is complete at most molarities (until you start reaching extremely high molarities - i.e. $\text{18 M}$). So sulfuric acid is a strong acid.

$\text{2)}$ The second ionization (i.e. the ionization of the hydrogen sulfate ion, $\ce{HOSO3^-}$, can be complete at low starting concentrations of hydrogen sulfate ion. This is because $K_a(\ce{HOSO3^-}) = 1.2 \times 10^{-2}$ is rather sizable!

So we can generalize these two cases of sulfuric acid solutions:

For very dilute sulfuric acid solutions:

$[\ce{H3O^+}]=2M_i - [\ce{HOSO3^-}] = [\ce{HOSO3^-}]+ 2\times[\ce{SO4^{2-}}]$, in which we generally take $[\ce{HOSO_3^-}]$ as negligible (i.e. = $\epsilon \to 0$).

For non-dilute solutions of sulfuric acid, we only have these relationships:

$$[\ce{H3O^+}]=M_i +[\ce{SO4^{2-}}] = [\ce{HOSO3^-}]+2[\ce{SO4^{2-}}]$$

As we can see, the first relationship (one for very dilute solutions) follows from the second because in a very dilute solution, $[\ce{HOSO3^-}]\approx 0$ - i.e. it has fully ionized into $[\ce{SO4^{2-}}]$.


Interaction between energy and entropy in highly elastic polymers?

If I remember correctly, the relationships you have presented are related to the configurational entropy of the polymer chains between cross links of the polymer network.  In these equations, R is the spatial distance between cross links, N is the number of chain segments between cross links, and l is the length of each chain segment.  The smaller the value of R, the greater the number of configurations that the chain can exhibit (i.e., the greater the entropy).  Thus, I believe that there should be a minus sign in your equation for S, because, as R increases, the fewer the number of configurations that the chain can exhibit, and thus the lower the entropy.  The parameter you call U is, I believe the Helmholtz free energy, and is a measure of the stored elastic energy of the polymer network.  As the R gets longer, the polymer chains have been stretched more, and more elastic energy is stored in the chains.  Conceptually, the polymer chains are like springs between the cross links.


Why does silver (I) have a larger lattice enthalpy and hydration enthalpy than sodium?

The electron configuration of sodium ion is that of the inert gas neon (1s^2 2s^2 2p^6), a spherical shell with a positive charge at the nucleus. There are 10 electrons inside this shell.

The electron configuration of silver ion is that of krypton plus 4d^10, a spherical shell with a positive charge at the center. There are 46 electrons inside this shell.

While the slightly smaller sodium ion might have slightly more polarizing power than silver ion, silver ion has far more polarizability than sodium ion. Polarizability is the distortion of the electron cloud as a result of a point charge (or dipole). Dispersion forces are a form of polarizability where two polarizable entities polarize each other. https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Physical_Properties_of_Matter/Atomic_and_Molecular_Properties/Intermolecular_Forces/Specific_Interactions/Polarizability

The effects of dispersion forces due to the polarizability of the anion can be seen by comparing the enthalpies. The crystal structures for the fluorides, chlorides and bromides and NaI are FCC, rock salt. AgI is wurtzite. The difference in enthalpies for the fluorides is 43, but as the polarizability of the chloride anion increases, the attraction of dispersion forces increases the difference to 125, and for bromide, 148. Even tho the crystal structure of AgI is different, it still seems to have a polarizablity-polarizability effect (difference = 181).

This is not what you would call a predictable effect, but does a good job of explaining the trend. It is not a first-order attraction, nor a second-order polarizability, but a third-order polarizability-induced polarization effect.

And very perspicacious of you to have observed this trend.


Nitrogen fertilisation expressed as nitrogen equivalent

This is referring only to the mass of elemental nitrogen available, rather than the total mass of fertilizer. In your case, ammonium sulfate is 21% nitrogen (see the table here for other fertilizers and here for more information on fertilizer composition), which gives ~40.95g ammonium sulfate.


Solve this mole fraction calculation with incomplete information

I don't think you can solve that unless you know additional information like mass of water added or volume of water added, or the density of the final solution.

You could approximate by dividing mass of ethanol by density of ethanol to get volume of ethanol, and saying volume of water added is 50mL-(volume ethanol), but this is only an approximation because volumes are not additive when the two solutions are different.

The error could be estimated from this graph:

source: http://en.wikipedia.org/wiki/Partial_molar_property

or the data from the graph could be used to improve the accuracy of the answer

or use this density table: http://www.handymath.com/cgi-bin/ethanolwater3.cgi?submit=Entry


What causes Starch's lump creation when solvent (butter/water/oil) is added and how to prevent them

The trick here is that you're not actually dissolving the starch. Starch is made up of small semi-crystalline granules of amylose and amylopectin which are actually pretty much insoluble at room temperature. Some water can get in as amylose breaks up the crystallinity, but it doesn't really dissolve. When you drop it into water, it hydrates a bit on the outside, holding it all together, but water doesn't make it into the inside. If you stir or shake it, you can break the clumps up into smaller pieces, eventually getting at most individual hydrated granules.

It's not until you heat the mixture that the granules really break apart and allow water in, in a process called gelatinization. This is what makes starch useful in cooking as the starch can form a more diffuse network, thickening the liquid.


Search materials by HOMO, LUMO and work function

Indeed, Prof. Alán Aspuru-Guzik has placed his "clean energy project" database on the web in a searchable form.

At the moment, it covers ~2.3 million compounds.

  Use of the data requires citation of the appropriate literature sources. The data in this database is released under the Creative Commons Attribution ShareAlike license and therefore any academic publications associated with the CEPDB should be appropriately cited.

As you point out, the database only covers HOMO and LUMO orbital eigenvalues. While this may not be the most useful for experimental applications (i.e., you really want ionization potentials and first excited state energies or electron affinities) it should be a helpful "first screen" to find interesting lead compounds.


Byproducts of the neutralisation of hydrochloric acid with sodium hydroxide

Given a 1M concentration of $\ce{HCL}$ and $\ce{NaOH}$ here are the pHs:

  pH of $\ce{HCL}$: 1

  pH of $\ce{NaOH}$: 13-14

First, a 1 molar concentration of hydrochloric acid solution, $\ce{HCl}$, has a pH of 0.0 if we take ${-log[H_3O^+]=pH}$ to be true. (And it is mostly true for dilute solutions).

  and of course water has a pH of 7.

Pure water at 25 degrees Celsius has a pH of 7.0. Boil the water and the water will no longer have a pH of 7.0. Cool the water and pH will similarly change. The only condition for neutrality is that the concentration of hydronium ion equals the concentration of hydroxide ion. It just so happens that water at normal lab conditions has a hydronium ion and hydroxide ion concentration of ${1.0*10^{-7}}$ M, which corresponds to a pH of 7.0. Your statement only holds at a specific temperature. At other temperatures, this reaction - the auto-ionization of water - which controls the pH of pure water - will go to varying extents:

$\ce{2H_2O <=> H_3O^+ + HO^-}$

  Here is a reaction that is most likely to occur:

  $\ce{HCl + NaOH + H2O <=> H2O + NaCl(aq)}$

Since we are talking about solutions of HCl and NaOH, your equation isn't reflective of what's actually going in on the system. HCl and NaOH both dissolve in solution (unless the solution is saturated with either). So you don't have HCl or NaOH in solution, but rather hydronium ion and chloride anions in the case of HCl and sodium ions and hydroxide ions in the case of NaOH solution. A more accurate equation would be:

$\ce{H_3O^+ + HO^- ->2H_2O}$

Regarding:

  $\ce{NaCl(aq)}$

Sodium chloride has a very high solubility in water and dissociates unless the solution is saturated. So it would be more accurate to write simply sodium ions and chloride ions, instead of "aqueous" sodium chloride (unless you are referring to the ion-pair effect). Ion pairs often occur in solution; the name describes exactly what is implied - two oppositely charged ions are attracted to each other. For sodium chloride, ion pairing is likely minimal, but in other cases, such as between a carbocation and a halide anion, the ion pair effect has noticeable effects on chemical reactions (i.e. partial racemization rather than full racemization).

  Like for example theoretically sodium oxides could form from the
  $\ce{OH-}$ taking a proton from another hydroxide anion thus giving
  you $\ce{O^2-}$ which could react with the sodium to give you
  $\ce{Na2O}$ which is a sodium oxide.

So hydroxide ion reacts as an acid and abstracts a proton from another hydroxide ion.

This is at least theoretically possible, but thermodynamically unfavorable. The oxide anion is very unstable in water solution to the point of being able to react with the solvent (water) itself to a large extent.


Who developed the concept of ICE tables?

As expected the concept of equilibrium calculations is more than a century old idea. The algebra has not changed. Check internet archives for classical general chemistry textbooks to see how equilibrium calculations were done in 60s-70s. If you are specifically looking for the representation of "initial, change, equilibrium" tables, this is also four decades old. Books from the early 80s used this approach. See for example "Chemistry with inorganic qualitative analysis" by Thenard (1980). These days everyone likes to coin fancy acronyms and words. Gold colloids became gold nanoparticles, and "initial, change, equilibrium" became ICE!


Equilibrium constant of a reaction whose order of forward and reverse reaction is not same

I'll try and answer your question, but a "full" answer would take a book.
Given the reaction:
$$\ce{aA + bB <=> cC + dD}$$
Then assuming an elementary reaction in the gaseous state the concentration equilibrium constant always has products over reactants and will be:
$$K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}$$
The above equation relies on two very specific assumptions:

The reaction is an elementary reaction which often isn't true and the actual coefficients must be determined experimentally since the reaction occurs in steps.
That concentrations can be used instead of activities.

Formally using activities explains part of your question immediately. For instance if A, B and C are gases and D is a liquid or solid, then the activity of D would be unity by definition.
Now breaking down the equilibrium expression into two rate equations we have:

$r_\mathrm{f}$ - Forward reaction rate
$r_\mathrm{r}$ - Reverse reaction rate
$\ce{a^*, b^*, c^*}$ and $\ce{d^*}$ = experimentally determined coefficients which may or may not be equal to the stoichiometric coefficients
$k_\mathrm{f}$ and $k_\mathrm{r}$ are constants for the forward and reverse reactions respectively.

Now using concentrations instead of activities:
\begin{align}
r_\mathrm{f} &= k_\mathrm{f} \ce{[A]^{a^*}[B]^{b^*}}\\
r_\mathrm{r} &= k_\mathrm{r} \ce{[C]^{c^*}[D]^{d^*}}
\end{align}
and at equilibrium by definition:
\begin{align}
r_\mathrm{f} &= r_\mathrm{r}\\
k_\mathrm{f} \ce{[A]^{a^*}[B]^{b^*}} &= k_\mathrm{r} \ce{[C]^{c^*}[D]^{d^*}}
\end{align}
so:
$$K_c = \frac{k_\mathrm{f}}{k_\mathrm{r}}= \frac{\ce{[C]^{c^*}[D]^{d^*}}}{\ce{[A]^{a^*}[B]^{b^*}}}$$
To determine the coefficients the experimentalist can manipulate the experiment to simply the kinetic expression. For example let $\ce{[A] \gg [B]}$ then $\ce{[A]^{a^*}}$ is essentially a constant.
Does this answer your question?


Why does the alkaline hydrolysis of benzamide yield ammonia?

If the reaction were simply having the form

$\ce{RC(O)NH2 + OH^- <=> RC(O)OH + NH2^-}$

then your reasoning might well apply.  But, in the real world, the carboxylic acid $\ce{RC(O)OH}$ deprotonates (more than the amide $\ce{RC(O)NH2}$) and the proposed, strongly basic amide ion of course protonates to ammonia.  These proton transfer reactions push the displacement forward giving the true reaction

$\ce{RC(O)NH2 + OH^- -> RC(O)O- + NH3}$


From this Wikipedia page:

  Kevlar is synthesized in solution from the monomers
  1,4-phenylene-diamine (para-phenylenediamine) and terephthaloyl
  chloride in a condensation reaction yielding hydrochloric acid as a
  byproduct. The result has liquid-crystalline behavior, and mechanical
  drawing orients the polymer chains in the fiber's direction.
  Hexamethylphosphoramide (HMPA) was the solvent initially used for the
  polymerization, but for safety reasons, DuPont replaced it by a
  solution of N-methyl-pyrrolidone and calcium chloride.

The structure of Kevlar is given from the same page:

However, Kevlar production on a small laboratory scale will likely be an expensive and difficult endeavor compared to purchasing Kevlar products, which are available in about any form you can imaging, many of them patented. Also, as Kevlar is water insoluble, concentrated sulfuric acid is required to keep it in solution during synthesis, molding, spinning, etc.  Good luck ;)


Conductivity and Molar Conductivity of Equimolar Solutions of LiCl and KCl

Since Li is highly solvated due to its small size, it's mobility is reduced and hence conductivity decreased.


What is the reaction between oxalic acid and potassium permanganate?

We had a titration of oxalic acid vs. potassium permanganate for an experiment in which we used concentrated $\ce{H2SO4}$.

Actually, the reaction requires an acidic medium i.e. $\ce{H+}$ is involved as a reactant... I suppose the $\ce{H+}$ released by the oxalic acid would be enough since you are studying the rate of the reaction though I am not sure.

The balanced ionic equation is:
$\ce{2MnO4- + 5H2C2O4 + 6H+ -> 2 Mn^2+ + 10CO2 + 8H2O}$

As you can see, $\ce{H+}$ is involved as a reactant and that is why you might have read that conc.$\ce{H2SO4}$ is necessary.


Why is the solubility of sodium acetate trihydrate so much lower than anhydrous sodium acetate?

The values of solubility for these salts are $100\%$ correct. Your understanding of solubility is also correct. What you didn't know (probably) is that solutions can exist in a non-equilibrium state. In case of anhydrous sodium acetate, it dissolves in water forming so-called supersaturated solution, which is metastable. Eventually, crystals of sodium acetate trihydrate will precipitate out of the solution, forming a system which is in the thermodynamic equilibrium.


Could Calcium Oxide be made from CaCO3 in a bonfire?

According to wikipedia, CaCO3 will turn into Calcium if heated above 825 °C (1,517 °F).

As for your fuel

Wood burns in several phases.

In the first phase, which can be referred to as dehydration, the heat produced by the wood's combustion drives of water from it in the form of vapour. It takes a lot of energy to do that which limits the fire's temperature to ~200-250°C.

The second phase, which may be referred to as pyrolysis, kicks in as the wood gradually becomes dehydrated. In that phase the fire's temperature will go from 250°C to topical peaks up to ~800°C.

As the wood further breaks down, you will encounter a third phase which is equivalent to the burning of charcoal. The temperature at the heart of your ambers will be at a minimum of ~700°C and could reach as much as 1300°C if you bring enough oxygen into it (by injecting compressed air into the heart of your ambers through a metal pipe for instance). This temperature would be fully sufficient to produce CaO from CaCO3 in a reasonably short amount of time.

At this point you could also add regular coal into the ambers, which requires a temperature above ~700°C to start burning. In ideal conditions, coal can burn up to ~ 2100°C which would significantly push the limits of what is possible for you.

If you would like to push these limits further, you could add coke on top of your regular coal, which would push the theoretical maximum temperature your fire could reach up to ~ 2500°C.

More practically

As Francis mentioned, there is a wide gap between theory in ideal conditions and practical reality.

The first aspect I already mentioned that will affect the practical temperature your chalk will be exposed to, is the amount of oxidising molecules that is brought to the heart of your fire. Injecting air through a pipe may be the safest option (though it is already significantly less safe than just making a regular bonfire, as it will significantly increase the combustion temperature and is at risk of propulsing fuel around if not adequately set up).
Theoretically, you could achieve a yet better result (at the cost of a lot of danger) by injecting pure compressed O2 into the heart of your ambers (which, for safety reasons, you should NOT do).

The second and very important aspect is insulation. The theoretical maximum temperature that is reached within your fire is of little practical importance. Insulating your fuel combustion zone without compromising the supply of oxidisers into it is key not only to increasing the maximum temperature that is locally reached inside of your fire, but more importantly to just keep calories inside of the combustion zone, which will significantly increase the average temperature inside of it (and tremendously lessen the required fuel expense for achieving any given amount of CaCO3 calcination). You can insulate your fire in numerous ways. Coal itself can act as a (burning) insulator if you place your CaCO3 at the heart of your combustion zone and cover it in a pile of coal/charcoal. Or you could build a structure around your fire: SiO2, Al2O3, CaO and MgO are particularly suitable materials for their resistance to high temperatures and for their refractive properties. You can look up their properties on wikipedia (MgO for instance melts at 2,852 °C, has a thermal Conductivity 45–60 W·m−1·K−1 and a Refractive Index of 1.7355).

Chemistry

My last point is that there might be the possibility for you to make things easier by adding some additional chemistry into the process (if you are inclined to do so). It would seem, for instance that Ca(OH)2 might turn into CaO starting around 512°C, which seems to be lower than CaCO3. So you might be able to lower your requirements on the combustion side by turning your CaCO3 into Ca(OH)2 prior to calcination. There are many ways to do this but one way would be to react CaCO3 with HCl to yield CaCl2 , which you'd subsequently react with NaOH (the very low solubility of Ca(OH)2 would drive the reaction forwards).


Why coefficient doesn't change amount of substance

When you say number of moles, it is a number to express the number of constituent species (atoms, in this case) in the given substance. It does not have any correlation to where that substance will be used and what will be its 'coefficient' in that reaction.

When you say 10g Mg, you mean 0.411 moles of Mg (and vice versa) irrespective of whatever you are going to do with those 10 grams.

Even if you were to use 10g Mg in a reaction such as $$\ce{ 2Mg + O2 -> 2MgO} $$, you would still have 0.411 moles of Mg as it is still equivalent to 10g. Nothing changes on the basis of the 'coefficient'.

You can further try to look up more information to get a better understanding about 'The Mole Concept' because that is where I think your problem lies.


Which unit cell do we obtain if spheres are arranged in ABABAB type in 2D and then that 2D layer is arranged in AAAAAA type in 3D?

The question is a little confusing, so I will interpret your 'ABABAB' type in 2D to mean a simple hexagonal layer (so, the same layer as used for your hcp and fcc packing). So, a stacking of a simple hexagonal layer, each layer directly above the other, will give a structure looking like right hexagonal prisms - or right triangular prisms if you look a bit more carefully.

Anyway, the resulting structure is the simple hexagonal Bravais lattice with a single atom basis. Hcp is also a simple hexagonal Bravais lattice, but with a two atom basis, one each from plane A and plane B. I do not know of any elements that crystallize in this crystal structure (but am open to corrections!).


Does pressure affect non-gaseous reactions?

Yes. Pressure, as a concept, is simply the force exerted by random molecular motion per unit area, and since all matter is composed of molecules, and all matter above absolute zero has some random motion, everything has molecules exerting some force over some area. (This pressure is the vapor pressure of the solid, which is typically extremely low. If the external pressure on the solid is higher than the vapor pressure of the solid, the solid molecules will not fly off and escape. Things with high vapor pressure exert a high "pressure" on their surroundings and thus escape easily - jet fuel or nail polish remover have very high vapor pressures.)

If you have a liquid reaction between molecules A and B, bringing the liquid to a higher pressure will increase the frequency of collisions between molecules of A and B (by the definition of pressure). Since there is a finite probability of the reaction proceeding for any given collision, increasing the collisions increases the rates.

Practically, for you to see that kind of effect in a laboratory would require a very pressure-sensitive reaction, large molecules whose shape changes with pressure, or a reaction with a large change in the number of moles. Biological chemistry is rich with examples pressure-sensitive non-gaseous reactions, mainly due to pressure changing the shape and behavior of large complex molecules.

There's also the effect of osmotic pressure, which is a pressure induced not by random molecular motion but by the tendency of the substance to seek a lower free energy state. Osmotic pressure will affect concentrations (the system will seek to reduce the concentrations of ions), which will affect reaction rates.

Increasing the pressure of solids and liquids can also change their physical properties (heat capacity and thermal conductivity are two excellent examples). These physical properties are innately tied to the reaction mechanism. Since most reactions occur in sequence, if the physical properties are different (and if they change over the course of the reaction), pressure will exert an effect there.

Finally, there is an effect technically excluded by your question, but important nonetheless, and that is the effect of pressure on solid-gas reactions (sublimation or deposition). The manufacture of integrated circuits involves depositing materials onto chips in layers, and pressure is an important way to control that rate.

In general, the more nonlinear the reaction mechanism, the more likely it is that the small effects that pressure creates will cascade into larger changes that are what you would call "significant."


What is the chemical structure of the cocoate ion vs. the olivate ion?

Note 1. Both coconut oil and olive oil are natural oils. The soap, produced from them, is mixture of several compounds.

The difference of olive and coconut oil may be easily found in wikipedia: http://en.wikipedia.org/wiki/Coconut_oil#Composition_and_comparison

In short, coconut oil contains mostly saturated acids esters, while olive contains mostly unsaturated ones. The unsaturated acid esters are responsible for olive oil ability to solidify on air. But here is

Note 2: In my knowledge, allergy is usually caused by carbohydrates or proteins. Fatty acids and their ions have too few binding sites to be recognized by immune system as something distinct from our own. The real source of reaction may be

Minor components of initial oil. There are quite many of them and removing them completely is not easy.
Other components of soap, such as perfume or pigment.
Soap solution pH, that may differs from soap kind to kind, but this is unlikely.


Chemical Free Energy or Gibbs Free Energy is best understood as potential energy for reactions.

If I have a lot of Gibbs Free Energy, then I have a lot of potential energy.
Kind of like if I were on top of a hill, it's very easy for me to roll down that hill.

Often what's talked about in books is dG or the change in Gibbs Free Energy. If we think of it is simply being on top of a hill (High Gibbs Free Energy/Potential Energy) G(initial)
 going to the bottom of the hill (Low Gibbs Free Energy/Potential Energy) G(final)

Then the change in Gibbs Free Energy will be negative for something spontaneous (like the rolling down a hill example).
Change is always calculated as: change = final - initial

Gibbs Free Energy is used in the same was as potential energy. But instead of translating to kinetic energy, it translates to Enthalpy (heat) and/or Electrochemical Energy (being able to move electrons). Both Enthalpy and Electrochemical Energy can then be converted to mechanical work.

The way to think about it is not so much that you can use Gibbs Free Energy, but rather it's just the innate property of whatever situation you're working with. Molecules have a tendency to be reduced or oxidized under different conditions, the Gibbs Free Energy just tells you which way is spontaneous.

dG = dH - TdS

This gets kind of complicated, since dG is affected by many factors, but it gives you an idea, that the Change in Gibbs Free Energy can become positive or negative under different conditions.

Keep in mind that Gibbs Free Energy isn't a real thing, but rather a concept that helps explain why things happen.
 Why do gases expand?
 I don't know, but the universe has a tendency to do this. So we say that it is spontaneous, just like how we would say things roll down a hill because of gravity.
 What's gravity?
 It's our way of explaining why things roll down hills. :p

I hope that helps a little bit.


H2O2 (Hydrogen Peroxide) catalysis in clear plastic bottle?

An easy test for this would be to react H2O2 with yeast or baking soda (sodium bicarbonate) before and after sun bathing. Or you could simply measure gas production in the sun vs in a box for say 24 hours.

An easy way to measure the O2 released is to use a closed vessel with a ballon, but the more scientific method uses an inverted, graduated column.

You can measure light intensity with a photometer.


Cis, trans diene reacting with trans dienophile in D-A reaction

Macrophage: I thought you did a fine job in analyzing the Diels-Alder reaction of the E,Z-pentadiene with fumaronitrile. To complement my comment above, I offer an illustration of the transition states of the two reactions leading to their respective products. Because fumaronitrile may add with equal facility (mirror image transition states) to either face of the achiral diene, both 2 and 4 are necessarily racemic. "Racemic" transition states represented by 1 are E-endo/Z-exo that afford racemic bis-nitrile 2. Alternatively, "racemic" transition state 3 is E-exo/Z-endo, which leads to bis-nitrile 4 as a racemate.


At some point in studying organic chemistry, you might have been taught that a nitrogen in an amide group is planar sp2, due to delocalization of its electrons into the carbonyl group. You also might have learned that nitrogen atoms in a pyrrole or imidazole ring are planar sp2 since they participate in the 6 electron aromatic pi bonding system. You would then conclude that all the nitrogen atoms in caffeine are coplanar. But without that essential background information, you might not have been expected to figure that out.


The optimized effective potential (OEP) is a natural connection between local density-functional theory and many-body perturbation theory, which is derived from the Sham-Schlüter equation. Converging the full set of OEP equations is quite a challenge. Therefore an approximation was made in order to reduce the integral equations to an analytically solvable one via a dominant orbital approximation, called The Krieger-Li-Iafrate (KLI) approximation.

$$\sum_{i,j\neq i}\frac{\phi_j^*(\mathbf{r})\phi_i^*(\mathbf{r})}{\epsilon_i-\epsilon_r}\left(f_i \langle \phi_i|v_x|\phi_j\rangle - \sum_{k, \alpha}\frac{d_{ik}^{\alpha}d_{kj}^{\alpha}}{2}(\epsilon_i-\epsilon_k)\left[\frac{f_i(1-f_k)}{\epsilon_i-\epsilon_k+\epsilon_{\alpha}}+\frac{f_k(1-f_i)}{\epsilon_i-\epsilon_k+\epsilon_{\alpha}}\right]\right) + c.c.$$

$$=\frac{1}{2}\sum_{\alpha,i,k}\omega_a\frac{\left|d_{ik}^{\alpha}\right|^2 f_i(1-f_k)}{(\epsilon_i-\epsilon_k-\epsilon_{\alpha})^2}\left(\left|\phi_k(\mathbf{r})\right|^2-\left|\phi_i(\mathbf{r})\right|^2\right)$$

Source: Maybe they can see comments above.


First of all, if in a certain condition it is said that mass (or any other parameter) is constant, you cannot just assume ${1gm}$ (or any other number). Mass being constant means $\Delta m =0$, not that it is ${1gm}$. You have no way of knowing what the mass is (unless stated in the question, of course). The correct approach is to assume the mass to be $'m'\space{units}$. Now, this mass $'m'$ can assume any value, according to the condition.

Moving on, you have made a mistake while comparing molar mass and volume.

The ideal gas equation gives us the relation,

$$PV=\frac{\text{Mass}}{\text{Molar Mass}}RT$$

For your question,

$$V=\frac{1}{\text{Molar Mass}}.\frac{(\text{Mass})RT}{P}$$

$$V\propto \frac{1}{\text{Molar Mass}}$$

Recheck your calculations.


Coordination number is the number of the nearest neighbours of an atom, and it is related to the Kepler's conjecture. For CN12 (12 is the 'kissing number' of a 3D sphere), there are two possible ways for the spheres to be 'packed' which are HCP and CCP. Space-filling efficiency they two are exactly the same, and it is why one coordination number can correspond to two crystal structures.
The crystal structure of metal may depend on many features, and the space-filling efficiency is just one of them. Even by the thermal expansion, the crystal structure can change. (Tin is the most familiar example.)


Magnesium Sulfate (MgSO4) back to Magnesium Oxide (MgO) + Sulfuric Acid (H2SO4)?

Yes, magnesium sulfate can be converted back to magnesium oxide by thermal decomposition or use of reducing agents.

Thermal decomposition

$$\ce{2MgSO4 ->[\Delta] 2MgO + 2SO2 ^  + O2 ^}$$

  The thermal decomposition of magnesium sulfate to produce magnesium
  oxide, sulfur dioxide and oxygen. This reaction takes place by heating
  to the temperature of 1137°C.(Chemiday 1)

Use of reducing agent

$$\ce{2MgSO4 + C ->[\Delta] 2MgO + 2SO2 ^  + CO2 ^}$$

  Magnesium sulfate react with carbon (as reducing agent) to produce
  magnesium oxide, sulfur dioxide and carbon dioxide. This reaction
  takes place at a temperature of 700-900°C.   (Chemiday 2)

The products other than magnesium oxide are gaseous and escapes leaving behind pure magnesium oxide.


Solubility curve for uranium hexafluoride (in any solvent)

So far I have found that it is soluble in:

Bromine Pentafluoride
Chlorotrifluoroethylene Liquid Polymer, Dichlorotetrafluoroethane, Perfluorotributylamine, Perfluoroisopropyl ether, 1,2-Dichlorohexafluoropropane, hydrogen fluoride and more.
chloroform,  carbon  tetrachloride  and  fluorocarbon  solvents;
liquid chlorine and bromine; dissolves in nitrobenzene to form a dark red solu-
tion that fumes in air

I think the resource given by Martin (2) is the best. Resource 1 gives a solubility curve, but resource 3 doesn't. However, I think there may be data floating around on compounds mentioned by 3.


Can a volume of gas be measured by change in PSI?

Volume is inversely proportional to pressure only when:

1) the number of moles of gas is held constant. We can say this is true in your situation because the repairman measured the pressure in the tank and I don’t think a significant amount of gas disappeared from the tank in the time it took for him to measure the pressure.

2) The temperature is held constant. Again, I don’t think temperature dropped 20 kelvin in the time it took for him to measure the pressure in the tank.

This follows from the ideal gas law equation – a good approximation most of the time:

$PV=nRT$.

So I say your repairman still has his wits about him and isn’t trying to rip anyone off.


A certain number of ions, atoms or molecules closely associate around a central atom leading to the formation of distinct entity called coordination complex.
The groups atoms or molecules linked to the central atom are said to be coordinated with the latter and their number gives the coordination number.

The non-ionisable units lie in the primary zone or sphere, whereas the  ionisable  units lie in the secondary sphere and appear as ions in the solution.

In hexammine platinic chloride $\ce{[Pt(NH3)6]Cl4}$  and in potassium ferrocyanide $\ce{K4[Fe(CN)6]}$ four chlorine ions of the former and four potassium ions of the latter ionize in the aqueous solution, whereas six ammonia molecules of the former and six cyanogen groups of the latter belong to the primary sphere or non-ionizable sphere and thus fail to respond to their characteristic tests or fail to ionize. Or you could also say that the chlorine ions in the former and the potassium ions in the latter occupy the primary valencies and hence are ionizable and the ammonia and cyanogen group occupy the secondary valencies  and are non-ionizable, while the $\ce{Fe}$ atom in the latter is present in the primary Sphere and thus fails to ionize.

As per your question during the formation of a compound say $\ce{[Pt(NH3)5Cl]Cl3}$, five molecules of ammonia and the chlorine ion (linked by secondary valencies) are present in the primary Sphere and the rest of the three chlorine atoms lie in the secondary sphere. Thus the coordination number of the compound would be  six as five ammonia molecules and one chlorine atom are attached to the central metal atom.

Now, every element tends to satisfy both its primary and secondary valencies. A negative ion tends to satisfy both these valencies i.e. primary as well as the secondary valencies.
The presence of a negative ion in the coordination sphere reduces the amount of charge on the complex ion by the amount of charge present on it while the negative ions in the primary sphere or the coordination sphere are not ionized.


Maximum hydrogen atom bonds per a single carbon atom

I would look at it like recursive problem. Starting with the simplest situation:

Carbon has 4 electrons available to form covalent chemical bonds (so 4 simple bonds for example). Hydrogen has one electron - 1 bond.

If you add one more carbon to the situation, you have to bond it somewhere - you need to use the first carbon. Now with this new carbon you added 3 more possible places to add 3 more elements (like hydrogens) but you canceled one hydrogen place on the first carbon. So every time you add another carbon, you can go up only for 2 hydrogen. It is actually called homologous addition or something like that. (Sorry for my english as well)

I ignored the oxygen whole time, cos as far as he is 2 bonded, he never adds anything he either substracts two hydrogen places, or you can instert it between a carbon and a hydrogen.


How to get $\Gamma_{vib}$ for hexafluoroethane?

To transcribe what is given in levineds's linked answer and remove the parts only relevant to linear molecules, here are the steps you can apply to determine the reducible representation of the vibrations.

Determine how the molecule is changed by all symmetry operations. If an atom is moved by a symmetry operation, it contributes it doesn't contribute to $\Gamma_\text{atoms}$. If the atom remains in place, add +1 to the column of that symmetry operation.

For hexafluoroethane, this gives:
$$\small \begin{array}{c|cccccc} \hline
D_{3\mathrm{d}} & E & 2C_3 & 3C_2' & i & 2S_6  & 3\sigma_d  & \\ \hline
\Gamma_\text{atoms} & 8 & 2 & 0 & 0 & 0 & 4 &\\
\end{array}$$

Multiply each column of $\Gamma_\text{atoms}$ by the trace of that operation's matrix representation.
$$\begin{array}{c|c}E&3\\\hline C_2&-1\\\hline \sigma&1\\\hline i&-3\\\hline C_n&1+2\cos(\frac{2\pi}{n})=1+2\cos\theta\\\hline S_n&-1+2\cos(\frac{2\pi}{n})=-1+2\cos\theta\end{array}$$

$$\small \begin{array}{c|cccccc} \hline
D_{3\mathrm{d}} & E & 2C_3 & 3C_2' & i & 2S_6  & 3\sigma_d  & \\ \hline
\Gamma_\text{cart} & 24 & 0 & 0 & 0 & 0 & 4 &\\
\end{array}$$

Use reduction formula to determine irreps. $n(i)=\frac{1}{h}\sum_R N\cdot\chi_r(R)\cdot\chi_i(R)$ where $n(i)$ is the number of the $i^{\text{th}}$ irrep, $R$ is a symmetry operation, $h$ is the order of the group, $N$ is the coefficient in front of the operation, and $\chi(R)$ is the character of the operation in the reducible/irreducible representation.

I won't go through the calculation (there are online calculators that you can use to do this), but this gives. $$\Gamma_{xyz}=3A_{1g}+1A_{2g}+4E_g+1A_{1u}+3A_{2u}+4E_u$$
which if we subtract off the translational and rotational irreps you have already obtained gives:
$$\Gamma_{vib}=3A_{1g}+3E_g+1A_{1u}+2A_{2u}+3E_u$$
(You had small error with your rot and trans, the $E$ irreps are degenerate and so count for 2 degrees of freedom. This means in each case, you would only have a single $E$ irrep.)


This question serves as a prime example of how much fake and scientifically incorrect information, or more famously, fake news is there on the web. This is why ordinary web should not be anyone's first priority to search about scientific topics. There is a famous saying that "paper never refused an ink" which applies to the web as well in the sense that anyone can write anything. Use Google Scholar, which only searches research level articles (relatively more authentic).
With this type of search one can arrive at Thermal stability of potassium carbonate near its melting point. This paper rules out the possibilities of (1) and (2). The authors note that there is significant vapor pressure of carbon dioxide near the melting point, and suggest $\ce{K2O}$ formation. They also advise the readers to consult the following as the most comprehensive study on alkali carbonates. I will let you try Google Scholar for this one.
References

A. Reisman, Reactions of the Group VB pentoxides with alkali oxides and carbonates. IX. A DTA study of alkali metal carbonates, J. Am. Chem. Soc. 80 (1958) 3558-3561.


In this specific case is the proton considered product or catalyst?

As you've illustrated with your figure from the Wikipedia article, a catalyst appears as the same compound on each side of a chemical equation. The short answer to this part of your question is that since the proton only appears on one side of the equation, it cannot be a catalyst. A hydrogen atom bound to a water molecule is not the same thing as a proton. Rather, it is the result, or product, of a chemical reaction. Also note that the reaction given in your question would be more correctly expressed as:

$$\ce{CO2 + H2O <=> H2CO3 <=> HCO3− + \color{red}{H+}}$$

In the second part of the question, that regarding the video excerpt, the proton from your reaction actually becomes a reactant in a second completely different reaction. Here, the proton reacts with hemoglobin resulting in two new products: the protonated hemoglobin an an oxygen molecule.

I hope that I have addressed the crux of your question. Please don't hesitate to ask for any clarifications in the comments below.


Hydrogen Bonding (with oxygen) - restricted to oxygen singly bonded to H or not?

Hydrogen-bonds require a $\delta^+$ hydrogen atom attached to an electronegative element, i.e. O, F, N, or Cl, and also an "active" lone pair on another electronegative atom.  An electrostatic attraction between these two is an extremely strong kind of intermolecular dipole-dipole interaction.

Compounds which have O-H or N-H groups are able to both form hydrogen-bonds through the $\ce \delta^+$H atom and accept hydrogen-bonds through the lone pair on the oxygen or nitrogen.  These compounds tend to have higher then expected melting and boiling points due to the hydrogen-bonding.  Compounds are even able to form internal hydrogen-bonds with other O or N atoms in the structure.

However hydrogen bonding can occur between substances that cannot form the bonds with their own molecules.  Because compound Y is an ether and can only accept hydrogen-bonds there are fewer bonds made overall between the mixture of substances, so the effect on the physical properties isn't so great.  Nonetheless water can form hydrogen-bonds through its proton and functional groups such as ethers, esters, amines, amides, alcohols and carboxylic acids can all accept those bonds through the relevant lone pairs.

Another example would be a tertiary amine such as $\ce{N(CH3)3}$, which is miscible with water because of the hydrogen bonds it accepts from the water molecules even though it has no N-H bonds itself.  Even $\ce{N(C2H5)3}$ is miscible with water at room temperature, and other tertiary amines continue to be soluble in water for the same reason: they are hydrogen-bond acceptors.

So the mark scheme is correct, and the teacher has shown a poor understanding of intermolecular interactions.


The answer is approximately 6.88. Therefore a $1\  M$ solution of glycerol in water will be ever so slightly acidic (considering that the hydroxyls are much weaker bases than acids, i.e. that the equilibrium constant for the reaction $\ce{C3H7O2OH + H2O <=> C3H7O2O^- + H3O^+}$ is much larger than the constant for $\ce{C3H7O2OH + H2O <=> C3H7O2OH2^+ + OH^-}$, which may or may not be true. Inclusion of the second equilibrium will drive the true pH even closer to 7)

For very weak or very dilute acids/bases, solving a dissociation problem is a bit more tricky because the self-dissociation equilibrium of water must be taken into account. It is often a valid approximation to forget about it, as water has only a small tendency to self-dissociate and so its effect on equilibrium concentrations of $\ce{H^+_{(aq)}}$ and $\ce{OH^{-}_{(aq)}}$ is often negligible, but this is not your case; glycerol in water seems to be about as weak an acid as water itself.

The best way to solve equilibrium problems is to set up a large equation including all factors by combining smaller equations. We need to make an equation that takes into account all the species in the medium ($\ce{H^+_{(aq)}}$, $\ce{OH^{-}_{(aq)}}$, $\ce{C3H7O2O^{-}_{(aq)}}$ and $\ce{C3H7O2OH_{(aq)}}$ {for convenience I shall label the latter two $\ce{A^-}$ and $\ce{HA}$}) and the related equilibrium constants ($K_a$ and $k_w$). The smaller equations are found by balancing charges (one equation) and the amount of matter of each substance (one or more equations depending on the problem).

Charge balance: $[H^+]=[OH^-]+[A^-]$ (I)

Matter balance: $[HA]+[A^-]=C_{acid}=1\ M$ (II)

The equations for the equilibrium constants are:

$K_a=\frac{[H^+][A^-]}{[HA]}=10^{-14.15}$ (III)

$k_w=[H^+][OH^-]=1\times 10^{-14}$ (IV)

To solve the problem, we need to play with the equalities until we get a single equation with a single variable. It's convenient to obtain an equation in $[H^+]$ because we are able to find the pH directly by applying $-log$ to the answer. Notice that

$[OH^-]=\frac{k_w}{[H^+]}$ (V)

$[HA]=\frac{[H^+][A^-]}{K_a}$ (VI)

Substituting (VI) in (II) yields, after some algebra:

$[A^-]=\frac{C_{acid}K_a}{[H^+]+K_a}$ (VII)

Now insert (V), (VI) and (VII) in (I) so that you get:

$[H^+]=\frac{k_w}{[H^+]}+\frac{C_{acid}K_a}{[H^+]+K_a}$

A little bit of persistence will get you to the following polynomial:

$[H^+]^3+K_a[H^+]^2-(C_{acid}K_a+k_w)[H^+]-K_ak_w=0$

Replacing the values of $K_a$, $k_w$ and $C_{acid}$, the only positive root for the equation is $[H^+]=1.30688\times 10^{-7}$, which after applying the antilogarithm results in $pH = 6.88$.


Are precipitation and crystallization both analogous between Chemistry and Meteorology?

In chemistry, on Earth in gravity, if a material from solution conformally, over all surfaces vertical and horizontal (e.g. the sides as well as the bottom of a container) is that still sometimes considered precipitation? Or is that crystallization from solution, and analogous to frost formation but not snow or rain?

Precipitation, as typically used an understood in chemistry, is really a very very old terminology (~ several hundred years). It refers to the formation of a solid by adding or mixing chemical reagents or even by cooling. There is no such requirement that the precipitate has to settle to the bottom. For example, colloids may not settle for ages. Precipitation involves crystallization, but not all precipitates are crystals (organic substances can be precipitated out by adding suitable solvents to a solution).
Now there are special ways to generate precipitates in solution. There was an experiment which we taught in classical analysis. The experiment goes like this. Add urea to an iron salt. No precipitate forms and there are no seeds to begin with. Let it boil slowly. With the passage of time, the entire solution becomes basic and it would uniformly produce precipitate iron hydroxide.  Since it was a in situ chemical decomposition of urea to produce iron hydroxides in solution, this process occurs very uniformly in the solution i.e., there is no localization where two reagent were being mixed.  This yielded uniformly sized particles. Such a process is called homogeneous precipitation in solution. Contrast this situation where we would add NaOH solution to iron salts. It will be impossible to avoid localized mixing.
Meterologists use the term precipitation as well, but as per the unabridged Oxford Dictionary (can't provide a link as it is by subscription), the full word is atmospheric precipitation.
In case you are interested, the etymology is

Etymology: < classical Latin praecipitāt-, past participial stem (see
-ate suffix3) of praecipitāre to throw or cause to fall headlong, to ruin, destroy, to fall headlong, to suffer ruin, come to grief, to
hasten, to rush, in post-classical Latin also to cause to be deposited
as a solid from a liquid solution (a1490, 1652 in British sources) <
praecipit- , praeceps (adjective) headlong, sheer (see precipe n.).
Compare Middle French, French precipiter , précipiter precipit v.


Why is the work done by a gas during free expansion zero and not infinity?

The external pressure in space is 0 (well, depends where and not quite, but for the discussion let's say that it is). So $P \cdot \Delta V$ is also 0. Gas expanding from a hole in a spaceship is not an irreversible process. Such processes are very slow and gradual, such that the external pressure has time to equilibrate with the internal pressure.

On the contrary - you have the most irreversible process in here. Gas leaking from a hole in a spaceship into a medium (space) that will never equilibrate with the internal pressure.

Think about it from an intuitive point of view. Work done by an expanding system is there because it moves other things around. In space there is nothing to move, so no work done.


Will SnO2 or SnO form from this reaction?

The actual reaction of tin (II) chloride with water in the presence of air is given here.

$$\ce{6SnCl2 + O2 + 2H2O → 2SnCl4 + 4Sn(OH)Cl}$$

  Tin(II) chloride react with oxygen and water to produce tin(IV) chloride and hydroxotin(II) chloride. Interaction with oxygen and air humidity. This reaction takes place slowly.

As Mithoron noted, the reaction takes place very, very slowly and thus get hydrolysed partially to form the basic salt, $\ce{Sn(OH)Cl}$ instead of forming the hydroxide or the oxide.

The basic salt will also form when tin (II) chloride reacts with water in the absence of air, thanks to Gannex.

$$\ce{SnCl2 + H2O ⇄ Sn(OH)Cl + HCl}$$

  Tin(II) chloride react with water to produce hydroxotin(II) chloride
  and hydrogen chloride. This reaction is proceeds by diluting the
  solution tin(II) chloride. (source)

As noted earlier, the reaction is very, very slow and maintains an equilibrium between the reactants and the products.

However when tin (II) chloride is present in a concentrated, acidified solution, it will form a complex.(source)

$$\ce{SnCl2 + H2O → [Sn(H2O)Cl2]}$$

  Tin(II) chloride react with water to produce dichloridoaquatin.
  Tin(II) chloride - concentrated solution. pH < 7.


To understand logic and relation of biochemical cycles

tl;dr: You can't reasonably 'propose' a biochemical pathway on paper the same way you might be able to for an organic reaction.

Biochemical pathways are highly complex and interwoven. The Krebs cycle doesn't act in isolation, but rather is part of a far larger network which provides it with the components it needs (enzymes, cofactors, oxidatants/reductants etc.) and the regulation required to ensure it works correctly.

The figure below demonstrates a more elaborated Krebs cycle, complete with all of the relevant interactions with various other pathways:[1]

Given their complexity, the discovery of the true nature of a pathway takes a long time, and uncertainty often remains due to the difficulty in directly observing the reactions taking place (though with modern genomics/sequencing its becoming easier to see what might be taking place).

In the case of the Krebs cycle, the key discovery was made by Gyorgyi who won the 1937 Nobel Prize in Physiology for 'his discoveries in connection with the biological combustion processes, with special reference to vitamin C and the catalysis of fumaric acid'.

Using isolated cells from pigeon breast tissue, Gyorgyi was able to identify the importance of succinic acid and fumaric acid in the pathway, noting that the rate of respiration was increased significantly if the tissue was incubated in their presence (as measured by gas evolution, which was already known at the time to be a consequence of respiration).

Building on this discovery, Krebs published the first full account of cellular respiration in his 1937 paper Metabolism of Ketonic acids in animal tissues.[2] In this paper he outlines all of the key steps needed, but even at this stage many of the important details were missing (many of the other necessary pathways weren't known, so Krebs had no idea where the succinate was coming from in the body, for example).

[1]: Gerhard, M.; Schomburg, D. Biochemical Pathways- An Atlas of Biochemistry and Molecular Biology; Wiley: New Jersey, 2012

[2]: Krebs, H. A.; Johnson, W. A. Biochem. J. 1937, 31, 645


Why is pyrimidine less basic than pyridine?

It is not the number of lone pairs that in any way explains basicity. Take a random sugar and it will have ten times the number of lone pairs (albeit on oxygen, not on nitrogen) without being significantly basic.

The problem is electronics. Nitrogen, being a rather electronegative atom, is able to draw the π electrons towards it well — to the extent that pyridine derivatives are termed extremely electron poor and thus very slow in electrophilic substitution reactions.

Going from pyridine to pyrimidine we need to double that. Thus, there is a perceived severe shortage of electrons in that ring since both nitrogens are trying to draw them towards themselves.

Now assume you protonate that species. We now have two nitrogens of which one is positively charged. Immediately, that positively charged nitrogen multiplies its electron-withdrawing force. Equally immediately, the other nitrogen feels electron density leaving its vicinity. The whole system is much less favourable than the pyridine system with only a single nitrogen.

Rephrasing that argument: Basicity is increased when electron-donating neighbours increase the electron density (that equals a partial negative charge) on an atom. In pyrimidine’s case, we have an electron-withdrawing neighbour that reduces the electron density (giving a partial positive charge) and making protonation (i.e. more positive charge) less favourable.


Does MOPAC allows calculation of Standard Enthalpy of reaction in this way?

The free energy change of a reaction occurring in solution can be computed using a thermodynamic cycle:
$$\Delta G^\circ_{r,soln} = \Delta G^\circ_{r,gas} + \Delta\Delta G^\circ_{r,solv} $$
$\Delta G^\circ_{r,soln}$ can also be written as
$$\Delta G^\circ_{r,soln} = G^\circ_{soln}(B)- G^\circ_{soln}(A)$$
where
$$G^\circ_{soln}(X) = G^\circ_{gas}(X) + \Delta G^\circ_{solv} $$
and
$$G^\circ_{gas}(X) = E_{gas}(X) + G^\circ_{gas,RRHO}(X)$$
The solvation energy is usually computed using a continuum as the difference in electronic energy with and without the continuum.
$\Delta G^\circ_{solv}(X) = E^\prime_{soln}(X)-E^\prime_{gas}(X)$
The parameters in the continuum model are usually optimized for a particular level of theory, which may not be appropriate for the computation of $\Delta G^\circ_{r,gas}$, so $E_{gas}(X)$ is not necessarily the same as $E^\prime_{gas}(X)$

However, if $E_{gas}(X) = E^\prime_{gas}(X)$ then
$$G^\circ_{soln}(X) = E_{soln}(X) + G^\circ_{gas,RRHO}(X) $$

The parameters in the continuum model are usually determined using the gas phase geometry $(X)$.  But if $E_{gas}(X) = E^\prime_{gas}(X)$ then many people often choose to use the geometry optimized in solution
$G^\circ_{soln}(X_{soln}) = E_{soln}(X_{soln}) + G^\circ_{soln,RRHO}(X_{soln})$.

For semiempirical calculations, the electronic energy corresponds to the enthalpy of formation at 298K, so if one simply makes the substitution
$$E(X) \rightarrow \Delta H^\circ_{f,298}(X)$$
then (at 298K)
$$G^\circ_{soln}(X) = \Delta H^\circ_{f,298}(X) -TS^\circ_{gas,RRHO}(X)$$ $G^\circ_{soln}(X)$ doesn't correspond to a measurable free energy because $S^\circ_{gas,RRHO} \ne S^\circ_{f}$ but it can still be used to compute $\Delta G^\circ_{r,soln}$ since the missing terms cancel.

For temperatures other than 298K
$$G^\circ_{soln}(X) = \Delta H^\circ_{f,298}(X) + (H^\circ_{gas,RRHO}(X,T)-H^\circ_{gas,RRHO}(X,T=298K)) -TS^\circ_{gas,RRHO}(X)$$


For this experiment, you'll be seeing an interesting color change when Iodine is oxidized at the anode to form the brownish $\ce{I2}$ on the electrode and yellow $\ce{I3-}$ in solution.

The reduction of $\ce{H+}$ to form $\ce{H2~(g)}$ and $\ce{OH-}$ will turn the indicator pink at the cathode.

The cool thing about your experiment is that you will not see any redox reaction at all (at either electrode) until iodide ions diffuse to the anode! That is, unless you use a high voltage battery, $\gtrapprox2.0~\mathrm{V}$, you will not have sufficient energy to oxidize $\ce{SO4^2-}$ to $\ce{S2O8^2-}$.

This makes the experiment an exercise in patience, and then sudden color display! Measuring the time it takes can yield data that can help you calculate the diffusion coefficient of $\ce{I-}$ if you wanted to take it another step!

In summary: Soaking the filter paper with KI would result in the electrodes producing color changes immediately. I suspect the sulfate was used in this experiment to illustrate the diffusion of dissolved ions, because the redox reaction would not take place until the iodide ions reached the anode.


How to form an acetal from an enol ether?

Steps (Source: from comment made by Zhe)

Tosic acid protonates alkene to form oxocarbenium through resonance.
One of the alcohols from ethylene glycol attacks the activated carbonyl and toluenesulfonate deprotonates the oxygen.
Methoxy group is protonated by the regenerated torsic acid.
Methanol is eliminated via SN2 with the other hydroxyl group of the ethylene glycol.
Tosic acid is regenerated by deprotonating the 1,3-dioxolane group.


Calculating charge of amino acids (am I doing it right?)

I second Lighthart's "Looks good to me"

  Let's break out the Henderson-Hasselbalch equation to be sure.

The Henderson-Hasselbalch equation is usually used to determine the pH of a buffer, but we can work backwards to get the ratio of an acid to its conjugate base at any pH.

$$\ce{pH}=\ce{p}K_a +\log{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}$$
$$\log{\left(\dfrac{[\ce{A-}]}{[\ce{HA}]}\right)}=\ce{pH}-\ce{p}K_a$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}=e^{\ce{pH}-\ce{p}K_a}$$

Let's take the three possible cases:

  $\ce{pH}>\ce{p}K_a$

If the pH is greater than the pKa, then

$$\ce{pH}-\ce{p}K_a>0$$

$$e^{\ce{pH}-\ce{p}K_a}>1$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}>1$$
$$[\ce{A-}]>[\ce{HA}]$$

For OH and SH functional groups in proteins, when $\ce{pH}>\ce{p}K_a$, then the group is deprotonated and has a negative charge.

For amino groups, $\ce{HA}$ is $\ce{RNH3+}$, so for amino groups at $\ce{pH}>\ce{p}K_a$, they are not protonated and neutral.

  $\ce{pH}<\ce{p}K_a$

If the pH is less than the pKa, then

$$\ce{pH}-\ce{p}K_a<0$$

$$e^{\ce{pH}-\ce{p}K_a}<1$$
$$\dfrac{[\ce{A-}]}{[\ce{HA}]}<1$$
$$[\ce{A-}]<[\ce{HA}]$$

For OH and SH functional groups in proteins, when $\ce{pH}<\ce{p}K_a$, then the group is protonated and neutral.

For amino groups at $\ce{pH}<\ce{p}K_a$, they are protonated and positively charged.

  $\ce{pH}=\ce{p}K_a$

In this case you will find that:
$$[\ce{A-}]=[\ce{HA}]$$

Half of the groups are protonated and the other half are not.


Does bleach turn a substance white or colorless?

First, let's think about what makes a substance coloured: a substance will appear to be coloured if it absorbs light in the visible spectrum. Beta carotene:

absorbs blue/green light (400–500 nm), so it looks red-orange when white light is shined on it as the blue/green light is subtracted from the white and what remains is what reaches our eyes. The details aren't dreadfully important, but you'll notice that the molecule has a lot of conjugated double bonds. This is what allows the molecule to absorb visible light and this conjugation is what produces the colour in many things that might stain clothing or the like. The size of this conjugated system affects how it absorbs light. In general, the more conjugation, the more absorption occurs and at longer wavelengths (towards red).

Substances used as bleaches tend to be oxidizing agents. Again, the details aren't too important, but one thing oxidation can do to these conjugated molecules is break them apart or convert double bonds to other functionalities. By disrupting the conjugated system, one changes the absorption properties of the molecule. Breaking the molecule apart or removing double bonds shrinks the extent of conjugation, reducing absorption and shifting it to shorter wavelengths. If the wavelength is short enough, the absorption is the in the UV range, which is invisible to the human eye, making it colourless.

Now, to your questions:

  From what I understand, bleach works to weaken the ability of a substance to absorb light.

This is as described above.

  Thus this seems to be "whitening" i.e. reflecting all light.

Reflective/scattering properties are distinct from absorption. e.g. if I add bleach to an aqueous solution of a dye, it may go from a coloured clear solution to a colourless clear solution, but if I bleach a shirt stained with a dye, the shirt doesn't become transparent—the reflective properties are determined by the fabric, not the dye. Bleach simply changes the structure and thus, absorption of the dye, it doesn't render them reflective.

  So is 'colorless' a misnomer when used to describe bleach?

Colourless is the correct term to describe the effects of bleach—it makes coloured substances colourless.

  Also, to double check when you bleach something, is it still there? Only that it is now white??

The coloured molecules undergo a chemical change so they aren't the same substances with which you started. Whether the products of the reaction are still there depends on what they are. It may very well be that the oxidation products of a dye are more soluble and easier to remove when you wash a piece of clothing. The bleaching itself implies a chemical reaction that alters the absorption of coloured substances, not their removal.


Calculating time taken for a set percentage of a reaction to occur based on half-life

Ryan showed that the rate constant is given by $k=\frac{ln(2)}{t_{1/2}}$.  So the concentration at time t is given by
$$C=C_0\exp(-kt)=C_0e^{-ln(2)\frac{t}{t_{1/2}}}$$But,
$e^{-ln(2)}=\frac{1}{2}$.  Therefore, $$C=C_0\left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}$$Substituting $n=\frac{t}{t_{1/2}}$ gives
$$C=C_0\left(\frac{1}{2}\right)^n$$


The maximum no. of isomers (including stereoisomers) that are possible on mono-chlorination of the compound: 3,3-dimethylpentane is?


Judging from the color, this clearly looks like more or less hydrated $\ce{Fe(OH)3}$ or $\ce{FeO(OH)}$, and it could also contain dissolved $\ce{FeCl3}$ formed by the reaction of iron with $\ce{Cl2}$ evolved at the anode. The fact that it smells like chlorine when you heat the gelatinous precipitate would support the latter assertion, but the smell could also arise from residual $\ce{Cl2}$ which was still dissolved in the water of the jelly. Heating $\ce{Fe(OH)3}$ or $\ce{FeO(OH)}$ intensively above 200 °C will give you $\ce{Fe2O3}$ by thermal decomposition (dehydration).

\begin{aligned}
\ce{2Fe(OH)3 ->& Fe2O3 + 3H2O} \\
\ce{2FeO(OH) ->& Fe2O3 + H2O} \\
\end{aligned}

During electrolysis, the zinc layer on the electrode is degraded by pitting corrosion of $\ce{Cl-}$ and oxidation in the presence of oxygen. When the coating is completely removed by the corrosion in some places, the exposed iron reacts with oxygen and chlorine to the products mentioned above. On anodes made of pure zinc, $\ce{Zn(OH)2}$ will likely be the primary product, which will form a transparent to white, jelly-like precipitate. Zinc hydroxide formed by oxidative degradation of the $\ce{Zn}$ layer would also be a constituent of the precipitate on a zinc-coated iron anode.


Computational Efficiency of Periodic Boundary Conditions

The application of periodic boundary condition does not mean that the system is explicitly replicated. You only deal with the particles within the unit cell (the box that you replicate), but you let them interact with the replicated ones outside of this unit cell.

Consider a simple 1-dimensional case with three atoms in the unit cell: $\ce{A-B-C}$, and lets assume we only need to treat nearest-neighbor-interactions. So the "monomer" has only 2 interactions: AB and BC.

One option to simulate the macroscopic system would be to explicitly deal with $\ce{A-B-C-A-B-C-A-B-C-A-B-C}$ (or even longer), but this requires a lot of atoms with accordingly many interactions between them. And even then both ends of the chain would be missing one neighbor.

Using periodic boundary conditions, one explicitly treats $\ce{A-B-C}$ only. But this time we will allow $\ce{C}$ to interact with $\ce{A}$, since the periodic boundary conditions implicitly tell us, there is another $\ce{C}$ to the left. Similar, there is an implicit $\ce{A}$ to the right of $\ce{C}$. Compared to the monomer from above, we have again 3 atoms, but now there are 3 interactions to be considered: AB, BC and AC.


Based on what you have in the question, you are good up until formation of the carbocation by loss of a water leaving group. It looks like you are missing one of two things:

The carbocation is resonance stabilized.
One of the carbon-carbon double bonds in the product is new.

The scheme below has two of the major resonance contributors to that carbocation. Perhaps one of give you some insight on how to show the formation of the second alkene.


Is there anything in the literature that supports revolutionary claims for this "new" sulfuric-acid battery?

There is nothing in the chemistry of what they describe that is different in any way to a conventional Lead-Acid battery. What they really seem to claim is improvement in materials in the battery not to change the electrochemistry but to improve other aspects of battery operation. For example:

MEG's E-SOURCE™ power cell technology is based on a lead-fiberglass composite material that is one-third the weight and 50 times the strength of metal alloys used in conventional batteries. To date, more than $100 million has been invested in research, development, manufacturing, and market testing of this technology in the United States and several international markets.

This suggests they have built a better lead-acid battery rather than something more radical. This is not ridiculous. They claim the following advantages:

Extraordinary manufacturing and user cost savings
Significant weight reduction compared to conventional lead-acid    batteries
Exceptionally long service life
Dramatically fast recharge speeds at all operating temperatures
Safe, simple, and low-cost maintenance

It is not impossible to imagine clever overall design and improved microstructural components and electrode materials could deliver these sorts of improvements and it is certainly far easier than inventing a whole new battery electrochemistry. For example, charging speed is often dominated by the physical material properties of the electrodes and not the underlying electrode chemistry, so clever design can mitigate the physical limits on charging speed such as by increasing the effective surface area.


Why do larger alkyl groups show a greater inductive (+I) effect?

Qualitatively:

If a methyl group is more electron donating than a hydrogen, then an ethyl group would be more donating that a methyl group by the same reason.

The effect drops off with distance, and generally a pentyl group is considered no more donating than a butyl group.

As for why a methyl group is more electron donating than a hydrogen, the carbon has a higher electronegativity than hydrogen (slightly), polarizing the bond (slightly).  This increased electron density near the carbon, which increases (slightly) its electron donation (I+) ability.


The entire process has two parts: Excitation and relaxation. The first is clear enough, we excite an electron into a higher energy level.

Now this electron wants to fall back down. There is no roadsign or intrinsic selection rule that requires the electron to immediately go back to the ground state. Any state in a lower shell will be less energetic and by chance electron 1 happens to stop in state A while electron 2 of a different atoms stops in state B. Any electron that is not in a ground state will remain there — probably for a duration of about a few femtoseconds or less — and then relax further. The universe is not deterministic so nothing is telling the electron it has to take the fastest path.

Note that this gedankenexperiment does not make any prediction which path is most likely. You would need more theory (that DavePhD is likely to give) to explain that.


Suppose you have $n_1$ mole of A and $n_2$ mole of B. If they are at temp. $T_1$ and you want to raise the temp. to $T_2$ then the amount of heat you need to add is
$$\Delta H~= n_1 \times c_{pA} \times \Delta T + n_2 \times c_{pB} \times \Delta T$$
$$\Delta H~= (n_1 \times c_{pA} + n_2 \times c_{pB}) \times \Delta T$$
$$\Delta H~= (n_1 +n_2)\times \frac{(n_1 \times c_{pA} + n_2 \times c_{pB})}{(n_1+n_2)} \times \Delta T$$
$$\Delta H~= n_{total} \times(y_1 \times c_{pA} + y_2 \times c_{pB}) \times \Delta T$$
$$\Delta H~= n_{total} \times c_{p,avg} \times \Delta T$$
So basically you are not losing any information in case of ideal situation where molecular interaction of two different species has no effect on each other's heat capacity. But from statistical mechanics we know,
$$C_v=\frac { <E^2>-<E>^2}{k_B \times T^2}$$
Now if we solve the schrodinger equation for individual species, we must include an extra potential energy term to account the interaction from other species, which in turn will result in fluctuation of $C_v$ from it's previous value.

Slightly off topic: When I see the question first things that came to my mind is kopp's rule. It's used to find heat capacity of molecule using atom's heat capacity or to find alloy's heat capacity using the constituents heat capacity.


Find moment of inertia around two axes of sulphur dioxide

Without explaining some more its hard to help, but the most likely error is in using the wrong units. The rotational constant is usually given in wavenumber units (cm$^{-1}$), e.g. 2.02736 cm$^{-1}$. In a linear molecule the equilibrium value is

$B_e = \frac{h}{(8\pi^2cI)}$

with $B_e$ in wavenumbers and c is the speed of light in cm/sec. In your case since the SO$_2$ is not linear (its planar bent at the S atom, C$_{2v}$ point group) there are three rotational constants usually called A$_0$, B$_0$, C$_0$.  (You can check the units by converting Joules in to base units, m, kg, s etc . The moment of inertia is in kg m$^2$ and $10^{-46}$ is of the correct order of magnitude. Seven decimal places seems too many as the rotational constants are not usually known to more than 5 places.)


What is it meant by two different groups in stereoisomerism?

Stereoisomers are different due to the spatial arrangement of their atoms.

The picture that you gave has two $\ce{CH3}$ groups on one side. That is called cis - as they are on the same side. There is another isomer, called trans. That is when one of the $\ce{CH3}$ is on the top and the other $\ce{CH3}$ is at the bottom. The arrangements of the $\ce{CH3}$ and $\ce{H}$ is different in cis and trans.

This link explains it well: Chem.Libretexts


Why don't all electrodes produce the same results in the electrolysis of potassium iodide?

To distill down your observations: you always saw pink at the cathode, but only some materials produced a brown color when used as the anode. This means that the anodic materials that didn't produce brown didn't oxidize the iodide to iodine.

So, the question is what was happening at the anode when the iodide was not being oxidized. The answer is simply that the anode material itself was easier to oxidize than the iodide in those cases, so the anode itself was oxidized and the iodide was not, so there was no brown color produced.


Storing oxygen generated via water electrolysis

I think the simplest approach is to fill a test-tube with your electrolyte solution, seal the open end with your latex-glove covered thumb, then immerse the sealed end in your cell. If you can tilt your anode several degrees from the perpendicular, you should get a stream of oxygen bubbles that you can then catch with the test tube. You certainly won't get them all but it shouldn't be a problem to get 3/4 of them anyway. If you have trouble inverting the test tube without having a little air left, just slide some flexible tubing to the top and suction out the air. Take care to watch the level of your electrolyte as the solution that is displaced from your test tube will of course be filling your cell.

After you have collected all of the oxygen you need, seal the open end of the test tube with a piece cut from a latex glove and a rubber band. You will then have a test tube filled with oxygen gas and any residual electrolyte solution.


Why does (atmospheric) pressure have an effect on solubility?

It comes from a combination of Henry's law and Le Châtelier's principle. Henry's law is
$$p=k_hc$$
where $p$ is the partial pressure of the gas, $k_h$ is the Henry's law constant and $c$ is the concentration of the gas (because liquids and solids are not affected as much as gases, if at all). This means that with a greater pressure comes a greater concentration, and with a lower pressure comes a lesser concentration, and vice versa

Le Châtelier's principle states that if the pressure, concentration, temperature or volume of a system changes, something else will change to compensate and swing the whole thing back into equilibrium. If there is a greater pressure, there will therefore be a grater concentration. A greater concentration means there is a greater solubility.

The UC Davis Chem Wiki (I didn't see that DrMoishe Pippik had linked to it until after I posted this!) is rather helpful - much more helpful than Wikipedia, in this case.