How do I setup the electrodes for cyclic voltammetry on a capacitor made on dry copper sheets?

The reason reference electrodes are needed in 3 electrode solution cells is that you're trying to control the interfacial potential of the working electrode (basically from the surface to a point very near the electrode in solution). The wiki's kinda lousy, but the diagram explains essentially what the potentiostat is doing: It's controlling the counter electrode (the op-amp output), to set the potential from the working electrode (ground) to a point just into the solution and the reference electrode is used to read the potential and provide feedback to the amplifier to correct the output. This way, any extra solution resistance between the counter and reference electrodes doesn't reduce the interfacial potential (known as iR drop in electrochemical circles) because you're measuring the potential near the surface. In other words, we're interested in applying a potential across $Z_2$ precisely, but we don't care about what's happening with $Z_1$.

Now, since you have a capacitor, what you're trying to do is control the potential of one plate versus the other plate, so you want the point where you're getting feedback to be one plate, and the other plate is ground—one plate is the "reference electrode" and the other is the "working electrode". Your capacitor is $Z2$ and there's no solution, so there's no $Z1$. (In some cases, it still makes sense to use a separate cable for the reference input, but it's probably fine to just short them together at the inputs) Many potentiostats will work with the counter electrode and reference electrode inputs shorted together (which is how one typically connects a 2 electrode cell), but a resistor between them can help prevent it from oscillating by increasing the time constant of the system.

You're not really using a reference electrode in this case because you don't care about measuring the applied potential versus some known potential, just in controlling the potential across the capacitor. In a cell in which we're interested in studying redox reactions and such, it's important to know how the potentials relate to one another and to compensate for the various potentials that arise when electrodes are dipped in electrolyte solutions.

If you're really interested in the details of potentiostats, Bank has a good primer with lots of details.


Using molecular sieves comes to my mind, they're easy to use and efficient. You can find some practical information about how this is done for example here http://www.klmtechgroup.com/PDF/Articles/articles/Molecular-Sieve-Applications.pdf


Chemical water treatment: Acid Base reaction

The question is long and detailed, and deserves a long answer, but I can only hit on the highlights, and urge others with some expertise in this area (which I lack) to weigh in.

First, the science of water treatment is mature and has addressed many of the issues raised here, primarily through biological and chemical processes which are very good at breaking down and dissolving most of the different sized particles you refer to.

As far as the composition of sewage is concerned it contains a lot more water than human waste, and a lot more normal waste (urine and fecal matter) than vomit, which, since most people are sick only a tiny fraction of their lives, and nauseous an even smaller fraction, likely is a vanishingly small contributor to the content of sewage. The composition of sewage at the treatment plant intake varies much less, and more slowly, than you might think, since it is averaging the inputs from people all over a city or region. Still, it may vary seasonally due to changes in temperature or water usage, so the treatment must vary accordingly, as you have surmised.

Finally, you concern about neutralizing every last iota of HCl is misplaced. Adjustment of pH is commonly used in sewage treatment and is done using automated equipment and pH sensors that can withstand extended exposure to sewage and to pH fluctuations. The glass membrane electrode is a workhorse that is stable to extremes of pH well beyond what would occur in sewage.

But I conclude my answer by saying that some of your ideas are very interesting, and there may be many applications of nanotechnology in water treatment along the lines you mentioned, that could have practical benefits for peoples' lives. I urge you to continue your studies of wastewater chemistry with your full vigor and an open mind.


Are there any organic compounds that do NOT contain hydrogen ? And if so, what are they?

Distinguishing organic and inorganic substances is often very formal. We say CX4 are organic compounds, we should also say carbon does not form carbon halogenides, which are inorganic compounds.

One of more known halogen-hydrogen-free examples is carbon suboxide $\ce{C3O2}$,  $\ce{O=C=C=C=O}$, that is somehere on inorganic-organic border, being often considered as the second anhydride of malonic acide $\ce{HOOC-CH2-COOH}$, as it is prepared from it by $\ce{P4O10}$ aggressive dehydratation.

But the true malonic anhydride has 4-atom cycle.

One of less known examples is the anhydride of the mellitic acid, which could be formally considered as a carbon oxide with the summary formula $\ce{C12O9}$

  ... [mellitic anhydride] is one of the only four [oxides] that are reasonably stable under standard conditions.  It is a white sublimable solid, apparently obtained by Justus Liebig and Friedrich Wöhler in 1830 in their study of Mellite

As an interesting side info:

  Mellite, also called honeystone, is an unusual mineral being also an organic chemical.   Chemically identified as an aluminium salt of mellitic acid; that is, aluminium benzene hexacarboxylate hydrate, with the chemical formula $\ce{Al2C6(COO)6 \cdot 16H2O}$


What you get is a polymeric (starch) composite, with salt as inorganic filler.

The salt particles give toughness to the resulting composite, while the starch phase, which is just the same as in your bread, minus the unnecessary blowing agent (baking powder, yeast, whatever), connects the particles in an elastic, flexible way. You get a material that is nearly as hard as a piece of rock salt, but much less brittle.

(Elastic here means more elastic than salt. Dry bread is quite flexible, compared to rock salt ;-) .)

See also https://en.wikipedia.org/wiki/Composite_material


Oxidation of sulfur dioxide to trioxide and its decomposition under same condition at higher temperature

There is an equilibrium between $\ce{SO3}$ and $\ce{SO2 + 1/2 O2}$, at least at high temperatures and with an appropriate catalyst.  We could write the equilibrium as
$$\ce{SO2(g) + 1/2 O2(g) <=> SO3(g)}$$
Just from writing this reaction this way we can glean several things:

The reaction combines 1.5 gas molecules into one gas molecule, a net reduction in the number of molecules.  Since fewer molecules (but the same number of atoms) means there are fewer translational degrees of freedom, the entropy will decrease the more $\ce{SO3}$ "product" is formed.  Higher temperatures will thus disfavor the right-hand $\ce{SO3}$ side of the equilibrium and favor the left-hand $\ce{SO2 + 1/2 O2}$ side.

Total pressure will also affect the equilibrium but in the opposite way.  Higher pressure (crudely speaking) is a force that squeezes things together, so at high pressures the right-hand side will be favored.

The other principle factor that influences the equilibrium would be the partial pressure of $\ce{SO2}$, of $\ce{O2}$, and of $\ce{SO3}$.  So for example  if you had 99.9 mol % oxygen gas and only a trace of $\ce{SO2}$, that would favor the conversion of nearly all sulfur to the LH sulfur trioxide side.

The exact balance will depend on the precise enthalpy change of the reaction, the partial pressure of each species, the temperature, and the total pressure.  You could modify the system to favor one side or the other, and could model this accurately and quantitatively with appropriate numerical data.


Do the spins of outer orbitals block measurement of spin of inner orbitals?

With regards to your first question, as you get to higher energy levels, the distance between each one increases and as a consequence the shells get closer together and begin to overlap. This is also seen with 5s and 4d iirc.


Why are enantiomeric excess % and specific rotation considered?

Enantiomeric excess is very important concept in organic chemistry because it clearly states the purity of one isomer over the other. For example, have a look at the structure of Thalidomide which is racemic (50:50 —R:S isomers) in nature. Between the two form, one is having very important medical property where as another form is sedative. In these cases, synthesizing enantiomerically pure (>99%) form is mandatory and in January 1996, the FDA announced it would consider further incentives for developing single isomer drugs, owing to their better pharmacokinetics prosperity, safety, and tolerability.
http://en.wikipedia.org/wiki/Thalidomide

These are the reason which makes chemists to target enantimerically pure form using latest available technologies like, Chiral GC or HPLC.

Specific optical rotation is very old concept to judge the enatiopurity of the compounds but its not accurate because it depends on various factors like concentration of the sample, solvent which we use, temperature, etc. Even small impurity in the sample gives drastic change in the rotation value. Thats why chemists adapt latest technique like HPLC of GC, nevertheless, chemists still carry out specific optical rotation to make sure that the enantiomerically excess compound has some rotation.


Why is F2 (fluorine) colourful despite being diamagnetic?

The colour of a molecule does not depend on the whether the molecule is diamagnetic or paramagnetic. The reason is that the colour is caused by absorption of a photon in the visible part of the spectrum. The photon has enough energy to excite one of the two (spin paired in a diamagnetic molecule)) electrons  in the highest occupied molecular orbital (HOMO) and into a higher energy unoccupied orbital, usually the lowest unoccupied molecular orbital (LUMO). From spin conservation rules the electron should not change spin. A spin change would correspond to a 'forbidden process' which has a very low chance of occurring so leads to exceptionally weak absorption.

The molecule now has one electron in the lower orbital and one in an upper orbital, these form the excited state, and if the spins are paired it is called a singlet state and the spin multiplicity is 1.

(multiplicity is 2*(total spin) + 1:  if the spins are paired the total spin quantum numbers are +1/2+(-1/2)=0. If the spins are unpaired the total spin is 1/2+1/2 and the multiplicity is 3, a triplet state).

In oxygen, which has a triplet ground state, on absorbing a photon an electron is still excited into an upper unoccupied orbital, the spin multiplicity of the excited state produced depends on the relative spins of all the unpaired electrons just as in the case of fluorine.

(As it happens fluorine seems to be a complicated  example as it has a low dissociation energy so absorption of a photon puts the molecule into a repulsive excited state leading to two F radicals as the molecule dissociates. It also absorbs weakly at longer wavelengths than needed to dissociate and this may be due to  a forbidden, spin changing, transition as mentioned above).


The food that you eat regularly is something that you are used to. You know exactly which smell combines with which taste ... but that does not imply that the chemicals you smell are the same that you taste.

If you had been eating your whole life long fragrant soap with apple smell, you would wonder about the taste of real apples as you wonder right now about the taste of apple soaps.

There are a lot of different substances that are responsible for the special taste of everything you eat. You can find more about flavor or aroma compounds on wikipedia.

In contrast to real fruits, fragrant soaps and other industrial products with a certain smell only contain those substances that are responsible for the characteristic smell of e.g. fruits (often only the main one or two aroma substances or even only synthetic aroma substances that smell similar to the natural products).

But as not all products that shall smell like certain thing also need to taste like those things you only smell the "apple" but you won't taste it.


What is the color and structure of CrO3?

According to its Wikipedia page:

Color

$\ce{CrO3}$ is a dark-purple solid under anhydrous conditions, that turns bright orange when wet.

Production

$\ce{CrO3}$ is generated, as you said, by the following reaction:

$$\ce{H2SO4 + Na2Cr2O7 -> 2 CrO3 + Na2SO4 + H2O}$$

Structure

Its monomeric structure is predicted to be pyramidal like so:

Its crystalline structure is:

Each chromium center shares two oxygen centers with neighbors, but two oxygen atoms are not shared, giving the overall stoichiometric ratio of 1:3.


Why is benzoic acid a stronger acid than 3-aminobenzoic acid?

The $\mathrm{p}K_\mathrm{a}$ of benzoic acid is $4.2$. The $\mathrm{p}K_\mathrm{a}$ of aniline (aniline's conjugate acid really) is $4.8$.

The two macroscopic $\mathrm{p}K_\mathrm{a}$s of 3-aminobenzoic acid (aniline-3-carboxylic acid) are $3.1$ and $4.8$.

When the microscopic $\mathrm{p}K_\mathrm{a}$s of two groups in a molecule are similar, the macroscopic $\mathrm{p}K_\mathrm{a}$s are not reflective of one particular group protonating/deprotonating. The neutral form will include both a zwitterionic and an non-zwitterionic form in the case of 3-aminobenzoic acid.

When the groups are close in space within the molecule, there will be electrostatic interaction. If the amino group is protonated, it facilitates deprotonation of the carboxylic acid group. If the carboxylic acid group is deprotonated, it facilitates protonation of the amino group.

Addition: There is a very relavent article: Carbon-13 NMR determination of acid-base tautomerization equilibriums J. Org. Chem., 1984, 49 (4), pp 691–696.  It reports that the neutral species is 59% zwitterion and 41% molecular (non-zwitterion).


What is the difference between fire and warm body?

You're really asking about the difference between a physical change and a chemical change.*
In a physical change, the chemical identity of the substance doesn't change.  In a chemical change, it does.
Suppose you take cube of sucrose (table sugar) at room temperature.  If you heat it up, it will become warmer.  But it will still be, chemically, sucrose.  It will just be sucrose at a higher temperature.  Since the chemical identity did not change, it is a physical change:
$$\ce{C12H22O11(s, room\,T)  -> C12H22O11(s, warm\,T)}$$
[$\ce{C12H22O11}$ is the chemical formula for sucrose; "s" means solid, "l" means liquid, "g" means gas.]
If you warm it more (to $170 \,^\circ \text{C}$) it will melt.  Melting is a physical change, because after the cube melts, it still consists of sucrose molecules; though these will now be in the liquid state:
$$\ce{C12H22O11(s)  -> C12H22O11(l)}$$
But suppose you instead flow enough energy into it to break the bonds that hold the sucrose together, causing it to combust, in the present of oxygen, into water and carbon dioxide.  That is what happens when it burns.  This is a chemical change because, as it is happening, the sucrose and oxygen are being changed into new chemicals: water and carbon dioxide:
$$\ce{C12H22O11(s) + 12 O2(g) -> 12CO2(g) + 11 H2O(l)}$$
This chemical change releases energy, which (in this case) takes the form of heat and light (which is why it feels hot, and you can see a flame).
In sum, a warming body is undergoing a physical change.  A burning body is undergoing a chemical change.
[*Yes, some might take issue with this distinction, but it is used in all first-year chemistry textbooks I've seen, indicating there is a consensus among educators in this field that this is a useful distinction for beginning students to learn.]


How does one find the ground-state term symbol for a configuration that is exactly half-filled?

The full calculation is laid out below. Start by calculating the spin, orbital and total of all angular momentum. Each electron has spin quantum number $s = 1/2$ and magnetic quantum number  $m_s = \pm 1/2$. Orbitals have angular momentum of $s=0, p=1, d=2, f=3$ etc.

The total spin angular momentum is the series of values
$$S = |s_1 + s_2|\cdots|s_1 − s_2|$$
where $s_1 , s_2 , s_3 =+1/2$  are the quantum numbers of each electron. The total angular momentum is
$$ L=|l_1+l_2|\cdots|l_1−l_2|$$
and total angular momentum
$$J=|L+S|\cdots|L−S|$$
Each of these are called Clebsch-Gordon series. The way to calculate them is to calculate the maximum and minimum values and make intervening values separated by $1$.

The term symbol has the form $^{2S+1}L_J$. The super-prefix is the spin multiplicity, for spin angular momentum S this is $2S+1$ or in general $2X+1$ for angular momentum X.

When there are three 3 electrons the the spin and orbital angular momentum terms have to be added in two parts. First as a pair, with the equations above then again with each of the values in the series with the last electron or orbital.

For the spin the S values are
$S = |s_1 + s_2|\cdots|s_1 − s_2| = |1/2+1/2|\cdots|1/2-1/2| = 1, ~ 0 $ .
We call these values $S_1$ and $ S_0$
To calculate the total with the third electron gives
$$S_{s_1}=|S_1+1/2|\cdots|S_1 - 1/2| = 3/2 , ~ 1/2$$
and for the $S_0$ there is one value $S_0=1/2$. Thus for the spin the values are $3/2$ and $1/2$. This means that the multiplicity produces quartet and doublet states, thus so far we have $^4L_J$ and $^2L_J$.

The same method is followed for the orbital angular momentum. The first two p orbitals give
$$ L=|l_1+l_2|\cdots|l_1−l_2|= |1+1|\cdots|1-1| = 2,~1,~0 $$
then for each of these values called $L_1,L_2,L_3$ we combine with the last orbital angular momentum, for example
$$ L_{L_1}=|L_1+l_1|\cdots|L_1−l_2|= |2+1|\cdots|2-1| = 3,~2,~1 $$
and so forth for the other values. The L values are $3,2,1,0$

Putting all these values into a table gives
$$\begin{matrix} L & S &  J &  \text{term symbol}\\
3 & 3/2 & 9/2, 7/2, 5/2, 3/2 & ^4F_{9/2, 7/2, 5/2, 3/2} \\
2 & 3/2 & 7/2, 5/2, 3/2,1/2 & ^4D_{ 7/2, 5/2, 3/2,1/2}\\
1 & 3/2 & 5/2, 3/2,1/2 & ^4P_{5/2, 3/2,1/2}\\
0 & 3/2 & 3/2 & ^4S_{3/2}\\
3 & 1/2 & 7/2, 5/2 & ^2F_{7/2,5/2}\\
2 & 1/2 & 5/2, 3/2 &^2D_{5/2, 3/2}\\
1 & 1/2 & 3/2, 1/2 & ^2P_{3/2, 1/2}\\
0 & 1/2 & 1/2 & ^2S_{1/2}\\
\end{matrix}$$
Finally to find the lowest energy term use Hund’s rules, i.e. of the terms given by equivalent electrons, i.e. having the same L values, the one with the greatest multiplicity is lowest in energy and of these the lowest is that one with the greatest L value. The only terms remaining after considering Pauli exclusion are $^4S$ which is lowest in energy and next $^2D$ then $^2P$


A question about rotational symmetry of organic molecules and in particular, the cyclic Isomers of C3H4X2

If the website diagram didn't clarify the situation immediately, the issue is not to answer questions about 1 to 7, but to understand why the question is wrong.
My organic chemistry course had us playing around with molecular models - 3D representations of reality, one type with a metal framework representing the multivalent nucleus and plastic tubes representing bond lengths, the other type with wooden balls and springs. They are a far cry from reality - but molecular models give an added dimension of understanding.
Lacking such 3D models, and lacking an answer sheet, one possibility would be to go to 3D images on the web: they are actually in 2 dimensions, but with perspective in the drawing, so it could be sort of 2.5 D, like this:

Only after knowing what is represented by the dotted lines and tilted bond can you understand what you are drawing. The proposed differences between structures 1, 2 and 3 indicate an unfamiliarity with what the bond representations actually mean. One look at the picture above should clarify everything. Perhaps one more comment should be added: the bonds between a carbon and its two substituents are never coplanar with the 3-carbon ring. Interestingly, I did not find any other similar images (of dichlorocylopropane). Everything was of the straight line or dashed line/tilted bond type. This just assumes that the lines/dashes have an obvious meaning. Sometimes a good picture is worth a thousand words. (This answer is worth 1254 words, if you include 1000 words for the picture.)


when using glass to glass stoppers, is it best to have them dry or use lubricant?

Sometimes dry glass-to-glass sealing of the stopper leads to seizing of the stopper and can become very hard to remove. Therefore vacuum grease is often applied lightly. However, if your material in the flask can dissolve the vacuum grease, the material can become contaminated by the vacuum grease. If your material is volatile, you will need the vacuum grease to ensure an airtight seal. Another method you can use is to use teflon tape between the stopper and the ground glass joint to make a seal that will not seize. But, vacuum grease will give you the best seal.


Which of these carbocations are more stable?

As we know (to some extent), resonance in a molecule occurs due to sideways overlapping of p-π orbitals.

Now in the molecule B, there is high steric hinderance near the carbocation due to which the plane of the "carbon with carbocation" changes, so that it can feel minimum repulsion from the other groups. This forces the carbocation to remain in the mutually perpendicular plane, which inhibits the resonance.

Thus, this makes A (the less hindered carbocation) more stable than B.


What is the difference between the units 'torr' and 'mm of Hg'?

The two units torr, and mm of Hg were the same until they were redefined.

The torr was named after the Italian Evangelista Torricelli. 1 atmosphere is $101325\ \mathrm{Pa}$. The torr is defined as $1/760$ of an atmosphere. This is equal to $133.322\overline{368421052631578947}~\mathrm{Pa}$, which periodically infinitely repeats.

The mm of Hg was defined as the pressure generated by a column of mercury one millimeter high. Since the pressure of mercury depends on temperature and gravity, it is now redefined as

$$\mathrm{0.001\ mL \cdot 13595.1~mg \cdot 9.80665\ m/s^2 = 133.322387415 ~Pa}$$

This has caused a difference between the units of $0.000015~\%$, so unless you require extremely high precision, you can assume that $1\ \mathrm{torr} = 1\ \mathrm{mmHg}$


Diffusion and potential gradients for two similar atoms

This answer has been edited quite a bit. New additions are put in blockquotes.

To avoid any potential (pun not intended) confusion, let me make one thing clear before I start: Any mention of "potential" in this answer refers to an electric potential, not a chemical potential. The definition of electric potential can be found here on Wikipedia.

Let's start off simple and say that we have a beaker with two compartments. The two compartments are separated by a membrane that is ONLY permeable to $\ce{Na+}$ ions. In the left compartment, we put 1 M of $\ce{K+}$ ions and in the right compartment, we put 1 M of $\ce{Na+}$ ions (with appropriate counterions). If diffusion of $\ce{Na+}$ entirely ignored the presence of $\ce{K+}$ as you say, then we would expect that after a while, we would have 1 M $\ce{K+}$ and 0.5 M $\ce{Na+}$ on the LHS, and on the RHS we would have 0.5 M $\ce{Na+}$.
This does not happen. In actual fact, there would be a concentration gradient that favours net movement of $\ce{Na+}$ from RHS->LHS, and there would be a electric potential difference that favours net movement of $\ce{Na+}$ from LHS->RHS. We could model the concentration at any point in time with:

LHS - 1 M $\ce{K+}$ and $x$ M $\ce{Na+}$
RHS - $0.5 - x$ M $\ce{Na+}$

where $x$ is between 0 and 0.5.
When $x = 0$, there is no potential difference since both sides have the same amount of charge (overall both sides are electrically neutral). However, there is a steep concentration gradient, so $\ce{Na+}$ starts diffusing from RHS-> LHS, causing $x$ to increase. Now as $x$ increases, the concentration gradient becomes gentler and the potential difference starts to become larger. At one particular value of $x$, the concentration gradient will cancel out the potential difference and there will no longer be any net movement of $\ce{Na+}$ across the membrane. This is the point where we could say that diffusion has stopped.
Now, let me change the scenario and say that the membrane is permeable to BOTH $\ce{K+}$ and $\ce{Na+}$ ions. In that case, clearly the equilibrium concentrations will be 0.5 M $\ce{K+}$ and 0.5 M $\ce{Na+}$ for both the LHS and RHS, because under these conditions there is neither a potential difference nor a concentration gradient.

It is important to note that the diffusion of $\ce{Na+}$ is NOT directly affected by the pre-existing concentration gradient of $\ce{K+}$. It is indirectly affected only because the concentration gradient of $\ce{K+}$ leads to a potential difference between the LHS and the RHS, which in turn affects the diffusion of $\ce{Na+}$. If we were to replace $\ce{K+}$ with an electrically neutral species, such as glucose, the concentration gradient of glucose would have absolutely zero effect on the diffusion of $\ce{Na+}$.

Now we can go on to explain the action potential in the neuron, which is not something I originally wanted to go into, but since the original question has been edited I feel that I have to address this now. Even though the original question now makes no reference to the action potential, it is describing exactly those events that occur during an action potential.
The inside and the outside of the neuron is separated by a cell membrane, which is selectively permeable to ions such as $\ce{Na+}$, $\ce{K+}$, $\ce{Ca^2+}$, and $\ce{Cl-}$. Of these four, sodium and potassium ions are the most important as they are directly involved in depolarisation and repolarisation when an action potential is generated.
The distribution of ions is not symmetrical across the membrane. When at rest, the extracellular concentration of $\ce{Na+}$ is much higher than the intracellular concentration of $\ce{Na+}$; the converse is true for $\ce{K+}$. Due to the membrane having different permeabilities to $\ce{Na+}$ and $\ce{K+}$ (for details of this, one can consult a neuroscience text), the resting membrane potential is -70 mV. Conventionally, the membrane potential is described as the intracellular potential with respect to the potential of the extracellular potential; this means that the inside of the cell is more negative.
Neurons react to certain stimuli by having $\ce{Na+}$ enter the cell, via various means which are not important to the current discussion. Since positively charged ions are entering the cell, this causes the membrane potential to rise. Once the membrane potential has risen to roughly -55 mV, voltage-gated $\ce{Na+}$ channels open, making the membrane permeable to $\ce{Na+}$.
At this stage, we have to consider the two factors that influence the diffusion of $\ce{Na+}$:

Since the potential difference is still negative, it favours influx of $\ce{Na+}$.
Since the extracellular concentration of $\ce{Na+}$ is larger than the intracellular concentration, the concentration gradient also favours influx of $\ce{Na+}$.

Hence more $\ce{Na+}$ enters the cell, leading to depolarisation of the membrane until the membrane potential reaches approximately +30 mV. At this stage, the voltage-gated $\ce{Na+}$ channels close and the voltage-gated $\ce{K+}$ channels open. Again, we need to consider the two factors that influence the diffusion of $\ce{K+}$:

The potential difference, which is now positive, favours efflux of $\ce{K+}$.
Since the intracellular concentration of $\ce{K+}$ is higher than the extracellular concentration, the concentration gradient also favours efflux of $\ce{K+}$.

Therefore $\ce{K+}$ leaves the cell, leading to repolarisation. There are further events such as hyperpolarisation and the refractory period, but they are not relevant to this discussion. After all this is said and done, the concentration gradients of both $\ce{Na+}$ and $\ce{K+}$ are not the same as they were at the very start of this discussion. They are restored by the $\ce{Na+}$/$\ce{K+}$ pump.
The most important thing to takeaway from this is that the diffusion of species X across a membrane is affected by two things: 1) the concentration gradient of X; 2) the potential difference across the membrane, if and only if X is a charged species. Other concentration gradients of other species, such as Y, Z etc. do not directly affect the diffusion of species X unless they contribute to the potential difference.
Prior to my edit you might have seen that I wrote that "diffusion does not ignore similarly charged species". This might seem to be a contradiction with what I say now, that the concentration gradient of Y does not affect the concentration gradient of X. The answer is that the concentration gradient of Y can only affect that of X when both X and Y are charged species, and this is only an indirect effect, in that the concentration gradient of Y affects the electric potential difference which in turn affects the concentration gradient of X. If Y was uncharged, its concentration gradient would not affect the potential difference; if X was uncharged, the potential difference would not affect its concentration gradient.
I think the marble analogy is inherently flawed in that there is no real analogous concept to a potential difference. You could modify it to have some kind of analogous concept, but in my opinion it wouldn't make sense any more.


Standardization of sodium thiosulfate using potassium dichromate

I think your procedure is incorrect. Typically you would mix $\ce{KIO3}$ (limiting reagent) with access of $\ce{KI}$. Then you add $\ce{HCl}$ to have a reaction:
$$\ce{KIO3 + 5KI + 6HCl -> 6KCl + 3I2 + 3H2O}$$
This is a way to generate a known amount of $\ce{I2}$.

You would then dissolve $\ce{Na2S2O3}$ in a buffer (sodium bicarbonate). This is done because $\ce{Na2S2O3}$ is not stable under acidic conditions. Then you can titrate $\ce{I2}$ in presence of starch with $\ce{Na2S2O3/NaHCO3}$. $\ce{NaI}$ and $\ce{Na2S4O6}$ are formed in this reaction.
In any case you cannot replace $\ce{HCl}$ with other acids. $\ce{H2SO4}$ and $\ce{HNO3}$ are oxidizers strong enough to turn some more $\ce{KI}$ into $\ce{I2}$ and mess up your measurement.


Detecting impurities in polyethylene glycol 3350

Extraction and GC-MS or maybe LC-MS, if you really need to measure these specific impurities
I don't know a lot about this area, but you are looking for very low amounts of impurities.  For ethylene oxide, for example, your desired detection limit of 0.07 mg in a 10 g sample corresponds to 7 parts per million, or 7 ppm.  Your limits for the glycols correspond to 100 and 400 ppm.
I'm not an NMR expert, but I think NMR will not be remotely  sensitive enough for this purpose.   Even if pure, NMR on a 0.07 mg sample is challenging and far from routine.
The best route forward might be a selective extraction of these impurities.  You should chose a solvent that does not dissolve polymeric PEG at all, or as little as possible.  This can be a challenge, as PEG is soluble in water, alcohols, and even dichloromethane.  Ethylene oxide is a relatively nonpolar gas, so you may be able to dissolve it in hexanes or a similar nonpolar solvent.  Unfortunately the glycols are unlikely to be soluble in hexanes.
So for ethylene oxide, say you extract 10 g of sample with X mL of hexanes, and say X is ~100 mL.  The resulting solution would be 700 ug/L of ethylene oxide at your desired detection limit.  That should be detectable by GC-MS.
For the glycols, supposing it is somehow possible to find a solvent that selectively dissolves the impurities but not the polymer, you could then derivatize the solution of glycols with a reagent like TBDMS and analyze them by GC-MS.  Alternatively, you may be able to analyze them without derivatization by LC-MS, but sensitivity of electrospray-based LC-MS for aliphatic alcohols is limited.  GC-MS is probably the way to go...if you can find a solvent for the selective extraction.
If no selective solvent is available, you might be able to analyze the glycols by directly derivatizing the solid PEG sample with a reagent like TBDMS in a liquid solution of some (non-selective) solvent.  The derivatization will convert the glycols into trialkylsilyl ether derivatives.  These may be lower polarity than PEG or more volatile than PEG such that they could be selectively removed and analyzed after derivatization.
Do you really need these specific impurities?
Maybe thermogravimetric analysis (TGA) would work for you.  There are variants of TGA where a mass spectrometer samples the gas phase above a material while it is slowly heated and decomposed.  If you have a sample of PEG you know is "pure enough", you could use that as a reference material, and then spike it with known amounts of contaminants to verify that these contaminants result in a perturbation to the TGA signal, or to the TGA-MS signal.
You may or may not get a precise numerical estimate of the mass fraction of your target impurities, but you will probably get a sense of whether a given 10 g sample of PEG is pure "enough" as measured by similarity to known reference material.
Analytical chemistry is hard
No matter what, you'll need to do a lot of methods development work to verify the appropriateness and robustness of your method.  E.g. what's the best value of X to use for selective extraction?  Can you for sure detect the impurities at your target levels with your chosen procedure?  You'll probably need to spike reference material with known amounts of impurities to make sure.  And is the response of your analysis method or instrument linear with respect to impurity mass fraction?  If so, over what range?
These are things that analytical chemists spend a great deal of time fretting about.  It's not easy.
Good luck!


Expectation value of an observable in quantum system

Roughly, the significance is that before measurement the system can be in a superposition of states, the wavefunction of which can be represented by a linear combination of the individual eigenfunctions

$\Psi = \sum_i c_i{\psi_i}$

However the very act of observing the system reduces the wavefunction to a single eigenstate due to interaction with the external world

For most cases, the expectation value is just the quantum mechanical way of calculating properties. Remember, the wavefunction is not observable! Only properties like density, dipole moment, etc are and the expectation value allows us a way to go from wavefunctions to properties via operators.

These other answers may also be helpful

https://physics.stackexchange.com/questions/67756/quantum-collapse
https://philosophy.stackexchange.com/questions/48425/is-collapse-of-the-quantum-state-conceptually-misnamed
https://physics.stackexchange.com/questions/151457/how-does-a-wavefunction-collapse


What are the ideal pair of (everyday) elements to pair in a thermocouple

My book has a few metals/alloys listed in a Seebeck Thermoelectric Series:

  Antimony, Nichrome, Iron, Zinc, Copper, Gold, Silver, Lead, Aluminum, Mercury, Platinum-Rhodium, Platinum, Nickel, Constantan, Bismuth

There's also the Seebeck formula for emf of a thermocouple:

$$E_{\text{AB}} = a_{\text{AB}}\theta + \frac{1}{2}b_{\text{AB}}{\theta}^2$$

Here, $a_{\text{AB}}$ and $b_{\text{AB}}$ are constants that depend on the pair of metals $\text{A}$ and $\text{B}$ chosen. $\theta$ is the temperature difference between the junctions of the thermocouple.

There's another defined quantity called thermoelectric Power, which is defined as $\frac{dE_{\text{AB}}}{d\theta}$.

The graph of $E_{\text{AB}}$ with respect to $\theta$ would be a parabola. If $a_{\text{AB}}$ and $b_{\text{AB}}$ are of opposite signs, then the parabola crosses the $x$-axis twice, and thus, the thermo-emf becomes zero for two values of $\theta$ and maximum for one value of $\theta$.

From the above thermoelectric series, a iron-nickel might seem like a good option to consider building a thermocouple with, considering they're quite far away in the series and are easily accessible. For this couple, we get the values of thermoelectric constants as $a_{\ce{Fe-Ni}}=\pu{2.5 \mu V/°C}$ and $b_{\ce{Fe-Ni}}≈0$.Thus, you get:

$$E_{\ce{Fe-Ni}}=2.5\theta$$

If you're using your temperature range of 300°C-500°C for the hot junctions and room temperature (25°C) for the cold junction, you would get your thermo-emf in the range $\pu{0.687-1.1875mV}$.


How to visualize the rearrangement of a pentacyclic triterpene alcohol?

You are correct about the beautiful chain of eight Wagner-Meerwein rearrangements. The stereochemistry of each of these is given by the fact that this arrangement is suprafacial and proceeds under retention, thus a group on the top of the molecule will stay above the ring plane and a group below will stay below. During the rearrangement process, since each intermediate cation is planar, the rings A to D (from where the hydroxy group was to where the double bond will be) will undergo a ring flip each, preserving the general trans-decalin structure for each pair of rings.

Once the final rearrangement has occurred, we arrive at that position in the molecule which features a cis-decalin system: note how in your drawing of the starting material the final ring E is not aligned with all the others but pointing downward. Assuming that final proton, which is eliminated, would actually migrate, we would generate the final carbocation at the cis-decalin junction. The methyl group between the eliminating proton and carbon A is in the wrong orientation to migrate; the carbon belonging to ring E has a much greater tendency because it is better aligned (anti to the proton). That migration would lead to a 6,5-spiro compound and a trisubstituted double bond. That isn’t bad in itself, but the pathway asked for in the question leads to a trans-decalin system by elimination, which features much less strain than a spiro compound. Additionally, a tetrasubstituted double bond is also more stable.

To arrive at the all-trans-polcyclic system, the proton on the junction of rings D and E must eliminate which it is happy to do. Ring D flips, which now turns the cis-decalin system of DE into a trans-decalin system, ring strain is released and the reaction terminates.


(I'm no chemist, but here's my take on it)

Let's see: I’m an alcoholic, and I really hate having to dole out large amounts of cash to purchase your usual "drinking"-ethanol.

So I get this amazing idea to separate the ethanol by distillation (methanol has a lower boiling point than methanol, so what I'll really be doing is distilling out the methanol, leaving behind ethanol and water). Pretty smart, right (for someone who's still inebriated)? Thus invigorated, I sally forth... and return with a dozen gallons  of industrial (denatured) ethanol. However, as I slowly get sober, I begin to see two major obstacles to my plan:

It’s going to be very, very difficult to maintain a constant temperature of exactly 64.7 °C (boiling point of methanol). Now, even though ethanol boils at around 78 °C, trying to maintain the denatured alcohol at under (say) 70 or 75°C isn’t really an option either. Remember, what I’m trying to do here is criminal (where I'm from at least), plus I don’t really have a Nurdrage-worthy laboratory. If say, I were to simply buy a large metal vat with a distillation apparatus clamped on the top and light a fire below, at such a large scale (a dozen gallons in my case) and with the non-professional equipment I use, it’ll be impossible to ensure that the vessel is uniformly heated at a specific temperature. So what’ll happen is, I’ll end up distilling out a mixture of ethanol and methanol, and the ethanol that remains will also have significant amounts of methanol in it (10ml of methanol's enough to make you go blind… just to give you an idea about what I mean by “significant”). Regardless, I wouldn't want to engage in something that involves producing methanol vapors (be it in a proper laboratory or otherwise)

Say I did manage to get hold of the required laboratory apparatus (I, uh, happen to know this cancer-stricken, bald, high-school Chemistry teacher who was willing to lend me them), and I were to distill the alcohol in small batches to ensure I evenly heat the solution at the desired temperature, but the whole process would be so time-consuming and labor intensive that it would probably just be better if I just went and bought "drinking"-ethanol in the first place.
(Should have mentioned this first, but since you were talking about methanol, ah well…) To begin with, I probably wouldn't know what the manufacturers actually used to denature the alcohol. They could have used methanol, pyridine, or some other foul-tasting, foul-smelling additive for all I know, or maybe even mixed all of them together. So trying to get ethanol out of that mess, is going to be a HUGE pain in the neck.

Also you might want to read up on denatured alcohol here.

I wasn't able to find anything on ethanol-methanol azeotropes in this Wiki page, so if anyone knows where I can find something on it, gimme a shout! :-)


Suggest a method, other than spectrophotometry, for determining the reaction rate

Spectroscopy is often the best method, since it allows for (almost) continuous monitoring of the concentration.

If that is not wanted, ask yourself:

What is released in the hydrolysis of the complex?

Chloride would be a good guess.

Can I take samples and perform a wet chemical analysis to determine the concentration of new species?

Looking at the graph in your other related question, the answer is yes. The hydrolysis is slow enough.

Which methods can I possibly use to determine the concentration $\ce{Cl-}$ in solution?

An argentometric titration, based on the low solubility of $\ce{AgCl}$ is a good idea. What about the Mohr titration, using $\ce{K2CrO4}$ as an indicator?


Does a timed reaction "ready indicator" paper exist?

I'm not aware of any commercially available examples, but I know that the speed of a fluid wicking through a paper microfluidic device has been used for timing. I'm also reminded of pencil detonators, dating back to WWII, where the time taken to dissolve a lead wire is used for timing. Since electronic timing is now so cheap, easy, accurate, and reliable, I'm not surprised that chemical timers have pretty much died out. There are a number of chemical phenomena that could be exploited for timing, but unless you need a huge number of cheap, one time use timers with relatively poor accuracy, there are probably better choices.


No, according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), numbering of spiro compounds starts in the smaller ring.

  P-24.2.1 Monospiro alicyclic ring systems

  Mono spiro parent hydrides consisting of two saturated cycloalkane rings are named by placing the nondetachable prefix ‘spiro’ before the name of the unbranched acyclic hydrocarbon with the same total number of skeletal atoms. The number of skeletal atoms linked to the spiro atom in each ring is indicated by arabic numbers separated by a full stop, cited in ascending order and enclosed in square brackets; this descriptor (called in these recommendations the ‘von Baeyer spiro descriptor’) is placed between the spiro prefix and the name of the acyclic alkane. Numbering starts in the smaller ring, if one is smaller, at a ring atom next to the spiro atom and proceeds first around that ring, then through the spiro atom and around the second ring.

The fixed numbering of the spiro hydrocarbon ring is not modified by the introduction of heteroatoms, double bonds, or functional groups. However, low locants must be attributed if there is a choice.


Like aliphatic cleaning solvents such as hexane, both liquid and supercritical $\ce{CO2}$ are effective solvents for fatty stains. According to Green America, though, "[L]iquid CO2 is pumped into the mix. Clothes are rotated in a cycle that lasts five to 15 minutes at room temperature."

The phase diagram for $\ce{CO2}$ shows that it is a supercritical fluid above ~300 K, just above room temperature, at ~80 kg/cm2, so that it may be either liquid or supercritical during the cleaning. However, at the end of the cycle, "[T]he liquid CO2 is pumped back into the storage tank, to be reused again, if possible. The remaining CO2 is released in the air." That implies a bit of a pressure drop at the end, with concomitant adiabatic cooling.


Are there any full worked examples of DFT calculations?

The XC potential for DFT actually consists of two terms: $V_x$ and $V_c$. Depending on which XC functional you choose, the exchange part is either computed exactly (using Hartree-Fock) or using fitted parameters or in a combination of both. In the case of B3LYP the exchange is 20% Hartree + 8% Slater + 72% Becke88. The correlation functional consists of three (3) different terms (LYP) in a different form. If you take a different functional, say M06-2X, it has different functional type (54% Hartree exchange, 200+ parameters!).

Coming to the next part: You use a basis set only to compute the kinetic energy and exchange terms in the system. Let me walk you through a HF routine first:

The Hamiltonian consists of four terms(nuclear kinetic energy is zero because of Born-Oppenheimer Approximation) : $T_{elec} + V_{nn} + V_{en} + V_{ee}$. The nuclear-nuclear potential energy is constant (because they are 'clamped'). The first and third terms are fairly trivial, and are computed with relative ease. The electron-electron potential term consists of Pauli's exchange term, and the coulomb repulsion term. They are the computationally expensive part of the HF calculation, and the eigenvalues (energies) are solved self-consistently.

So, when you obtain the HF energy of a system, it is obtained variationally with respect to all the energy components in an interacting system of electrons and nuclei. Coming to DFT:

The first term you mentioned $V_{ext}$ is the external potential, which is usually the nuclear potential for an unperturbed system. The second term is equivalent to Coulomb repulsion. The third term is evaluated using the XC potential of your choice(every DFT method differs only at this part). You evaluate the energies for each and every orbital using the effective potential $V_s$ which you mentioned. Note: You evaluate energies for every orbital separately.

This is because of the KS theorem that equates the density created by non-interacting particles to be the same as that of interacting particles. So, we solve Kohn-Sham equation instead of Schrodinger equation, under a constraint that the sum of square of all the orbitals gives the density:

$$\rho(\mathbf r)=\sum_i^N |\phi_{i}(\mathbf r)|^2$$

This means that the energy computed using DFT is minimum with respect to a non-interacting system of electrons. This is the biggest assumption of DFT, which is quite valid too. As a result, DFT provides a good account of correlation energy at a very minimal computational expense (HF accounts for ZERO correlation)!!!

I took a computational chemistry course, where DFT was a small part, so I am not able to explain more in detail. I used Prof. Kieron Burke's ABC of DFT book, which he has uploaded online in his site for free use. It is a good source to start. There is also a video lecture of Prof. David Sherrill on Youtube: https://www.youtube.com/watch?v=5orzn-XA29M .

I missed the last question: For any molecule:

Compute the kinetic energy (using the KE functional)
Compute the external potential ($V_{ext}$)
Compute the Coulomb repulsion using an initial guess density ($n(r)$)
Compute the XC potential from the XC functional
Use Kohn Sham equations to find out orbital energies (under the constraint that the density ($n(r)$) is equal to the density of interacting particles (electrons). In this case, the density must be equal to the total number of electrons.
Using the newly obtained density, recompute the effective potential ($V_s$)
Perform until self-consistency is achieved. Then total energy is the sum of occupied orbital energies times the occupancy number.

Hope this helps!


Sephadex resins swell as they absorb buffer, so the amount of buffer you need to add is dependent on the amount that the specific resin you are using is expected to swell. It is also dependent on the final thickness of slurry that you want to achieve. Too thick - the resin will not be suspended well enough and will pack poorly. Too thin - the resin at the bottom will settle before enough buffer has flowed through to allow you to add more, and again you will not get good packing (and it will take a lot longer to pour).
Sigma Aldrich provides technical documents online with recommendations for preparing Sephadex slurries from dry powder, which you can find here.
Here are some the the key points. The expected swelling of the Sigma Aldrich resins is provided in units of mL per 1 g dry resin:
Sephadex G-10 Medium: 2-3 mL/g
Sephadex G-25 (all grades): 4-6 mL/g
Sephadex G-50 Fine: 9-11 mL/g
In all cases, add buffer and allow the resin to swell (1-3 hours at room temperature), then top off the buffer so that the settled swollen resin is ~75% of the total volume. That will give you a slurry of appropriate thickness when stirred.


Chemicals used as explosives are those which, by some reaction/state change, can produce a large pressure (and thus usually volume) change very quickly.

While dry ice (solid carbon dioxide) is much more dense than in its gaseous phase, at room temperature the change happens very slowly, and is just unstable rather than stable but reactive, so can't be left and triggered later.

This makes it pretty useless as a working explosive.

Since it can exert a significant pressure while it sublimates, however, it can make other things explosively fail -- like the dry ice in the soda bottle demonstration. A quick search suggests that soda bottles are rated to withstand a pressure difference of around 100 psi (~6.8 atm, 0.69 MPa), and the increasing pressure won't stop the carbon dioxide subliming/boiling (since at that point it'll be a liquid) until ten times that.

Now, there is a market for things that slowly expand with a lot of force ("non-explosive demolition agents"), but they're solids that you mix into a slurry and pour into a cavity, since these can trivially exert orders of magnitude more pressure -- a common brand-name one, Dexpan, is rated at 18000 psi -- and they also turn from a slurry or solid into another solid. If you tried this operation with something that turned into a gas, you'd need to seal the cavity to stop the gas just escaping, whereas with solid-to-solid reactions, gravity and the solid itself will do the job for you.


Colloidal solutions may contain particles made of liquids or of solids. The particles are liquid in substances like mayonnaise or moisturizer, whose particles are droplets of oil (in water) for mayonnaise or droplets of water (in oil) for moisturizer. They can be solid in substances like acrylic paints, where the particles are made of semi-solid paint dispersed in water. The great advantage is that these paints can be diluted with water. Fresh spots are easily removed. Colloidal gold does exist. The colloidal particles can even be gas in whipped cream, froth and foams.


Looking at what has been published about hydrolysis of polycarbonate resins to recover bisphenol A, typical conditions involve water, heat, optional base catalyst (or more rarely, acid, which probably further degrades BPA), and very often a cosolvent such as tetrahydrofuran that itself is stable to hydrolysis. So your intuitions about whether polycarbonate should be susceptible to acid hydrolysis are correct.

There has been a lot of attention paid to whether polycarbonate in household items (e g. baby bottles) releases bisphenol A into foods, causing health concerns- which it certainly can.

On the practical side, such as preparing dilute aqueous acids or bases in a polycarbonate container at ambient temperature (e.g.for lab use), degradation would be very slow due to the density, hydrophobicity,  and perhaps cross-linking of polymer chains into a tight matrix. This prevents the aqueous acid or base from physically contacting the bulk polymer, although slow surface degradation is still likely.


If cis is what's naturally found, by default it's unsaturated fat isn't it?

Both cis and trans are unsaturated.  "Unsaturated" means having at least one C-C bond that is not a single bond (but instead a double or triple bond).  The terms cis and trans refer to the geometry of the molecule about the double bond.

Trans fat also occurs naturally in milk, for example fats of vaccenic acid

Also, since a fatty acid can have more than one double bond, it can be both cis and trans, for example fats of Rumenic acid found in milk.

  When disqualifying trans fat, are we qualifying cis-fat as healthy?

Cis-fat can also have negative health consequences.  For example, according to the review article Dietary fat, insulin sensitivity and the metabolic syndrome Clinical Nutrition vol. 23 pages 447-456.

  monounsaturated and ω-6 polyunsaturated fatty acids have less negative effect on insulin sensitivity than saturated fat

In other words, even consumption of too much unsaturated, naturally occuring, cis fat can have negative health consequences, such as insulin resistance.

Your question says:

  Because everything they adore seems to contain trans-fat from breakfast cereals, syrups, chips...

but is this really true?

From what I've seen, cereals have little or no fat of any kind, let alone trans-fat.

Similarly, maple syrup has no fat.

Even potato chips don't necessarily contain any trans fat:

  So we gotta explain why they should go for trans fat free, unsaturated food or in the worst case just choose a saturated food instead of a trans-fatty thing

The amounts of partially hydrogenated trans fat in foods has been greatly reduced in the last 10 years.  Milk and meats naturally contain trans fat.

I wouldn't want you telling my kids they can't drink milk or eat a hamburger.  Sure it's bad to eat too many hamburgers.  The focus should be a balanced diet.


Where do their electrons go?

IIRC, in the gas phase, the energy required for heterolysis is usually much greater than that required for homolysis. Consequently, usually, each of the atoms of a diatomic molecule retains one of the originally bonded electrons on dissociation.

  Which quantum mechanical technique can decide the question -
  Hartree-Fock?

In principle, one could scan the potential energy surface1) by elongating the bond using some quantum chemical method to look on what actually  happens. Beaware though, that the most used variant of the Hartree-Fock (HF) method, the so-called restricted Hartree-Fock (RHF) method is know to be ill-suited for such calculations: it is known to fail completely when the homolytical dissociation of a closed-shell molecule into two open-shell fragments is considered.

Such behaviour is usually refered to as the Hartree-Fock "dissociation catastrophe", although, strictly speaking, only the restricted variant of the Hartree-Fock method is a subject to it. The unrestricted Hartree-Fock (UHF) method does not suffer from the "dissociation catastrophe" and tends to give at least qualitatively correct dissociation limit. So, use UHF at least. Even better, use some correlated method.

1) Simply a curve in this case, since there is only single degree of freedom (bond length).


As we are discussing acid-base neutralisation reactions they will be in solution (mixed with water).

First consider LiOH when mixed with water:
$$\ce{LiOH -> Li+ + OH-}$$

As we can see $\ce{OH-}$ ions are formed, meaning this solution is a base.  This means we need an acid to neutralise it.  (Acids and bases react to neutralise each other).

Lets say we use $\ce{HCl}$ as our acid.  Consider it mixed with water:
$$\ce{HCl -> H+ + Cl-}$$

When we mix these two solutions together we get:
$$\ce{Li+ + Cl- + H+ + OH- -> LiCl + H2O}$$

Notice how the $\ce{OH-}$ and $\ce{H+}$ react to form $\ce{H2O}$.  The $\ce{LiOH}$ is now neutralised.

This can be written in one equation like this:
$$\ce{LiOH + HCl -> LiCl + H2O}$$

Other acids could also be used to neutralise the $\ce{LiOH}$ such as $\ce{H2SO4}$:
$$\ce{2LiOH + H2SO4 -> Li2SO4 + 2H2O}$$

I am uncertain of your second question.  There is no way of neutralising $\ce{MgCl2}$ as it is a neutral salt (no $\ce{OH-}$ or $\ce{H+}$).  Solubility tables (at this level) also do not affect the pH of your solution.  If you haven't learned about it yet then don't attempt to now, you will likely just confuse yourself for your test tomorrow.

I hope this has been at an appropriate level for you.


Why is the nitroso group a deactivating group for electrophilic aromatic substitution?

Let's look at nitrosobenzene.

In order to be delocalized (participate in resonance), the lone pair on the nitrogen needs to be in a $\ce{p}$ orbital. As I have draw it above, the lone pair is in a $\ce{sp^2}$ orbital, and the $\ce{N=O}$ pi bond is conjugated with the ring:

If the lone pair on the nitrogen atom is delocalized, then yes the nitrogen atom is $\ce{sp}$ hybridized. As a consequence, the $\ce{N=O}$ pi bond is perpendicular to the pi system. It is no longer conjugated:

Either the lone pair or the $\ce{N=O}$ pi bond is conjugated, not both.

If the lone pair is conjugated, the nitroso group can be electron donating by resonance:

If the $\ce{N=O}$ bond is conjugated, the nitroso group is electron withdrawing by resonance:

Finally, because the nitrogen is more electronegative than carbon, the nitroso group is electron withdrawing by induction:

It is more likely that the $\ce{N=O}$ bond is conjugated because that gives a higher degree of conjugation: 8 atoms versus 7 atoms.


Are the following two compounds isomers or not?

You are right, these two molecules are enantiomers. If you rotate the bottom molecule 180 degrees, you can see that it is the mirror image of the top molecule. Enantiomers are stereoisomers, so the only reason your book would say they are not isomers is that it (unfortunately) contains a typo.


Why does the reduction of Dichromate to Chromium 2 need an acid?

A big factor is kinetics.  A lump of zinc or even zinc powder sits there undissolved in a neutral solution and does not provide good contact with the chromium species with which you want the zinc to react.  Adding acid creates nascent hydrogen in the solution, and this is what really does the job of reducing the chromium species.  This source, drawn from a related answer by Nilay Ghosh, explores the matter more fully, and I recommend going there.  In addition to the nascent hydrogen, the acid enables keeping the zinc ions in solution as a salt, where otherwise they would form a hydroxide precipitate that interferes with the reaction.

Not every acid is equally good.  To get the most nascent hydrogen you need a strong acid and, even more important, one that does not eat up its hydrogen with other oxidizing functions (nitric acid, you're out!).  Add the fact that cheap is also good, and hydrochloric and sulfuric acids are the usual choices.


Alphanumeric preference between hexyl and heptyl

According to the IUPAC nomenclature, preference for placing prefixes is given in accordance with alphabetical order (just like in a standard English dictionary). It has nothing to do with the number of letters in a prefix. Since "P" in heptyl comes before "X" in hexyl, heptyl will be placed before hexyl in IUPAC nomenclature.
Hence, the correct name will be the one where heptyl is placed before hexyl (8-heptyl-10-hexylheptadecane).


The configuration for the free atoms is:

Ba - $\mathrm{[Xe]\ 6s^2}$
Br - $\mathrm{[Ar]\ 3d^{10} 4s^2 4p^5}$
S - $\mathrm{[Ne]\ 3s^2 3p^4}$
Si - $\mathrm{[Ne]\ 3s^2 3p^2}$

Thus:

Barium has no unpaired electrons.
Bromine has one unpaired electron in 4p subshell.
Sulfur has two unpaired electrons in the 3p subshell.
Silicon has two unpaired electrons in the 3p subshell.


What does the plus-minus sign with a value in parentheses mean?

It's simply describing the absolute error of the relevant quantities.

So if this precedes a table of melting points, you can expect the true value to vary by 1 degree, e.g.

$\ce{H2O}: 0.0 \pm 1~^{\circ}\mathrm{C}$

means the melting point is between $-1$ and $+1~^{\circ}\mathrm{C}$.


$\ce{Na2SO4}$ will dissolve and dissociate into separate ions. And there will be some protonation $\ce{H+ + SO4^{2-} -> HSO4-}$.


Selection of parent chain (IUPAC nomenclature)

The most important simplified criteria for the choice of a principal chain are:

greater number of substituents corresponding to the suffix (principal characteristic group)
longest chain
greater number of multiple bonds
lower locants for suffixes
lower locants for multiple bonds
greater number of prefixes
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding wording of the rules taken from Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) is as follows.

P-44.1 SENIORITY ORDER FOR PARENT STRUCTURES
When there is a choice, the senior parent structure is chosen by applying the following criteria, in order, until a decision is reached. These criteria must always be applied before those applicable to rings and ring systems (see P-44.2) and to chains (see P-44.3). Then criteria applicable to both chains and rings or ring systems given in P-44.4 are considered.
P-44.1.1 The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).
(…)
P-44.3.2 The principal chain has the greater number of skeletal atoms [criterion (b) in P-44.3].
(…)
P-44.4.1 If the criteria of P-44.1 through P-44.3, where applicable, do not effect a choice of a senior parent structure, the following criteria are applied successively until there are no alternatives remaining. These criteria are illustrated in P-44.4.1.1 through P-44.4.1.12.
The senior ring, ring system, or principal chain:
(a) has the greater number of multiple bonds (P-44.4.1.1);
(b) has the greater number of double bonds (P-44.4.1.2);
(…)
(h) has the lower locant for an attached group expressed as a suffix (P-44.4.1.8);
(…)
(j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);
(…)
P-45.2.1 The preferred IUPAC name is based on the senior parent structure that has the maximum number of substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.2 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes (other than ‘hydro/dehydro’) to the parent structure.
P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.
(…)

According to Rule P-44.1.1, the priority list of substituent groups has to be considered first in order to decide the parent chain for this compound. Apparently, you have already done that in both of your suggested possibilities: The principal characteristic group that is expressed as a suffix is the aldehyde group ($\ce{-CHO}$).
The next rule (P-44.3.2) is looking at the longest chain. Therefore, the correct answer is B; the principal chain contains six carbon atoms and the hydroxyl group. Thus, the complete name is 2-ethynyl-5-hydroxyhexanal.

The next rule (P-44.4.1.1) would be looking at the multiple bond, but it is not needed since we already found the principal chain.
Note, however, that the IUPAC recommendations on this matter have been changed. In the current version, the order of seniority between unsaturation and length of chain given in earlier recommendations is reversed.

(…), the first criterion to be considered in choosing a preferred parent acyclic chain is the length of the chain; unsaturation is now the second criterion.

If your book is still using old nomenclature rules, it would expect the other answer A to be correct.


What should be the order of specific conductance (κ) of floride, chloride and bromide ions?

ppm is ambiguous quantity, it can be v/v, w/v,w/w,n/n.

By other words, all 1 ppm very probably is not the same molar concentration, unless it is n/n ppm, relating itself to the molar amount of the solution.

$\Lambda = \kappa /c$ is correct, but $c$ must be the molar concentration.

$$\Lambda[\pu{Sm^2mol^-1}]=\frac{M[\pu{g/mol}]}{c[\pu{ppm as g/m^3}]}\cdot \kappa[\pu{S/m}]$$


How do phase transitions occur in the solid phase?

Phase transitions in solids occur in the same manner as in liquids or gases, through nucleation and growth. Yes, those processes require atomic motion. Yet,atoms diffuse in solids all the time, through various mechanisms. While potentially slower, the thermodynamics of solid phase transitions are easily seen in calorimetry measurements for example.

Nucleation and growth of precipitates is a key part of many engineering alloys. On the other hand, some phase changes (like diamond to graphite) are very hard to make happen at near-human-tolerable temperatures.

The solid phase also sees the martensitic phase transitions, which are second-order diffusionless atomic rearrangements. The name comes from the martensite phase observed in steels. Shape memory alloys are another example of martensitic phases.

A classic materials science text would be 'Phase Transformations in Metals and Alloys", by D.A. Porter and K.E. Easterling. In my opinion, the second edition is vastly preferable to the more recent third edition where the changes mainly seemed to be reintroducing all the errors that were removed in the second edition (sigh).


In carbon compounds Coulson's Theorem can be used to relate bond angles to the hybridization indices of the bonds involved.

$$1+\lambda_{i} \lambda_{j} \cos(\theta_{ij})=0$$

where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.

Let's look at some examples.

In the case of methane, each $\ce{C-H}$ bond is $\ce{sp^3}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt3$. Using Coulson's theorem we find that the $\ce{H-C-H}$ bond angle is 109.5 degrees.
In the case of the methyl carbocation ($\ce{CH3^{+}}$) the $\ce{C-H}$ bonds are $\ce{sp^2}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt2$.  In this case, using Coulson's theorem we find that the $\ce{H-C-H}$ bond angle is 120 degrees.
Finally, in the case of acetylene the $\ce{C-H}$ bond is $\ce{sp}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt1$.  In this case, using Coulson's theorem we find that the $\ce{C-C-H}$ bond angle is 180 degrees.

Conversely, if the bond angle is known, the hybridization can be determined.


Why does the PH3 geometry deviate more from the trigonal planar one, than does NH3?

Apparantly it was community consensus that this question be clear enough to stay open.

Neither $\ce{PH3}$ nor $\ce{NH3}$ have any incentive whatsoever to assume a trigonal planar geometry. The most stable ground state of both nitrogen and phosphorus atoms is $n\mathrm{s^2},\ n\mathrm{p^3}$ where $n = 2$ for nitrogen and $n=3$ for phosphorus. Since the p-orbitals are higher in energy per se, it makes sense to use them for bonding and lower their energy — which, as a side note leads to much more efficient overlap than using s-orbitals for bonding, since p-orbitals have a direction in space.

Therefore, the most stable geometry for both would be trigonal pyramidal with $90^\circ$ angles between the substituents. This would allow all-p bonding orbitals and an all-s lone pair on the central atom. If you check out $\ce{PH3}$, you will find almost exactly that geometry save a little distortion.

The distortion you see in $\ce{PH3}$ is of the same type as the one which moves the hydrogen atoms much further apart in $\ce{NH3}$: Steric stress. The hydrogens themselves repel each other; they require a certain space. Around a large phosphorus atom there is enough space for them to crowd at the ideal bonding angle while the small nitrogen does not offer that space. The final bond angle of $\approx 107^\circ$ is a fine-tuned result of trying to keep the hydrogen atoms as far away from each other as necessary and keeping the bonding angle as close to $90^\circ$ as possible.

$\ce{BH3}$ is an odd one out in this comparison. Boron does not have a lone pair and thus we expect isolated $\ce{BH3}$ to assume trigonal planar geometry. However, we also expect isolated $\ce{BH3}$ to have a low stability since boron does not have an octet. Hence it dimerises to $\ce{B2H6}$ with two $\ce{B-H-B}$ banana bonds of the 2-electron-3-centre type. The result can be described as a distorted tetrahedron around boron.


The capacitance sensor itself will certainly not present an ignition risk to your system. The sensor will be on the outside of the container, and it does not introduce any electrical potential that would cause a spark.

The way the device works is to measure the change in the sensor's capacitance depending on the dielectric effect of the liquid, which will increase as more liquid enters the electric field of the sensor (a capacitor).


Are hydrogen atoms in acetone molecule acidic?

You’re not going wrong anywhere. Acetone can indeed be deprotonated. The resulting structure is called an enolate.

$$\ce{H3C-C(=O)-CH3 <<=> H3C-C(-O^-)=CH2}$$

When compared with truly acidic compounds such as $\ce{HCl}$ and even $\ce{NH4+}$, acetone is a very, very weak acid, though. That does not mean it should be considered aprotic. Given a strong enough base — e.g. $\ce{HMDS}$ ($\ce{(Me3Si)2N-}$) — it can be deprotonated. Weaker bases will allow for partial (but probably hardly verifiable) deprotonations.


Why are excited Slater determinants used to describe electron correlation?

It perhaps isn’t a very satisfying answer, but mathematically the reason is that this is our only option!

The atomic spin-orbitals form a complete basis set: we are guaranteed to be able to represent the true wavefunction as some linear combination of them. Under Hartree-Fock, as you point out, this is a single Slater determinant. If we want to do better than this, the extra chemistry that we want to add is electron correlation. The extra maths we need to add to cope with this is more basis elements… and the way to add these is to mix in more Slater determinants. Which is why I say this is our only option: if we start with the HF ground state, and want to mix in more basis functions still, then by necessity we will have to mix in some HF excited states.

(There is one extra point that might help in understanding the Kümmel description, which is that if we could get a lower energy for the ground state by mixing in a single-electron excitation, then the HF procedure would already have done this. So we need to consider two-electron excitations, which as Kümmel says can be interpreted as interactions between these two electrons that raise them above the HF ground state.)


How to determine omitted units in a publication

These are degrees Celsius.

To prove that, you could search JACS for the publications of the same year/issue for the temperature-related physical constants such as melting point:

Douglas et al. [1] lists m.p. of $\ce{LiF}$ as $\pu{848 ± 1 ^\circ}$, whereas currently Wikipedia lists $\pu{845 °C}$.
Yankwich et al. [2] lists m.p. of malonic acid as $\pu{140 ^\circ}$ whereas currently Wikipedia lists $\pu{135 to 137 °C}$.

You may find further examples, but even at this point it's fairly obvious what those temperature units are.

References

Douglas, T. B.; Dever, J. L. Lithium Fluoride: Heat Content from 0 to 900°, the Melting Point and Heat of Fusion. J. Am. Chem. Soc. 1954, 76 (19), 4826–4829. https://doi.org/10.1021/ja01648a016.
Yankwich, P. E.; Promislow, A. L. Intramolecular Carbon Isotope Effect in the Decarboxylation of Liquid Malonic Acid Near the Melting Point. J. Am. Chem. Soc. 1954, 76 (18), 4648–4651. https://doi.org/10.1021/ja01647a050.


electron - nuclei recombination within stars

Nuclear fusion in stars produces huge amounts of energy that raise the temperature of the star (e.g. our sun has a surface temperature of about 6000K). At such high temperatures, a lot of atoms possess enough energy to become ionized. However, what percentage of atoms at a particular temperature has what amount of energy is determined by the Boltzmann distribution. So, there will always be some atoms which are not completely ionized (i.e. maybe the electrons are excited from the ground energy level, but they have not completely 'left' the nucleus).
The upshot of all of this is that as the temperature rises, more and more atoms get ionized. Atoms with lower ionization energies like Na, Mg and other metals lose their electrons first, whereas atoms with high ionization energies like He hold onto their electrons until the temperature reaches ~30,000K. Elements like C, N, O, H come somewhere between them. Also, the Boltzmann distribution means that as the temperature rises, more number of a certain atom become ionized. (This page contains a nice diagram showing which ions form at which temperature)
Conversely, if you lower the temperature, you start to form more neutral atoms, and even molecules after a point. However, that happens only below ~3000K, where it starts to look less and less like a star, and more and more like a huge gas giant planet (see Brown dwarf).
As you lower the temperature even more, you can get big complicated molecules like the ones we have on Earth. All of this happens outside of stars, of course*. (*It requires elements like C,N,O,P all of which were synthesized in a pre-existing star, see how planets form )

As an aside, there is another way electron and nuclei can 'recombine' in stars. At the end of the life of a heavy star (> 8 times the mass of sun), the nuclear fusion (that provides an outwards radiation force) suddenly stops. The star then collapses inward on itself due to the heavy gravitational force. This force is so massive that it crushes the electrons and protons together to form neutrons and neutrinos and a huge amount of energy (a type of nuclear reaction). That type of rapid nuclear reaction causes the star to detonate in a supernova.


I think, first I should clarify what causes the osmotic pressure:
Osmosis occurs when two solutions of different concentrations are separated by a membrane which will selectively allow some species, e.g. the solvent, through it but not others, e.g. the solute.
So, there is a concentration gradient between the two solutions which would lead to a diffusion from the side with the higher to the side with the lower concentration.
But since the solute cannot pass through the membrane the diffusive flow cannot even the concentration differences out and thus constantly presses against the membrane exerting a force on it which leads to a pressure.
The pressure originating from this tendency for osmotic flow to occur is called the osmotic pressure $\Pi$ of the solution.

So, the question is: Is the theory behind diffusion applicable to colloidal solutions?
To reason about this, I will first address the matter of whether colloids can be considered "real" solutes or not. Colloidal particles are usually some polymer particle coated with a tenside layer. So, it is the molecules of those tenside layers that interact with the solvent molecules and they are chosen in such a way that the colloidal particles neither lump together (coalesce) nor sink to the ground or float to the top - that keeps the colloidal solution stable. So, a colloidal particle is surrounded by solvent molecules which solvate the tenside layer. Of course, the situation is a bit different from a "normal" solution since the colloidal particle is a lot bigger than, say, an alcohol molecule and thus the the ratio between solute particle size and the size of its solvation sphere is much smaller. But, in essence, a colloidal particle and an alcohol molecule are solvated in pretty much the same way.
Furthermore it can also be assumed that interactions (e.g. electrostatic forces) between the colloidal particles are weak.
This means that colloidal particles are thus not subject to external forces which influence their movement. They simply move through the solution by Brownian motion. This is important because then their motion can mathematically be described by a stochastic process which is the basis for the thermodynamic description of diffusion.
And, as I described above, the reason for diffusion to happen is the same as the origin of the osmotic pressure.

So, to answer your last question: Yes, I think, the description for osmotic pressure caused by colloidal particles should be the same as for "normal" solutes.


how protons(H+) react with matter and cause the solution to be acid?

Well HCl won't desolve all plastics. In case of your hand there are complex proteins involved containing a variety of elements e.g, carbon, nitrogen, oxygen..., when your hand comes in contact of a proton, the proton is a electrophile and seeks negative charge centers, where ever on a protein it finds the negative charge center e.g, on oxygen atom in a hydroxyl(-OH) group in some amino acid in some protein in your hand, it will attach itself to the oxygen and eventually causing it(oxygen) to loose off from the protein as water, because the ion H20^+ isn't stable and the newly formed Oxygen Hydrogen bond isn't easy to break, so the part breaks off as water. millions of such reactions or even trillions happen eventually corroding off the hand or al large part of hand exposed to protons in a mobile medium.
An acid burn isn't actually a burn but corrosion of proteins, and it doesn't either melt things away rather disintegrates them.


Nomenclature of unusual nitrogen functional group

According to Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name of the compound $\ce{H2N-CH=NH}$ is methanimidamide. The name formimidamide may be used in general nomenclature. The name formamidine is no longer recommended.
The letter locants ‘N’ and ‘N'’ are used to differentiate among the different nitrogen atoms. Thus, the compound $\ce{R-NH-CH=N-OH}$ would be named as N'-hydroxy-N-…methanimidamide.


What do you think is a good way to explain electronic configurations?

As an undergrad student that has recently learnt that the Bohr model is just an approximation at best, and mostly incorrect, I think I am well equipped to answer this.

When I first learnt about the structure of an atom, I was told that "electrons move around the nucleus in fixed paths having fixed energy, called orbits. The angular momentum of the electrons is an integer multiple of $h/2\pi$."

When I was first taught the electronic configurations and the quantum mechanical structure of an atom, my teacher explained it thus (and it was easily understood by most of the class): he made an analogy between a person's location and the location of an electronic.

Just how a person can be located if we know which country they're in, which city they're in, which locality, which room, and finally which area in the room; an electron can be approximately located if we know its principal quantum number, its azimuthal quantum number and its magnetic quantum number. Another quantum number, i.e., spin quantum number tells us which direction it is spinning in.

After this, a brief historical understanding of why the Bohr model was discarded and this model was proposed was given to us. The Heisenberg's uncertainty principle, and the Schrodinger equation were introduced, as the uncertainty principle explained why Bohr's model was incorrect, and the Schrodinger equation explained how to find the probability density of the presence of the electron.

This method helped me understand as it felt like I was transitioning through the same phases that all the people of the scientific community had to when the quantum mechanical model was on its way in, and the Bohr model on its way out. This really felt like a fluid transition, rather than "Bohr was wrong, this is right", which would've caused a lot of confusion (at least for me).

Finally, there isn't a one size fits all answer to this, as every student is different. I am just sharing my experience.


Relative strengths of methyl, primary, secondary, and tertiary C–H bonds

$\ce{X–H}$ bond dissociation energies usually correspond inversely to $\ce{X^.}$ radical stability, so another way of looking at it is to recall that the stability of radicals increases in the order methyl < primary < secondary < tertiary.

The root cause of this is hyperconjugation (from $\sigma_\ce{C–H}$ to singly filled $\mathrm{p}$ orbital on carbon), analogous to the stabilisation of carbocations. So, it is just the second side of the same coin, really.


The heterocycle in this question is indole and is aromatic. This means that the N lone pair is delocalised and not readily available for nucleophilic attack. Think of it as similar in reactivity to a secondary amide nitrogen RCONHR. Generally you need to formally deprotonate to functionalise, though there are some interesting techniques using carbonyl azoles catalysed by DBU[1] and others using aldehyde and alcohol substrates.[2] Note that 3-unsubstituted indoles react with acyl halides by F-C acylation at the 3 position.[3]

References:

Heller, S. T.; Schultz, E. E.; Sarpong, R. Chemoselective N-Acylation of Indoles and Oxazolidinones with Carbonylazoles. Angew. Chem. Int. Ed. 2012, 51 (33), 8304–8308 DOI: 10.1002/anie.201203976.
Maki, B. E.; Scheidt, K. A. Single-Flask Synthesis ofN-Acylated Indoles by Catalytic Dehydrogenative Coupling with Primary Alcohols. Org. Lett. 2009, 11 (7), 1651–1654 DOI: 10.1021/ol900306v. PMID: 19320508 (with free text available).
Okauchi, T.; Itonaga, M.; Minami, T.; Owa, T.; Kitoh, K.; Yoshino, H. A General Method for Acylation of Indoles at the 3-Position with Acyl Chlorides in the Presence of Dialkylaluminum Chloride. Org. Lett. 2000, 2 (10), 1485–1487 DOI: 10.1021/ol005841p.


NaHCO₃ + HNO₃ → NaNO₃ + H₂O + CO₂ ... but something weird happens to me in practice

Let's look at the equation first:

$$\ce{NaHCO3 + HNO3 \to NaNO3 + H2O + CO2}$$

In other words, when added in equal parts, you'd ideally expect the bicarbonate anion to be protonated by the nitric acid, forming weaker carbonic acid (which will dissociate into water and carbon dioxide under temperature or even agitation), and the remaining ions will pair up into sodium nitrate (soda niter) solution in water.

However, nitric acid is tricky, and doesn't like to act "ideally". First, you already know that nitric acid decomposes into nitrogen dioxide and water; this happens whether you want it to or not. Highly exothermic reactions (most nitric acid reactions, including this one, are so) ramp up this inevitable side reaction with the extra heat. This will reduce your yield, but I'd expect it to go the other way than you apparently did; 1:1 mixtures of acid and bicarbonate will inevitably leave unreacted bicarbonate in the solution (not acid) along with some dissolved nitrogen dioxide (aka nitrite hydrate, $\ce{H2NO3}$).

In addition, "pure" WFNA is defined as anywhere north of 98% nitric acid, so pretty much any quantity of nitric acid you could get is likely to have impurities. So, I think your first mistake was to keep adding sodium bicarbonate; if you mixed a mass of "pure" (>98%) WFNA (even if you made it from scratch) with equivalent molar weight of sodium bicarb, you'll have a small excess of bicarb due to impurities in the nitric acid (pre-existing and created), but the solution will still read acidic (because of dissolved $\ce{CO2}$ and $\ce{NO2}$) after you've added more bicarb than will dissolve in the solution.

Now, heating the equilibrium solution to obtain the dry solid salts will remove the water from the dissolved nitrogen dioxide, and it will start acting like the nitrite ion it is, reacting with any unreacted bicarbonate to form yellow sodium nitrite, water and more $\ce{CO2}$. This sounds like exactly what happened to you, resulting in a more yellowish solid than the pure white soda niter you were expecting.

If you're not careful with the heat while evaporating off the water, parts of the reaction, namely the precipitating solid collecting on the bottom of the flask, can be superheated beyond boiling, and reach the temperature at which both of the niter salts will decompose into sodium oxide and more oxides of nitrogen. This is typically accomplished only at temperatures in excess of those capable with the average benchtop rig, but it becomes more thermodynamically favorable in the presence of excess sodium (which you have in the sodium bicarbonate). I'm not saying this happened - it's more likely that the nitric acid decomposed more than you might have wanted - but it's a theoretical possibility.


Nitrogen molecule vs carbon-carbon triple bond stability

You might already know that elements like to follow the octet rule, which means that atoms tend to combine in a way that each atom can have eight valence electrons. Nitrogen can exist diatomically because through the triple bonds, each nitrogen will have one lone pair and three shared pairs. As seen in the picture, each nitrogen satisfies the octet rule.

Carbon, on the other hand, cannot satisfy the octet rule diatomically. Since carbon has one less valence electron than nitrogen, a triple bond would not work because the carbon does not have enough electrons to bond and satisfy the octet rule. It would leave each carbon atom with three bonded pairs of electrons and only one lone electron, giving each atom a configuration of only seven electrons per atom, not eight. Sorry I don't have a picture of a triple carbon bond. Hope this answers your question though!


Why is a weak base more able to deprotonate a strong acid than a weak acid?

The key is that water is the great equalizer and mediator. Let's deal with weak and strong base first: You are right that a sufficient quantity of a strong base would have the same effect. In this particular case, the hydrogencarbonate will separate into $\ce{CO2}$ and water on protonation and we can assume that most of the carbon dioxide will bubble out. So it will functionally not make a difference whether you started with $\ce{NaOH}$ (a strong base) or $\ce{NaHCO3}$. This is a part of the equalizer part: in an aqueous solution, you cannot have a stronger base than $\ce{OH-}$, because any stronger base will just deprotonate a readily available water molecule.

To some extent, even weak bases will do so in an equilibrium. To understand how a proton gets from an acid to a base, we have to understand the structure of water: A water-based solution like the one from the question is a huge network of water molecules engaged in hydrogen bonding with each other and mainly electrostatic bonding to ions. A when a proton "moves" through this network, it will basically jump from one water to the next, and that water will give up a proton to the next etc. (This is the reason for the huge mobility of protons in water as measured by conductivity experiments.) Thus, an acid and a base such as $\ce{RCOOH}$ and $\ce{HCO3-}$ do not have to come close to each other to effectively transfer a proton between them.

Thus the question "How does the base know to deprotonate the strong acid instead of the weak acid?" loses its basis. In an equilibrium, the stronger acids and bases will deprotonate and protonate more, respectively, than their weaker counterparts because that's what acid/base strength means.


The importance of the cold chain in the food and the pharmaceutical industry

Avoiding cooling  again after keeping it ( for non critical products ) at room temperature is mainly to prevent forgetting it was not cold all the time and that it may not last as long as expected.
Taking food from the fridge, you may not remember after a week or two that it was outside longer than it should have been. It can violate the safety by growth of microbes, which does not fully stop when it gets cold again.
For drugs or similar critical products, it violates stability and guarantee of usability, as thoroughly described by @Julian answer. I did  related analysis within drug stability studies in past as well.
Food freezing is another case. Industrial freezing of food is done by shock freezing using liquid nitrogen. This leads to quick forming of small ice crystals that do not puncture the cell walls, so after melting food before usage, it is OK.
But when refrozen back e.g in a freezer, freezing is slow, leading to growth of large ice crystals puncturing the plant/animal cells. After another melting, cell liquids leak out, cells lose their tonus and the food gets unpleasant mechanical and sensory state, affecting the final taste. Additionally, what was told above about the fridge applies here as well.
Presence of free ice along with frozen vegetables is a sign the content was refrozen.
Slow melting of previously shock-frozen stuff leads to growth of bigger crystals, similar to recrystallization, as bigger crystals are thermodynamically preferred due to lower surface energy. So some cell wall tearing and subsequent deterioration of the content happens even during melting and continues until content is frozen again. Also, shock freezing still induces some stress on the product, even if lower than slow freezing.
So the final effect of

fast freezing -> slow melting -> slow freezing -> melting

can be much worse than just

slow freezing -> melting

as the former has more time and opportunity to cause damage.


Calculating the heat capacity of a calorimeter

$12.5\ \mathrm{kJ}$ of heat was absorbed by the surroundings.

I found this by using the mcat formula and the specific heat capacity of water (4.18 J/(g °C)):

$Q=m\cdot c\cdot\Delta T$

$Q = 950\ \mathrm{g} \times (4.18\ \mathrm{J\cdot g^{-1}\cdot {^\circ C^{-1}}}) \times (23.25\ \mathrm{^\circ C} - 20.10\ \mathrm{^\circ C}) = 12508.7\ \mathrm{J}$

If you wanted to use this whole formula for solving the calorimeter's specific heat capacity, you would need to know the mass of the calorimeter as well, which is not given.

What your book is probably asking is for what is called the "calorimeter constant". This is given in units of $\pu{J/^\circ C}$ notice that it does not include mass.

Note: Sometimes "the calorimeter's specific heat capcity" is used instead of referring to the calorimeter constant, but in this case we cannot find a value which will include mass in the units, so I think it is more clear to use the term "calorimeter constant."

You can determine the constant by this formula: $$Q_\text{cal} = C_\text{cal} \times \Delta T_\text{cal}$$

Where $Q_\text{cal}$ is the energy absorbed, $C$ is the constant and $\Delta T$ is the same as the change in temperature of the water.

You may calculate $Q_\text{cal}$ by using this formula: $$Q_\text{cal} = -(Q_\text{water} + Q_\text{glucose})$$

It may also help to think of $Q_\text{water}$ = $Q_\text{surroundings}$ and $Q_\text{glucose}$ = $Q_\text{system}$

To find $Q_\text{glucose}$ I did: (glucose has lost energy, it is negative value)

$-2820\ \mathrm{kJ} \times 0.007\ \mathrm{mol}$ and $Q_\text{water}$ is simply the $12508.7\ \mathrm{J}$ positive because $\Delta T$ is positive for the surroundings (the system/glucose lost energy)

$Q_\text{cal} = -(12508.7\ \mathrm{J} + (-19740\ \mathrm{J}))$

So my final answer is then: $2.3\times10^3\ \mathrm{J/{^\circ C}}$

It is important that heat capacities are positive, think about what it would mean if this were a negative value.

In the laboratory, it is necessary to do a calculation such as this one before using a calorimeter for anything. Normally it can be done by heating a piece of nickel or something, recording the temperature of the metal and the water, and then dropping the metal into the calorimeter to find the final temperatures, and then calculate the calorimeter constant. You can then proceed with further experimentation using that calorimeter, but only after this constant has been found can you find the specific heat capacity of other materials.


How do chemists find a procedure to produce a given substance?

Chemists use many sources of information to do this.

Part of being trained as a chemists is learning, in general terms, what kinds of reactions are known and what their best uses are. That training can go 5-10 years beyond a bachelors degree and is very intensive.

Beyond what you can keep in your head, one of the most important information sources is the "primary literature", consisting of:

Patents
Journal publications

As an example, see this patent about ethanol manufacture.

Alternatively, see this very recent journal publication dealing a new catalytic reaction.

Having all of this information out there is a start, but it's a vast collection spread out over many publishers and 100 years or more. To find what you need, it's essential to have a good indexing system. This allows you to search large databases of reactions and molecules for references that exactly match, or which are close to what you need.

This last part - being able to reason by analogy - is critical to being able to get the most out of the primary literature. Maybe I don't find an exact match for the exact reaction I want to do. But I find something pretty close. Here's where my training and reasoning ability come in to either decide to adapt what I've found or keep looking for something better.

An example of one of these search systems is Chemical Abstracts Service, but there are others.

As a professional chemist, you're expected to keep up with all of the main journals in your field so you stay up to date. You do this by reading articles, talking to people, giving talks, and going to conferences.

There's another category of information consisting of 'reviews'. These are articles that summarize a large area of chemistry - for example, recent large-scale methods for the production of ethanol.

The discovery/invention of new reactions is a very active area of current chemical research.


You are indeed right, I confirmed that it is the fault in the diagram itself. See the diagram by wikipedia:


Why does radioactive matter decay in series of half-lives?

There are fewer decays because there are fewer atoms to decay

The simple reason why the number of decays (strictly, the number of decays per unit time) decreases in simple radioactive decay is because there are fewer atoms left to decay.

Nuclear decay is probabilistic. The probability of any given unstable atom decaying is constant (independent of time or the environment). So the reaction rate (the total number of decays per second) per mole is constant. If this were a chemical reaction we would call it a first-order reaction (since it isn't a chemical phenomenon we can call it a first order process but the mathematics are the same).

In any first order process there is a characteristic time when half the original material will have gone (we call this the half life of the process). So, if we measure time in half-lives at t=1 we will have, say, 1 mole of the material; by t=1 we will have 0.5 moles; at t=2 we will have 0.25 moles of material. If we count the number of emissions per second they will follow the amount of material left. So the radioactivity will decay at the same rate as the material.

We don't need to invoke any interactions between the atoms of the material to explain this.

PS it gets a little more complicated if one radioactive compound decays into another. Then the total radioactivity will depend on two competing decays with different half-lives (but the radioactive emissions may well be different as each has a characteristic energy and expensive instruments can tell them apart giving two separate emission counts). But the mathematics remains the same.


The one on the left (RR) and the right (SS) are not superimposable mirror images of one another and are thus enantiomers. In your head, try flipping one of them. For instance, if you flip the RR molecule on the left as I drew it, it will look something like this:

The two pictures I posted are depicting two same molecules, except the one on the left was flipped around an axis in plane of the screen. With the second picture, it's easier to see that these two molecules are non-superimposable mirror images. Also notice how neither of these molecules have a plane (or point) of symmetry.

I recommend trying to build this with a model kit if you can't visualize it in your head easily.


For A: The pyridine ring would be very strained. The lone pair on the nitrogen is also in the plane of the ring and would hardly be able to form a double bond this way.

For E: Putting a positive charge on an sp hybridized atom is never good. The more 's' character you have, the closer the charge is to the nucleus (since s orbitals are closer to the nucleus). This is higher in energy. On top of that, putting a negative charge on a 'sp3' carbon is not particularly favored either.