As in aqueous solution:
$$\ce{NH3+CH2COOH_{[b]}<=> NH3+CH2COO-_{[z]} + H+_{[x]}\\NH3+CH2COO-_{[z]} <=> NH2CH2COO-_{[a]} + H+_{[x]}}$$
So,
$$10^{-2}=\frac{[z][x]}{[b]}\\10^{-10}=\frac{[a][x]}{[z]}\\\implies 10^{-12}=[x]^2$$
Notice that a=b as the solution is always neutral; also z is a neutral compund
As,
$$[H^+]=10^{-6}\implies p^H=6$$
$$\implies[b]=100[z][x]=10^{2-2-6}=10^{-6}$$

  You should have done $p^H=-log[H^+]$, your [b] is correct.

For part (b), yes, using Henderson-Hasslebalch equation:
$$p^H=10+\log\left(\frac{1.05\times10^{-4}}{10^{-6}}\right)\approx 10.97$$

In basic solution, the second equilibrium shifts well(as a result the first one too) to the right.As NaOH is a strong acid, we can neglect any other bases.In presence of NaOH,
$$\ce{NH3+CH2COOH_{[b]} + 2NaOH_{[k]}<=> 2NH2CH2COONa_{[a]} + 2H2O_{[y]}}$$
In millimoles:
$$\begin{array}{c|c|c|c|c}
&entity&b&k&a&y\\\hline
&initial&10&.1&0&0\\\hline
&final&9.95&0&.1&.1+110(original)\\\hline
\end{array}$$
Now you can similiarly apply agin the Henderson-Hasslebalch equation:
$$p_H=12+log(.1/9.95)\approx 10.01$$

Note that I am treating the compund as a monoprotic acid of $pK_a =10(-NH_3^+)+2(-COOH)=12$


Why are indicators used only in the form of dilute solutions?

The most simple and general answer is: because we can! However, it may require explanation.

Indicators will always affect the observable of the solution they are measuring. For example, an acid-base indicator is an acid or a base itself (depending on the pH of the solution). It will take at least one indicator-equivalent of acid/base to protonate/deprotonate the indicator and induce the colour change. (Some indicators may even require two or more equivalents.) Until that full equivalent is supplied, we cannot witness the new colour. Thus, we want the indicator concentration to be as minute as possible because that will give us the lowest indicator error possible.

All other properties and effects of indicators are secondary to the major point above. For example, indicators are typically chosen to be very intensely coloured so that the colour may be observed even at very low concentrations; this allows us to reduce indicator loading and error and thereby increase accuracy.
Synthesising these indicators may be time-consuming and they may be expensive. This further strengthens our wish to use as little as possible.
The colour change of an indicator is ideally abrupt and strong, again meaning we can reduce the loading a lot.
And so on and so forth.

Thus, because we can use indicators at these low concentrations, we do.


For bond dissociation energies (BDEs) of double and triple bonds, is number shown in tables for that second or third bond? Or all of them?

They're for the entire bond (so for a triple bond, all 3).

From Wikipedia, it can be defined as

  The standard enthalpy change when [a bond, be it single, double, or triple] is cleaved by homolysis to give fragments A and B, which are usually radical species

So you get two fragments as your final state.


Acidic nature of benzene and other alkenes

A sp2-hybridized carbon is more electronegative than an sp3-hybridized carbon, and you would expect it to be more acidic.

Experiments support this prediction.  Benzene has a pKa of ~ 43, compared to alkanes, which have a pKa of ~ 50.  You can prepare lithioaromatics from bromo/iodoaromatics and butyllithium via metal-halogen exchange.

The situation with alkenes is slightly more complicated; the pKa of vinylic protons is in the same ballpark as of aromatic protons, but allylic protons are slighty more acidic.

Consult any of the usual pKa tables on the web.


I don't know if this would be very helpful or not but as i had some free time on hand I got a list of Madelung Constant Values from the Internet

In the list most of values were reported to 3-4 decimal places so I have Compared only the first decimal places as once the second decimal place was used none of the values matched. Maybe this would be a helpful start.

$$M_a$$ of ZnS (Wurtzite Hexagonal) is $10.7153$ (nearby the 12th and 18th entry in your table)

As Oscar Lanzi has commented $1.386$ is approximately $$2\log(2)$$ which is the Madelung constant for a one-dimensional line of alternating charges

As the question has been edited to add further details maybe the second one is a coincidence.

Only Nearby Matches So Far


Why is partial pressure proportional to mole fraction in Henry's law?

1st, Henry's law of Partial Pressure 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."

Quoting from WikiPedia here:

Henry's law is a limiting law that only applies for 'sufficiently dilute' solutions. The range of concentrations in which it applies becomes narrower the more the system diverges from ideal behavior. Roughly speaking, that is the more chemically 'different' the solute is from the solvent.
For a dilute solution, the concentration of the solute is approximately proportional to its mole fraction $x$, and Henry's law can be written as:

$$p = K_{\rm H}\,x$$
,where $p$ is the partial pressure of the gas above and in equilibrium with the solution, and $K_{\rm H}\,$ is the proportionality constant called the Henry's Law constant.
This is a conceptual conclusion as steeper the concentration gradient, the rate of diffusion (and hence the amount of diffused substance) will be greater. Here, the concentrations of the "solute" is represented by partial pressure of the gas in the gaseous state and by it's corresponding mole fraction in the solution state.
What you have supposed stands only for a closed system. As in the second quote from WikiPedia, you understand that the law is applicable only for conditions in which the behavior/condition which you supposed is either not relevant or is negligible.
If you go through the linked article, you'll find that Raoult's Law for Vapor Pressure of Solutions appears to be a special case of Henry's Law for more concentrated solutions, and can be used as such.
Note that both Raoult's Law and Henry's Law are limiting laws and they are applicable each as when:
Henry's Law: $x$ → $0$
Raoult's Law: $x$ → $1$


Establish the formula of an oxide of chromium

Rounding judiciously, the mass of chromium is 52 amu and oxygen is 16 amu.

Our compound is CrxOy and has 4 atoms.  The sum of those 4 atoms' mass is 100 amu.

x + y = 4

52x + 16y = 100

Solve the system of equations to find the answer.  Alternatively, you can just guess and check.

Any species with 2 chromium atoms must have a mass greater than 100 amu, since Cr2 = 104 amu and we haven't included any oxygens.  Thus we know for certain that there is only 1 chromium atom in the compound.  The remaining three must be oxygen.  I leave it to you to check that the answer is correct.

I think your confusion stems from assuming that something called "chromium oxide" must have 1 chromium atom for every 1 oxygen atom, and that isn't the case.


Converting mol files to InChiKey in large batches

This is exactly why I wrote Open Babel.

obabel *.mol -oinchikey -m

As long as you're not exceeding the number of command line arguments, you're done.


What is the difference between X-ray photoelectron spectroscopy and X-ray absorption spectroscopy?

You could easily google this...

X-ray photoelectron spectroscopy uses low energy monochromatic x-rays to excite electrons from atoms near the surface of a sample. The kinetic energy of the electrons is then measured and then the binding energy of the electrons can be calculated.

X-ray absorption spectroscopy measure the attenuation of a x-ray beam as a function of x-ray wavelength (or energy). So what you are detecting is the absorption edges of the atoms in the sample.


Dependance of heat capacity at constant volume on temperature for solids and liquids

The molar heat capacity of classical ideal gases at constant volume is temperature independent, because there are not considered quantization steps of electronic, vibrational and rotational energy.
The molar heat capacity ideal gases in the context of quantum-mechanic-aware gas theory is not temperature independent anymore. Because the electronic, vibrational and rotational energy contribution depends on ratio of mean energy per degree of freedom $E=\frac{1}{2} \cdot k_\mathrm{B} \cdot T$ versus the energy of quantization steps.
The capacity raises as rotation energy contribution kicks in, and later vibration energy as well, and finally electronic energy, as their quantization steps grows in this order.
There is a similar temperature dependent effect for liquids and solids, where quantized vibration is essentional, especially for solids.

The classical model of ideal gas does not take into account quantum effects. By warming up of an ideal gas, the mechanical energy gets evenly distributed between energy of linear motion, bond vibration and molecular rotation, proportionally to absolute temperature.
Quantum effects, taking part in behaviour of real gases, liquids and solids affects energy distribution. Vibration, rotation or electron energy are allowed by laws of quantum mechanics to have just some energy levels.
Imagine you throw small stones upwards to a steep hill. Stronger you throw it, the higher it would stay. This is analogous to an ideal gas.
Now you try to throw such stones to open windows of a tall building. If you make weak throwing, it will not flow even to the nearest window and will fall back. Stronger and stronger throwing ( = higher temperature ) would eventually cause some stones would reach at least the 2nd floor. Later even 3rd one etc.  This is like quantization of possible energies for real gases and condensed phases.


Rutherford's Alpha Ray Scattering Experiment

Your assumption is correct. For alpha particles, the main contribution to the total stopping power can be attributed to the electronic stopping power, i.e. inelastic collisions with electrons. Only a small contribution comes from the nuclear stopping power, i.e. elastic Coulomb collisions in which recoil energy is imparted to atoms.

The stopping power of a material is defined as the average energy loss per path length that the alpha particle suffers when travelling through the material.

According to the International Commission on Radiation Units and Measurements (ICRU) Report 49 Stopping Powers and Ranges for Protons and Alpha Particles (1993), the contributions to the total stopping power for alpha particles in gold are as follows.

Typical low-energy alpha particles with $E=1\ \mathrm{MeV}$:

Electronic stopping power: $3.887\times10^2\ \mathrm{MeV\ cm^2\ g^{-1}}$
Nuclear stopping power: $8.394\times10^{-1}\ \mathrm{MeV\ cm^2\ g^{-1}}$

Typical high-energy alpha particles with $E=10\ \mathrm{MeV}$:

Electronic stopping power: $1.650\times10^2\ \mathrm{MeV\ cm^2\ g^{-1}}$
Nuclear stopping power: $1.315\times10^{-1}\ \mathrm{MeV\ cm^2\ g^{-1}}$

Since the energy that is required for excitation or ionization is only a few $\mathrm{eV}$, an alpha particle with an initial energy of a few $\mathrm{MeV}$ can liberate many electrons on its path. When the alpha particle has sufficiently slowed down due to the stopping interactions, it can finally catch two electrons to form a neutral helium atom. Hence, you can release and detect helium when you dissolve old minerals of uranium or thorium.

In the Rutherford experiment, however, the used gold foil was very thin ($<1\ \mathrm{\mu m}$; note that gold is very malleable and can be beaten thin enough to become semi-transparent).
Even for low-energy alpha particles of only $E=1\ \mathrm{MeV}$, the range in gold is still $r/\rho=3.974\times10^{-3}\ \mathrm{g\ cm^{-2}}$, which can be calculated using the continuous-slowing-down approximation (CSDA), i.e. by integrating the reciprocal of the total stopping power with respect to energy. Considering a density for gold of $\rho=19.3\ \mathrm{g\ cm^{-3}}$, the range $r$ can be calculated as

$$\begin{align}
r&=\frac{3.974\times10^{-3}\ \mathrm{g\ cm^{-2}}}{19.3\ \mathrm{g\ cm^{-3}}}\\[6pt]
&=2.06\times10^{-4}\ \mathrm{cm}\\[6pt]
&=2.06\ \mathrm{\mu m}
\end{align}$$

I.e., the experiment was deliberately designed so that the alpha particles were not completely stopped and converted to neutral helium atoms within the gold foil.


As the $\mathrm{s}$ characteristic of an atom increases, so too does its electronegativity. Proof of this can be seen when comparing the $\mathrm{p}K_\mathrm{a}$'s of 2-carbon hydrocarbons. The $\mathrm{p}K_\mathrm{a}$'s of $\ce{CH3 -CH3}$, $\ce{CH2 =CH2}$, and $\ce{CH ≡CH}$ are $51$, $44$, and $25$ respectively.

When considering phenyl group for what it is, a ring of six $\mathrm{sp^2}$-hybridized carbons, it is easy to see why it shows a strong electron withdrawing effect, and a weak electron donating effect. The electronegativity of the carbons make benzene a nucleophilic structure that does not easily give up its electron cloud.


How does one calculate chemical shift in a gauge independent manner?

The references below describe the answer.

Prog Nucl Magn Reson Spectrosc. Author manuscript; available in PMC 2012 May 1.
Published in final edited form as:
Prog Nucl Magn Reson Spectrosc. 2011 May; 58(3-4): 176–201.
Published online 2010 Dec 15. doi: 10.1016/j.pnmrs.2010.10.003
PMCID: PMC3058154
NIHMSID: NIHMS248338
Chemical shift tensors: Theory and application to molecular structural problems
Julio C. Facelli

Nucleic Acids Res. 2014 Dec 16; 42(22): e173.
Published online 2014 Nov 17. doi: 10.1093/nar/gku1006
PMCID: PMC4267612
Accurate ab initio prediction of NMR chemical shifts of nucleic acids and nucleic acids/protein complexes
Andrea Victora,1 Heiko M. Möller,2 and Thomas E. Exner1,3,*

A comparison of models for calculating nuclear magnetic resonance
shielding tensors
James R. Cheeseman, Gary W. Trucks, Todd A. Keith, and Michael J. Frisch

J. Chem. Phys. 104 (14), 8 April 1996


Why can we not compare the coupling constants to chemically equivalent (but not identical) protons to determine magnetic equivalence?

The problem is that from the perspective of $\ce{H_{d}}$ and $\ce{H_{c}}$, $\ce{H_{a}}$ and $\ce{H_{b}}$ are different (you can tell which is which by their coupling constants). This difference means that there is a magnetic inequivalence between the two hydrogens $\ce{H_{a}}$ and $\ce{H_{b}}$, so there will be coupling between them.


Why color of alkali metal peroxides, superoxides and ozonides deepens down the group?

Most likely, we are seeing the same band structure effects.

There is an additional trend, whereby the compounds become darker and more reddish as one adds more oxygen.  In oxygen-chain species such as $\ce {O2^{2-}} $, antibonding molecular orbital are occupied.  So The valence band containg these electrons is raised.  That lowers the band gap enabling absorption of more visible light.

Incidentally, the visual effects described here are limited to alkali metal (when oxygen is the electronegative element).  When we move to the corresponding alkaline earth oxide the monoxides are white and the peroxides ($\ce {Mg}$ and heavier) are white too.  Less strongly ionic bonding and a lower cation/anions ratio lead to less negative charge on the oxygen species, thus lower valence band energies and gaps  remaining large enough to remain uv-absorptive only.


Clarification on Recrystallization concept

You note that you are pretty close to the answer and you are. Let’s assume for a minute (I’m pretty sure the assumption is false, but it’ll get us a good way) that the solubility of the impurity is independent of the amount of desired product dissolved and vice-versa. Let’s also assume we have a mixture of $95~\%$ desired product and $5~\%$ side product which you want to purify by recrystallisation. The desired product has a solubility of $1\,\mathrm{\frac{mol}{l}}$ at $4\,\mathrm{^\circ C}$ and $10\,\mathrm{\frac{mol}{l}}$ at $80\,\mathrm{^\circ C}$; and the side product has the same solubility. And finally, let’s assume that you have $100\,\mathrm{mmol}$ total product (side and desired).

You then add $9.5\,\mathrm{ml}$ of your recrystallisation solvent and heat to $80\,\mathrm{^\circ C}$. That would be enough to dissolve both your $95\,\mathrm{mmol}$ desired product and your $5\,\mathrm{mmol}$ side product. You observe a clear (hopefully boiling because best solubility) solution.*

You then let your solution cool down and put it in the fridge overnight. Due to the decreased temperature, the solubility of both desired product and side product drop. The same volume of solution now dissolves only $9.5\,\mathrm{mmol}$ of both. Since the desired product is supersaturated ($95 > 9.5$), $85.5\,\mathrm{mmol}$ of the desired product will precipitate or crystallise. But the side product is still not supersaturated. We have $5\,\mathrm{mmol}$ of the side product dissolved, but the solvent could dissolve $9.5\,\mathrm{mmol}$. So all of the side product will remain in solution (ideally).

Next step is usually filtration and washing. The filtrate will now contain the saturated product solution which is undersaturated with respect to the side product. The side product didn’t disappear, it merely remained in solution. Wash your pure product, record an NMR spectra and rejoice upon its purity.

Of course, as I stated in the beginning, the assumption that solubilities be independent is wrong. It is also tendencially wrong to assume similar solubilities of desired product and impurity. One would attempt to select a solvent so that the product is not well soluble at low temperatures, but the impurity ideally is. However, the basic principle remains valid: The desired product will form a supersaturated solution and precipitate/crystallise while whichever impurity you have should still be soluble to not co-precipitate.

*: A note on how to do it practically: Usually, you would put your crude product into a flask into an oil bath with a refluxing condenser on top. You would then slowly, dropwise add solvent through the condenser until the solution is boiling and everything is dissolved. One wouldn’t go about saying ‘I should have $x\,\mathrm{mmol}$ so I should need $y\,\mathrm{ml}$ of solvent.’


Triclosan (2,4,4′-trichloro-2′-hydroxydiphenyl ether) is a nonionic, broad spectrum, antimicrobial agent that, because of its favorable safety profile, has been incorporated into a variety of many personal care products, including deodorant soaps, underarm deodorants, shower gels, and health care personnel handwashes. Triclosan exhibits a moderate degree of substantivity to the skin, and, in many products, it imparts a remnant antimicrobial effect.Here is the link.

U.S. Food and Drug Administration (FDA) states that "Triclosan is not currently known to be hazardous to humans."

Despite having significant uses, it also posses little amount of adverse effects on prolong use.It impairs muscle function and skeletal muscle contractile property.This research has only been conducted in mice and results were not good; when mice were exposed to triclosan, heart muscle function was reduced by 25 percent and grip strength was reduced by 18 %.
So result has concluded that it might cause skeletal and heart muscle failure in humans as well.Here is the link.

Exposures to triclosan can also cause dermatitis, or skin irritation, and an increase in allergic reactions, especially in children.Due to its link with dioxin, it has been suspected to cause other health problems like weakening of the immune system, decreased fertility, miscarriage, birth defects, and cancer.Here is the link

But to eradicate these effects,FDA has been engaged in an ongoing scientific and regulatory review of this ingredient. FDA does not have sufficient safety evidence to recommend changing consumer use of products that contain triclosan at this time.

So we can conclude that, it cannot be considered entirely harmful for human body.


How well (or poorly) does natural leather insulate?

Interestingley enough, there are papers entirely devoted to the subject of measuring the properties of living skin. The paper linked gives a value of $7 \times 10^{-4}~\mathrm{\frac{cal}{cm\, s\, °C}}$ for human skin, or in everyday units used nowadays: $0.29 ~\mathrm{\frac{W}{m\, K}}$. This is only for the dermis, the part of skin that then gets transformed into leather.

Now, this website has a comprehensive list of many different thermal conductivity values, and dry leather clocks in at $0.14 ~\mathrm{\frac{W}{m\, K}}$, which is about 2 times less the conductivity than that of living skin. What a difference the tanning process can make!

Given the similarity  observed in the review paper between man, beef and porcine tissue, I think it is fair to say that the value for leather will not change greatly across species.


Can Carbon Form bonds without Hybridization?

Hybridisation is a mathematical concept. It is one way of deriving chemical bonds from elements but not the only way. A different one (that computational chemistry typically uses) would be to just plant atoms at certain positions in space, and then mix their unhybridised atomic orbitals with each other to see what comes out. One can then back-calculate this to localise the molecule-spanning orbitals to single bonds and thereby deduce a hybridisation that would have worked to build up the molecule. Take the following example of methane (taken from Professor Klüfers’ internet scriptum for basic and inorganic chemistry at the university of Munich):

On the left you have four hydrogen s-orbitals, on the right you have the orbitals of carbon. As you see, there is no hybridisation assumed at all and the ground state is taken to be a fully populated $\mathrm{2s}$ orbital and three $\mathrm{2p}$ orbitals populated with two electrons amoung them for a total of $\mathrm{2s^2~2p^2}$. The hybrid orbitals are gained from symmetry considerations (in a tetrahedron, the p-orbitals transform as $\mathrm{t_1}$) and hydrogen orbitals were selected as was seen fit. The electrons are added a posteriori. Spin conservation would have produced a twice-excited state with very high total energy, so immediate spin pairing is assumed.

To localise the bonds, linear combine all four orbitals. If you take every orbital with factor $+1$, you arrive at the $\ce{C-H}$ bond to the upper right hydrogen. By shifting around a factor of $-1$ among the three $\mathrm{t_1}$ molecular orbitals during linear combination, you can address every other $\ce{C-H}$ bond. The resulting bonding orbital looks like n $\mathrm{sp^3}$ orbital mixed with a hydrogen $\mathrm{s}$ orbital.


SN2 nucleophilic attack with alpha unsaturation

As Ben Norris said, the aspects of nucleophilic and electrophilic attacks are not relevant in the second case.

Talking about the compound given in question 1(b), the first step of nucleophilic attack is opposed by the conditions you mentioned, i.e., of the electron withdrawal effect of the double bond and replusion of nucleophile by $\pi$ electron cloud.

But talking in terms of yield of the product, it is necessary to consider two things:

Chlorine is a good leaving group.
The stability of the carbocation intermediate formed by $SN_1$ mechanism (steric factors don't favour $SN_2$). The carbocation formed will be very stable due to conjugation with the double bond. Hence, the yield of the product is not much low.

While solving the above question, we should look at the stability of carbocation intermediate formed,which is a more dominating effect. Though Bromine is a better leaving group, and there is decrease in tendency for nucleophilic attack due to $\alpha $-unsaturation; both factors which support higher nucleophilic attack for the first compound, stabilisation of carbocation by resonance (conjugation) is highly stabilising, and this shadows the opposing factors.


How do I know how many molecules a solute will split into when it dissolves in a solvent?

You need to know the type of reaction that is taking place and what types of bonds are being made and/or broken to determine how many ions the given molecule will split into.


Why is it that increasing the molar mass generally increases the Entropy?

The allowed energy levels for a particle of mass $m$ in a box of length $X$ are given by
$$E_i=\frac{h^2}{8mX^2}i^2\quad i=1,2,\ldots$$

(Already this equation gives a hint that an ideal gas of high molar mass has more thermally accessible translational states than a gas of low molar mass. Therefore, the molar entropy of a gas of high molar mass is greater.)

In this case, the partition function for translational motion in the $x$-direction is approximately
$$z=\left(\frac{2\pi mkT}{h^2}\right)^{1/2}X$$

The partition function for translational motion in three dimensions is
$$\begin{align}
z&=\left(\frac{2\pi mkT}{h^2}\right)^{3/2}XYZ\\
&=\left(\frac{2\pi mkT}{h^2}\right)^{3/2}V
\end{align}$$

For a system of $N$ particles

$$\begin{align}Z&=\frac{z^N}{N!}\\&=\frac{1}{N!}\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}V\right]^N\end{align}$$

The statistical entropy is
$$\begin{align}
S &=\frac{U-U_0}{T}+k\ln Z\\
&=\frac{U-U_0}{T}+k\ln \frac{z^N}{N!}\\
&=\frac{U-U_0}{T}+kN\ln z-k\ln N!\\
&\approx\frac{U-U_0}{T}+kN\ln z-kN\ln N+kN\\
&=\frac{U-U_0}{T}+nR\ln z-nR\ln nN_\text{A}+nR\\
&=\frac{U-U_0}{T}+nR\ln\left[z\cdot\frac{1}{nN_\text{A}}\cdot \operatorname{e}\right] \\
&=\frac{U-U_0}{T}+n R \ln \left[\left(\frac{2\pi m k T}{h^2}\right)^{3/2} \cdot \frac{\operatorname{e}V}{nN_\text{A}}\right] \\
\end{align}$$

The internal energy of a monoatomic ideal gas is

$$U=U_0+\tfrac{3}{2}nRT$$

which leads to the Sackur–Tetrode equation:

$$\begin{align}
S &=\frac{U-U_0}{T}+nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]\\
&=\tfrac{3}{2}nR+nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]\\
&=nR\left\{\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2}\cdot\frac{\operatorname{e}V}{nN_\text{A}}\right]+\tfrac{3}{2}\right\}\\
&=nR\ln\left[\left(\frac{2\pi mkT}{h^2}\right)^{3/2} \cdot \frac{\operatorname{e}^{5/2}V}{nN_\text{A}}\right]
\end{align}$$

This equation implies that the entropy $S$ of a monoatomic ideal gas depends on the mass $m$.


Purification/Extraction of Sodium Permanganate

You are correct. First step produces $\ce{K2MnO4}$, then you need to oxidize it.

My first guess was that adding $\ce{H2SO4}$ will cause disproportionation of $\ce{K2MnO4}$ into $\ce{KMnO4}$ and $\ce{MnO2 + K2SO4}$. Filtering removes $\ce{MnO2}$. Crystallization will give you pure product. Or you can use a mix of
$\ce{KMnO4 + K2SO4}$ for many purposes.

Industry uses more efficient method: electrolysis of $\ce{K2MnO4}$.


Why are hydrogen-bonded compounds (such as NH3) excellent solvents for most lab reagents?

Recall that hydrogen bonding interactions result from the polarized nature of bonds between hydrogen and nitrogen, oxygen, and fluorine. This polarized bond results in a hydrogen atom with very low electron density—a property that allows the atom to interact very strongly with negatively charged particles.

Thus, anions and electronegative atoms in polarized molecules have very strong electrostatic interactions with the low-electron-density hydrogen atom in $\ce{NH3}$ or $\ce{H2O}$, for example. Cations and positively charged elements in polarized molecules can also be stabilized by the electronegative atoms in molecules exhibiting hydrogen bonding ($\ce{N}$ and $\ce{O}$ in this case).

These strong interparticular interactions ultimately result in polarized or ionic solutes generally being very soluble in solvents with hydrogen bonding such as $\ce{NH3}$ or $\ce{H2O}$.


Mechanism for reaction of Tollens' reagent with aldehydes

Here are the two half reactions:

$$\begin{align}
\ce{[Ag(NH3)2]+ + e- &-> Ag^0 + 2NH3} \\
\ce{RCHO + 3OH- &-> RCO2- + 2H2O + 2e-}
\end{align}$$

which together yield the overall reaction

$$\ce{2[Ag(NH3)2]+ + RCHO + 3OH- -> 2Ag^0 + RCO2- + 4NH3 + 2H2O}$$

Here is a diagram of the reaction mechanism. The carbonyl group is oxidized in the process and the $\ce{Ag^+}$ is reduced. The resultant oxidized aldehyde (now a radical cation) reacts with hydroxide to form a tetrahedral intermediate. A gem-diol like intermediate is formed via a hydrogen shift, which then continues on to the final carboxylate anion.


Is phosphorus a stereogenic centre in this phosphate ester?

In compound A, the negative and double bonded oxygens bound to the phosphorus are equivalent:

$\hspace{5.1cm}$

In compound B, they are not equivalent:

$\hspace{7.5cm}$


How does an additive like aluminum oxide make a material like polyurethane harder/more resistant to wear?

Alumina is extremely hard and very wear resistant, likely harder than anything which would come in contact with the floors. Dirt for example would mainly contain silica based abrasives which are far softer than alumina and hence alumina additives, of sufficient size would provide scratch resistance when present in sufficient quantities.

If you want the exact details, they are going to vary from one manufacturer to another. If you do a search for patents you can see various procedures some of which are pre-mixed, some of which require onsite mixing, for example :

Surface coverings containing aluminum oxide

That kind of post-mixing is very common in concrete, though there the aggregate usually isn't alumina (could be just decorative as well or dual purposed as is common locally) and is far bigger. In general you want the aggregate to be at least as large as, and ideally bigger than the particle which is trying to cut it. If the aggregate is much smaller it won't resist wear strongly as it will be plowed out with the substrate.

Effect of abrasive particle size on abrasive wear
of hardfacing alloys

If you are curious about a specific product then just call the manufacturer and ask to speak to the engineering department. However as they are all likely patented products, you can even search manufacturer specific and turn up results.


How do you determine how many electrons occupy the anti bonding orbitals?

This is the MO diagram for Hydrogen. As you can see, for hydrogen atom, there is only 1 electron in the 1s orbital. Two hydrogens could come together to form a sigma bond, resulting in lowered energy. This doesn't mean that there is no sigma antiorbital, but rather due to the fact that hydrogen only contributes one electron each leading to a sum of two electrons, there are not enough electrons to fill the antiorbital, hence it is unfilled.

Now, imagine if it was helium instead. Helium has two 1s electrons, therefore, if two He atoms form a bond, 4 electrons has to be placed into the molecular orbitals. Since every orbital can hold a maximum of 2 electrons, two electrons would be in the sigma orbital, and the other two would be in the sigma antiorbital. An orbital and antiorbital would effectively cancel out, therefore, you see Helium exist in the monoatomic state.


After further reflection I realized an initial comment I made was flawed*. To quote the Wikipedia:

  Isenthalpic processes on an ideal gas follow isotherms, since $\mathrm dh = 0 = c_p\,\mathrm dT$.

That means that for an ideal gas isenthalpic is synonymous with isothermal, and this implies $\Delta U = 0$ and so $q = -w = P_\mathrm{ext}\Delta V$.

For the isobaric process described in the problem, the work done by the gas is $\pu{500 J}$, and therefore the heat expected for an isothermal process is also $\pu{500 J}$.

* The source of my error is simple: for an irreversible process the heat is not equal to $\Delta H$. In fact, since the external pressure is not balanced by that of the system, $P_\mathrm{ini}\neq P_\mathrm{fin}$ for the system (the process is not strictly isobaric for the system). For a reversible isobaric process $q_p = \Delta H$.


Why are the vinylic bonds in alkenes slightly polar?

Maybe because of resonance:

$H_{2}C=CH_{2}$    $\leftrightarrow$    $H_{2}C^{-}-C^{+}H_{2}$


$\ce{CF3+}$ is more stable than $\ce{CH3+}$. The comparison between -I and +R effect is made while talking about activation or deactivation of halogens when substituted on benzene. Fluorine always stabilises a carbocation to a large extent because of a very good overlap of $\ce{2p -~2p}$ orbitals.


Why don't ions form crystal lattices in water and other polar solvents?

They actually form: ions can be dissolved up till a certain concentration, and beyond that, they form ionic lattices just as you predict.

The easiest way to describe the scenario as an equilibrium between ions-in-water vs ions-in-lattice:

If the ion-water interaction is very strong, the ions can be very stable in water, even more than in an ionic lattice, so ions will be in the solution.
If the ion-water interaction is weak compared to the solid, then the solid phase is much more stable, and only a very low concentration of ions can be in the solution.

Looking at only interaction energies is a very simplistic view as it does not consider entropic factors, yet can describe at least part of the story. Entropy is generally a huge driving force for any kind of dissolution and allows some ions to be in solution even if the solid state energetically more stable.


Methylene blue used as funguscide in cichlid reproduction - concentration measurement?

The concentration of an aqueous solution of methylene blue can be determined using a UV-VIS spectrometer.

Measure the absorption at $\lambda = 664\, \mathrm{nm}$, use the molar extiction coefficient of $95000\, \mathrm{dm^{3}\cdot mol^{-1}\cdot cm^{-1}}$
(Reference) and calculate the concentration using the Lambert-Beer equation.


Using the general case equation
$$\frac{P_1V_1 }{T_1} = \frac{P_2 V_2} {T_2}$$
and given that pressure is constant at $\pu{1  atm}$ in your experiment, the equation simplifies to
$$\frac{V_1 }{T_1} = \frac{ V_2} {T_2}$$
We know $V_1=\pu{960 \mu L}$ and it appears that the experiment is being run at $\pu{(273+23)K}$, so we need to calculate what the volume ($V_2$) would be at standard T ($\pu{273 K}$).  This is given by
$$V_2 = \frac{V_1 T_2}{T_1}$$
if we next divide $V_2$ by $\pu{22.423E6 \mu~L}$, we now know how many moles of $\ce{O2}$ we have.


There is no sense of comparing the reaction rate when one of the compound doesn't show SN2 mechanism. And, that compound is $\ce{CH2=CHBr}$, due to the fact that it is a vinyl halide.

$\ce{CH2=CHBr}$ will hinder the approach of nucleophile due to the presence of pi-electron cloud around double bond.

And, therefore the obvious answer is $\ce{CH3-CH2Br}$


Short answer:
No. No. Yes. Yes. No.
Next time, try searching before posting.
Longer answer:
It is just a sale trick. Chemistry community has no direct influence on used trademarks of commercial products. Some may call whiskey "liquid gold", but nobody expects there is any gold there.
The product is called "liquid chlorine" just because it shares some of characteristics of chlorine dissolved in water/water based solutions. As in both cases , there are some equilibrium concentrations of hypochlorous acid, free chlorine and atomic oxygen.
The "liquid chlorine" is basically a solution of sodium hypochlorite, sodium chloride and sodium hydroxide. There is the answer directly on the photo: "130 g/L available chlorine(Cl) present as sodium hypochlorite". I.e. chlorine is not present in its free form, but in chemically alternative form of hypochlorite, that can release chlorine at different than storing conditions.
Sodium hypochlorite is produced by dissolution of chlorine in solution of sodium hydroxide:
$\ce{Cl2(g) + 2 NaOH(aq) -> H2O + NaCl(aq) + NaClO(aq)}\tag{1}$
If acidified, e.g. by sulphuric acid, the reaction (1) is reversed:
$\ce{H2SO4(aq) + NaCl(aq) + NaClO(aq) -> Na2SO4(aq) + Cl2(g) + H2O}\tag{2}$
If dissolved by about neutral solutions, reaction (2) happens in lower extent, as there occurs also
$$\ce{ClO-(aq) + H2O <=> OH-(aq) + HClO(aq)\tag{3a}}$$
$$\ce{HClO(aq) -> H+(aq) + Cl-(aq) + O(aq)\tag{3b}}$$
The atomic oxygen is reportedly the true desinfecting agens, rather then chlorine itself.


How to synthesize 4‐[(4,4‐dimethylpentyl)oxy]‐2,2,8,8‐tetramethylnonane starting with 4,4‐dimethylpent‐1‐ene?

Divide your alkene starting material in two portions - one twice the amount of the other.
Take the larger portion and do an anti-Markovnikov addition of HBr to it using peroxide or light reference to give the terminal bromide (1-Bromo-4,4-dimethylpentane).
Take the smaller portion and ozonolyse it, working up with Dimethyl Sulfide to give the aldehyde (3,3-dimethyl butanal).
Take half the bromide material and form the Grignard with Mg metal. Add the Grignard to the aldehyde. Work up and isolate the resulting secondary alcohol. Dissolve the secondary alcohol in DMF at 0c and form the alkoxide with NaH, add the second portion of the bromide to the alkoxide solution. Work up, isolate and purify to obtain the desired product.


The mathematical description of the cell constant ($l/A$) does not include factors like fringing fields (the ionic current does not travel precisely in a well-defined path between the electrodes) and the actual surface area for rough surfaces.
The cross section of a round electrode may have different fringing fields depending on its separation from its companion. And if the two electrodes are of different geometry, you have to decide how to average the areas out. I think there was more standardization in the choice of electrode materials, size and spacing than has been carried over into the beautifully simple formula $l/A$, so when the formula is used outside of certain narrow bounds (and I don't know what they are), it's only an approximation, and resistance or conductance can be measured more precisely than the cell constant from geometry. So you might as well calibrate against a known cell or solution.


Why cannot nitrosonium ion attack aniline in electrophylic aromatic substitution?

What actually happens is that the nitrosonium ion always adds faster to the amino nitrogen.  But that reaction is rapidly reversible unless a proton can dissociate from the amino nitrogen to give a stable N-nitroso product.

A primary or secondary amine has the required N-H proton, and in the case of a primary amine the presence of a second N-H proton enables further reaction making a diazonium ion (as shown for aniline by the OP) or even molecuar nitrogen.  But a tertiary amine has no N-H proton for the elimination step.  A tertiary amine then undergoes no net nitrosation reaction at the nitrogen atom, but if it's aromatic then EAS emerges as an alternative.


Nitrobenzene and other monosubstituted benzenes 1H vs 13C NMR assignments

As I wrote in the comments, there is not really a good way of arriving at these conclusions for a general aromatic molecule. The energy differences at play in these systems are so small that qualitative arguments will often be insufficient to get the order right. The diamagnetic term, which can be qualitatively rationalized by arguments based on electron density, for non-H atoms, is dominated by the paramagnetic term, which is, in some sense, an orbital response term due to the application of the magnetic field of NMR. Qualitatively reasoning about the paramagnetic term is possible in some special systems like PhLi or pyridine, but the accuracy of these qualitative arguments is insufficient for general substituted benzenes. The paramagnetic term can be calculated by electronic structure packages, but I assume that is not really the question.

I wrote some lecture notes a few years ago for a lecture I gave on the subject of diamagnetic and paramagnetic terms in NMR. I tried to pare it down a bit but they are still rather long. I will start with the qualitative quasiclassical conclusions and then have the derivation.

Qualitative Explanation of Diamagnetic and Paramagnetic Terms

The diamagnetic and paramagnetic terms in NMR shifts arise from considering how the electron motion is affected by an applied field. The terms diamagnetic and paramagnetic only indicate the sign of the magnetic response, opposed to the field and with the field, respectively (it does not have anything to do with unpaired electrons, which is the place most chemists have heard these words before). The diamagnetic term can be thought of being like Lenz's Law in classical electromagnetism; applying a field induces a current which opposes the field. This is sort of (with excessive handwaving) why electron density correlates with how shielded the nuclei are: with less electron density around, the amount of current is less, the induced field opposing the applied field is less, and the effective field is larger.

For the paramagnetic term, think of an electron moving (remember this a quasiclassical picture, but it's not wrong) in a $p$ orbital. It has angular momentum $l=1$, including an $m_l$ value of $+1$ which would generate a huge field and an $m_l$ value of $-1$ which generates the same field and in the opposite direction (also $0$ but that's not important here). Hence, in the absence of a field there is no field due to this orbital angular momentum being ``quenched'', that is $m_l=\pm 1$ are represented equally in the ground state wavefunction and exactly oppose each other. If a field is applied though, the $m_l$ aligned with the field (let's say $+1$) is stabilized while the $m_l=-1$ is destabilized and so there is more of one than the other in the ground state wavefunction at that angular momentum is unquenched. This results in the addition of a field in the same direction as the applied field, making it larger. This is the origin of the much larger range of shifts for $\ce{C}$ as opposed to $\ce{H}$. Calculating the magnitude of the paramagnetic term is not super straightforward and so qualitative arguments to try to rationalize the order of simple aromatic molecules is hard to do (examples like PhLi or pyridine are possible because there is an obvious paramagnetic perturbation, the lone pair). The amount of paramagnetic shifting tells us something about the low-lying vacant orbitals of the system and so can be a useful tool for predicting reactivity in some cases, but predicting those shifts without calculating them explicitly is not readily possible.2

In hydrogen, which is essentially only an $s$ orbital (which has an angular momentum of 0), there is no paramagnetic term (because there is no orbital angular momentum to unquench) and all of the shifts are due to the diamagnetic term. For heavier elements, like C, where $p$ orbitals get involved, there is a paramagnetic term and it is dominant.

Derivation of the Diamagnetic and Paramagentic Terms

Classical electromagnetism tells us that the magnetic field $\textbf{H}$ generated by a charge $q$ moving with velocity $\mathbf{v}$ at a point $\mathbf{r'}$ away is

$$\textbf{H}=\frac{q}{c}\frac{\mathbf{v}\times\mathbf{r'}}{r'^3}$$

Since we care about the field at the nucleus, we set the coordinate system origin at the nucleus and consider the field at the origin of a charge a distance $\mathbf{r}$ away. Then $\mathbf{r'}=-\mathbf{r}$ and we have

$$\textbf{H}=\frac{q}{c}\frac{\mathbf{r}\times\mathbf{v}}{r^3}=\frac{q}{mc}\frac{\mathbf{r}\times\mathbf{mv}}{r^3}=\frac{q}{mc}\frac{\mathbf{L}}{r^3}$$

We'll need this later. Let's consider though: the above implies that $\textbf{H}$ is on the order of $\approx \beta/r^3$, where $\beta$ is the Bohr magneton and $(1/r^3)$ for a 2p orbital is about 0.25 Å and so $H$ is about 600,000 gauss, which is massive! This should bring up the question: if non-zero $\mathbf{L}$ leads to huge magnetic fields why don't these fields completely destroy our NMR signal? Are there magnetic fields present in diamagnetic molecules when there are no magnetic fields present? I've sort of already hinted at the answer. No, there aren't such huge fields. Loosely, this is because the field generated by the electrons circulating in the $m_L = +1$ level exactly cancels the field of the electrons in the $m_L=-1$ because both are equally represented in the ground state (as they must be without a magnetic field present). This quenching of angular momentum occurs in all cases where the orbitals are real valued, such as if they are solutions of a Hamiltonian in the absence of a magnetic field (so nothing with unpaired spins, which carry a magnetic field). If we turn on an $\textbf{H}_0$ field though, one of the directions of circulation will be favored over the other (the $m_L=-1$) and the wavefunction will adjust so that the ground state has a slight net circulation in the favored direction. In this way, an applied magnetic field unquenches a small amount of the orbital angular momentum. This gives rise to the paramagnetic contribution.

Let's develop this mathematically. First, we should show that our semi-classical picture is a valid way to consider this quantum mechanical system. Fundamentally, a magnetic field $\textbf{H}$ is given by the curl of some vector potential $\textbf{A}$. We ultimately have two fields to consider: $\textbf{H}_0$, the applied field, and $\textbf{H}_n$, the field generated by the nucleus of the atom, each with their associated vector potentials.

\begin{align}
\textbf{H}_0&=\nabla\times \textbf{A}_0\\
\textbf{H}_n&=\nabla\times \textbf{A}_n
\end{align}

We're going to neglect the magnetic fields of the nucleus for this discussion. Obviously, they are important, since the perturbation of the nuclear spin energy levels are mediated by the coupling of the nuclear magnetic fields to the electronic magnetic fields, but we first need to determine what those electronic fields are.

As you may recall from classical electromagnetism, there is more than one vector potential that produces a given field. Specifically, if $\textbf{H}=\nabla\times\textbf{A}$, another vector potential $\textbf{A}' = \textbf{A}+\nabla\phi$, where $\phi$ is a scalar function, gives the same field, since the curl of a gradient of a function is always 0. The transfomation $\textbf{A}\rightarrow\textbf{A'}$ is called a gauge transformation. We need all physical results that we calculate to be gauge-invariant.

So now, how does an applied magnetic field show up in our Hamiltonians. It appears by changing $(\hbar/i)\nabla\rightarrow(\hbar/i)\nabla-(q/c)\textbf{A}$. So the Hamiltonian becomes

$$\mathcal{H} = \frac{1}{2m}\left(\mathbf{p}-\frac{q}{c}\textbf{A}\right)^2+V$$

If we were to calculate expectation values of $\hbar/i\nabla$ with different gauges, we would find that they aren't equal and so $m\mathbf{v}$ is no longer $\hbar/i\nabla$ but is $\hbar/i\nabla-\frac{q}{c}\textbf{A}$. Similarly, $\mathbf{L}$ is now

$$r\times\left(\frac{\hbar}{i}\nabla-\frac{q}{c}\textbf{A}\right)$$

We define something that will be useful to us throughout this discussion, the current density $\mathbf{j}(\mathbf{r})$.

\begin{align}\label{current}\tag 1
\mathbf{j}(\mathbf{r})=\frac{q}{2m}\frac{\hbar}{i}[\psi^*\nabla\psi-\psi\nabla\psi^*]-\frac{q^2}{mc}\textbf{A}\psi^*\psi
\end{align}

Note that it is a function of position and real-valued. It is also $q$ times the quantum mechanical probability current and gauge-invariant. If we assume $\psi$ is a solution of the time-dependent Schrodinger equation, then

$$\text{div} \mathbf{j}+\frac{\partial \rho}{\partial t} = 0$$

where $\rho = q\psi^*\psi$ (that is, the charge distribution). For pure states, $\psi^*\psi$ is time-independent and so $\text{div} \mathbf{j}=0$, and hence behaves as a classical current density. One can do some more math and show that it is possible to express the chemical shift, a quantum mechanical phenomenon, entirely classically as the coupling of the the nuclear magnetic moment to the effective magnetic field at the nucleus due to the circulating density $\mathbf{j}_0$.

\begin{align}\label{chemshiftE}\tag 2
E_\text{pert}=-\mathbf{\mu}\cdot\left[\frac{1}{c}\int{\frac{\mathbf{r}\times\mathbf{j}_0(\mathbf{r})}{r^3}d\tau}\right]
\end{align}

So if we can calculate $\mathbf{j}_0$ we have done the hard part of finding the chemical shift. Let's turn to that now.

Now, the Hamiltonian for our electron is

$$\mathcal{H} = \frac{1}{2m}\left(\mathbf{p}-\frac{q}{c}\textbf{A}_0\right)^2+V$$

We multiply out the binomial and rearrange to get

$$\mathcal{H} = \frac{p^2}{2m}+V-\frac{q}{2mc}(\mathbf{p}\cdot\textbf{A}_0+\textbf{A}_0\cdot\mathbf{p})+\frac{q^2}{2mc^2}A_0^2$$

Let's assume we have the eigenfunctions of $\mathcal{H}_0=\frac{p^2}{2m}+V$ in the absence of the field (i.e. $\mathcal{H}_0\psi_n=E_n\psi_n$). Our goal is to calculate $\mathbf{j}_0$, which has a term linear in $\textbf{A}_0$, which looks like

$$\textbf{A}_0=\frac{1}{2}\textbf{H}_0\times\mathbf{r}$$

That's one choice of gauge, but in any gauge $\textbf{A}_0$ is proportional to $\textbf{H}_0$. So if we want $\mathbf{j}_0$ right to first order, we need $\psi$ at least accurate to first order in $\textbf{H}_0$ (the $\psi^*\psi$ in the last term of $\mathbf{j}_0$ can be the unperturbed wavefunction since that term already has a $\textbf{A}_0$). So let's do some first-order perturbation theory! In case you forgot your first-order PT, we first need to define the perturbing Hamiltonian. We'll use the linear part of our Hamiltonian above.

$$\mathcal{H}_1=-\frac{q}{2mc}(\mathbf{p}\cdot\textbf{A}_0+\textbf{A}_0\cdot\mathbf{p})$$

Our first-order corrected ground-state wavefunction (this is just standard first order PT) is

$$\psi'_0=\psi_0+\sum_n{\frac{\langle n|\mathcal{H}_1|0\rangle}{E_0-E_n}\psi_n}$$

where $E_0$ is the ground state energy, $E_n$ is the energy of the $n$th excited state $\psi_n$. Let's define

$$\epsilon_{n}=\frac{\langle n|\mathcal{H}_1|0\rangle}{E_0-E_n}$$

So that

$$\psi'_0=\psi_0+\sum_n{\epsilon_{n}\psi_n}$$

Substituting this $\psi'_0$ into the equation for $\mathbf{j}$ (equation \ref{current} above), we obtain

$$\mathbf{j}_0(\mathbf{r})=\frac{\hbar q}{2mi}(\psi_0^*\nabla\psi_0-\psi_0\nabla\psi_0^*)+\frac{\hbar q}{2mi}\sum_n\left[{(\psi_0^*\nabla\psi_n-\psi_n\nabla\psi_0^*)\epsilon_n+(\psi_n^*\nabla\psi_0-\psi_0\nabla\psi_n^*)\epsilon_n^*}\right]-\frac{q^2}{mc}\textbf{A}_0\psi^*\psi$$

The first term

$$\frac{\hbar q}{2mi}(\psi_0^*\nabla\psi_0-\psi_0\nabla\psi_0^*)$$

is the current flowing in the absence of a magnetic field. If our angular momentum is quenched as we described above (as in closed-shell molecules), then this is 0. If we take the $\psi_n$ as real, then

$$\mathbf{j}_0(\mathbf{r})=\frac{\hbar q}{2mi}\sum_n\left[{(\epsilon_n-\epsilon_n^*)(\psi_0\nabla\psi_n-\psi_n\nabla\psi_0)}\right]-\frac{q^2}{mc}\textbf{A}_0\psi^*\psi$$

We're gonna see that the first term is the paramagnetic current, related to $\epsilon_n$, which recall is the matrix element coupling the ground state to excited states by the angular momentum operator, and the second term is the diamagnetic current, always present and independent of $\textbf{L}$. Let's go back to that choice of the vector potential $\textbf{A}_0$ we made before

$$\textbf{A}_0=\frac{1}{2}\textbf{H}_0\times\mathbf{r}$$

Subbing that into the perturbing Hamiltonian gives

\begin{align}
\mathcal{H}_1&=-\frac{q}{2mc}(\mathbf{p}\cdot\frac{1}{2}\textbf{H}_0\times\mathbf{r}+\frac{1}{2}\textbf{H}_0\times\mathbf{r}\cdot\mathbf{p})\\
&=-\frac{q}{2mc}(\textbf{H}_0\cdot(\mathbf{r}\times\mathbf{p}))\\
&=-\frac{q}{2mc}(\textbf{H}_0\cdot\mathbf{L})
\end{align}

Where that $\mathbf{L}$ is in the absence of a field. Then we have that

$$\epsilon_{n}=\frac{\langle n|\mathcal{H}_1|0\rangle}{E_0-E_n}=\frac{qH_0}{2mc}\langle n|\mathbf{k}\cdot\mathbf{L}|0\rangle$$

where I assumed that $\textbf{H}_0$ is some magnitude $H_0$ along some unit vector direction $\mathbf{k}$. Now, going back to our chemical shift energy (equation \ref{chemshiftE}), we want the field that this current $\mathbf{j}_0$ generates. This is

$$\textbf{H}=\frac{1}{c}\int{\frac{\mathbf{r}\times\mathbf{j}_0(\mathbf{r})}{r^3}d\tau}$$

Substituting $\mathbf{j}_0$ and $\mathcal{H}_1$ in gives

$$\textbf{H}=H_0\frac{q^2\hbar}{2m^2c^2}\sum\frac{\langle 0|\sum_j\frac{\mathbf{L}_j}{r_j^3}|n\rangle\langle n|\mathbf{k}\cdot\mathbf{L}|0\rangle+\langle 0|\mathbf{k}\cdot\mathbf{L}|n\rangle\langle n|\sum_j\frac{\mathbf{L}_j}{r_j^3}|0\rangle}{E_n-E_0}-\frac{q^2}{2mc^2}H_0\langle 0|\sum_j\mathbf{r}_j\times(\mathbf{k}\times\mathbf{r}_j)|0\rangle$$

where $\mathbf{k}$ is the direction of the applied field and the $j$ subscripts denote electron number in the wavefunction.

For an $s$ orbital, all matrix elements with $L$ are 0 and so the first term is 0. This leaves only the second term which is opposed to the applied field. For anything with higher angular momentum than $s$ orbitals, the first term will be nonzero and aligned with the field. This term arises due to unquenching of the orbital angular momentum in the presence of a magnetic field and hence the direction of the paramagnetic field was determined by the projection of angular momentum which aligns with the field, rather than arising out of Lenz's Law. Said yet a different way, the orbital was already a magnet, it's just that we couldn't see it before. The applied field does not induce the quantum mechanical current flow or paramagnetic field (in the sense of classical EM), merely reveals it by mixing more of the aligned magnet into the ground state wavefunction than the anti-aligned one.

Putting some reasonable numbers into these equations gives that the paramagnetic term is larger at distances less than 1 Å from the nucleus, while the diamagnetic term dominates outside this range. As we mentioned above, the distance important for chemical shifts (average $r^3$ for a $2p$ orbital) is about 0.25 Å and so the paramagnetic term will be larger for NMR related problems.  On the other hand, these paramagnetic terms play practically no role in macroscopic magnetic moments of molecules.

References:

Slichter, Charles P. Principles of Magnetic Resonance (Harper and Row Publishers, 1963).
Gordon, C. P., Yamamoto, K. , Liao, W., Allouche, F., Andersen, R. A., Copéret, C., Raynaud, C., Eisenstein, O. Metathesis Activity Encoded in the Metallacyclobutane Carbon-13 NMR Chemical Shift Tensors. ACS Cent. Sci., 2017, 3 (7), 759–768. DOI: 10.1021/acscentsci.7b00174


How many grams of solid sodium bicarbonate will be consumed?

$$\ce{Cu(NO3)2_{(aq)} + 2NaHCO3_{(s)} -> CuCO3_{(s)} + 2NaNO3_{(aq)} + H2O_{(l)} + CO2_{(g)}}$$
The equation is telling us that for every $1$ mole of $\ce{Cu(NO3)2}$ consumed, there are $2$ moles of $\ce{NaHCO3}$ correspondingly consumed in the reaction.

So, let us first find how many moles of $\ce{Cu(NO3)2}$ is consumed. Here,
$$\text{Molarity} = \frac{\text{#Moles of } \ce{Cu(NO3)2}}{\text{Volume of Solution}}\\ \implies \text{#Moles of } \ce{Cu(NO3)2} = 0.240 \text{ mol.L}^{-1}\times 510  \text{ mL} \times \frac{1\text{ L}}{1000 \text{ mL}} = 0.1224 \text{ mol}$$

Now, think.

$$1 \text{ mole } \ce{Cu(NO3)2}  \iff 2 \text{ moles } \ce{NaHCO3}\\
0.1224 \text{ moles } \ce{Cu(NO3)2} \iff 0.2448 \text{ moles } \ce{NaHCO3}$$

$\text{Weight of } \ce{NaHCO3}= 0.2448 \text{ mol} \times 84.007 \text{ g mol}^{-1} = 20.5649136 \text{ g}$


Wouldn't the partial pressures of all the reactants change by the same factor, so Q wouldn't change?

If the number of gas particles is the same for reactants and products, then Q would not change. However, if you have a reaction like

$$\ce{2N2O(g) <=> N2O4(g)}$$

even though the reactant and product partial pressure changes by the same factor, the changes don't cancel out because the coefficients are different.

$$ Q = \frac{[\ce{N2O4}]}{[\ce{N2O}]^2}$$


Ascorbic acid and aldehydes: Reaction and Influencing Metals

You would need very strongly reducing compound to be able to reduce ascorbic acid. Aldehydes are not the case.
Ascorbic acid would not reduce aldehydes like ethanal either, as can be seen by comparison of their respective redox potentials below:

Redox system
$E^{\circ} [\pu{V}]$

Ascorbic/dehydroascorbic  acid
+0.35

Ethanal/ethanol
-0.197

Transition d-orbital  metals with multiple possible oxidation numbers very frequently act as redox catalysts. Both neighbour oxidation numbers may be common ( $\ce{Fe^{+III}/Fe^{+II}}$, $\ce{Mn^{+III}/Mn^{+II}}$,
$\ce{Mn^{+IV}/Mn^{+III}}$ ), or one may be uncommon ($\ce{Cu^{+III}/Cu^{+II }, Ag^{+II}/Ag^{+I}}$)
Catalytic effect is often related to the opposite charges of reagents and catalysts. The typical example is reaction of oxalate and permanganate anions, catalyzed by manganese cations.
A good metal catalyst has standard redox potential comparable with respective redox potentials of the oxidant a/o reductant.
E.g  $\ce{Ag+}$ may act as a catalyst during oxidation by peroxodisulphate, as $\ce{Ag^2+/Ag+}$ has the similar redox potential. Similarly, $\ce{Mn^2+}$ ions act as the catalyst during  oxidation of oxalic acid  by permanganate in acidic solutions, as $\ce{Mn^2+/Mn^3+}$ has similar redox potential as $\ce{MnO4-/Mn^2+}$.


What causes the yellow colour in the Briggs-Rauscher oscillating reaction?

The color is in fact caused by the elemental iodine. At the 'yellow' stage of  reaction the concentration of iodide is low, so the triodide anion (yellow at low concentration and red-brown at high) does not form, hence there is no formation of the blue triodide-amylose complex.

The OP wonders how the iodine concetration can be high enough to give the solution its yellow coloration, given the low solubility of elemental iodine in water?

First let me make a remark that crystalline iodine is very sluggish to dissolve in water to reach its maximal concentration. Even when making potassium iodide - iodine solutions - it takes a lot of stirring, anyone who worked in an analytical lab knows that.

But this is not the answer to the question. You have to consider that the iodine in solution does not come from a solid phase going to the liquid phase - it is generated in a homogenous chemical reaction. Therefore, it can reach concentrations far beyond the maximal solubility, forming a metastable supersaturated solution. This stage of the oscillation does not last long enough to allow nucleation and precipitation of the elemental iodine. The subsequent increase of iodide concentration causes formation of triodide, which along with starch gives the blue color and so on.


Calculating pH for titration of weak base with strong acid

First of all, when you titrate  a weak base (methylamine) with a strong acid, the equation of titration is:$$\ce{H+(aq) + CH3NH2 -> CH3NH3+ }$$The reaction of titration, as you can see is total and quantitative. The constant of the aforementioned equilibrium is:  $$ K=\frac{K_\mathrm b}{K_\mathrm w}= 5\times 10^{10} \gg 10^3$$
According to the stoichiometry of the neutralization equation, at the equivalence point: $n_{\mathrm{acid}}=n_{\mathrm{base}}$ $$ C_{\mathrm{acid}}.V_{\mathrm{acid}}=C_{\mathrm{base}}.V_{\mathrm{base}}$$

As the initial concentration of the base equals the initial concentration of the acid, the volume of the acid at the equivalence point equals the volume of the base:  $$ V_{\mathrm{acid}}^{equiv.}=V_{\mathrm{base}}$$
You can see that the total volume is doubled.

Now, let's calculate the $\mathrm{pH}$ at the equivalence point: We have a weak acid at the initial concentration $$C_{\ce{CH3NH3+}}=\frac{C_{\mathrm{base}}.V_{\mathrm{base}}}{V_{\mathrm{base}}+V_{\mathrm{acid}}}=\frac{C_{\mathrm{base}}.V_{\mathrm{base}}}{2V_{\mathrm{base}}}=\frac{C_{\mathrm{base}}}{2}= \frac{0.13}{2} = \pu{0.065 M}$$
It's in equilibrium with its conjugated base:
$$\ce{CH3NH3+<=>H+(aq) + CH3NH2}$$ The acid is partially dissociated:$$ [\ce{CH3NH3+}]= 0.065-x$$ $$ [\ce{CH3NH2}]=x$$ $$ [\ce{H+(aq)}]=x$$
Let's write the constant of this equilibrium:
$$K_\mathrm a= 0.2\times 10^{-10}= \frac{x^2}{0.065-x}$$
We solve this equation of second order to find $x$, the concentration of ion hydronium.
We find: $\mathrm{pH}= 5.94$


Air recycling system(CO₂-->O₂) or (CO₂-->fuel)?

This is quite a hot topic at the moment - the current buzz-phrase is power to gas.

There are quite a few conference papers and technical papers, and at least one dedicated conference, for it.

And it's too soon for there to be a known "best" process.

Audi and others are using a modified Sabatier process.

$$\ce{CO2 + 4 H2 -> CH4 + 2 H2O }$$

Others are looking at a modified Fischer-Tropsch process.

$$\ce{(2n +1) H2  +  n CO -> C_{n} H_{2n +2}  +  n H2O}$$

(image courtesy of the RWE power to Gas Project)

Essentially, you've got to prise the Oxygen off the Carbon, which is an energy-intensive process; and then attach the Hydrogen, which is not. But the Hydrogen typically comes from electrolysis of water, which is energy-intensive.

Not only are the processes being prototyped high-energy, but they're high-exergy too, requiring very high heat and / or electricity.

So it's unlikely that this is a better option than using solar power for solar-thermal or photovoltaics, unless the application is long-distance travel, which would require an energy source with a very high energy density. In which case, there are several renewable-fuel synthesisers being prototyped at the moment, but it's too early to tell which is the best.


What does "parallel in-register" mean for a protein structure in a fibril?

There is a clear explanation and figure showing this in the following reference:

Anton Gorkovskiy, Kent R. Thurber, Robert Tycko, Reed B. Wickner, PNAS 2014, 111 (43) E4615-E4622.

A parallel beta sheet is one where the direction from N- to C-termini on adjacent strands run in a parallel direction (rather than antiparallel). In-register means that each residue in one molecule is aligned with the same residue in the neighboring molecule, producing lines of identical residues along the long axis of the filament.


Why stepwise dissociation for acids but not hydroxide bases?

This is not a general rule. Much like with acids for some bases we write progressive capture of one proton at a time. It all depends on the nature of the acid/base.

In the case of $\ce{Ca(OH)2}$ you don’t have a molecule like $\ce{HO-Ca-OH}$ or something like you can assume for $\ce{HO-SO2-OH}$ (sulfuric acid). Rather, you have an ionic compound composed of two $\ce{OH-}$ ions per $\ce{Ca^2+}$ ion. Once we dissolve this, all the ions are released at the same time resulting in $2~\mathrm{mol}$ of $\ce{OH-}$ per mole of $\ce{Ca(OH)2}$ dissolved.

As a different example take sodium oxide $\ce{Na2O}$. If you slowly add water to that compound the oxide ion will grab one proton at a time resulting in $\ce{2 NaOH}$ first before a second set of protons will generate water. This is much the same mechanism as for diprotic acids. (Note, however, that the oxide ion is such a strong base that it is immediately protonated with a single equivalent of water — much like sulfuric acid is such a strong acid that it will immediately release one proton on contact with water.)


A square planar geometry is somewhat constrained, if you wish. There need to be compelling electronic reasons to move four coordinating anions (especially large ones like bromide) closer together than tetrahedral — which is the case in square planar, cf the smaller bond angles. Usually, this geometry is associated with $\mathrm{d^8}$ ions but it can happen for other compounds, too.

Another thing to consider is the field split. Its size is influenced by the metal’s charge. Higher charge typically means greater field split. Square planar complexes typically require a large field split to stabilise their low-spin state.

Manganese(II) is a $\mathrm{d^5}$ ion. Since it only has a low charge, it is very likely to be in the high-spin state. The high-spin state of a $\mathrm{d^5}$ ion does not stabilise any geometry whatsoever. Thus, no reasons exist to go square planar and the tetrabromidocomplex must be tetrahedral.


How does adding lemon juice to sugar make better caramel?

The citric acid in lemon juice hydrolyzes the sucrose molecules into glucose and fructose as you heat the sugar. The water in the lemon juice is also part of this process. This mixture of monosaccharides is much less crystalline in the same way that most mixtures are less crystalline than pure substances. Wikipedia has a good article on this process under the title "Inverted sugar syrup". The syrup is "inverted" in the sense that the sign of its optical rotation has changed because sucrose has a large positive optical rotation while fructose has a large negative optical rotation. Glucose has a positive optical rotation, but of a lower magnitude than sucrose.

The reaction is:
$$\ce{C12H22O11}(sucrose)+\ce{H2O -> C6H12O6}(glucose)+\ce{C6H12O6}(fructose)$$
The reaction can occur to a small degree by simply heating sugar with water in absence of acid catalysis. I assume the acid is necessary to get good conversion before you burn your sugar or vaporize all of the water.

Honey is a naturally occuring inverted sugar syrup of this type. Honey is resistant to crystallization because it is a mixture of gluclose and fructose. Honey is also slightly acidic.


Why can potassium hydroxide not be used to remove atmospheric carbon dioxide?

$\ce{KOH}$ is produced by electrolysis of aqueous $\ce{KCl}$ solutions. Taking into account that the electrical energy still mostly comes from power plants that burn fossile fuels, it doesn't look like the most clever idea to release $\ce{CO2}$, just in order to capture it from the atmosphere again.

While it doesn't make sense to tackle the global $\ce{CO2}$ problem with $\ce{KOH}$ or other hydroxides, they do nevertheless bind $\ce{CO2}$ just as you said. Consequently, these hydroxides were and are still used for the individual protection against $\ce{CO2}$ (self)poisoning in emergency cases and/or in confined spaces. Rebreathers used by fire brigades, divers or miners are typical examples. Initially, $\ce{KOH}$ was indeed used, but nowadays, these $\ce{CO2}$ filters mostly contain mixtures of sodium hydroxide and calcium hydroxide.


What you are measuring is not the potential $Fe^{2+}/Fe^{3+}$. It is the potentiel of the couple $[Fe(CN)_6]^{4-}/[Fe(CN)_6]^{3-}$ which is equal to +0.36 V in the table.


Explain why alumina-containing zeolites possess chemically active surfaces

Alumina containing zeolites have composition that can be represented as

where M is metal ion with m charge, and so on, respectively.
Meaning you have $[SiO_4]^{4-}$ and $[AlO_4]^{5-}$ tetrahedra which make up negatively charged framework. Positive ions compensate for this negative charge. There is a rule from 1954. called Löwenstein rule, which states that x/y has to be less than 1, which in turn implies there are no Al-O-Al bonds, as two tetrahedra have Al atoms in their tetrahedral positions. This gives really great internal specific area and resistance to high temperature, which makes this type of zeolite a great candidate for catalysis (heterogenous).
The following picture nicely represents the active sites in zeolites - they act as molecular sieves, as only the specific molecules fit in the pores:


Quadrupole MS: mass as a function of voltage

These parameters arise from non-dimensionalization of the Mathieu equations for ion motion in exogenous, periodically varying electric fields (i.e. oscillating potentials).  You can't calculate $m/z$ directly from $a$ and $q$.  For a fixed $U$ and $V$, you can regard $a$ and $q$ as non-dimensional parameters that correspond to $m/z$.  (Conversely, at fixed $m/z$ you can regard $a$ and $q$ as non-dimensional versions of the electrical potentials $U$ and $V$.)  The reason that those parameters don't tell you everything is that you still have to solve the Mathieu equations.

An analogy might be to the Reynolds number for characterizing fluid flows.  It, like $a$ and $q$, is a non-dimensional number that depends on properties of the molecules under study and on the system geometry.  Knowing just the value of the Reynolds number is useful for understanding some of the general characteristics of a flow, but if you want to compute fluid motion, i.e. the velocity profile in a pipe, you still have to solve the Navier-Stokes equation.

For quadrupole mass analyzers, only certain regions of $a$-$q$ space result in stable ion trajectories (i.e. where ions don't collide with the surfaces of the four rods in the quadrupole).  This map shows which values for $a$ and $q$ result in stable solutions to the Mathieu equations, and thus is called the Mathieu stability diagram.  The same document you linked to shows an example of such a diagram:

Thus, to compute $m/z$, we need to impose the criterion that $a$ and $q$ are in a stable region of the stability diagram, more specifically in the region "A" shown in the diagram.  By rough eyeballing, the "middle" of this diagram corresponds to about $q = 0.75$ and $a = 0.07$ or so.  You can use either of those values to estimate $m/z$, but a more precise estimate, including line-widths etc., would require knowing both.


Why does copper gain electrons in redox reactions, when it instead has a tendency to lose them?

As you already said it, it only gains electrons when it's oxidized, not like a Cu(-II) would form or so. Therefore it's just regular redox-chemistry. Copper is somewhat noble, or at least it has a mediocre redox-potential. When dissolved in acids it only reacts with oxidizing ones like nitric acid. On the other hand if you have Cu(II) and add an iron nail, the redox-potential of iron is much lower making it a good reducing agent. But that's basically the same for every element. Once you have an oxidizing agent it will be oxidized, if you add a reducing agent it will be reduced. Gold would do the same, sodium as well if you used lithium to reduce although gold doesn't readily give off it's electrons and sodium will do it quite easily. It really depends on the other redox-system you add.

You may also take a look at the redox-series:

Here the stronger reducing ones are on top (but depending on the artist it may also be the other way). Here it means the upper one will go from left to right if the lower one goes from right to left. To take an example:

Zn will form Zn(II) and Cu(II) will form Cu(0) in a system where both are mixed. But Cu will form Cu(II) and Pt(II) will form Pt(0) if those are mixed.


How does boiling water extract dyes from vegetables?

These dyes (NOT pigments$^*$) are highly water soluble. As you pointed out, boiling causes cell walls to rupture, releasing the cell plasma into the water. Boiling of the plasma fluid causes pressure build up, something the cell walls aren't designed to withstand.

You can also extract them with cold water by mashing up the raw material in a food blender or grinding it in a mortar and pestle. This has the same effect of rupturing the cell walls.

An interesting fact is that many of these dyes are pH sensitive: their colour is different in acid or alkaline conditions. Notably red cabbage extract changes from wine red to greenish blue when you add some sodium bicarbonate (baking powder) to it, because 'Bicar' is a (weak) alkali. Then adding a lot of vinegar (a weak acid) will change it back to its original colour.

$^*$ pigments are inorganic substances (many are refined minerals) and mostly insoluble in water.


How to write correct NMR integrals in a formula?

You assume the total number of protons summed over reactants and products is constant during the course of the reaction, which is perfectly reasonable, and that all monomer protons are incorporated into the polymer (perfectly reasonably), but also that the total signal is a constant. Note that because polymer signal will relax more rapidly by $T_2$ relaxation, usually, signal ratios are not expected to be strictly proportional to concentration, but we'll ignore that point, assuming this is a solution experiment and the polymer is highly dynamic.

Therefore, you could write

$ f_{conv}= (s_A-s_C)/(s_A+s_C) = (s_{A,norm}-1)/(s_{A,norm}+1)$

where $s_C$ is the isolated signal from one type of monomer proton and  $s_B=s_C$, the other monomer signal, overlaps with the polymer signal to generate $s_A$. Then you define units for the area $s_B=s_C=1$ according to the expression on the right. This is explicit in the way I have written the equation. You should explain the meaning of the symbols in the text, of course.


Does ionization energy have anything to do with how reactive an element is?

This comes about as a consequence of the effective nuclear charge on the atom or Z. eff, and the size of the atom. Z effective if you are not familiar is a concept usually touched upon in chem 2 as a periodic trend. It is a measure of the charge felt by the outer most electron. It is calculated by accounting for the electrons on the inner shells essentially canceling out or shielding the outer electron from feeling all of the charge from the positive nucleus. As you could expect this make the electron easier or harder to take away causing ionization depending on the Zeff. A small atomic radius causes the energy barrier to be higher to remove an electron, because the electron is closer to the nucleus. If the atom is larger it will be easier to remove the outer electron because it's further away from the nucleus. As you move along the periods of the periodic table from left to right and low to high, the effective nuclear charge increases. As you move along the periodic table from top down and right to left, the atomic radius increases.

We find that fluorine is at the top right of the periodic table, where the effective nuclear charge is high, and the radius of the atom is small. This means that it has a small condensed positive charge permeating through the inner shells. This causes the reactivity of fluorine to other atoms due to it's electron affinity from the unshielded nucleus. It is hard to ionize compared to sodium because of its smaller than sodium, and it's z eff is higher. These both cause the electron to be attracted to the nucleus more, causing a higher energy barrier to take it away.


Solution of a first- and second-order consecutive reaction scheme

Look for equation solvers in Numpy/ Scipy and load at the top of your script scipy.integrate import odeint and import numpy as np . There are examples of how to numerically solve differential equations in the examples and on line. The Master equation approach does not work for second order steps. There is also the Gillespie method which is a sort of Monte-Carlo method see this answer which has an example code Probability of the single molecule reaction to happen


Increasing the length of a copper wire via electrolysis

The two arrangements would produce identical results, because the copper wire conducts electricity so much better than the electrolyte solution.
However, the second arrangement is more practical because no joins are needed so it would be more reliable.
Generally, the surface area of the copper anode should be similar to the cathode which is to be coated to avoid concentrating the current unevenly, so increasing the anode area would probably help shorten the coating time somewhat, as it would avoid oxygen being evolved from the anode which can happen if the anode area is too small.


Does benzene structure stand for a single resonance form or the whole molecule?

The answer is you are referring to neither of them. That is because resonance structures don't actually exist in reality. We only use them to give us a rough idea what the actually molecule and bonds look like. A common way to explain resonance structures is this:

An explorer from a far distant land travels to a new continent and sees a strange animal that he has never seen before, a rhinoceros! When he comes back home, not really sure how to describe this strange animal, he tells everyone that it looked like an unicorn and a dragon mixed together.

So what the explorer did was he tried to explain some weird, not easily imagined object (a rhinoceros) by describing it as a mixture of two mythical, non-existent objects, but which are easily imagined by people.

That is exactly what the resonance structures of benzene (also for any other molecule) are. They don't actually exist, but rather benzene exists as a weird mixture of these resonance structures. The reason why we use resonance structures is that they give us an idea how to imagine what benzene looks like.


Yama Cold Brew Tower: what are these lab parts / components?

I think that you can make your own cheaper Yama cold brew tower however what they are selling to you is the design! What you see in the picture is not chemistry glassware but if you want to make your own with chemistry glassware (this will cost more money because good chemistry glassware is purer and must have many expansive features that normal glassware don't have) I've written the analogous components in the figure. You need an open-top cylindrical dropping funnel with a stopcock valve.

See here and here for reference. You can use a normal funnel and attach to it a stopcock valve if you want. Then there is something similar to a Buchner funnel, I imagine that there is a filter in the bottom, may be included into the funnel, in this case, a 90-150 micron filter should be sufficient. See here for more info. Then there is a glass coil spiral, up to my knowledge this has no practical use is only beautiful and geeky but will cost a lot. Then you have a normal carafe.
(probably a piece of filter paper is placed over the grounded cafe to allow the diffusion of water in the whole volume of the cylinder and not only in the centre).

With some filter paper and a couple of half cut plastic bottles you can do your Yama tower with less of 5 $: you can make a little hole in the screw cap of your first bottle to make drip the first funnel, and use a piece of filter paper in the second.  If you want to use glassware don't use chemistry glassware you don't need it really, try to find normal glassware to adapt to your project and you can make your own cheaper and nice Yama tower.


What solvent would have the fewest ions?

2-Methylpropane has a $\mathrm{p}K_\mathrm{a}$ of 53 ($\mathrm{p}K_\mathrm{a}$ table).  That sounds pretty non-ionic.  Can't think of any common all-carbon (no substituents of any kind) liquid compounds.


The "Ostwald Isolation Method": How does this work?

The powers $x$ and $y$ are intrinsic to the reaction (at least if the reaction pathway remains the same) and will not vary with reaction parameters, so that adding them then poses no difficulty.

More specifically, it is presumed that the reaction rate is given by an equation of the form $$r = c\ce{[A]}^x\ce{[B]}^y;$$ the Ostwald isolation method finds the powers $x$ and $y$ by flooding the system with $\ce{B}$ and $\ce{A}$ respectively in excess in two separate experiments, so that in the former case we have $$r = (c\ce{[B]}^y)\ce{[A]}^x \approx k\ce{[A]}^x$$ and in the latter $$r = (c\ce{[A]}^x)\ce{[B]}^y \approx j\ce{[B]}^y.$$

Adding the powers $x$ and $y$ gives the order of reaction by definition.

Incidentally, Wikipedia lists this method as the 'method of flooding', here.


Where should I publish my formula related to Aufbau Principle?

Have you searched the Journal of Chemical Education? There are so many articles on this topic you are interested in. If you don't have access to formal databases, start from Google Scholar and read abstracts.

Journal of Chemical Education is the right platform for such "teaching" tricks. However, you have to justify with some theory as to why your idea works. What is the logic behind it? As far as I can recall (>10 years ago), there was an article by a high-school student in that journal. So, students can also publish but you have to be quite good at that subject - similar to a professional.

However, sometimes I receive such articles for review but the tricks or analogies are so horrible that there is no option but to reject them. Similarly, make sure you know where your approach fails. Can it predict the electron configuration of all the elements? If you are a student share the idea with a chemistry teacher. See their reaction (no pun intended). Do they find it useful or have they seen it before? If yes, certainly consider writing a short note.


Why are porphyrins colored when porphyrinogens are colorless?

The precursor porphyrinogen does not have a conjugated pi system that spans the macrocyclic ring. On oxidation to porphyrin, however, the macrocyclic ring becomes fully "conjugated", i.e. connected by sp2 hybridized atoms. The HOMO/LUMO spacing in such systems is much narrower than in non-conjugated systems, which means the photon energy required to excite electrons into the LUMO is lowered from "deep" UV to the visible region, resulting in strong absorbance of light in the visual frequency range.


Is CH₃CH₂CH₂Cl saturated or unsaturated?

The workbook must be wrong, it is a saturated molecule.


I will try to make the answer as non-technical as possible, to make it understandable for you.

Some substances are composed of atoms, like $\ce{H}, \ce{O}, \ce{C}, \ce{Fe}$, etc. Each of these atoms actually exist out of even smaller particles, called protons, neutrons and electrons. Protons are positively charged ($+1$), neutrons are, as you might have guessed, neutral, and electrons are negatively charged ($-1$).

The protons and the neutrons are together in the core of the atom, the electrons are in a cloud around it. Because atoms are not charged, an atom has equally many protons as neutrons. A picture will help: (Source: User:Fastfission at Wikipedia)

However, in some cases, those atoms take an extra electron or give a electron to another atom. They are then called ions. For example, a normal natrium atom $\ce{Na}$ has 11 protons and 11 electrons, but a natrium ion $\ce{Na^+}$ has 11 protons and 10 electrons. Similiarly, a normal chloride atom $\ce{Cl}$ has 17 protons and 17 electrons, but a chloride ion $\ce{Cl^-}$ has 17 protons and 18 electrons.

Now we know enough to define what a salt is: A salt is a praticle that is is composed of one or more positively charged ions and one ore more negatively charged ions. For example $\ce{Na^+Cl^-}$, more commonly just written as $\ce{NaCl}$, is common salt, i.e. the kind of salt people do on some of their food (at least in some countries).

There are two special ions, $\ce{H+}$ and $\ce{OH-}$. If you have a liquid, then these two are inside. Now, we call something an acid if there is more $\ce{H+}$ inside than $\ce{OH-}$, and we call something a base if there $\ce{OH-}$ inside than $\ce{H+}$.

I'd also like to note that the book definition is dangerous: An Acid is a chemical compound that is sour in taste. Acids, or chemical compounds in general, are of course not things you should taste to test what they are! They might be poisonous.


What are the meanings of dotted and wavy lines in structural formulas?

I was wondering what the wavy and the dotted line represent?

A dashed line indicates that the bond is extending behind the plane of the drawing surface
A bold-wedged line indicates that the bond is protruding out from the plane of the drawing surface
A solid line indicates that the bond exists in the plane of the drawing surface.
A wavy line indicates that the stereochemistry of the bond is unknown.
A dotted line indicates that the bond is not a full bond, it is only a partial bond as in a hydrogen bond or a partially formed or broken bond in a transition state. Also, as Mrigank Pawagi notes in the comments, the dotted line can be used along with the solid line to denote bonds that have partial double bond character due to resonance.
An arrow indicates a dative bond in which both electrons in the bond originate from one atom (not common)


Does ammonium react with water in an ammonia solution?

Is it possible for the ammonium to simultaneously establish another reaction with water?

It might help to add these two reactions to see what the net effect is:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$
$$\ce{NH4+ + H2O <=> NH3 + H3O+}$$
$$+\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_ $$
$$\ce{2H2O <=> H3O+ + OH-}$$

So yes, this is the autodissociation of water, and this reaction goes on at the same time. The three equilibrium constants are linked (K1 x K2 = Kw), so there is never a situation where two of these reactions are at equilibrium and the third one isn't, causing an endless change in concentrations. In fact, for reactions in solution, there is always a state where all of the reactions are at equilibrium, and if all reactions proceed, they will all reach equilibrium eventually.

You can take any reaction that includes hydronium (or hydroxide) ions and formulate another reaction that includes hydroxide (or hydronium) ions instead by combining it with the autodissociation reaction.

  If so, we would enter a cycle where the additional concentration of ammonia resulting from the second reaction would, in turn, shift the first equilibrium to the right according to Le Châtelier's principle.

There is no cycle, unless you call reactants making products and products making reactants at the same time a cycle. The latter is a feature of all reactions that are at equilibrium. If you label a hydrogen, it will forever cycle between being part of ammonia, ammonium, hydroxide, water, and hydronium. There are some more complicated systems of reactions where the path toward equilibrium involves oscillations of concentrations, but in most cases, concentrations approach the equilibrium concentration in a monotonic way.


Lagrange Multipliers for the derivation of Maxwell-Boltzmann distribution

Ignoring multiplicative constants, the required integral is (in general form)

$$ \int \sqrt{x} e^{ax} \mathrm{d} x = \frac{1}{a}\sqrt{x}e^{ax}+\frac{i\sqrt{\pi}}{2a^{3/2}} \mathrm{erf}(i\sqrt{ax})$$

where $$ \mathrm{erf}(x) = \frac{2}{\sqrt{\pi}} \int_0^x e^{-t^2}\mathrm{d}t$$

Setting $x=E$, $a=\beta$, and the integration limits as $0$ and $\infty$, gives

$$ \int_0^\infty \sqrt{E} e^{\beta E} \mathrm{d} E = \frac{1}{\beta}\sqrt{E}e^{\beta E}\Bigg|_{0}^\infty + \frac{i\sqrt{\pi}}{2\beta^{3/2}} \mathrm{erf}(i\sqrt{\beta E})\Bigg|_{0}^\infty = \frac{1}{\beta}\sqrt{E}e^{\beta E}\Bigg|^\infty +\frac{i\sqrt{\pi}}{2\beta^{3/2}} \mathrm{erf}(i\sqrt{\beta E})\Bigg|^\infty $$

since $\mathrm{erf}(0)=0$.
To make further progress, consider that the function $\mathrm{erf}(x)$ must have a real input $x$ (otherwise it will "blow up" to $\infty$), which implies that $\beta$ must be negative, or, explicitly, that  $\beta = -|\beta| $ (we could also argue that the second term on the r.h.s must be real and come to the same conclusion). With negative $\beta$ we have that

$$\mathrm{erf}(i\sqrt{\beta E})=\mathrm{erf}(i\sqrt{-|\beta|E})=\mathrm{erf}(-\sqrt{|\beta|E})=-\mathrm{erf}(\sqrt{|\beta|E})$$

With this condition we also see that the first term on the r.h.s. is zero and as a result that

$$ \int_0^\infty \sqrt{E} e^{-|\beta| E} \mathrm{d} E =  \frac{\sqrt{\pi}}{2|\beta|^{3/2}} $$


How can a methyl halide react by intramolecular SN2?

I agree. There is no way that a one-carbon and one-heteroatom compound could react by intramolecular substitution. All intramolecular nucleophilic substitutions will require some ring-like transition state. The smallest ring you can build is a three-membered ring — but bromomethane does not have three atoms that could build up that ring (hydrogen does not participate).


What is the number of stereoisomers for hex-4-en-2-ol?

The approach is right, and your answers are right, but there are a couple of things to look at in your explanation.

First, the presence of a chiral centre does not in itself make a compound chiral.

It is in this case, but a molecule is chiral if, and only if, it is not superimposable on its mirror image.  You could have a molecule with more than one chiral centre, which was superimposable on its mirror image, and was therefore not chiral.

Second, a chiral molecule is not necessarily optically active.  Its specific rotation may be zero (or at least unmeasurable), in which case it is said to exhibit cryptochirality.

Thirdly, the R- and S- form of each of the geometric isomers are enantiomers (ie non-superimposable mirror images) of one another, not diastereomers.  Diastereomers occur when a molecule has (for instance) two chiral centres and rather than inverting both (giving the enantiomer, its mirror image) you only invert one, which is actually a different compound, with different physical and chemical properties.

Personally I think using the term stereoisomer to refer to the cis and trans versions is misleading, even if strictly geometric isomers are a subset of stereoisomers.  I'd prefer to say that the R- and S- forms of each geometric isomer are enantiomers of one another, and leave it at that, rather than lumping all four together as stereoisomers.  But you don't get to choose the wording of the question, I guess.


This is a bit late, but I'll add an answer anyway. XPS is already a method that can be used for the speciation, so you should be able to differentiate between the different the different oxides of the same metal or metalloid. This is because the work function to eject an electron from a shell of the metal centre is dependent on its chemical environment.


I don't think HF will react with ethanol under normal conditions. Nucleophilic substitution could be a possibility, but $\ce{F-}$ is worse nucleophile than $\ce{OH-}$, so this won't happen.


Biologically inert water mixture (or other fluid) for heat-exchange purposes

Apparently an acceptable choice of additive is silverised hydrogen peroxide. Only a few 100ppm should be necessary to keep the water sterile and the half life is about 5 weeks.

I'll go with this and update this answer if there are any issues.

PS

This article on overclocking (http://www.overclockers.com/pc-water-coolant-chemistry-part-ii/) has a number of suggestion on water additives to improve the thermal conductivity and  keep the system clean.


I got the answer on a Pymol listserve:

Yes, it can be done in Pymol.

Choose the residue you want to mutate from cytosine to 5-methylcytosine.
Use Action -  Hydrogens - Add function to add hydrogens.
Change mouse selection to choose atoms.

Click on the hydrogen you want to replace.

Click builder (there is more work to do if you're on a Mac, see below) to open builder window.

Click C

If you're on a Mac, you won't have Builder functionality unless you change your MacPyMOL.app file name to PyMOLX11Hybrid.app


What can I use to remove print from hard surfaces?

Tritan is a co-polyester. It is prone to stress cracking in the presence of a number of common solvents - cyclohexanone and tetrahydrofuran are used as surface bonding agents.

Acetone can induce stress cracking, as can IPA.

I would suggest overprinting if you want to disguise the logo.

As for the aluminium item - take your choice - cyclo, THF, acetone, xylene - lots of options.


What source of error in an acid-base titration could lead to a higher reading of the acid concentration?

A Excess water in the conical flask will have no influence whatsoever* (in first order approximation). Since you are determining the the amount of substance of methanoic acid with the titration, the volume is irrelevant. The volume you use to determine the concentration is from when you take the aliquot with the pipette.
This option is therefore incorrect.

B You are correct, the excess water left in the burette will dilute you potassium hydroxide solution, causing you to use more of it, and therefore also reading a higher amount of substance of methanoic acid than would be correct. Obviously this leads to a higher concentration.
This is a correct option.

C Rinsing the pipette with water will cause you analysis solution to be diluted. Therefore the calculated concentration would be lower than the actual concentration.
This option is therefore incorrect.

D Rinsing the conical flask with potassium hydroxide solution will lead to having some of the methanioc acid already neutralised. You will use less of your analysis solution, therefore calculating a lower concentration.
This option is also incorrect.

In order to achieve the highest accuracy, you should

rinse the burette with the potassium hydroxide solution you use to analyse.
rinse the conical flask with plenty of deionised water (to remove any residues).
rinse the pipette with the solution which is to be analysed.

* If it is too much water, then the volume significantly changes. This will cause the indicator substance to react ever slightly differently, which might lead to you not seeing the equivalence point early enough. This could lead to reading a higher concentration.
However, you eyes are a larger source of error if it is a few drops of water.


1) solubility. Typical liquid fuels consist entirely of hydrocarbons and are non-polar and thus cannot dissolve polar molecules like ascorbic acid.

2) The molecules mentioned in the provided quote can form stable radicals on loss of $\ce{OH}$ group hydrogen, i.e. act as a radical trap, transforming active radicals into non-active particles. It is a typical mechanism for antioxidants to work.

3) Strongly depends on the definition of effectiveness employed. Differen applications demand different properties.

4) Because in industry people typically do not bother with isolation of individual products if a mixture can work just as well. Sure, one can isolate individual products of phenol alkylation, but it is a tiresome work, so why bother, if the mixture works just fine?


Are the bonds within bismuth pentafluoride (BiF5) covalent or ionic?

For all nomenclature purposes, one should refer to the official IUPAC recommendations, which in this case are laid down in Nomenclature of Inorganic Chemistry — IUPAC Recommendations 2005, the ‘Red Book’. The relevant section for this question is section IR-5: Compositional Nomenclature, and Overview of Names of Ions and Radicals
The introduction of this section (IR-5.1) states:

The simplest type of compositional name is a stoichiometric name, which is just a reflection of the empirical formula (Section IR-4.2.1) or the molecular formula (Section IR-4.2.2) of the compound. In stoichiometric names, proportions of constituent elements may be indicated in several ways, using multiplicative prefixes, oxidation numbers or charge numbers.

The following details on Stoichiometric Names of Elements and Binary Compounds (section IR-5.2) detail how to derive a stoichiometric name:

The stoichiometric name of the compound is then formed by combining the name of the electropositive constituent, cited first, with that of the electronegative constituent, both suitably qualified by any necessary multiplicative prefixes (‘mono’, ‘di’, ‘tri’, ‘tetra’, ‘penta’, etc., given in Table IV). The multiplicative prefixes precede the names they multiply, and are joined directly to them without spaces or hyphens. The final vowels of multiplicative prefixes should not be elided (although ‘monoxide’, rather than ‘monooxide’, is an allowed exception because of general usage). The two parts of the name are separated by a space in English.

Thus, the stoichiometric name of $\ce{BiF5}$ is bismuth pentafluoride.[1]
In the later section Generalized Stoichiometric Names, guidance is given on how to name compounds based on one constituent’s charge or oxidation number (IR-5.4.2.2).

It is possible to provide information on the proportions of the constituents in names by using one of two other devices: the charge number, which designates ionic charge, and the oxidation number, which designates oxidation state. […]
[…]
The oxidation number (see Sections IR-4.6.1 and IR-9.1.2.8) of an element is indicated by a Roman numeral placed in parentheses immediately following the name (modified by the ending ‘ate’ if necessary) of the element to which it refers. The oxidation number may be positive, negative or zero (represented by the numeral 0). An oxidation number is always non-negative unless the minus sign is explicitly used (the positive sign is never used). Non-integral oxidation numbers are not used for nomenclature purposes.

Thus, according to this nomenclature, $\ce{BiF5}$ (determination of bismuth’s oxidation state is unambiguous) is bismuth(V) fluoride.

These two different styles of naming binary compounds are equivalent. In no way do they include any statement concerning the bonding situation. In fact, these names explicitly do not contain any information concerning structure or connectivity: dinitrogen tetraoxide is the correct name for both $\ce{O2N-O-N-O}$ and $\ce{O2N-NO2}$ — and the generalized name nitrogen(IV) oxide is also correct for $\ce{NO2}$.
Notably, usage of one or the other nomenclature cannot be used to distinguish between covalent or ionic bonding. As a case in point, $\ce{CCl4}$ — a rather clearly covalently bonded molecule — can be referred to as:

carbon tetrachloride
carbon(IV) chloride

And vice-versa, $\ce{BaCl2}$ — a very ionic compound — can be referred to as:

barium dichloride
barium(II) chloride
barium chloride[1]

Therefore, neither bismuth pentafluoride nor bismuth(V) fluoride make any reference to the type of bonds present in the compound; it is up to the chemist to use their previously acquired knowledge to realise whether bonds are predominantly covalent or predominantly ionic or something else.

Note:

Multiplicative prefixes need not be used in binary names if there is no ambiguity about the stoichiometry of the compound […]. The prefix ‘mono’ is, strictly speaking, superfluous and is only needed for emphasizing stoichiometry when discussing compositionally related substances […].

(Section IR-5.2)
Therefore, bismuth pentafluoride is permitted alongside monobismuth pentafluoride; and barium chloride is permitted, since only barium dichloride is reasonable.


Diluting a sodium chloride solution with specific glassware

Assuming you can use a beaker (or just one of those 50 ml volumetric flask for mixing), you could first dilute some of the 0.3 M solution to 0.1 M by adding 5 ml of stock to 10 ml of water. Now you have a 0.1 M solution of NaCl which you can calculate how to dilute using the M1*V1=M2*V2 equation. Doing this, I calculate that to make 50 ml of 0.01 M solution, you need 5 ml of the 0.1 M solution you made, which you can do with a pipette.


Can chemical energy from food be stored in a battery?

Three ways to turn chemically stored energy into electricity:

a fuel cell: While that works fine for hydrogen or methanol, I'm not sure if sugars are possible, and general "food" won't work. Of course you can chemically convert food into hydrogen or alcohol, or into sth. very specially designed to power a biofuel cell.
dry, burn, run a turbine or steam engine with attached generator, very obvious and simple.
direct conversion into mechanical energy via an artificial muscle, which powers a generator. Horribly complicated (bio-)chemistry, theoretically possible, because our body does it, too.


Is it possible to export a rotating .gif format picture from chemapps.stolaf.edu?

As of now, Java applets only work with Firefox ESR 32-bit. Assuming that OP is using Chrome and Windows 10, in my opinion the best solution would be to use a screen capture software, like ScreenToGif, which is FOSS, has nice interface and is also quite intuitive:

And here is the result:

I must remind though that the major limitation of GIF is that you are stuck with 256 colors. Structures displayed with GIF will never look as slick as those demonstrated in browser with JavaScript.