Yes, this really happened, probably (it's not a peer-reviewed effort).

The reactor is a Farnsworth fusor, a very low-density, low-rate inertial-confinement reactor. They're surprisingly easy to build and get working; many people have done so, and verified their function by measuring neutron output. An accelerating voltage of 10 or 20 kV produces particle energies equivalent to over 100 million degrees C.

There's active debate about whether a fusor of this design could ever produce more power than it consumes, but there's been some interest in using them as compact neutron sources (that you can actually switch on and off).

You asked about fusing hydrogen-1 to hydrogen-2. These experimenters don't attempt that; they just buy deuterium, which is also readily available.


What type of glass are Pasteur pipettes made of?

This page from the glass shop in the Department of Chemistry and Biochemistry at the University of Delaware lists the properties of several common glasses that are used in laboratory equipment.

The two most common types of glass for less expensive applications are soda lime glass and borosilicate glass. The anneal point (or glass transition temperature) can be thought of as the temperature above which the glass can start to have some molecular motion, but it is not a liquid. The soften point is probably what you are looking for.

$$
\begin{array}{lcc}
\hline
\text{Glass} & \mathrm{Anneal\ Point\ ^\circ C} & \mathrm{Soften\ Point\ ^\circ C} \\
\hline
\text{Soda Lime} & 545 & 724 \\
\text{Borosilicate} & 560 & 821 \\
\hline
\end{array}$$

Soda lime glass is cheaper, and is generally used when you want something very inexpensive or disposal. Borosilicate glass has a lower thermal expansion coefficient so it is less likely to break. If you want to know which type of glass your pipets are made from , look at the wide end from an angle. If there is a slight greenish cast to the curve of the glass, you have soda-lime glass. Or you could look at the box they came in.


Short answer is: Yes and no. It depends on what acid-base theory is applied.

Redox reactions like $\ce{Mg -> Mg^2+ + 2 e-}$  are not usually considered acid-base reactions, but  over-generalized acid-base theories like Usanovich acid-base theory treat them as a special case of acid-base reactions.
Within this theory, $\ce{Mg}$ is relatively strong Usanovich base, $\ce{Mg^2+}$ is very weak Usanovich acid.

If we consider typical reactions by classical Broensted-Lawry theory:
$\ce{H2SO4 + H2O -> HSO4- + H3O+},$
or $\ce{NH3 + H3O+ -> NH4+ +H2O}$,
or by another classic - Lewis theory:
$\ce{NH3 + BF3 -> H3N-BF3}$,
then atoms involved in acid-base interactions have the same oxidation number before and after the reaction.
$\ce{Mg^2+(aq)}$ is the conjugate acid to the conjugate base $\ce{MgOH+ (aq)}$.
$$\ce{[Mg(H2O)6]^2+ + H2O <=> [Mg(H2O)5(OH)]+ + H3O+}$$
$\ce{MgOH+ (aq)}$ is is then the conjugate acid to $\ce{Mg(OH)2 (aq)}$.
$$\ce{[Mg(H2O)5(OH)]+ + H2O <=> [Mg(H2O)4(OH)2] + H3O+}$$
$\ce{\pu{Mg}}$ has in all 3 species the oxidation number +2.


Ammonia water, AKA ammonium hydroxide ($\ce{NH4OH}$), reacts with formaldehyde to form hexamine, a comparatively innocuous solid (though it is flammable when dry).

Safer might be Neutralization of Formaldehyde Gas by Ammonium Bicarbonate and Ammonium Carbonate, which would greatly reduce the flammability of the product.


Has a Gaussian expansion of STOs of higher quantum numbers (for example, 2s or 3p STOs) been performed

I am unaware of such fitting. Typically, basis sets using higher angular momenta (cc-pV$n$Z, def2-$n$ZVP, pc-$n$) are fitted on total energies or correlation energies. See https://www.basissetexchange.org/ for examples and references.

  Are there any GTO integration schemes for Cartesian GTOs of higher principal quantum numbers?

Yes, there are. I have seen the following approach in action before:

hand-crafted algorithms are used for common, low-angular-momentum cases
more general schemes, but still optimized schemes for higher cases
a catch-all, not very efficient, unspecific algorithm above a certain threshold


Why do we use phenolphthalein as an acid-base indicator?

If you analyse a titration curve you will immediately notice the pH shooting up near equivalence point. For this reason, there is very little difference to the amount of base added for pH 7 and pH 8 (due to the logarithmic nature of pH). So, it's not entirely accurate but certainly good enough (for strong acid - strong base and weak acid - strong base), if you want something better use an electronic pH tester


Normal phase

Normal phase HPLC systems are similar to the flash-column chromatography that you might be familiar with.

A silica stationary phase is eluted with a non-polar solvent such as hexane, or a fairly non-polar solvent mixture such as 2-propanol in hexanes. In normal phase chromatography, only organic solvents are used.

In the normal phase, polar molecules elute slowly, and non-polar (greasy) molecules elute quickly.

Reverse phase

Reverse phase is essentially the opposite of normal-phase.

A non polar stationary phase (often silica in which the free hydroxyl groups are end-capped with something greasy, C18 chains are common but many many variants are possible) is eluted with a polar solvent such as acetonitrile/methanol, or a fairly polar solvent mixture (acetonitrile water mixtures are common, or methanol water mixtures).

In the reverse phase, polar molecules elute quickly, and non-polar (greasy) molecules elute slowly.


Isomers of Octahedral Coordination Compound with Two Bidentate Ligands

Note: I am answering this as a homework question, i.e. not actually giving a definite answer.

As a first idea, reduce the degrees of freedom in your complex. You could start by assuming what the number of isomers is for $\ce{[CoCl2(en)2]}$, in which the two non-ethylenediamine ligands are identical. And you can further reduce that complex’ complexity by going to something like $\ce{[CoCl2(NH3)4]}$, in which the ethylene bridges have also been removed. (It may have made more sense to stick with $\ce{NH3}$ ligands in the first step, but then I cannot simply substitute for ammine in the second step, so I chose something entirely different there. What you choose is up to you, as long as you don’t create ambiguity.) For the tetraammine complex, it should be clear that there are two different isomers: cis and trans. Upon readding the ethylene bridge, these two isomers still exist. Do your four structures account for this isomerism?

Next up, ask yourself whether any of the isomers of $\ce{[CoCl2(en)2]}$ can be optically active, that could give you another set of isomers. This is important because of the next step.

This next step involves readding the difference between the two monovalent ligands, i.e. going back to $\ce{[Co(NO2)(en)2(NH3)]+}$. Remember to consider the topicity of the two chlorides you had before in case you run into enantiotopic or diastereotopic chlorides that would give you a new set of isomers.

Finally, remember that nitrite has two or three possible means of attaching to a metal centre: κN, κO and κO,O'. Remember to consider whether all theoretical possibilities are actually possible.

The final answer I came up with is given in a spoiler tag.

   The trans-isomer has a high symmetry of $D_\mathrm{2h}$ if I did the transforming correctly, meaning that it is does not have enantiomers. Also, it renders the two chlorides homotopic, meaning that we get one trans isomer.

   There are two cis-isomers that behave as enantiomers of each other: a Δ and a Λ isomer. These two each have homotopic chlorides again, meaning we get a set of two cis-isomers. To be honest, I only saw that when I tried it out with a modelling kit.

   This means that substituting one of the two chlorides with something else will result in identical compounds, giving us three different isomers. (Note that you have drawn two pairs in which one transforms into the other by simple rotation. You only have two different complexes drawn in the question.)

   The chelating nitrite ion is not possible in your complex since you have a rather small metal centre that only allows octahedral coordination. Therefore, each of the three isomers we established earlier can exist in a κN and a κO form. That gives us a grand total of six possible isomers whereof two pairs are enantiomers of each other and three are constitutional isomers of the other three.


We want to perform the reaction you mentioned:
$$N_2+3H_2\longrightarrow 2NH_3$$

At the beginning we have certain concentrations of reagents, say $[N_2]_{in}$, $[H_2]_{in}$ and $[NH_3]_{in}$. The reason why I am assuming we have some ammonia is just a way for not dealing with infinite values of energy as it will be clear soon. Since Chemists often work at constant temperature and pressure we will follow the spontaneity of this reaction following the changes in the Gibbs Free energy $G_r(T,P)$ of reaction. A model describing the Gibbs free energy changes is the following:
$$\Delta G_{r}(T,P)=\Delta G_r^0(T,P=1bar) + RT \ln\biggl(\frac{[NH_3]^2}{[N_2][H_2]^3}\biggr)$$
The reason we have chosen not to have a zero concentration of ammonia is because in that case we would get $\Delta G_{in}=-\infty$ and this is not sensible. It is important to note that $\Delta G^0$ only depends on the temperature since the pressure has been fixed at 1 bar. Another aspect to note is that the concentrations in square brackets are the concentration at a certain point of the reaction. Now we are ready to find the answer to the question. What happens at the equilibrium? By definition the equilibrium is reached when the Gibbs free energy is minimum, thus when $\Delta G_r =0$. Taking the above equation and setting this condition, isolating the concentrations from the logarithm, we get that the concentrations of our reagents/product at the equilibrium are
$$\frac{[NH_3]_{eq}^2}{[N_2]_{eq}[H_2]_{eq}^3}=\exp\biggl(-\frac{\Delta G_r^0}{RT}\biggr)=K_{eq}$$
Clearly the equilibrium constant depends only on the temperature and not on the pressure since in $\Delta G_r^0$ pressure has been fixed at 1 bar by definition. Changing the total pressure does not effect the amount of ammonia you get as you can see from a "mathematical" point of view. However in order to increase the product concentration we can increase the concentration of the reagents (in this case the partial pressure since we are in gas phase) since what is really fixed at the equilibrium for a given temperature is $\frac{[NH_3]_{eq}^2}{[N_2]_{eq}[H_2]_{eq}^3}$ and not the concentrations taken singularly.


Is nitrogen of aniline sp² or sp³ hybridized?

The nitrogen in aniline is somewhere between $\mathrm{sp^3}$ and $\mathrm{sp^2}$ hybridized, probably closer to the $\mathrm{sp^2}$ side.  We are correctly taught that the nitrogen in simple aliphatic amines is pyramidal ($\mathrm{sp^3}$ hybridized).  However in aniline, due to the resonance interaction between the aromatic ring and the nitrogen lone pair, considerable flattening of the nitrogen occurs (if it were completely flat it would be $\mathrm{sp^2}$ hybridized).

We can assess the nitrogen hybridization by measuring its barrier for pyramidal inversion. If a trigonal nitrogen is $\mathrm{sp^2}$ hybridized, the barrier will be zero. On the other hand, in aliphatic amines where the nitrogen is $\mathrm{sp^3}$ hybridized the inversion barrier is typically around $\pu{4-5 kcal/mol}$.

(pyramidal inversion diagram)

In aniline this barrier is very low, somewhere around $\pu{1-2 kcal/mol}$. This indicates that the nitrogen in aniline is not quite planar, but is much closer to being  planar ($\mathrm{sp^2}$) than pyramidal ($\mathrm{sp^3}$).


Short answer: Yes.
Longer answer:
Isotopes are atoms of the same element that have different numbers of neutrons in their nuclei. Everything else about them is the same. The numbers of protons and electrons in a neutral atom is governed by the atomic number.
So, for example, chlorine has two naturally occurring isotopes:
$$\begin{array}{|c|c|c|c|c|c|c|}\hline
\mathrm{isotope} & Z & A & p & n & e & \mathrm{abundance}\\ \hline
\ce{^35_17Cl} & 17 & 35 & 17 & 18 & 17 & 75.78\%\\
\ce{^37_17Cl} & 17 & 37 & 17 & 20 & 17 & 24.22\%\\ \hline
\end{array}$$
Where:
$$\begin{aligned}
&Z=\mathrm{atomic\ number}\\
&A = \mathrm{mass\ number}\\
&p = \mathrm{number\ of\ protons}\\
&n = \mathrm{number\ of\ neutrons}\\
&e = \mathrm{number\ of\ electrons}\\
\end{aligned}$$
Isotopes generally behave the same chemically1 since almost all chemistry is governed by the number of electrons ($Z$). Thus, anything that contains chloride ions contains a roughly 3:1 ratio of $\ce{^35Cl-}$ and $\ce{^37Cl-}$. We could call these ion isotopes or isotope ions, but in most all cases, we do not distinguish between the two in reactions or compounds involving chloride.
1 There is one caveat in the "isotopes behave the same chemically" rule of thumb that is worth mentioning. Since isotopes have different masses, they can produce kinetic isotope effects. Given the $KE=\frac{1}{2}mv^2$, two isotopes with different masses will move at different velocities and have the same kinetic energy. Heavier isotopes generally move more slowly.
This effect is most pronounced in hydrogen, where the second most common isotope $\ce{^2H}$ ($\ce{D}$) is twice as massive at the most common isotope $\ce{^1H}$ ($\ce{H}$). If two atoms have the same kinetic energy, then:
$$\begin{aligned}
m_1 v_1^2 &=m_2 v_2^2\\
\dfrac{m_1}{m_2} &=\dfrac{v_2^2}{v_1^2}\\
\sqrt{\dfrac{m_1}{m_2}} &=\dfrac{v_2}{v_1}
\end{aligned}$$
For the isotopes of hydrogen, the heavier isotope of hydrogen moves at only 70% of the velocity of the lighter isotope. This kinetic isotope effect can be used to probe reaction mechanisms. If the rate decreases dramatically when $\ce{H}$ is swapped for $\ce{D}$, then that hydrogen atom is involved in the rate determining step.
$$\begin{array}{|c|c|c|c|}\hline
\mathrm{isotope} & A & \mathrm{rel.\ mass} & \mathrm{rel.\ velocity}\\ \hline
\ce{^1H} & 1 & 1 & 1 \\
\ce{^2H} & 2 & 2 & \dfrac{1}{\sqrt{2}}\approx 0.707\\ \hline
\end{array}$$
For chlorine, the outcome is a little different. Since the ratio of the masses of the two isotopes is close to 1, the ratio of the velocities is also close to 1.
$$\begin{array}{|c|c|c|c|}\hline
\mathrm{isotope} & A & \mathrm{rel.\ mass} & \mathrm{rel.\ velocity}\\ \hline
\ce{^35Cl} & 35 & 1 & 1 \\
\ce{^37Cl} & 37 & 1.06 & 0.97\\ \hline
\end{array}$$


What about Miller Indices of planes passing through origin?

Since a comment does not allow the inclusion of a picture, perhaps this way is helpful:

Have a look at the following picture, derived from an illustration about the Miller Indices on wikipedia.  It depicts a simplified situation of a unit cell which may occur if your crystal is of the cubic lattice system, i.e., the three vectors defining the unit cell are orthogonal to each other, and all are of same length. In the drawing, the author choosed $\vec{a_1}$, $\vec{a_2}$, and $\vec{a_3}$ for the unit cell's dimensions.

Not only the plane closest to the coordinate system's origin drawn is a valid Miller plane, equally so the second, farther one drawn.  For each set of Miller indices $(hkl)$, the corresponding Miller planes are parallel to each other.  The distance between these planes depends both on the indices, as well as on the symmetry of the crystal.

Since a crystal consists of many unit cells adjacent to each other, it is perfectly reasonable to draw Miller planes either in "positive direction" along vectors $\vec{a_1}$, $\vec{a_2}$, and $\vec{a_3}$ -- as shown here -- or in opposite direction along $-\vec{a_1}$, $-\vec{a_2}$, and $-\vec{a_3}$ with the same set of $(hkl)$.


Trying to understand the chemistry in an iron carbon battery

Ions in solution

  [OP] I understand that ferrous chloride dissociates in water to $\ce{Fe^2+}$ and $\ce{Cl-}$ ions.

Yes. The same goes for the iron(III) species.

  [OP] I'm told that iron(II) chloride loses electrons to the positive electrode and becomes iron(III) chloride.

No, iron(II) loses electrons and becomes iron(III). The chloride is less relevant, and could be exchanged for a different ion.

  [OP]  How does this happen if the iron(II) chloride has dissociated in to ions, there is no iron(II) chloride anymore?

The reactant is $\ce{Fe^2+(aq)}$, not $\ce{FeCl2(s)}$.

Ion movement

  [OP] During charge the $\ce{Fe^2+}$ ions are attracted to to the negative iron electrode and the $\ce{Cl-}$ ions are attracted to the positive carbon electrode.

Yes, but a small fraction, only. Otherwise the repulsion between like ions would be too great.

  [OP] My problem is understanding what happens at the positive carbon electrode. I'm told that iron(II) chloride loses electrons to the positive electrode and becomes iron(III) chloride.

Just $\ce{Fe^3+(aq)}$. There might be a double layer, but no solid iron(III) chloride.

  [OP] Fine, but how does this happen if all the $\ce{Fe^2+}$ ions are at the negative electrode?

There are plenty of cations in the bulk of the solution, and plenty near either electrode.


Why does acetylene, having an acidic proton, not react with strong bases like NaOH and KOH?

The $\mathrm{p}K_\mathrm{a}$ of acetylene is $25$. It is very weakly acidic, and hence, $\ce{KOH}$ is not sufficiently alkaline to react with it. Even methanol, with a $\mathrm{p}K_\mathrm{a}$ of $15.5$ does not react much with hydroxide ion. On other hand, $\ce{NaH}$ or $\ce{NaNH2}$ have more basic anions than hydroxide, and weak acids such as acetylene or alcohols can react with them.

Predicting the relative basicities can be done qualitatively by invoking the Brønsted conjugate acid-base theory. The rule to remember is, a base gives its corresponding conjugate acid in solution, and stronger the conjugate acid, weaker its conjugate base (and vice versa). This can be understood from the following

$$\ce{HA<=>H^+ + A-}$$

Here, $\ce{A-}$ is the conjugate base of the acid $\ce{HA}$. Now, if $\ce{A-}$ is highly unstable, it will have a stronger tendency to react with a proton and get converted to its conjugate acid which is relatively more stable. Since the equilibrium lies more to the left in that case, we say that the conjugate base is strong (and consequently the conjugate acid is weak). Now consider the following anions for the sake of example.

$$\ce{HC#C-} > \ce{NH2-} > \ce{OH-} > \ce{F-}$$

These are listed in decreasing order of stability, and hence also their strength as bases. This is because as the electronegativity increases from carbon to fluorine, the negative charge becomes increasingly stable on the atom. More electronegative elements are increasingly stable with the extra electron due to higher effective nuclear force of attraction. Correspondingly, by the dictum "stronger base is weaker conjugate acid", the order of the acidity is:

$$\ce{HC#CH} < \ce{NH3} < \ce{H2O} < \ce{HF}$$


How do the stored chemicals in batteries drive electronic current from their terminal to outside devices?

A battery, at its heart, is what we call a redox cell: some compounds are stripped of their electrons, and others gain those electrons. The movement of electrons between these two compounds can be harnessed to provide electricity, but that's taking us a step too far. What we want to understand here is why the electrons move. To do this, we'll have to understand the redox cell. As much as I hate to assume things, you're going to have to understand a bit about ions and equilibrium to understand this explanation.

(Note: the explanation I'm about to use is heavily taken from the excellent Chemguide site, run by Jim Clark. I'll include a link at the end.)

Let's say I place a bar of magnesium in a water bath. I'll take a look at the magnesium bar a short while after I place it in the water. If I could visualize atoms and electrons, this is what I'd see:

Some of the atoms of magnesium leave the bar and become ions in the solution, solvated by water (and other things, but that's not crucial to this explanation). Since they become ions, they leave electrons on the metal bar. Because the ions are positively charged, they are attracted to the negative charge on the metal bar, and tend to hang around there.

If I wait a little longer, I'll see more and more ions coming off the bar. As they start to build up in the water around the metal, every once in a while, one of the ions will hit the bar and reclaim its electrons. This is what we'll see then:

If I wait long enough, eventually, the number of atoms coming off the metal will be equal to the number of ions reabsorbing onto the metal. If I take out the "bites" in the metal bar to simplify the situation, this is what I see:

Notice how there's quite a buildup of electrons on the magnesium. This is because magnesium is a very active metal, like lithium. What if we used a much less active metal, one that doesn't have a large propensity to lose atoms into solution? An example of such a metal would be copper. Here's a side-by-side of the two:

Because the magnesium is a lot more active, it tends to lose ions more, and so it has more electron buildup. If we connect the two with a wire, we give the electrons a way to flow from the high concentration to the low concentration. We can then place something along the wire in the middle (say, a lightbulb) and harvest some energy!

This, of course, is a hilariously oversimplified scheme. For one, as the electrons go to the copper, more and more magnesium ions break off to replace them. At some point, there are so many positive ions in the solution that they draw the electrons back toward them, through electrostatic attractions (which is a fancy way of saying that positive and negative charges attract each other). To solve this, we'd put a salt bridge between the solutions that lets ions migrate back and forth.

There are other problems too: for example, this setup wouldn't work in real life because there are no ions in solution: real cells use salt solutions of the metal e.g. $\ce{MgCl2}$ or $\ce{CuSO4}$. There's also the issue that this battery is liquid and wouldn't do so well if you tipped it over, the fact that the voltage (which is really the difference between the charge on the metal bars) is only about 1.1 volts, yada yada. That's not central to the explanation.

The fact that certain compounds are more likely to gain electrons and some are more likely to lose them is the central reason why things like batteries can exist. If you'd like to read more about it, I pulled this explanation from chemguide.co.uk, run by Jim Clark. I am currently ready to graduate with a B.Sc in chemistry, and his website is still one of the best resources for conceptual learning that I've come across---as a matter of fact, I learned about things on that website (eutetic mixtures and azeotropes) that I still have yet to come across in four years of university-level study. If you'd like to read more into redox reactions, the redox equilibria page is at this link right here.


Which is the correct name: 2,3-dimethyl cyclopentene or 1,5-dimethyl cyclopentene?

The correct name is 1,5-dimethylcyclopent-1-ene. The double bond has the highest priority and so these two carbons must be numbered 1 and 2 if possible (yes in this case because the molecule is cyclic). We then have to choose between numbering the methyl substituents 1,5 or 2,3. The lowest locant sum rule that you mentioned doesn't actually exist, even though it is often taught. It is a agglomeration of several rules which works most, but not all of the time. The actual rule we are interested in is the first point of difference rule which states:

  2.2 - The longest chain is numbered from one end to the other by Arabic numerals, the direction being so chosen as to give the lowest numbers possible to the side chains. When series of locants containing the same number of terms are compared term by term, that series is "lowest" which contains the lowest number on the occasion of the first difference. This principle is applied irrespective of the nature of the substituents.

Therefore the locant set 1,5 is lower than the locant set 2,3 since at the first point of difference, 1 is lower than 2.


Actually, there are three reasons why BF3 is more acidic than AlCl3.

BF3 is smaller in size and can easily attract the incoming pair of electrons.
Fluorine is more electronegative than chlorine. So boron has less ionisation potential as compare to aluminium thus forms anion easily.
Boron is more electronegative than aluminium, therefore boron has more tendency to attract electrons. It means that BF3 behaves more like Lewis acid than AlCl3.

This photo has been taken from this article.


How many molecules are there typically in items made of plastics?

The short answer is a very large number and far more than you could reasonably count

Polymerization reactions typically produce very large molecules but not so large that a single molecule would make up as much material as a whole plastic bag or item of any significant size. This source on LDPE tells us that the typical molecular weight of the industrial product is 90,000 (some polymers are much bigger than this).

For non-chemists molecular weights are (simplifying slightly) the weight relative to a single hydrogen atom. Carbon is 12 on this scale and ethylene (the monomer from which the plastic is made which consists of 2 carbons and 4 hydrogens) is 32. This mean that a typical molecule of the plastic is (crudely) made from a little under 3,000 ethylene molecules. The other fact you need to know is that there are about 6 x 1023 hydrogen atoms in a gramme of hydrogen (see the wikipedia definition of Avogadro's number). Simple mathematics tells us that a gramme of LDPE will contain roughly 6.7 x 1018 molecules so that is the order of the number in small plastic item which, I expect, weighs just a few grammes.

So, the bottom line is that even in a small plastic item consisting of a long chain polymer (LDPE is relatively short but other polymers might have chains 10-100 times more at most) there will be something like 1016 to 1018 molecules of polymer, which is a lot.


How to calculate the Gibbs free energy from the equilibrium constant for the decomposition of sodium bicarbonate?

For your reaction:

$$\ce{2 NaHCO3(s) ⇌ Na2CO3(s) + H2O(g) + CO2(g)}$$

$$K=P_{\ce{CO2}}\cdot P_{\ce{H2O}}$$

However, $K\ne0.545\text{ atm}$ because the sum of the pressures is $0.545\text{ atm}$ and the equilibrium constant is the product of the pressures.

$$P_{\ce{CO2}}+P_{\ce{H2O}}=0.545\text{ atm}$$

Since you started with no gas, and the stoichiometric coefficients of both gases are equal, then

$$P_{\ce{CO2}}=P_{\ce{H2O}}=\dfrac{0.545\text{ atm}}{2}$$

  What is the value of $K$ now?


Why does 1,2-dichloro-4-nitrobenzene undergo SNAr para and not meta to the nitro group?

You can obtain an answer by looking at the electronic structure.  $\ce{NO2}$ is a meta director for electrophilic aromatic substitution (EAS).  The question asks about nucleophilic aromatic substitution (NAS). Aromatic substitutions involve a free pair of electrons resonating around a structure. Electrophiles may add at any resonance site where the pair of electrons are. Nucleophiles may inversely add to any site where electrons are not present shown in the picture below. Therefore if a functional group is meta directing for EAS it will be ortho/para directing for NAS and vice versa.


Pepto-bismol molar bismuth concentration

Your calculation is correct. You are not missing anything obvious. Your set up, which converts mL to L and then mg to mmol is correct and you get the correct units.

$$\require{cancel}
\mathrm{\dfrac{17.5\ \cancel{mg}}{\cancel{mL}}\times \dfrac{1000\ \cancel{mL}}{L}\times \dfrac{mmol}{364.10054\ \cancel{mg}} =48.2\ mM}$$

This concentration is not "too high". It is what it is. 48.2 mM is 0.0482 M, which is a pretty low concentration.

For comparison, your 17.5 mg per mL is at most 2% by mass (low-balling the density of the suspension to be around 1). 1 M hydrochloric acid (which perhaps seems like a high concentration) is around 3% by mass. Concentrated HCl is 36% by mass and about 12 M.


Can anyone explain the reaction mechanisms of these synthesis?

The ammonium chloride, calcium hydroxide, ammonium sulfate, and sodium hydroxide are all ionic compounds, and when dissolved in water they dissociate into $\ce{NH4+}$, $\ce{Cl-}$, $\ce{Ca^2+}$, $\ce{SO4^2-}$, and $\ce{Na+}$ ions. The $\ce{NH4+}$ ammonium ion can donate a proton to the $\ce{OH-}$ ion, creating $\ce{NH3}$ and $\ce{H2O}$. The other ions don't really do anything, so these spectator ions remain in solution but the charges are still balanced. If you removed the ammonia and water, which is easy with a vacuum, you'd be left with the $\ce{CaCl2}$ and $\ce{Na2SO4}$ salts. These reactions will have some equilibrium, with $\ce{H2O}$ donating a proton back to $\ce{NH3}$ to form $\ce{NH4+}$ and $\ce{OH-}$ again.


Yikes.

As far as I'm aware, a thorough application of ordinary soap or washing-up liquid should be able to wash this off using the normal surfactant processes (binding to organic components with lipophilic regions of the molecule). If you're experiencing long-lived after-effects (say, more than a week maybe), you might want to see a doctor -- it's not impossible you could have had some sort of allergic reaction. Oh, and if you're going to wash hair with soap or washing-up liquid, I'd suggest using conditioner afterwards, to reduce the long-term damage incurred by stripping the natural fatty acids and oils off.

I'm not a medical professional or a tricologist, though, so please don't take this as Official Medical Advice.


Changes in defects in a material due to heat treatment

It seems, that your oxide can have different crystal polymorphs and probably also other non-crystalline phases.

You synthesize your material as mixture of those, where the composition depends on the preparation process.

Upon heating and slow cooling you probably recrystallize the species to more thermodynamically stable polymorph (new peaks c,d,e) on the expense of less stable one (a).
It can also depend on the speed of cooling, so you could obtain some metastable phases, etc...

Even then, treat your material always as an unkonwn mixture of uknown number of polymorphs and let the spectra tell you in which way the composition shifted.


What are computed energies in DFT and FF methods relative to?

Start by looking at the Hamiltonian for a molecular system
\begin{equation} \label{eq:coulomb_hamiltonian}
    \hat{H}
    = - \sum\limits_{α=1}^{ν} \frac{1}{2 m_{α}} \nabla_{α}^{2}
    - \sum\limits_{i=1}^{n} \frac{1}{2} \nabla_{i}^{2}
    - \sum\limits_{α=1}^{ν} \sum\limits_{i=1}^{n} \frac{Z_{α}}{r_{αi}}
    + \sum\limits_{α=1}^{ν} \sum\limits_{β > α} \frac{Z_{α} Z_{β}}{r_{αβ}}
    + \sum\limits_{i=1}^{n} \sum\limits_{j > i}^{n} \frac{1}{r_{ij}} \, ,
\end{equation}
where $m_{α}$ is the rest mass of nucleus $α$ and $Z_{α}$ is its atomic number, $r_{αi} = |\vec{r}_{α} - \vec{r}_{i}|$ is the distance between nucleus $α$ and electron $i$, $r_{αβ} = |\vec{r}_{α} - \vec{r}_{β}|$ is the distance between two nuclei $α$ and $β$, and $r_{ij} = |\vec{r}_{i} - \vec{r}_{j}|$ is the distance between two electrons $i$ and $j$.

Now, thinking classically, the total energy is equal to zero when all the terms in the Hamiltonian are equal to zero, so that the zero of the total energy corresponds to the case when, (again) classically speaking, particles are infinitely far away from each other and not moving. Obviously, the total energy of any stable molecular system should be lower than that, i.e. negative, for otherwise the system will be unstable with respect to dissociation towards particles. That is the way it works for any method of computational chemistry that solves the Schrödinger equation or its relativistic counterpart is some way or another (HF, post-HF, DFF, etc.).

But force field methods and energies are completely different beasts, because the meaning and the zero of energy in this realm depends on the functional form of potential energy and parameter set used in a particular force field. So, in molecular mechanics in general absolute energies have no meaning, only relative energies (energy differences) should be used for any kind of analysis.


An ideal gas is undergoing an isothermal expansion against a constant external pressure. Which of the following statements is true?

If the gas is ideal, for an isothermal expansion

$$p=\frac{nRT}{V}=\frac{\textrm{constant}}{V}$$

As the volume of the system changes, so must the pressure.

Now a condition of reversibility is mechanical balance between the system and surroundings, that is, $p=p_{\textrm{ext}}$. However, if a process is performed at constant external pressure and temperature, then as shown above the condition of mechanical balance cannot be satisfied, since $p$ will change during the expansion whereas $p_{\textrm{ext}}$ is constant. Therefore the process must be irreversible.


The issue is that basicity has a couple of meanings and you have to be able to tell from the context what it means.

The first is logical, and is used to describe how basic a base is. That is the trend that you mentioned about the Group 15 hydrides. (In my experience, this is by far the more common usage of the word.)

The second is somewhat confusing, and it is used to describe how many protons of an acid can be removed by neutralisation with base. For example, $\ce{HCl}$ has one acidic proton, so it is monobasic; $\ce{H2SO4}$ has two acidic protons, so it is dibasic.


Quick and simple explanation of molar mass, molecular mass and atomic mass

The quantity molar mass (quantity symbol: $M$) for a pure sample is defined as
$$M = m/n$$
where $m$ is mass and $n$ is amount of substance.
The dimension of the molar mass is
$$\dim M = \mathsf{M}\;\mathsf{N}^{-1}$$
The coherent SI unit for molar mass is ‘kilogram per mole’ (unit symbol: $\mathrm{kg/mol}$).
(The usually used unit is gram per mole ($\mathrm{g/mol}$) rather than kilogram per mole ($\mathrm{kg/mol}$).)

The quantity relative molecular mass (quantity symbol: $M_\mathrm{r}$) is defined as the ratio of the average mass per molecule or specified entity of a substance to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$; i.e. the relative molecular mass is the ratio of the mass of a molecule to the unified atomic mass constant:
$$M_\mathrm{r} = m_\mathrm{f}/m_\mathrm{u}$$
where $m_\mathrm{f}$ is mass of entity (molecule or formula unit) and $m_\mathrm{u}$ is the unified atomic mass constant.
The relative molecular mass is a quantity of dimension one (for historical reasons, a quantity of dimension one is often called dimensionless):
$$\dim M_\mathrm{r} = 1$$
The coherent SI unit for relative molecular mass is the unit one (symbol: $1$).
For historical reasons, the IUPAC accepts the use of the special name ‘molecular weight’ for the quantity relative molecular mass. The use of this traditional name is deprecated.

For atoms, the relative molecular mass $M_\mathrm{r}$ is the relative atomic mass, and the quantity symbol $A_\mathrm{r}$ may be used. The quantity relative atomic mass is defined as the ratio of the average mass per atom of an element to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$:
$$A_\mathrm{r} = m_\mathrm{a}/m_\mathrm{u}$$
where $m_\mathrm{a}$ is the atomic mass (see below) and $m_\mathrm{u}$ is the unified atomic mass constant.
The relative atomic mass is a quantity of dimension one:
$$\dim A_\mathrm{r} = 1$$
The coherent SI unit for relative atomic mass is the unit one (symbol: $1$).
For historical reasons, the IUPAC accepts the use of the special name ‘atomic weight’ for the quantity relative atomic mass. The use of this traditional name is deprecated.

The quantity atomic mass (quantity symbol: $m_\mathrm{a}$) is defined as rest mass of a neutral atom in the ground state.
The dimension of the atomic mass is
$$\dim m_\mathrm{a} = \mathsf{M}$$
The coherent SI unit for atomic mass is ‘kilogram’ (unit symbol: $\mathrm{kg}$).


What is the standard heat of formation of the nitrate ion, based on the standard enthalpy of formation of silver nitrate?

Silver nitrate, $\ce{AgNO3}$ is a strong electrolyte. Based on the information given, I expect you are meant to assume that it dissociates completely when dissolved in water:

$$\ce{AgNO3(aq) -> Ag+(aq) + NO3- (aq)}$$

Thus, we can make the assumption that $$\ce{AgNO3(aq)} \equiv \ce{Ag+(aq) + NO3- (aq)}$$

Then, we can state:

$$\Delta_\mathrm{f} H (\ce{AgNO3(aq)}) = \Delta_\mathrm{f} H (\ce{Ag+(aq)})+\Delta_\mathrm{f} H (\ce{NO3- (aq)})$$

$$\implies \pu{-99kJ/mol} = \pu{+105kJ/mol} +\Delta_\mathrm{f} H (\ce{NO3- (aq)})$$

Thus $$\Delta_\mathrm{f} H (\ce{NO3- (aq)})=\pu{-204kJ/mol}$$


What are the pros and cons of using Gilman reagent over Grignard reagent to synthesize a hydrocarbon?

I am not sure if Gilman reagents are harder to make than Grignard but they are softer nucleophiles and so they are used to perform 1,4-(conjugate) additions on $\ce\alpha,\beta-$unsaturated electrophiles. Grignard reagents will generally go for 1,2-attack instead. See this Wikipedia article for more info:


Why does the quantum number n limit the value of l, and why does l, likewise, limit the value of m?

The shapes of the orbitals, i.e. the solutions to the Schrödinger equation for the hydrogen atom $\hat{H}\Psi_i = E_i\Psi_i $, are primarily due to the requirement that all orbitals are orthogonal: $\langle\Psi_i|\Psi_j\rangle=0$.  Orthogonality is made possible by the fact that the orbitals have "positive" and "negative" regions separated by nodes.

One what to think about $n$ is that $n-1$ is the number of nodes in $\Psi_i$. So there can be no $1p$ orbitals since $p$-shaped orbitals have at least one node. Similarly, there can be no $2d$ orbitals, and so on.

Because $s$ orbitals have a spherical shape, there is only one way to orient them, i.e. $m=0$ if $l=0$. $p$-shaped orbitals lie along one axis and there are three Cartesian axes, so there are three possible - orthogonal - orientations (I'm sacrificing a bit of accuracy here for clearness).

It's a bit harder to visualize why there should be 5 and only 5 $d$-shaped orbitals, but the answer is that there are only 5 ways to construct 3-dimensional shapes with (a minimum of) 2 nodes that are orthogonal to the $s$- and $p$-shaped solution as well as the other $d$-shaped solutions.


No, most semiconductors are ionic.

I'm not sure I've seen the term "covalent semiconductor" before. Wikipedia has an article on covalent superconductors.

I think the main point would be that many if not most semiconductors are inorganic solid-state compounds. Consequently, most of these (e.g. III-V semiconductors are ionic solids, not really covalent ones. Consider GaAs or InP, which are typically classified as ionic.

That is the solid state electronic structure can be considered mainly composed of cations and anions (unlike organic chemistry where the electronic structure is considered mainly of neutral atoms).

(Incidentally, defining whether some of these compounds have purely ionic or purely covalent interactions is somewhat futile, since most bonding interactions are both somewhat ionic and covalent.)


Software recommendations for overlaying molecular structures

You can use the "compare" command in the Jmol program, either online or as free-standing program. A tutorial on how to use the program for this purpose is here, http://proteopedia.org/wiki/index.php/Superposition_with_jmol. It outputs the root mean square of the pairwise differences (RMSD) as well as the superposed coordinates.


Why are arenes with conjoined benzene rings drawn as they are?

Questions about why we represent things in one way or another are hard to answer in general, because they appeal to one’s sense of taste… de gustibus non est disputandum. Here, I want to state the position and the reasoning of IUPAC on this matter. The relevant publication on this are the IUPAC 2008 Recommandations, titled Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008) [1].

Section GR-6 of the report deals with “Aromatic rings and other types of electron delocalization”. I quote part of it (emphasis is mine):

  GR-6.5 Curves should only be used when delocalization is being represented

  Curves represent delocalization, yet lots of structures have some delocalized elements. Although all such elements could be depicted with curves, there is little gained by doing so. On the contrary, the arbitrary use of curves can draw the viewer’s attention to insignificant portions of a structural diagram, and away from areas that are chemically more important such as an active site or a reactive group. Accordingly, curves should only be used when the delocalization is specifically being highlighted as an important feature of the structure.

  When curves are not used, any alternating configuration of double bonds is acceptable within the further constraints discussed in GR-3.5.

The text then goes further to note:

  It is generally not acceptable to use curves in two adjacent fused rings, since such diagrams are at best ambiguous in terms of the character of the shared fusion bond between the two rings.

Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008), Pure Appl. Chem., 2008 DOI: 10.1351/pac200880020277

PS: hmm, the leader for the IUPAC report works at CambridgeSoft…


What is the energy of the London dispersion force holding together two atoms of the same element?

Polarisability as used here has dimensions of volume. The London formula is $$U \approx -\frac{3}2\frac{I_aI_b}{I_a+I_b}\frac{\alpha _a \alpha _b}{r^6}$$  for two species a and b so you just need to square $\alpha$.


Coloration of substances because of F-Centers

For a simple explanation, we could explain the coloration of the substance by the absorption of the energy of the electrons in the void, which leads to the color of the compliment color (as is suggested by http://rspa.royalsocietypublishing.org/content/204/1078/406)

However, a more rigorous explanation could be found here: http://ptp.oxfordjournals.org/content/4/2/181.full.pdf

For the second question, it is written in Ceramic Materials: Science and Engineering that "the term color center is now applied to any defect, including an impurity, that produces color in an insulator". It also notes that the original observation was made by the production of F centers, which suggests that the modern interpretation of the phenomenon applies to the defects which product color in an insulator.

I'm not sure if I answered you question exactly as you hoped, but if you look into the resources, there is a wealth of knowledge there.


Difference between Resonance Effect and Mesomeric Effect

The IUPAC gold book (IUPAC are the organisation which 'define' standards for things such as chemical nomenclature) describes them as being essentially synonymous. The following definitions are given (link):

Resonance  In the context of chemistry, the term refers to the
representation of the electronic structure of a molecular entity in
terms of contributing structures. Resonance among contributing
structures means that the wavefunction is represented by 'mixing' the
wavefunctions of the contributing structures. The concept is the basis
of the quantum mechanical valence bond methods. The resulting
stabilization is linked to the quantum mechanical concept of
'resonance energy'. The term resonance is also used to refer to the
delocalization phenomenon itself.
Mesomerism Essentially synonymous with resonance. The term is particularly associated with the picture of π-electrons as less localized in an actual molecule than in a Lewis formula. The term is intended to imply that the correct representation of a structure is intermediate between two or more Lewis formulae.

There does seem to have been some historic difference that has largely been reconciled into a single definition however one use of Mesomerism that still gets commonly used is the definition of groups as having a positive (+M) or negative (-M) effect to describe the effect of a group on a conjugated system.

The Mesomeric Structures are all the compounds with all possible arrangements or the pi bonds drawn separately. The Resonance Structure is one diagram that uses dashes and lines to denote the partial double bond characters in the compound.

This statement is not entirely correct.  The 'Mesomeric' structures you talk about are usually refered to as resonance structures/forms. The depiction with the dashed lines and partial bonds is a resonance hybrid.


Does Methyl Iodide (CH3I) react with water (H2O) and if it does what would be its byproducts?

Methyl iodide is hydrolyzed by water in an $\ce{S_{N}2}$ reaction ($\ce{S_{N}1}$ is disfavored by the instability of the primary methyl cation), which yields methanol and hydroiodic acid as products.

$$\ce{CH3I + H2O <=> CH3OH2^+ + I-}$$
$$\ce{CH3OH2^+ + H2O <=> H3O+ + CH3OH}$$

A possible side reaction is the formation of iodine and methane by the redox reaction

$$\ce{CH3I + H3O+ + I- <=> CH4 + I2 + H2O}$$

However, this reaction becomes only important when a large concentration of methanol pushes the hydrolysis equilibrium to the educt side. It is negligible in pure aqueous solution (reference).


What type of bond(s) are present in the borohydride ion?

No, this is simply a bond between four hydrogen atoms ($\ce{H}$) and a boron anion ($\ce{B-}$). These bonds are identical to and indistinguishable from the other $\ce{B-H}$ bonds. They are all covalent bonds of length $1.21\:\mathrm{Å}$.

Your idea of $\ce{H-}$ completing $\ce{BH3}$'s $2\mathrm{p}_z$ orbital isn't quite right either. Atom's individually have atomic orbitals, but when they form bonds these atomic orbitals combine to form molecular orbitals. This is picture more accurately demonstrates the bonding between $\ce{B-}$ and the $\ce{H}$ atoms.

[Photo Source]


How to simulate a Lorentzian broadened spectrum in MATLAB?

Update : In the given data by the OP, the x scale is not evenly spaced so the discrete convolution approach will not work to generate a spectrum until and unless your data is evenly spaced! The trick mentioned by EdV is a very nice shortcut. Here, if you really want to play around with convolution via MATLAB, then read below.
Solution: You can satisfy yourself as follows:

Make sure you have a even spacing of the x-axis. Your uneven axis threw me off, and it took a while to discover why it is not working. I wish you mentioned it earlier.
Your sticks are basically scaled delta functions as Prof. Ed said.
Generate a vector for a single Lorentzian peak, whose center is zero, which is also evenly spaced. The x-axis of the Lorentzian would be symmetric [-length (wavenumber/2):Sampling rate:length(wavenumber/2)-Sampling rate].
Perform convolution of your spectrum and this Lorentzian.

Let us start with a Dirac delta centered at point $x.$ I don't know how to draw a displace Dirac delta in MATLAB. I manually made it in an Excel file. If you convolute it with a zero centered Lorentzian, then you will get a Lorentzian centered at $x.$ This is called the sifting property.
Basically, all you have to ensure is that your Lorentzian is centered at zero. First, play with convolution using single stick. Once you correct the code, so that the line position does not change, apply it on the entire spectrum.

A great book is Bracewell's Fourier Transform And Its Applications. It is all there. From MIT 2.14 / 2.140 Analysis and Design of Feedback Control Systems, Spring 2007 handout:

1.1 The “Sifting” Property of the Impulse
When an impulse appears in a product within an integrand, it has the property of ”sifting” out the value of the integrand at the point of its occurrence:
$$\int_{-\infty}^\infty f(t)δ(t - a)\,\mathrm dt = f(a)\tag{5}$$
This is easily seen by noting that $δ(t − a)$ is zero except at $t = a,$ and for its infinitesimal duration $f(t)$ may be considered a constant and taken outside the integral, so that
$$\int_{-\infty}^\infty f(t)δ(t - a)\,\mathrm dt = f(a)\int_{-\infty}^\infty δ(t - a)\,\mathrm dt = f(a)\tag{6}$$
from the unit area property.


If an action increases the reaction rate in a particular equilibrium mixture, does it mean that both forward and backward rates are increased?

The forward and reverse reaction rates both increase, but the reverse rate increases by more. By Le Chatelier's principle, we know that $K_{eq}$ for an exothermic reaction will decrease when the temperature is raised. We also know that $K_{\text{eq}}=\frac{k_{\text{forward}}}{k_{\text{reverse}}}$ and that $k=Ae^{\frac{-E_a}{RT}}$ (The  Arrhenius equation). Since this is an exothermic reaction, $E_a(\text{forward})<E_a(\text{reverse})$, which means any increase in $T$ will increase $k_{\text{forward}}$, but it will increase $k_{\text{reverse}}$ by more. This lead to a decrease in $K_{\text{eq}}$, consistent with what was predicted by Le Chatelier's principle.


When writing a paper should I describe the procedure thoroughly?

Use the following criterion in mind when you write a method in a paper: If you were reading this synthesis procedure from a book or a paper, could you replicate it or not? Do you feel anything is missing for proper duplication? When I review a methodology section of a papwe, I keep this criterion in mind: "Using this written information, can I reproduce his/her work?", otherwise I ask for more information.
Since you mentioned synthesis, a good sample procedure could be from Organic Synthesis http://www.orgsyn.org/
See how they write good procedures, and these methods are verified by independent person. Additionally, some journal put restrictions on word count, but still you want more details. For that, Supporting Information file exists on most good quality journals. I once saw someone even added pictures of their synthetic procedure/apparatus. Nobody will mind but it is not common.
Odd cases: Mentioning the source of glassware might be important if glassware surface were affecting the reaction. I "re-discovered" this only once. Sometimes the age of a solvents might affect the reaction. Tetrahydrofuran when used a solvent can have traces of peroxide which can show new unexpected products.
Now, there is a trend these days to intentionally hide some information so that competitor's cannot catch up with your work or write the method in a sloppy bare minimum way. This is really sinister but it happens and such practices must be curbed. Ayway scientists are not saints-they are human beings like everyone.


Is the end of a cigarette on fire when lit?

A lighted cigarette is burning, technically smoldering (no flame).  The stuff inside the paper bears only a passing resemblance to tobacco,

http://en.wikipedia.org/wiki/Cigarette#Construction


How to determine the mole fraction of the liquid phase from a given mole fraction of the vapour phase for a benzene/toluene mixture?

The question gives you the mol percentage of benzene ($53 \%$) in the vapour phase, and not in the liquid phase.

It's always good to write down what you have first. (note: I use $y$ to denote mole fractions in the vapour phase and $x$ to denote mole fractions in the liquid state)

First of all, we have $y_\text{benzene} = 0.53$. Additionally, we have been given the vapour pressure of pure benzene and toluene (denoted with a superscript $*$):

$$p^*_\text{benzene} = 0.125\ \mathrm{atm}  \qquad p^*_\text{toluene} = 0.037\ \mathrm{atm} $$

and we require $x_\text{benzene}$

We also, know $p_\text{benzene} = x_\text{benzene}\ p^*_\text{benzene}$, and the total vapor pressure is given by $$P = p_\text{benzene} + p_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene} + x_\text{toluene}\ p^*_\text{toluene} = x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}$$

Finally,

$$y_\text{benzene} = \frac{p_\text{benzene}}{P} = \frac{x_\text{benzene}\ p^*_\text{benzene}}{x_\text{benzene}\ p^*_\text{benzene}+ (1- x_\text{benzene})\ p^*_\text{toluene}}$$

It is now, trivial to solve this equation. Since I am lazy, and was not in the mood to write down two lines of algebra and reach for a calculator, I chucked it into Mathematica, and the result is:

Thus, $x_\text{benzene} = 0.25$, which is indeed the result given in your answer key.


Why do some covalent compounds form lattice while others do not?

See, as far as I know, their is no fixed way to compare catenation properties of elements randomly without any context. Some general pointers however, to keep in mind are-:
Carbon has one of the highest catenation powers, as soap molecules like sodium stearate can contain up to 17 carbon atoms.
After carbon, some other elements having significant catenation tendencies are silicon, sulphur, boron etc. (roughly in that order). I will not be elaborating further here as I'm not very sure myself about this except perhaps a few rough pointers such as medium-ish size of atom and the presence of group 13 and 14(which have the maximum number of covalent sigma bonds formed without expansion of octet)
The thing you really need to understand is that the ability to form multiple bonds is more important here to predict the bonding nature of the crystal lattice. Oxygen and nitrogen can form multiple bonds, which would allow them to kind form cross links while satisfying the valency of the central atom. While on the other hand, fluorine will finish it's covalency in satisfying the needs of the central atom (Si) only, so the tetrahedral unit formed will majorly interact via Van der Waals forces (checking it with the facts, this is a pretty good prediction). The exact crystal structure of SiO2 and ALN are quartz and wurtzite respectively, which would be difficult to predict simply by making the covalent structures. However, a fair idea can be taken about the kind of interactions which will prevail in the crystal lattice, provided it is also covalent.


I would just try soap and water first. If you want to try any kind of solvent like alcohols, you really needn't be concerned with lingering smell, but I would be concerned with the (slight) possibility of destroying the foam.

If soap and water doesn't work, just try placing a drop of isopropanol on both the top and bottom of the insert and let it dry. If there is any change in the consistency or feel to it I would stop there. If not, then progress to cleaning a small area, and if that does no damage go ahead and soak the whole thing if needed.


How do electronegativity and lone pairs affect bond angles?

The electronegativity of the central atom does not really influence bond angles in any meaningful way. The size of atoms and the orbitals used and useable are much more important.

Sometimes, students may be taught concepts such as electronegativity or ‘lone pairs require space’ to explain why $\ce{H2O}$ has a smaller bond angle ($105^\circ$) than $\ce{NH3}\ (107^\circ)$ or similar. However, careful analysis of the molecular orbitals reveals that sterics are a much more important reason than ‘lone pairs requiring more space and thus making tetrahedral $\mathrm{sp^3}$ contract’. Indeed, as soon as a lone pair exists on a central atom, there is a tendency to have the lone pair occupy an orbital with an s-contribution as high as possible and to have bonding orbitals with a maximised p-contribution. p-orbitals, having a direction, can form stronger σ bonds due to better overlap.

This is why the bond angle drops quickly when comparing second-period to third-period central elements to close to $90^\circ$ but then hardly changes down the group. On the other hand, the reduced bond angles observed from ammonia to water are mainly due to more steric stress on nitrogen — which needs to accomodate an additional hydrogen and thus needs to space them out more.

Indeed, if you compare the bond angles of different $\ce{OX2}$ compounds, the only general trend is ‘the larger the second element’s atoms, the greater the bond angle.’ See the table below (all values taken from Wikipedia).

\begin{array}{ccc}\hline
\ce{X} & \angle(\ce{X-O-X}) & \angle(\ce{X-N-X})\\ \hline
\ce{H} & 104.5^\circ & 107.8^\circ \\
\ce{F} & 103^\circ & 102.5^\circ \\
\ce{Cl} & 110.9^\circ & -/- \\
\ce{Br} & 112^\circ & -/- \\ \hline \end{array}

The drop in bond angles from hydrogen to fluorine is sufficiently explained by the much longer bond lengths of the $\ce{OF2}/\ce{NF3}$ species when compared to the corresponding hydrogen species due to fluorine using p-orbitals for bonding where hydrogen uses its 1s orbital.

While the bond lengths also increase when going from fluorine to chlorine and bromine, the angle also does. Note especially $\ce{OBr2}$ whose bond angle is larger than the tetrahedral bond angle. Any argument involving electronegativity should assume a smaller angle here since the electronegativity of the partners drops from fluorine to bromine.


Ground state elevation for phase transition

In order to make the phase transition, you need to supply your sample molecules with an additional amount of kinetic energy (i.e., the heat of fusion or vaporization) to overcome their intermolecular bonding. This however means that after the transition, your system contains more internal energy than before. That additional energy is encoded in the vertical offset on the energy scale.

Note that if you look at each phase on its own, you could set the their ground states as the zero point on your energy scale. However, if you want to compare the system among different phases, then you need to take that offset into account. Otherwise you'd assume that your system can change from solid to liquid to gas without any effort.


What is the Major product in free radical halogenation of Alkanes. (Between 3°,2°,1°)

It depends on what halogen you use.

Chlorine has a reactivity ratio for $1^o$:$2^o$:$3^o$= 1 : 3.5 : 5.

To calculate the relative yield at that position, multiply the number of Hydrogens at that position to the reactivity at that position. This will give you a comparison of the yield at particular positions.

But in the case of bromine, the reactivity order is 1 : 250 : 6300. Therefore, without any further thinking, the bromine radical will always attack at the tertiary position if available.

In general, reactivity towards tertiary carbon increases down the group as in the case of halogens.


Building an average model from MD snapshots

This is possible with the Python package MDTraj. Assuming you have your trajectory saved as trajectory.dcd and topology in topology.pdb:

import mdtraj as md
from mdtraj.geometry.alignment import compute_average_structure

traj = md.load('trajectory.dcd', top='topology.pdb')
average = compute_average_structure(traj.xyz)

I believe it will RMSD align the trajectory first, but you may need to specify that explicitly first.

You might also want to check out MDAnalysis as well, it may have similar functionality. I've used both packages and both are still actively maintained.


Clarification on Oxidation Number and Ionic Charge

For simple mononuclear ions in ionic compounds, as you have in $\ce{NaCl, Al2O3, ZnS}$ or $\ce{GaN}$, the oxidation state of each element will always equal its ionic charge: $\ce{Na+}$ has $\mathrm{+I}$, $\ce{Al^3+}$ $\mathrm{+III}$, $\ce{Zn^2+}$ $\mathrm{+II}$, $\ce{Ga^3+}$ $\mathrm{+III}$, $\ce{Cl-}$ $\mathrm{-I}$, $\ce{O^2-}$ $\mathrm{-II}$, $\ce{S^2-}$ $\mathrm{-II}$ and $\ce{N^3-}$ $\mathrm{-III}$.

Something like the oxalate anion $\ce{C2O4^2-}$ is a multinuclear anion. Its ionic charge is $2-$, as evident by the superscript. However, you cannot always determine the constituent elements’ oxidation states a priori — the only thing you do know is that the sum of the oxidation states must equal the charge number. In oxalate, carbon has $\mathrm{+III}$ as you correctly mentioned and oxygen has $\mathrm{-II}$, as it should be of no surprise to you. Added up, this leaves us with:

$$2 \times (+3) + 4 \times (-2) = +6 + -8 = -2$$

And $2-$ is the ionic charge as we already know. This rule goes both ways, so you can use the (known) ionic charge of a multinuclear ion to determine an element’s oxidation state if the other oxidation states are known: in $\ce{SO3^2-}$ oxygen has $\mathrm{-II}$, so sulphur can only have $\mathrm{+IV}$.


Whats the difference between entropy and the (dis)order of a system?

Briefly, spontaneous processes tend to proceed from states of low probability to states of higher probability. The higher-probability states tend to be those that can be realized in many different ways.

Entropy is a measure of the number of different ways a state with a particular energy can be realized. Specifically, $$S=k\ln W$$ where $k$ is Boltzmann's constant and $W$ is the number of equivalent ways to distribute energy in the system. If there are many ways to realize a state with a given energy, we say it has high entropy. Often the many ways to realize a high entropy state might be described as "disorder", but the lack of order is beside the point; the state has high entropy because it can be realized in many different ways, not because it's "messy".

Here's an analogy: if energy were money, entropy would be related to the number of different ways of counting it out. For example, there, there are only two ways of counting out two dollars with American paper money (2 1-dollar bills, or 1 two-dollar bill). But there are five ways of counting out two dollars using 50-cent or 25-cent coins (4 50-cent pieces, 3 50-cent pieces and 2 quarters, and so on). You could say that the "entropy" of a system that dealt in coins was higher than that of a system that dealt only in paper money.

Let's look the change in entropy for a reaction $\rm A\rightarrow B$, where A molecules can take on energies that are multiples of 10 energy units, and B molecules can take on energies that are multiples of 5 units.

Suppose that the total energy of the reacting mixture is 20 units. If we have 3 molecules of A, there are 2 ways to distribute our 20 units among energy levels with 0, 10, and 20 units:

If we have 3 molecules of B, there are 4 ways to distribute 20 units among energy levels with 0, 5, 10, 15, and 20 units:

The entropy of B is higher than the entropy of A because there are more ways to distribute the same amount of energy in B than in A. Therefore, $\Delta S$ for the reaction $\rm A\rightarrow B$ will be positive.


How do I convert 4-(Methylthio)benzaldehyde to 4-(thioacetate)benzaldehyde?

You could react 4-(methylthio)benzaldehyde with $\ce{HBr}$ to form a thiol, and then react the thiol with acetyl chloride in triethylamine to form 4-(thioacetate)benzaldehyde.


Validity of Henry's Law for an Oxygen/Bitumen System?

Regarding Q1, Henry's law states:

  At a constant temperature, the amount of a given gas that dissolves in
  a given type and volume of liquid is directly proportional to the
  partial pressure of that gas in equilibrium with that liquid.

A1 is answered by the definition itself: the system should be at equilibrium.
A2 is straightforward: if the gas reacted, you would not have a constant pressure as a certain (if not all) quantity of it would be removed from the system[1]. This would also keep your system from being at equilibrium, defying the previous assumption.
A3 and A4 both depend on the fact that – as the concentration or the partial pressure rise – the system deviates from an ideal model where the solute is
infinite-diluted and in such system Raoult's Law should be used
instead[2]. So you need a low concentration and partial pressure of your gas.

To answer to Q2, Henry's Law can be applied to any solvent, whether liquid, vapour or solid[3], [4]. It has been used, within certain limits, to study absorption of gas in zeolites[5] or hydrogels, but I couldn't find any precise experimental data on it. Only that in such systems it is only a limited law, adequate at low mole fractions, but useful for practical purposes when high precision is not required.

    1 Henry's Law, UC Davis University ChemWiki
    2 North Carolina State University, CH 431 / Lecture 14
    3 Francis L. Smith and Allan H. Harvey (September 2007). "Avoid Common Pitfalls When Using Henry's Law".
    4 Rosenberg, Robert M., Peticolas, Warner L., Journal of Chemical Education, Volume 81, 11, pp 1647-1652
    5 Eric G. Derouane, Zeolites as solid solvents, Journal of Molecular Catalysis A: Chemical, Volume 134, 1–3, pp 29-45


How to identify the most stable Lewis adduct of boron trichloride?

You have to consider also the inherent strength of the covalent bond.  A strong covalent bond can entice an electronegative atom to accept some positive charge or give up some negative charge.

Roughly speaking, hydrogen and period 2 nonmetals form much stronger bonds among one another than with other elements, and this tends to override the electronegativity difference.  For instance, fluorine is much more electronegative than chlorine, but fluorine bonds so much more strongly with hydrogen that hydrofluoric acid is only a weak acid in water whereas hydrochloric acid is strong.  Note that the dissociation of hydrogen chloride in water creates hydrogen-oxygen covalent bonds.

Here, boron binds especially strongly with the period 2 nonmetals nitrogen and oxygen.  So (a) and (c) have this advantage, and after that (c) wins between these two by electronegativity difference.


Balancing redox equations in acidic media: to use H+ or H3O+?

This is a question of "style" so there is no correct answer.  In fact, just as $\ce{H^+}$ doesn't really exist in solution, neither does $\ce{H3O+}$ exist in solution.  In reality higher hydrates of $\ce{H^+}$ are what exist in solution. See the first paragraph in this Wikipedia article.  Again, both $\ce{H^+}$ and $\ce{H3O+}$ are just shorthand notations for reality, choose the style that you like best.


Semi-Infinite Potential Square Well: Negative Potential

Answering your quantum mechanics question seems to become a hobby of mine :). First let me say that there is nothing wrong with shifting the potential up so that $V_{0} = 0$. This choice would indeed simplify the problem as it gets rid of negative energy values and thus avoids the following point of confusion: when $E < 0$ the wave function of a particle in region $\text{II}$ is not a plane wave $\mathrm{e}^{\mathrm{i} k x}$ ($k \in \mathbb{R}$) like that of a free particle moving in vacuum but a decaying exponential $\mathrm{e}^{-\kappa x}$ (with $\kappa \in \mathbb{R}$). So, the choice $V_{0} = 0$ is advisable. Shifting the potential back later is always possible and might be easier.

Nevertheless, of course the problem is solvable irrespective of the choice of $V_{0}$. I think your problem in this case is that you enforced an incorrect boundary condition. To avoid confusion I will start from the beginning:

I will only consider the case $E < 0$ in the following, so $E$ is always a negative number between $-V_{0}$ and $0$. For region $\text{II}$ you have to solve the Schroedinger Equation:

\begin{align}
  - \frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &= E \psi_{\text{II}}(x) \\
  \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &= - \frac{2mE}{\hbar^{2}} \psi_{\text{II}}(x) \ .
\end{align}

Ok, here comes the first "problem" with the choice of $V_{0}$: In your solution you chose $\frac{2mE}{\hbar^{2}} = k^{2}$ as you would naturally do for a free particle. But since $E$ is negative $k$ will be a complex number (and not a real number as it would be for a free particle) and so this differential equation does not have an oscillatory solution $\psi_{\text{II}}(x) = A \sin(kx) + B \cos(kx)$ with $k \in \mathbb{R}$. Rather you get a decaying exponential function by doing the following: $- \frac{2mE}{\hbar^{2}}$ will be a positive number so you simply set $- \frac{2mE}{\hbar^{2}} = \kappa^{2}$ to get

\begin{align}
  \frac{\mathrm{d}^{2} \psi_{\text{II}}(x)}{\mathrm{d} x^{2}} &= \kappa^{2} \psi_{\text{II}}(x) \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ .
\end{align}

Now, this differential equation has a solution of the form

\begin{align}
  \psi_{\text{II}}(x) = A_{\text{II}} \mathrm{e}^{\kappa x} + B_{\text{II}} \mathrm{e}^{-\kappa x} \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ .
\end{align}

At this point you can enforce the boundary condition that the wavefunction $\psi_{\text{II}}$ must be square integrable over the whole interval $a < x < \infty$ (Note that you don't have to consider the interval $-\infty < x < \infty$ but only the interval on which $\psi_{\text{II}}$ is defined). Since $\lim_{x \to \infty} \mathrm{e}^{\kappa x} = \infty$ this mean that the prefactor $A_{\text{II}}$ must be equal to zero, so

\begin{align}
  \psi_{\text{II}}(x) = B_{\text{II}} \mathrm{e}^{-\kappa x} \qquad \quad \text{with} \qquad \kappa \in \mathbb{R} \ .
\end{align}

So much for region $\text{II}$. Let's move on to region $\text{I}$: Here you have to solve the Schroedinger Equation:

\begin{align}
  - \frac{\hbar^{2}}{2m} \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} - V_{0} \psi_{\text{I}}(x)  &= E \psi_{\text{I}}(x) \\
  \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} &= - \frac{2m(E + V_{0})}{\hbar^{2}} \psi_{\text{I}}(x) \ .
\end{align}

Again, care must be taken: Since $-V_{0} < E < 0$ the term $(E + V_{0})$ is positive and so $- \frac{2m(E + V_{0})}{\hbar^{2}}$ is a negative number and you can set $- \frac{2m(E + V_{0})}{\hbar^{2}} = -k^{2}$ which leads to

\begin{align}
  \frac{\mathrm{d}^{2} \psi_{\text{I}}(x)}{\mathrm{d} x^{2}} &= - k^{2} \psi_{\text{I}}(x) \qquad \quad \text{with} \qquad k \in \mathbb{R} \ .
\end{align}

This differential equation has a solution of the form:

\begin{align}
  \psi_{\text{I}}(x) = A_{\text{I}} \mathrm{e}^{\mathrm{i} k x} + B_{\text{I}} \mathrm{e}^{-\mathrm{i} k x} \qquad \quad \text{with} \qquad k \in \mathbb{R} \ .
\end{align}

(You could also formulate an equivalent solution in terms of $\sin$ and $\cos$ functions using Euler's formula if that is more to your liking. But I prefer plane waves ;))
This is the point where you made your main mistake: You discarded one of the terms in your $\psi_{\text{I}}$ wavefunction because it blew up for $x \to \infty$. But $\psi_{\text{I}}(x)$ is only defined on the interval $0 \leq x \leq a$ and within this interval both exponential terms are perfectly square-integrable and you can't discard any of them.

Finally, you just have to remember that $\psi_{\text{III}}(x) = 0 \ \forall \ x$ and enforce the boundary conditions given by the system:

\begin{align}
  \psi_{\text{I}}(0) &= \psi_{\text{III}}(0) \quad &&\Rightarrow \quad A_{\text{I}} \mathrm{e}^{0} + B_{\text{I}} \mathrm{e}^{0} = 0 \quad &&\Rightarrow \quad B_{\text{I}} = -A_{\text{I}} \\
  \psi_{\text{I}}(a) &= \psi_{\text{II}}(a) \quad &&\Rightarrow \quad A_{\text{I}} \mathrm{e}^{\mathrm{i} k a} - A_{\text{I}} \mathrm{e}^{-\mathrm{i} k a} = B_{\text{II}} \mathrm{e}^{-\kappa a} && \\
  & &&\hphantom{\Rightarrow} \ \quad A_{\text{I}} \underbrace{\bigl(\mathrm{e}^{\mathrm{i} k a} - \mathrm{e}^{-\mathrm{i} k a} \bigr)}_{= \, 2 \mathrm{i} \sin(k a)} = B_{\text{II}} \mathrm{e}^{-\kappa a} \quad &&\Rightarrow \quad B_{\text{II}} = 2 \mathrm{i} \sin(k a) \mathrm{e}^{\kappa a} A_{\text{I}}
\end{align}

Edit:

Concerning your comment: You can indeed enforce another boundary condition. The probability
flux is required to be continuous across any surface, since otherwise the surface would
contain sources or sinks. This entails the requirement that both $\psi$ and $\nabla \psi$ must be continuous. The continuity condition $\psi_{\text{I}}(a) = \psi_{\text{II}}(a)$ has already been enforced but the one for the gradient is still left:

\begin{align}
  \frac{\mathrm{d} \psi_{\text{I}}(x)}{\mathrm{d} x} \bigg|_{x = a} &= \frac{\mathrm{d} \psi_{\text{II}}(x)}{\mathrm{d} x} \bigg|_{x = a} \ .
\end{align}

From the above discussion we know that:

\begin{align}
  \psi_{\text{I}}(x) &= A_{\text{I}} \underbrace{\left(\mathrm{e}^{\mathrm{i} k x} - \mathrm{e}^{-\mathrm{i} k x} \right)}_{= \, 2 \mathrm{i} \sin(k x)} = 2 \mathrm{i} A_\text{I} \sin (kx) = C_\text{I} \sin(kx)
\end{align}

and

\begin{align}
  \psi_{\text{II}}(x) &= \underbrace{2 \mathrm{i} A_{\text{I}} \sin(k a)}_{= \, \psi_{\text{I}}(x=a)} \, \mathrm{e}^{\kappa a} \mathrm{e}^{- \kappa x} = \psi_{\text{I}}(x \! = \! a) \, \mathrm{e}^{- \kappa (x - a)}
\end{align}

with $C_\text{I} = 2 \mathrm{i} A_{\text{I}}$, so

\begin{align}
  \frac{\mathrm{d} \psi_{\text{I}}(x)}{\mathrm{d} x} \bigg|_{x = a} &= \frac{\mathrm{d} \psi_{\text{II}}(x)}{\mathrm{d} x} \bigg|_{x = a} \\
  k C_\text{I} \cos(ka) &= - \kappa C_\text{I} \sin(ka) \\
  \kappa &= - \frac{k}{\tan(k a)}  \ .
\end{align}

While this equation looks innocent enough it is not easily solved for $E$ and you will need the assistance of something like Mathematica for that.


How to prepare a solution of accurate concentration?

This is how I did it in my analytical practicals, works for primary standards with a decent accuracy, works for everything else with a larger error. (You'd want to check concentrations with a primary standard then.)

Determine the mass of the empty volumetric flask
Prepare approximate mass of the substance you want the solution of in a beaker or watch glass or similar
Transfer the prepared substance into the volumetric flask
Determine the mass of the filled flask; the difference will give you the mass of the substance
Fill the flask with the solvent.
Given the precision of the instruments used, you can calculate the concentration quite accurately. If necessary check the concentration with another method.


History and because of how pressure works

The original reason why mm of Hg are used is because of how chemists first started to measure and understand atmospheric pressure. The idea of the vacuum and atmospheric pressure is surprisingly new. Otto von Guericke demonstrated the power of atmospheric pressure by showing that two horses could not separate a brass sphere containing a partial vacuum only in the 17th century (remember a vacuum doesn't suck the atmosphere pushes).

This kicked off a vast amount of experimentation. Around the same time scientists like Torricelli and Pascal realised that they could measure atmospheric pressure by filling a closed tube with mercury and inverting it in a bath of mercury (you can do this with any fluid but mercury is about 13 times denser than water so the tube only needs to be a meter or so long; with water it needs to be ~10m long which is an inconvenient length in an indoor laboratory). The height of the mercury in the tube is a measure of the external pressure as a vacuum is formed above it in the closed tube. This simple instrument was used by a friend of Pascal to show that atmospheric pressure drops with height when he took a barometer up France's Puy de Dome mountain.

The diameter of the tube is irrelevant as pressure is weight/area (do the calculations and tube diameter cancels out).

We now call a tube like this a barometer. And the height the mercury rises to is a convenient measure of atmospheric pressure and can be read off using a simple ruler. We could use mm of water, but would need a much longer tube and an unusually inconveniently sized ruler. Hence why mercury is the fluid and a convenient way of measuring pressure is mm of Hg.

We tend to use modern SI units for pressure these days, not least because the wide use of mercury has been banned and calculations in consistent nits are far easier.


1) Yes that is correct.  Generally, when the intensity of the light transmitted or reflected is low, then the absorbance is high.

2) It is hard to say just what "most" means, but if we are speaking of visible light (remember absorbance is wavelength dependent), there are a LOT of colorless materials that do not absorb in the visible region.

3) Yes, more molecules will increase the probability of light being absorbed by the solution, but as you noted, the absorption coefficient is a "molar" absorption coefficient.  Therefor when using the molar absorption coefficient, the absorbance is adjusted for the number of molecules in the solution.


Both your paragraphs are valid, as all present components are balanced by 3, mutually linked equilibrium reactions, with the respective equilibrium constants:
$\ce{HF + H2O <=> F- + H3O+}$, $K_\mathrm{a}$
$\ce{F- + H2O <=> HF + OH-}$, $K_\mathrm{b}$
$\ce{2 H2O <=> H2O+ + OH-}$, $K_\mathrm{w}$
with the relation $K_\mathrm{a} \cdot K_\mathrm{b} = K_\mathrm{w}$
$\mathrm{pH}= \mathrm{p}K_\mathrm{a} + \log{\frac{\ce{[F-]}}{\ce{[HF]}}}$


I am sure the book, if it is any decent, should have said something like this already. $K'$ is not a measure of the forward reaction, nor is it a measure of the reverse reaction. It is a measure of the equilibrium position.

Equilibrium occurs when the rate of the forward reaction is equal to the rate of the backward reaction. By the laws of thermodynamics, this occurs at one special point where the instantaneous concentrations of all the compounds satisfy the equation

$$\frac{[\ce{CaO}][\ce{CO2}]}{[\ce{CaCO3}]} = K'$$

$K'$ for a reaction is a constant. Clearly, a large $K'$ indicates that when equilibrium is reached, there is a lot of $\ce{CaO}$ and $\ce{CO2}$, and only a little bit of $\ce{CaCO3}$. We say that the equilibrium position lies to the right. Conversely, a small $K'$ means that the equilibrium position lies to the left, i.e. reactants predominate.

The equilibrium constant is properly defined in terms of activities of the compounds, not concentrations. This means that for the reaction you gave,

$$K = \frac{(a_{\ce{CaO}})(a_{\ce{CO2}})}{a_{\ce{CaCO3}}}$$

The activity of a pure solid is defined to be equal to $1$, which means that you can simplify this by removing all the terms corresponding to pure solids:

$$K = a_{\ce{CO2}}$$

The "equilibrium constant" that you are learning that involves concentrations is a simplification.


Most of $\mathrm{pH}$ calculations make simplifying assumptions and one must determine if they are justified for the particular case.
Simplification for strong acids  $$\mathrm{pH}=-\log{[\mathrm{acid}]}$$
implies $\pu{[\mathrm{acid}] >> 10^{-7} mol/L}$.
Because water dissociation constant $K_\mathrm{w}=\ce{[H+][OH-]}=\pu{10^{-14} mol^2/L^2}$,
neutral water contains $\pu{10^{-7} mol/L}$ of $\ce{H+}$.
That means the very most $\ce{H+}$ ion must come from acid dissociation and not from water dissociation.

If OTOH very most $\ce{H+}$ are from water dissociation as in your case, another simplification says $\mathrm{pH}=7$, no matter how much of acid you added.

For general case:
$$c_{\mathrm{acid}}=\ce{[A-]}=\ce{[H+]} - \ce{[OH-]} = \ce{[H+]} - \frac{K_\mathrm{w}}{\ce{[H+]}}$$
$${\ce{[H+]}}^2 - \ce{[H+]}\cdot c_{\mathrm{acid}} - K_\mathrm{w}=0$$
$$\ce{[H+]} = \frac{c_\mathrm{acid} + \sqrt{{ c_\mathrm{acid}}^2 + 4\cdot K_\mathrm{w}}}{2}$$
The limit case for $c_\mathrm{acid} << \sqrt{K_\mathrm{w}}$ is $$\ce{[H+]} = \sqrt{K_\mathrm{w}}$$
$$\mathrm{pH}=\frac 12 \mathrm{p}K_\mathrm{w}$$
The limit case for $c_\mathrm{acid} >> \sqrt{K_\mathrm{w}}$ is $$\ce{[H+]} = c_\mathrm{acid}$$
$$\mathrm{pH} = -\log{(c_\mathrm{acid})}$$

Simplification for weak acids $$\mathrm{pH}=0.5 \left(\mathrm{p}K_\mathrm{a}-\log {\ce{[acid]}}\right)$$
implies negligible water dissociation and also negligible relative acid dissociation $\ce{[HA]} >> \ce{[A-]}$.
So one can assume:

The acid concentration $\ce{[HA]}$ is not affected by acid dissociation.
$\ce{[A-]} = \ce{[H+]}$


The nucleus is the very dense region consisting of protons and neutrons at the center of an atom.
The diameter of the nucleus is in the range of 1.75 fm ($1.75×10^{−15}$ m) for hydrogen (the diameter of a single proton)2 to about 15 fm for the heaviest atoms, such as uranium. These dimensions are much smaller than the diameter of the atom itself (nucleus + electron cloud), by a factor of about 23,000 (uranium) to about 145,000 (hydrogen)

Source:Wikipedia


Regioselectivity of electrophilic addition of Halogen and Water (Halohydrin) on Alkenes?

This is not a 100% answer, but it was too detailed to fit in the comment section.
Following up from the comment, here's a reference from Inorganic Chemistry by Morrison and Boyd:

There is evidence, of a kind we are not prepared to go into here, that these compounds are not formed by addition of preformed hypohalous acid, $\ce{HOX}$, but by reaction of the alkene with, successively, halogen and water.

Halogen adds (Step 1) to form the halonium ion; this then reacts, in part, not with the bromide ion, but with water (Step 2) to yield the protonated alcohol.
Propylene gives the chlorohydrin in which chlorine is attached to the terminal carbon. This is typical behaviour for an unsymmetrical alkene; orientation follows markovnikov's rule, with positive halogen going to the same carbon that the hydrogen of a protic reagent would. Now, this orientation would be perfectly understandable if the intermediate were an open carbocation: the initial addition of halogen yields the more stable carbocation - secondary, in the case of propylene. But the stereochemistry indicates that the intermediate is not an open carbocation, but a cyclic halonium ion.

And that's all. Quite anticlimatic, as I was expecting a reason for the regioselectivity. Most other references and papers I have seen online also do not have a concrete explanation for the regioselectivity, and they resort to hand-wavy explanations involving resonance for the same. This reaction mechanism is also quite similar to Oxymercuration-demercuration, whose regioselectivity is again explained by the "three resonating structures". I could not find diagrams of these resonating structures, and while this question asks the same, the resonance structures or the intermediate are not properly explained in the answer.
My best suggestion for you would be to accept the explanation in your textbook as that is adequate for now. Until someone more knowledgable posts a better answer with references, that should suffice.


is this the proper identification of this cyclic molecule?

It's correct, as already mentioned in a comment.

Some might find more or less helpful to redraw the structure to:

Your additional question can be tricky, understood a different way. It's possible to exchange H and Cl positions, or rather axial to equatorial bonds and the other way around – by switching to the other boat conformation, i.e. without changing the configuration:

(The two structures are identical, one of them can be rotated to superimpose the other.)


I'm going to use Theta-Solvent and Theta-temperature since that is what Wikipedia uses, and Wikipedia usually has the preferred names.

In order to get a polymer to the Theta-Condition two factors must be considered.

(1) Solvent

A Theta-Solvent is a poor solvent for a polymer is one such that the interactions between the polymer segments and the solvent are minimized. This results in polymer segment interactions to be maximized.

A good solvent for a polymer is one such that the interactions between the polymer segments and the solvent are maximized. This results in polymer segment interactions being minimized.

(2) Temperature

The Theta-temperature is the temperature for the particular theta-solvent
which minimizes interactions between the polymer segments and the solvent.

Point

The combination of using a theta-solvent and being at the theta-temperature satisfies the theta-condition.

This makes experimental determination of important quantities such as the root mean square end-to-end distance or the radius of gyration much simpler.

Note

This assumes that polymer will self-coil into its lowest energy configuration. For polymers with same unit repeating this is not unlikely.

For proteins in biological systems this self-arrangement is unlikely. Proteins do not have a "repeating unit." Therefore biological systems have a mechanism to fold a protein in a certain way.


Ultra Met uses tantalum for corrosion resistant coatings and has a list of corrosive substances that won't dissolve tantalum, comprising of hydrochloric, hydrobromic, and hydroiodic acids, nitric acid, sulfuric acid, acetic acid, hydrogen sulfide and even pirahna.

So unless you plan on using aqua regia, hydorfluoric acid, hot chlorine or other hot halogen gas, (none of which are safe for most metal jewelry) your ring should survive—at least the tantalum part will.


Viscosity of liquid interrupts accuracy of micropipette?

I think you try to use glass pipette instead. It may decrease viscosity effect.


How to specify a different isotope in the ORCA quantum chemistry program?

ORCA cannot read isotope information from XYZ files -- the OpenBabel specification does not include it and so ORCA doesn't look for it in the geometry file.

You need to copy-paste the geometry into your input file:

* xyz 0 1
  ...
  H   0.55630745024474     -2.41808278363585     -1.72738492075924 M 2
*

Alternatively, if you are only interested in the frequencies, normal modes, and/or IR spectrum, you can edit the atomic masses in the .hess file after the run is completed and use the orca_vib utility program to re-run the analysis with the substituted isotopes. See pp. 201-202 of the ORCA v3.0.3 manual for more details.


Does this mean that the higher the concentration of bicarbonates, the better the quality for drinking?

Not necessarily. This is more a subjective question involving what tastes good to you.

  Should we choose the bottle of water with the highest concentration of these anions?

Assuming the water is actually monitored for safety/quality, the upper limit on caustic (hydroxide) alkalinity is likely too low to do any damage. And carbonate alkalinity will be limited by pH and equilibrium with atmospheric carbon dioxide. So: If it tastes good to you, then drink it.

  And also, when boiling the water we loose bicarbonates, so is boiled water unhealthier?

When you boil water, you are increasing the concentration of dissolved species that do not boil off at that temperature. You can, for example, increase levels of nitrate and nitrite in water by boiling it, which is a bad idea if your water is contaminated by large amounts of nitrate/nitrite. Conversely, you can kill some if not all harmful bacteria in water by boiling it. And I don't think that boiling water results in the loss of bicarbonate in any measurable way, at least not on timescales we're interested in (like cooking).


Constitutional isomers of cis-1,2-dibromocyclopentane

If I understood you correctly, your objection is to the relative cis- designation in the name of the compound. As far as I can tell, the use of any stereochemical designation within the context of that question is either deliberately misleading, incorrect, or simply pointless. Per the definition of constitutional isomerism, the only criteria that the two molecules in your answer must satisfy is that they have (a) the same molecular formula, and (b) different structural connectivity. Stereochemistry, whether relative or absolute, is totally irrelevant. Note the wording in the IUPAC Gold Book entry for constitution (emphasis mine):

  The description of the identity and connectivity (and corresponding bond multiplicities) of the atoms in a molecular entity (omitting any distinction arising from their spatial arrangement).

Hence, any molecule with the formula $\ce{C5H8Br2}$ that differs in connectivity from a cyclopentane with two vicinal bromine atoms should be a valid answer. Two such isomers are given by simply permuting the positions of the bromines (i.e., 1,1-dibromocyclopentane and 1,3-dibromocyclopentane). Of course, there are numerous other possible answers.


Is it possible for Hydrogen to lose its electron?

Hydrogen can lose an electron meaning it can be in the +1 oxidation state. However, just like any other cation or anion it never occurs free in condensed matter, it always is in contact with solvent and/or anions. Moreover, because of extremely small size of proton, it is an extremely powerful Lewis acid. Consequently, in common conditions proton would react with first electron pair it comes in contact with, up to and including inert gas electron pairs and covalent bond pairs.

On the other hand, hydrogen ions are quite easy to generate in electric discharge and/or under extreme heating. In fact, producing and confinement of super-hot plasma, consisting of hydrogen ions and electrons, is an area of active research for several decades and, well, producing and confinement of relatively cold plasma is not a problem. Confinement of several billions Kelvin hot plasma, however, is still a problem.


Haemoglobin literature spectra does not match

As stated by Scott Prahl :

  One of the most confusing things about looking at hemoglobin (Hb)
  spectra is that the values are typically tabulated in equivalents. The
  term equivalent is used to indicate the amount of hemoglobin which
  contains 1 gm atom of Fe and combines with 1 gm molecule of O2 or CO.
  One equivalent of hemoglobin is assumed to be 64,500/4 or 16,125 gm. A
  concentration of 10$^{-6}$ equivalent is 16.125mg of hemoglobin per cc.

It seems that Assendelft and Zijlstra used the equivalent quantification of haemoglobin, meaning that all their extinction coefficients should be multiplied by 4. This leads to the following figure:

Sorry for not having read again Scott Prahl's description before posting. I read it several months ago, as I was gathering data from different source and forgotten it since...


Acidic strength of diethyl malonate vs ethyl acetoacetate

The hydrogens which you have chosen are correct but both the carbanions are NOT eqaully stablised via the resonance. In case of malonic ester, the elctrophilicity of the carbonyl(c=o) is reduced due to the resonance from the oxygen beside it. So the carbanion will not be pulled very effectively into resonance as the carbonyl becomes electron rich due to the oxygen of the ester.

However, in the ethyl acetoacetae only one of the carbonyl groups is affected by this reasonace. The other one is still highly elctrophillic. So it stablises the carbanion even more. So it has a lower pka(more acidic) than the malonic ester.


Due to the fact that starch (amylum) is a polysaccharide , a long chain compound consisting of glucose monomers it has poor solubilty in most solvents.

Some starches are water soluble because of the hydroxyl groups involved in the polysaccharide chain, some like the branched form amylopectin are more insoluble.

Starches form granules in solution with cold water, the granules break down as the water is heated and can be solubilized.

When ethanol is added, which is less polar than water, it lowers the solubility and potential hydrogen bonds of the starch , because the water is more attracted to the alcohol than the starch, the starch is dehydrated by the alcohol and precipitates.

So while polarity and temperature are variables to starches poor solubility, the critical variable in this case is the starches ability to hydrogen bond with the solvent.


What is the effect of substituting oxygen with sulfur in drug molecules?

In asking this question - Why do chalcogens (Group VI) stink so badly? - I received a couple answers and comments along the lines of "it's not quite so simple". The same would generally apply here; there isn't one change in behavior, or even a finite set of behavioral changes, that could be predicted by an oxygen-to-sulfur exchange.

Just as a basic example, compare methanol ($\ce{CH_3(OH)}$) to methyl mercaptan, ($\ce{CH_3(SH)}$). Methanol is a colorless liquid at RT with a sweetish odor. Methyl mercaptan, with a higher molecular mass due to the sulfur switch, is nevertheless a gas at RT and smells like rotting cabbage. Methanol metabolizes into formic acid and formaldehyde which in sufficient doses causes nerve damage, specifically to the optic nerve. Methyl mercaptan is pretty much its own metabolite; while also toxic in sufficient doses, it's readily filtered out by the liver and kidneys, as it's produced by bacterial breakdown of many of the foods we eat (especially highly-fibrous vegetables, legumes etc).

Perhaps the most common occurrence of sulfur in otherwise organic molecules is a sulfhydryl group (R-SH), forming a thiol. The alcohol-to-thiol switch is relatively common in human biochemistry and as such is also of interest in pharmacology. The Wikipedia page on thiols lists the following basic changes between an alcohol and its thiol analog:

More easily oxidized, producing disulfides, esters, ethers, etc
As thiols, more acidic than alcohols
Contrasting this, thiols dehydrogenate easily to form thiolate bases
Thiolates complex strongly to most metals

In general, you can expect the sulfur analog of pretty much any organic functional group (alcohols->thiols, aldehydes->thials, ketones->thioketones, esters->thioesters, ethers->thioethers, etc) to be more reactive. However, exactly how the sulfur switch will affect the actual reactions the compound will undergo is best studied case by case, as there is no one way that a particular sulfur-bearing functional group will behave when introduced into the complex chemical stew of human biology.

One near-constant; sulfur in small doses can smell pleasant, because sulfur is an essential not-so-trace element in humans (oranges and grapefruit, for instance, have some volatile sulfur compounds that contribute to their smell and taste). Sulfur in high concentrations stinks, because for the same reasons it is essential in biochemistry, too much of it is toxic (which is a pretty good illustration of the First Rule of Toxicology - every substance, in sufficient dosage, is toxic).


Regarding units of water potential and chemical potential

According to German Wikipedia water potential is defined as
$$\psi := \frac{\mu - \mu_0}{\bar{V}},\tag{1}$$
where $\mu_0$ is the standard chemical potential (usually pure water at atmospheric pressure at a specified reference height), and $\bar{V} \approx \pu{18 cm3 mol-1}$ is the molar volume of pure liquid water.
This means that water potential $\psi$ is defined as energy $E$ per volume $V$, which is equivalent to pressure $p$ if you perform dimensional analysis:
$$[\psi] = \frac{[E]}{[V]} = \frac{\mathsf{L}^2·\mathsf{M}·\mathsf{T}^{−2}}{\mathsf{L}^3} = \mathsf{M}·\mathsf{L}^{−1}·\mathsf{T}^{−2} = [p]\tag{2}$$
Megapascal $\pu{MPa}$ is usually used as the unit of measurement; tensiometers are usually graded in hectopascals $\pu{hPa}.$


Interesting question and this is a complicated topic. Actually it is not true that the wave and particle models are each valid 50 percent of the time. I think it is better to say that the wave model is always valid, but under some conditions the wave gets compressed down to a small area, so it looks like a particle. I hope that makes sense, please comment if it doesn't?

I am not a quantum chemist, but I think the normal conditions for electrons in atoms and molecules are not ones that make them act like particle. So therefore we should always think of them as waves. Luckily we can still talk about the "inner" and "outer" electrons while they are waves. It is not "inner" and "outer" exact position, it is an area occupied by the wave, the "inner" one is nearer to the nucleus, the "outer" one if further away. For atoms these are called atomic orbitals - if you look them up on Wikipedia it might be very interesting.

If you want to solve the energy, wavelength and position, it's quite complicated. I would recommend studying quantum mechanics. It will take a long time to get there, although it is very interesting. There might be a quicker but less detailed course just for chemists to do these calculation too. I hope that's helpful.