Understanding how mass spectroscopy works

Question 1: What’s the difference between an electric field and a magnetic field? And why does the magnetic field bend the ions while the electron field does not. Is this just a consequence of the shape of the apparatus? Could you use a magnetic field to accelerate a particle and a electric field to do the bending? I’m just really confused as to the difference between electric fields and magnetic fields.

Electric fields accelerate charged particles (linear acceleration). Magnetic fields deflect them (angular acceleration).

Charged particles respond to electric fields and magnetic fields differently. Electric fields accelerate charged particles linearly in the direction of the flux. For example, if an electric field exists between two plates, one with a buildup of positive charge and the other with a buildup of negative charge, then positively charged particles will accelerate from the positive part toward the negative plate. The acceleration of the particle ($\vec{a}$) depends on the strength of the field ($\vec{E}$), the mass of the particle ($m$), and the charge on the particle ($q$). Since the acceleration is a scalar multiple of the electric field, they have the same direction.

$$\vec{F}=q\vec{E}$$
$$\vec{F}=m\vec{a}$$
$$q\vec{E}=m\vec{a}$$
$$\vec{a}=\frac{q}{m}\vec{E}$$

Magnetic fields are orthogonal to electric fields. Magnetic fields alter the angular acceleration of charged particles (i.e., they deflect them). Thus, charged particles have a circular trajectory in a magnetic field. The radius of the circular path ($r$) is dependent on the strength of the magnetic field ($\vec{B}$), the charge of the particle ($q$), the mass of the particle ($m$), and the velocity of particle ($\vec{v}$). The radius of curvature is proportion to the mass to charge ration ($m/q$).

$$|\vec{F}|=m\frac{|\vec{v}|^2}{r}$$
$$|\vec{F}|=|q\vec{v}||\vec{B}|$$
$$m\frac{|\vec{v}|^2}{r}=|q\vec{v}||\vec{B}|$$
$$r=\frac{m}{q} \frac{|\vec{v}|}{|\vec{B}|}$$

The acceleration of the particle depends on the velocity, magnetic field, radius, and mass. Since acceleration in a magnetic field is a scalar multiple of the cross product of velocity and field strength, the direction of acceleration is always changing.
$$\vec{F}=q(\vec{v}\times\vec{B})$$
$$m\vec{a}=q(\vec{v}\times\vec{B})$$
$$\vec{a}=\frac{q}{m}(\vec{v}\times\vec{B})$$

  Question 2: I’ve come across a tutorial that said that what this detector is actually measuring is the mass to charge ratio. How is this calibrated and how does this step allow us to measure the mass of a single molecule. Won’t there be multiple particles hitting the detector at the same time, will this not affect the mass reading of a particular molecule? I'm just confused about this final step and how we can get an accurate mass reading for a particular molecule.

Now that we know the relationship between radius of curvature and mass-to-charge ratio, this question is easy. The detector counts the number of particles that hit it. That is all that it can do. That is all that it needs to do. The hard work of separating the ions by m/q is done by the mass analyzer, which is the magnetic field in curved tube. The magnetic field scans (quickly) over a range of $\vec{B}$ to alter the curvature of the ions by their m/q. Since the tube has a fixed curvature, only the ions with the correct m/q will pass to the detector. All others will hit the walls of the tube since their radius of deflection would be too large or too small. Thus, since the instrument controls the strength of the magnetic field, and physics determines what m/q will have the correct deflection at that field strength, there is no need to calibrate the detector.

More complex mass analyzers exist, for example the quadrupole mass analyzer. Instead of the typical bipolar magnetic field, it generates a quadrupolar magnetic field. Ions are thrown all over the place to hit a spherical detector. The physics is more complex, but the trajectory once again depends on m/q, so the detector only has to count hits in certain spots.


How to determine the equation for the reaction of europium with sulfuric acid based on experimental measurements?

Your equation isn't balanced; the oxidation state for europium is +3 and not +1; and europium sulfate does not exist in the solution you describe.
From this link:

Reaction of europium with acids
Europium metal dissolves readily in dilute sulphuric acid to form solutions containing the very pale pink aquated Eu(III) ion together with hydrogen gas, $\ce{H2}$. It is quite likely that $\ce{Eu^3+}$(aq) exists as largely the complex ion $\ce{[Eu(OH2)9]^3+}$

$$\ce{2Eu(s) + 3H2SO4(aq) -> 2Eu^{3+}(aq) + 3SO4^{2-}(aq) + 3H2(g)}$$


Why are branched alkanes used as fuels instead of straight chain alkanes?

The main reason for the high quantity of iso-octane is to prevent Engine Knocking. The Wikipedia page for Octane explains this quite well:

The octane rating was originally determined by mixing fuels from only heptane and 2,2,4-trimethylpentane (a highly branched octane), and assigning anti-knock ratings of zero for pure heptane and 100 for pure 2,2,4-trimethylpentane. The anti-knock rating of this mixture would be the same as the percentage of the latter in the mix. Different isomers of octane can contribute to a lower or higher octane rating. For example, n-octane (the straight chain of 8 carbon atoms with no branching) has a -20 (negative) Research Octane Rating, whereas pure 2,2,4-trimethylpentane has an RON rating of 100. Some fuels have an octane rating higher than 100, notably those containing methanol or ethanol.

A fuel made of only octane would have such a high enthalpy of combustion, that it would combust spontaneously in the cylinders, causing knocking, as seen by the RON number of -20 for it. Thus, iso-octane content is high because it reduces knocking and prevents engine wear.


The reactivity of NHS activated esters has mostly been desribed quite anecdotically. They are susceptible to hydrolysis in water, the kinetics of which is dependent on the nature of the NHS activated group (ester vs carnonate vs carbamate) as well as on the pH of the reaction buffer1, which is one way to show that primary amines are not the only nucleophile which can react with NHS-activated esters.

As for the chemical groups that have been observed to react with NHS activated esters, I did find a couple of examples upon looking up this issue. For instance, the Zenobi group have attempted a quite systematic study on the groups that can react with doubly NHS activated cross-linkers. Citing the abstract2 :

  As soon as additional cross-linkers were attached or loops were formed, other amino acids were also involved in the reaction. In addition to the primary amino groups, serine, threonine and tyrosine showed significant reactivity due to the effect of neighboring amino acids by intermediate or permanent Type-1 cross-link formation. The reactivity is highly dependent on the pH and on adjacent amino acids.

In an earlier work, the reactivity of a biotinylation NHS-ester based reagent was assessed on a single peptide,  [D-Lys(6)]gonadotropin releasing hormone. A similar reactivity of -OH groups was observed, with in addition that of the arginine guanidinium group3:

  In addition to the O-acylation of Ser(4) and Tyr(5) in this peptide, we have also identified a novel biotinylation of the Arg(8) side chain.

Other groups have also observed some reactivity with cysteins (sulfhydryl groups), but in a situation where the reagent is included within a ligand of the target protein. One can assume, as do the authors, that specific positionning of the ligand within the binding site can lead to binding with a cystein group4,5.

From these earlier work, one cas conclude, in answer to your question that although primary amines are the most reactive group for NHS-ester derivatization, other groups present in peptide side-chains (-OH for tyrosine, serine, threonine, guanidinium for arginine and sulfhydryls for cysteins).

The last issue raised in the question remains that of the secondary amine groups. It is widely accepted that pH influences strongly the kinetics of NHS derivatization, mostly through protonation of the amine groups which cannot act as nucleophiles in protonated state. This has been put to use in controlling the buffer pH in order to favor N-terminal reaction vs lysine side chain reaction. Although, in my experience, it is not at all a straightforward experiment and it is highly dependent on the nature of the derivatization group. One could hypothesize that since secondary amines are more basic then primary amines, the reactivity of secondary amines towards NHS should be reduced compared to primary amines, but I have no experimental or litterature proof for this part of the answer.


Will we ever run out of gold, silver, copper and other important conductors?

How long will these reserves last?

Short answer: longer than you think and will outlast me, you, our children and our grandchildren. Possibly last forever.

Long answer: This is a fairly complex issue. There was a book published about this topic several years ago claiming that we are running out of metals. Turns out that the book had a major flaw in its reasoning, so the same publisher followed up with a second book refuting the first one. Both books are open access and freely available so I recommend having a look.

The problem lies in the definition of the word reserve.

What are reserves?

A reserve is an ore deposit (that is, a concentration of a certain metal) that we know exactly how much is in there to a good degree, and we have detailed plans on how to extract it. This means, we have plans to build a mine, we know the cost of running the mine, etc. In some countries (for example, Canada and Australia) you are not allowed to call something a reserve unless you have the plans. This is a legal definition.

What are resources?

On the other hand, if you know how much metal is in the deposit, but you haven't made the plans to extract it yet, it is a resource. You are not legally allowed to call it a reserve.

The work required to develop a resource into a reserve costs money. You have to employ mine engineers, you need to get environmental approval, you have to do a lot of lengthy and costly things. Therefore, you don't "declare" a reserve until you have to. If you are a mining company and you have reserves of 40 years, why bother converting more of your resources into reserves? And this is exactly the problem: alarmists and the media consider the reserves, without the resources. Of course we have limited reserves! Because it's pointless to turn the resources to the reserves if we still have reserves!

Quoting Antonio Arribas, an exploration geologist, from an enlightening article published in the SEG Newsletter:

  One was published in issue 2605 of the international science magazine
  New Scientist in May 2007 (Fig. 5). In it, the author claims that,
  based on data by researchers from Yale University and the University
  of Augsburg, “if predicted new technologies appear and the population
  grows, some key resources will be exhausted” within what is
  graphically shown as an alarmingly short time frame (Cohen, 2007).
  Among the mineral resources covered, platinum, indium, and copper are
  particularly illustrative of the errors of interpretation contained in
  the article and the overall fallacy that this report represents.

  According to the article, humankind should be running out of platinum
  within four years (15 years from 2007). However, despite production of
  2,000 t of platinum between 2007 and 2017 and a metal price drop of
  26%, the global reserves of platinum group metals have decreased only
  by 2.8%3 (data from USGS, https://minerals.usgs.gov/
  minerals/pubs/mcs/). Likewise, according to this report we should have
  run out of indium sometime between 2012 and 2017. Never mind that, as
  a key component of indium tin oxide (ITO), this metal has been present
  as a transparent thin film conductor in the touch screen displays of
  every smartphone and tablet manufactured since then—including the 1.2+
  billion iPhones sold by Apple alone since the product’s release in
  June 2007, barely a month after publication of the New Scientist
  article.

  The second article and infographic selected were titled
  “Global resources stock check” and appeared in a June 18, 2012,
  article from BBC Future (www. bbc.com/future/story/
  20120618-global-resources-stock-check). In this report, the world
  supplies of indium, antimony, and copper, just to choose three, were
  projected to last through 2024, 2020, and 2044, respectively. I would
  argue that nothing but common sense is needed to believe that by the
  time of the Paris Olympiad in 2024, ITO-based transparent thin film
  conductors (or a new, and likely superior, substitute) will continue
  to be present in digital displays and thin-film photovoltaic cells,
  among many other applications. As for antimony, how do we reconcile
  the prediction that we should have only two years left of the metal
  now when, after production of
  1.1 Mt between 2011 and 2017, global reserves remain unchanged within the
  1.5- to 2.1-Mt range that has existed every year since 1997, when the U.S. Geological Survey started to record reserve estimates? In other
  words, for the time being and for the foreseeable future, the reserves
  of this metal are controlled by economics, technology, geopolitics,
  etc., and not by absolute limitations of availability in nature.


Structure of benzene dimer in an vibrationally excited state

The binding energy of the benzene dimer is really quite small. Ref. [1] reports the parallel-displaced and T-shaped dimers to have a binding energy at about 2.6 kcal/mol. This is about 900 cm$^{-1}$.

So, it is quite likely that if any vibrational mode with more energy than this is excited that upon redistribution of the energy, the dimer will fall apart.

On the other hand, excitation of vibrations less than the binding energy will likely lead to a very rapid interconversion between the T-shaped and parallel-displaced (or maybe sandwich) isomers. I have read that it is believed this happens basically no matter what.

Also, the sandwich isomer has a binding energy closer to 1.5 kcal/mol, so it is potentially possible to excite some low frequency mode which would cause the sandwich isomer to dissociate but not the T-shaped or parallel-displaced isomers.

Note that because this is such a weakly bound dimer, experimental measurement can only be made at very low temperatures. Ref. [2] describes one measurement of the conformers using supersonic jet expansion to form the dimers. This means temperatures close to milliKelvin.

References:

[1]:  Janowski, T., & Pulay, P. (2007). High accuracy benchmark calculations on the benzene dimer potential energy surface. Chemical Physics Letters, 447(1-3), 27-32.

[2]: Scherzer, W., Krätzschmar, O., Selzle, H. L., & Schlag, E. W. (1992). Structural isomers of the benzene dimer from mass selective hole-burning spectroscopy. Zeitschrift für Naturforschung A, 47(12), 1248-1252.


How do I find the final concentration after an additional dilution?

Check your math: It looks like you forgot to divide by $ 10\ \mathrm{ml} $ in finding $ c_2 $.

Also: the process you are describing in the first paragraph will not produced the same concentration that you started with. If you want $ 35\ \mathrm{ml} $ at the same concentration ($ 0.02\ \mathrm{g/ml} $) by taking some fraction of your original solution and diluting it, you will need to add more solute.

e.g.
$$ \rm(0.02\ g/ml) \cdot (35\ ml) = 0.07\ g $$
$$ \rm(0.07\ g + \mathit{m}) / 35\ ml = (0.02\ g/ml) $$
$$ m \rm = (0.02\ g/ml) \cdot (35\ ml) - 0.07\ g $$

The first expression is the concentration of your fraction (which doesn't change from the original solution, assuming a homogenous solution).

The second expression is equating the concentration that you want (conc. of the original solution, for example, with the concentration of your diluted fraction with $ m\ \mathrm{g} $ of solute added (using the mass we just found).

Then I rearranged this to express the mass of solute needed to obtain this concentration.


Stereochemistry of 1,4 disubstituted cyclohexane

Take the molecule on the left and ideally flip it by rotating 180 degrees along the vertical axis of the diagram. This inverts the cyclohexane without having to break or modify the molecule in any way


Your name, 1-chloro-2-methoxyethane, is correct.

Here is a link to the relevant page of the 1993 edition of the IUPAC Blue Book - Nomenclature of Organic Chemistry, which is published online by ACD Labs with permission of IUPAC.

Here are some example ethers and sulfides from that page, and we see a similar case to yours:

There is no difference in preference or priority between halo and alkoxy substituents (or any of the other prefix only substituents). Thus, in a case where there are two such groups and assignment of locants generates the same combination either way, you choose to assign the lower locant number to the alphabetically first.


Reason for negative Joule Thomson coefficient of Helium and Hydrogen at NTP conditions

Note: You can skip section I, and go straight to section II and/or the end of section III (specifically the conclusions subsection), if you are already familiar with the basic mathematical machinery/definitions

I. Preliminaries

Feel free to skip all of this if you are familiar with it

First, we have an adiabatic expansion, i.e $\mathrm{q} = 0 $.

Second, our process is carried out as follows: a gas at pressure $p_i$ flows into a region of lower pressure $p_f$ without significant change in kinetic energy. This is called the Joule–Thomson expansion.

Obviously, this process is inherently irreversible. Additionally, one can show that this expansion isenthalpic (i.e $H_i = H_f$). Unlike a free expansion, work is done, causing a change in internal energy; the exact specifics of this change will depend on the initial and final conditions and also on the nature of the working fluid.

What we measure experimentally is a change in temperature with respect to pressure at constant $H$, and we call it $\mu_{\mathrm{JT}}$ (Joule-Thomson Coefficient).

$$ \mu_{\mathrm{JT}} = \left(\frac{\partial T}{\partial P} \right)_H = \frac{V(T \alpha-1)}{C_p} \tag{1}$$

where $\alpha$ is the coefficient of thermal expansion $$\alpha = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_p$$

All real gases have an inversion point at which the value of  $\mu _{\mathrm  {JT}}$ changes sign. The Joule–Thomson inversion temperature, depends on the pressure of the gas before expansion.

$$\begin{array}{cc}
\hline
\text{Temperature} & \text{sign of}\ \partial P & \text{sign of}\ \mu _{\mathrm  {JT}} & \text{thus sign of}\ \partial T \\ \hline
\text{below  inversion temp.} & - & + & - \\
\text{above inversion } & - & - & + \\
 \hline
\end{array}$$

The final pressure is always lower than the initial pressure and thus the change is always negative. Depending on whether you are above or below the inversion temperature, your change in temperature can either be positive (warming) or negative (cooling). I would like to stress that the inversion temperature depends both on the temperature and pressure.

II. Qualitative Analysis

The arguments given below might seem a little "hand-wavy"/crude, but bear with me.

$$ H = U + pV$$
$$ \Delta H = \Delta U + \Delta (pV)$$

During a Joule-Thomson expansion (isenthalpic, adiabatic), the change in $pV$ represents the work done by the fluid.

If $PV$ increases,at constant enthalpy, then U must decrease as a result of the fluid doing work on its surroundings; the working material is using its own internal energy to do work. This produces a decrease in temperature and results in a positive Joule-Thomson coefficient.

Decrease in$pV$ means that work is done on the fluid and the internal energy increases. This can thus lead to an increase in temperature and the Joule-Thomson coefficient will be negative.

For an ideal gas, $pV$ does not change during a Joule-Thomson expansion.

As a result, there is no change in internal energy; also, there no change in temperature; and the joule-thomson coefficient is zero.

In real gases, $pV$ does change. The deviation from ideality (in general) is contained in $Z$, the compressibility factor. Thus, it would make sense if we could make some connection to it.

The ratio of the value of "real" $pV$ to that expected for an ideal gas at the same temperature is called the compressibility factor,

$$Z := \frac{pV}{nRT}$$

For a gas, this is typically less than unity at low temperature (attractive forces dominant) and greater than unity at high temperature.

At low pressure, the value of $Z$ always moves towards unity as a gas expands (tend towards ideality).

In general at low temperature, $Z$ and $PV$ will increase as the gas expands, resulting in a positive Joule-Thomson coefficient.

At high temperature, $Z$ and $pV$ decrease as the gas expands; if the decrease is large enough, the Joule-Thomson coefficient will be negative.

It follows, that for the $Z>1$, Joule-Thomson coefficient will be negative and the gas gets warmer. This is the case for the gases in question at the given conditions.

A graph from wikipedia showing how the Joule-Thomson Coefficient changes sign (for nitrogen):

Note: The question originally asked about gases, but I'll make a small point about condensed phases and supercritical fluids (since you can see them in the graph). At high pressures, these phases can't respond to an increase in pressure by reducing their volume, and $pV$ increases as pressure increases. The molecules are being forced together, and such the Joule-Thomson coefficient is negative.

Of course intermolecular forces, attractive and repulsive, and the limits in which they are dominant are key in explaining the changes of $Z$, and consequently of $\mu_{\mathrm{JT}}$. This point is further developed in the next section.

III. A slightly more quantitative, but ultimately Qualitative Analysis

To see how intermolecular interactions come into play, let's look at two model systems: an ideal gas (no intermolecular interactions) and a Van der Waals gas (both attractive and repulsive interactions).

Using equation (1) and the definition of $\alpha$ we can write

$$ \mu_{\mathrm{JT}} = \frac{V(T \alpha-1)}{C_p} = \frac{T \left(\left(\frac{\partial V}{\partial T}\right)_p -V\right)}{C_p} \tag{2}$$

For a perfect gast $\left(\frac{\partial V}{\partial T}\right)_p = \frac{V}{T}$ and so, $ \mu_{\mathrm{JT}} = 0 $

Since the Van der Waals gas equation is cubic in $V$, I will use the first few terms of the virial expansion and solve for $V$ to obtain $\left(\frac{\partial V}{\partial T}\right)_p$. I am omitting explicit computations, you may do them as an exercise.

$$Z = 1 + \frac{1}{RT} \left(b- \frac{a}{RT} \right)p + .... \tag{3}$$

Since $Z = \frac{p\tilde{V}}{RT}$ where $\tilde{V}$ is the molar volume of the gas; I am doing this for convenience's sake, I don't want to lug around a $n$ all over the place.

$$\tilde{V} = \frac{RT}{p} +  \left(b- \frac{a}{RT} \right) + .... \tag{4}$$

After differentiating we obtain

$$\left(\frac{\partial V}{\partial T}\right)_p = \frac{R}{p} +  \left(- \frac{a}{RT^2} \right) + .... \tag{5} $$

Substituting (4) and (5) into (2), and dropping all of the higher order terms.

$$ \mu_{\mathrm{JT}} \approx \frac{\frac{2a}{RT}-b}{C_p}$$

Now, consider the low temperature limit of this expression. The $\frac{2a}{RT}$ becomes large, and $b$ is negligibly small compared to it so

$$ \mu_{\mathrm{JT}} \approx \frac{2a}{RTC_p}$$

All the quantities on the right hand side of this equation are positive. so $\mu$ is positive. The attractive forces are dominant, since only the paramter $a$ appears in our expression.

Similarly, the high temperature limit is

$$ \mu_{\mathrm{JT}} \approx \frac{-b}{C_p}$$

Here, $\mu$ is clearly negative. The repulsive forces are dominant, since only the parameter $b$ appears in our expression.

Conclusions

The existence of an inversion temperature becomes obvious in a gas with both attractive and repulsive forces, since $ \mu_{\mathrm{JT}}$ changes sign from positive to negative.
The connection to attractive and repulsive forces also becomes more explicit.
In the low temperature limit, attractive forces are dominant and the gas needs to break free from these interactions as it expands. This requires energy, which has to be supplied by $U$ the internal energy of the gas. Thus the gas cools.
At high temperatures, the repulsions are dominant. As the gas expands, the over potential is lowered, and energy is released. Thus, the gas warms up.
Obviously, the exact magnitude of the high and low temperature limit depends on the nature of working fluid, since the parameters $a$ and $b$ are different for each gas. The gases you are talking about, have barely an attractive interactions.

Caveats

Our analysis based on the Van der Waals gas seems to suggest that the inversion temperature and $\mu_\mathrm{JT}$ is a function determined by temperature alone. This has experimentally proven to be false. This discrepancy arises because we dropped the higher order terms in our expansion which contained the pressure dependance.
This model is ultimately qualitative. It will give you the wrong numbers when you try to compute something, but it explains the general trends well.


From this paper, found the mechanism for neutralisation,

  Smoke particle charge interaction

  When smoke particles enter in the ionization chamber, they are
  subjected to the electric field generated by electrodes but also to the
  one generated by the charges. Close to the charges, the electric field is
  highly divergent resulting in dielectrophoretic force acting on smoke
  particles. The force tends to bind the smoke particles to the charges.
  When smoke particles are considered as isotropic neutral dielectric
  bodies linearly and homogeneously polarisable, dielectrophoretic
  force F can be expressed as

  $ F = \frac{1}{2} \alpha v \overrightarrow{grad} |E^{2} |$

  where $\alpha$ depends on smoke particle polarizability, $v$ is the smoke
  particle volume and $E$ is the electric field produced by the charges
  [11]. Coefficient $\alpha$ is typically of the order of $\epsilon_{0}/3$ [12] where $\epsilon_{0}$ is the
  dielectric constant of air. Volume $v$ is approximately $4 \mu m^{3} $ for a 1-$\mu m $
  diameter smoke particle.
  Amplitude of electric field $E$ produced by charges is given by
  coulomb’s law:

  $ E = \frac{1}{4\pi \epsilon_{0}} \frac{q}{r^{2}}$

  in which $r$ is the distance between the charge and the smoke particle.
  Due to the divergence of the field, the dielectrophoretic force
  decrease as $\frac{1}{r^{5}}$. If a smoke particle is close to a charge then they can bind. If the smoke particle is relatively far from the charge, the time
  of interaction may be relatively long and it is necessary that the
  charge drifts sufficiently slowly in the chamber to optimize the
  interaction. Indeed if the charge passes too rapidly there will be not
  enough time for the smoke particle to bind and thus nothing would be
  detected. In fact the capture radius reduces with the electric field and
  is of the order of 10 $\mu m $ for $E= 1  V/mm$.

  When the smoke particles attach to the free charges due to
  electrophoresis force, the charges become heavy bodies then the
  charge mobility $\mu$ decreases resulting in a drop of the electric current
  measured between the electrodes of the ionization chamber.
  Even if ionic detectors are very efficient, according to new
  standards they are no longer authorised in almost all of the developed
  countries because of the radioactive sources they include for
  generating the ions. Such radioactive sources are indeed difficult to
  collect and to recycle.


Here are four observations of the reactions of four metals

Before starting of with my reasoning , I must tell you that your answer ( whether guessed or logically reasoned ) looks correct according to me .

I am assuming that you are familiar with the Metal Reactivity Series ( if not look here )

We shall take use of a well presented reactivity series :

Now to crack the answer clue by clue :

1.

  Metal F and G react slowly with water.

Hence F and G are somewhere at the top.

2.

  Metal H displaces metal G from its chloride salt solution

Clearly H>G.

3.

  Metal E does not react with dilute acids

E will come somewhere at the bottom half so F,G>E (from 1) and we can further say H>G>E (from 2)

4.

  Metal H does NOT displace metal F from its sulphate salt solution

This gives F>H.

Combine last two clues' result to get F>H>G>E .


Grignard formation and vicinal dihalides

Taken from the Virtual Textbook of Organic Chemistry, which is maintained by William Reusch.

  If two halogen atoms are present in a given compound, reactions with
  reducing metals may take different paths depending on how close the
  carbon-halogen bonds are to each other. If they are separated by four
  or more carbons, as in the first example below, a bis-organometallic
  compound may be formed. However, if the halogens are bonded to
  adjacent (vicinal) carbons, an elimination takes place with formation
  of a double bond.

He does however not provide any explanation for this behavior. It could be that this result is largely empirically based, about some aspects of the Gringard reaction mechanism there is not much known about (or consensus among chemists). I looked at Clayden et al. and Vollhardt & Shore and neither give a satisfying answer. If I find any in an other perhaps more advanced textbook, I'll edit this answer.


How does voltage and resistance affect an electrolysis?

Voltage determines the kind of chemistry at each electrode.  Gold is purified by plating the anode to the cathode at minimum voltage for the half-reaction, so impurities either do not dissolve and do not plate out.  Copper electrorefining electrode mud contains valuable trace elements in concentrated from.

Current determines the amount of chemistry, ~96,500 coulombs/mole of electrons.

Cell internal resistance converts input energy, power multipleid by time, $\ce{I^2Rt}$, into useless heat.  You then want small electrode spacing and good electrolyte stirring.  You then need enough voltage to do the chemistry, plus more to compensate for resistance losses, but not so much voltage that undesirable chemistries are enabled.  Does cell resistance increase (ohmic) or decrease with temperature?


Can carbon dating tell how long a species would survive?

No, it can't be used for that. Read the mechanism of radiocarbon dating.

  Radiocarbon, often abbreviated as $\ce{^{14} C}$, is constantly being created in the atmosphere by the interaction of cosmic rays with atmospheric nitrogen. The resulting radiocarbon combines with atmospheric oxygen to form radioactive carbon dioxide. This is then incorporated into plants by photosynthesis, and animals acquire $\ce{^{14}C}$ by eating the plants. When the animal or plant dies, it stops exchanging carbon with its environment, and from that point the amount of $\ce{^{14} C}$ it contains begins to reduce as the $\ce{^{14}C}$ undergoes radioactive decay. Measuring the amount of $\ce{^{14}C}$ in a sample from a dead plant or animal such as piece of old wood or a fragment of bone provides information that can be used to calculate when the animal or plant died. The oldest dates that can be reliably measured by radiocarbon dating are around 50,000 years ago, though special preparation methods occasionally permit dating of older samples.


Yes, of course they affect the shape. But only slightly so. More precisely, bond angles can change, however this does not change the general shape predicted by VSEPR.

For example, if two atoms (one with a positive pole, and one with a negative pole)  are connected by another atom, then the two atoms would make an angle that is smaller than the angle predicted by VSEPR.

And also, polar bonds are capable of intramolecular bonding. Such as in protein-folding.  So they definitely contribute to molecular shape.


Inferring bond breakage in a reaction from InChI strings of reactans?

There is the Reaction Decoder Tool (RDT on github) that can - from the README - do:

  Reaction Annotator (Extract Bond Changes, Identify & Mark Reaction Centres)

EDIT: It potentially could have InChI support, but a quick glance at this file suggests that it currently doesn't...

In fact, that would require the CDK/RDT to implement the rInChI format

The input formats includes reaction smiles which should be possible to generate from your smiles strings. Looks like you just jam a ">" inbetween reactants and products with a "." separating the molecules in each set.

which sounds like what you want, I think. Also it should be able to handle those volumes as it has been developed to work with reaction databases at the EBI.

(Full disclosure : I worked on the project for a little while :) but it does have InChI support  - via the CDK).


What is the oxidation state of carbon atoms in Ethanoic acid?

Typically when figuring oxidation state of an atom, electronegativity of different atoms of same element is considered equal. This means that in thiosulphate oxidation state of central sulfur would be somewhere around 4.

However!
Let us condider a very tricky case: ozone. Technically it consists of atoms of same element and the oxidation state of oxygen atoms should be zero. But ouch, auch, the central atom bears full positive chanrge in all feasible structures.... So, should ozone, disregarding electron shifts in covalent bonds, have oxidation state of atoms +1 (for central atom) and -1/2  (for two other atoms)? But this doesn't make much sense. This is a place where oxidation state abstraction breaks.

So, there are situations when oxidation state abstraction doesn't make much sence.

Thiosulphate is somewhate similar. The central sulfur behaves very muc like sulfur in sulfate, so by analogy it should be in +6 oxidation state. However, formally it is not the case. Here a clear conflict between formal abstraction and real life occurs, so people cope with it variously. One possibility is to arbitrarily say that the central sulfur of thiosulfate is in +6 oxidation state (and it makes sence). Another is to go formally (and it also makes sense). I saw both options in the wild.

TL;DR : formally the central sulfur in thiosulphate is +4; but people sometemes disregard it as the ion is very similar to sulphate with central sulfphur atom being +6.


"No variables in Z-matrix" error in scan calculation of Gaussian

Ok, so I guess I'm on the right way now. Trying to fix such a complex system is always very difficult so what I did was using a similar system, I got one sodium atom instead of your NT and I'm moving the benzene away.

First of all: you got two different dummy atoms in there, atom 1 and atom 9, which I guess makes some problems. Also the extra lines between the sections are a problem.

So here's the input for this, for a first try just from 2.5 in 5 0.5 A steps which at least runs without any errors:

# opt=z-matrix hf/3-21g

Title Card Required

0 2
 Na                 0.00000000    0.00000000    0.00000000
X 1 R1
C 2 XC 1 A1
C 2 XC 1 A1 3 60.0
C 2 XC 1 A1 4 60.0
C 2 XC 1 A1 5 60.0
C 2 XC 1 A1 6 60.0
C 2 XC 1 A1 7 60.0
H 3 HC 8 A2 2 180.0
H 4 HC 3 A2 2 180.0
H 5 HC 4 A2 2 180.0
H 6 HC 5 A2 2 180.0
H 7 HC 6 A2 2 180.0
H 8 HC 7 A2 2 180.0

XC=1.39893
HC=1.09100
A1=90.0
A2=120.0
R1 2.5 S 5 0.5

So here's the point. The format basically needs to be exact like this. Everything else will confuse Gaussian. You can substitute the "Na" line with your cartesian coordinates, but then the numbering in the z-matrix section changes! you have to adopt those numbers. Oh and of course if you got a more complex system than one atom you want to not only define the distance of that dummy atom but also the angles, so it's in the right position. You might also need to adept the whole matrix so the benzene has the right orientation in relation to the NT, so at least three atoms in there need to be somehow defined by atoms in the NT.

If you want to have fixed coordinates (what you labeled as constants) in the z-matrix, you can add them after a blank line just below the last section.

  However, after a test run I'm not sure if it actually scanned that one
  coordinate, I just get one structure right now, I have to check how to
  do the scan correctly.

It works after adding that one 'S' in the coordinate which needs to be scanned.


Do we neglect the temperature changes during a reaction while calculating delta(H)?

Yes. By saying that $\Delta (pV) = \Delta (nRT) = RT\Delta n$ you have made the assumption that $T$ is constant. When you take anything out of a $\Delta$ sign, you are saying that the initial and final values are the same. Otherwise you would have to write $\Delta (nRT) = R \Delta (nT) = R(n_2T_2 - n_1T_1)$ - this only simplifies to $RT(n_2 - n_1)$ if you assume $T_1 = T_2 = T$.

By the way, I want to be picky about the formulae that you are using. Look at $$\mathrm{d}H = \mathrm{d}U + (c + d - a - b)RT$$

The problem with this is that $\mathrm{d}H$ and $\mathrm{d}U$ are infinitesimals, but $(c + d - a - b)RT$ is a finite quantity. Do not ever add or subtract these two in the same equation. It is like trying to add a mass to a length - what is $30 \mathrm{~kg} + 25 \mathrm{~cm}$? It doesn't make sense.

Same thing applies for $\mathrm{d}H = \mathrm{d}U + (pV)$.


In acidic media, the ester carbonyl will be protonated and the carbonyl carbon attacked by a water molecule. The intermediate will have a proton abstracted by a water molecule and the ethoxy group will be protonated (it can happen either by intermolecular reaction with an H3O+ or by an intramolecular reaction with a proton migration from the protonated intermediate after the water attack). The carbonyl is restored and an ethanol molecule leaves the structure. A subsequent proton abstraction by water forms the carboxylic acid, which will then suffer decarboxylation following the mechanism you presented above. (I'm sorry about my ChemDraw, I suck at drawing even on a computer).


Acid-base titration: Calculate pKa with only three values given

First, we note that this is not a strong acid, since when it is only partially neutralized, the $\mathrm{pH}$ has already risen to almost $5$.

At that point, we know that $\frac{10}{22.2}$ of the total acid has been neutralized. Therefore, the ratio

$$\frac{\ce{[A-]}}{\ce{[HA]}} = \frac{10}{22.2} \cdot \left(1 - \frac{10}{22.2}\right)^{-1} = \frac{1}{1.22}$$

Substitution into Henderson-Hasselbalch gives that the $\mathrm{p}K_\mathrm{a} = 4.73$.

Note that I assumed that the solution wasn't super dilute. If that were the case, I couldn't plug directly into the Henderson-Hasselbalch equation because the acid-base equilibrium would need to shift, but the change in the concentration of hydronium would be dominated by auto-ionization of water.

And in general, the total volume of solution does matter. You could consider a solution of hydrochloric acid that were dilute enough that even the first assumption of a weak acid were wrong. It's pretty easy to construct the solution such that the partially neutralized solution had a $\mathrm{pH}$ of $4.65$. I'll leave that as a follow-up exercise for you to explore.


How is the Maxwell-Boltzmann curve obtained?

There is, indeed, a function that plots the graph.

There are two common types of graph, one with kinetic energy as the $x$-axis, and one with speed. For the most part, they are qualitatively the same, although there is a slight difference in curvature near the origin. The speed function is much more commonly used, because one can more easily extract useful information from it, such as mean velocities, or mean free paths. It is:

$$f(v) = (4\pi v^2) \cdot \left(\frac{m}{2\pi kT}\right)^{3/2} \cdot \exp{\left(-\frac{mv^2}{2kT}\right)}$$

where $m$ refers to the mass of the particle, $k$ Boltzmann's constant, and $T$ the temperature.

As for the derivation, it is rather long and perhaps you were not really asking for it. If you are interested, I would recommend looking it up in a physical chemistry textbook, for example: Atkins' Physical Chemistry, 10th ed., section 1B(b), or Levine's Physical Chemistry, 6th ed., section 14.4. I am partial towards Levine because he gives a much more detailed explanation than Atkins, who is always very brief.


Most stable conformation of aldehydes and ketones

I found in Ian Fleming's Molecular orbitals and Organic Chemical Reactions, page 93-94 the answer (the aldehyde is propanal):

  The polarisation of the H-C bonds in the methyl group leaves a weak
  positive charge on the outside and this is attracted to the partial
  negative charge on the oxygen atom, lowering the energy of this
  conformation

(the difference in energy of the two eclipsed conformations of propanal is 8 kJ/mol).


Mechanism of a cyclization reaction - why does an enol ether form?

@Carbenoid: A year has passed since you requested a mechanism from @Mithoron. The statutes of limitation have run out. Your comment about why not do an aldol condensation (in base) is well taken. But the question was what is the structure of compound A.  I have no problem with the dienol ether 5 (aka A) in spite of many people's aversion to forming 8-membered rings.
Full stereochemistry of structure 1 could have been informative but not essential. I will forgo the mechanism of the acid-catalyzed aldol condensation and defer to @ringo's contribution. The bold bonds in structure 2 form a rigid framework so that the degrees of freedom are minimized in the formation of the hemiketal 3. Effectively, the cyclization is that of a much smaller ring. This ring closure is mechanistically more sound than the SN2 displacement with an enol particularly one with an (E)-geometry when at least the (Z)-geometry would be required. The red arrows terminate with dienolether 5. Aqueous acid and microscopic reversibility via the blue arrows return dienolether 5 to structure 2. In fact, structures 2 and 3 may well be in equilibrium with each other.


How do the toxicities of arsenate and arsenite compare?

I see now.

According to the same website with the reaction mechanism,

  Although the arsenite formed is more toxic than arsenate , it can be extruded from some bacteria by EC 3.6.3.16 , arsenite-transporting ATPase; in other organisms, arsenite can be methylated by EC 2.1.1.137 , arsenite methyltransferase, in a pathway that produces non-toxic organoarsenical compounds. cf EC 1.20.4.4, arsenate reductase (thioredoxin) .


In pig iron, from where does carbon in it comes from?

A blast furnace is made of successive layers of charcoal and iron oxide mineral. In the beginning, the lower layer of charcoal is lit and the fire is maintained by air being pressed through the burning matter. The combustion produces first $CO_2$. But as the temperature is high and the charcoal is in excess, $CO_2$ soon reacts with $C$  producing $2$ $CO$  according to the equations :$$\ce{C + O_2 -> CO_2}$$ $$\ce{CO_2 + C -> 2 CO}$$ Then the CO gas touches the bits or pieces of mineral, which produces the following equation :$$\ce{3 CO + Fe_2O_3 -> 2 Fe + 3 CO_2}$$ This produces some metallic iron which is liquid if the temperature is high enough. This liquid flows between the bits of burning charcoal, and dissolves some charcoal during this flow. The obtained pig iron contains about 5% carbon. which must be later on burned in special furnaces, to get steel (containing about 1% dissolved carbon)


Are the bonds in the monomers at the end of a polymer filled or unfilled?

You are asking about polymers made by radical polymerization.

  But in the monomers at the very ends of a polymer chain, there is an unfilled bond, a carbon atom with only three bonds filled.

During polymerization this is true, the radical will continue to react with monomer and propagate the polymerization, but the radical does not sit in the bulk. There are a number of ways which an additional atom amy bond with the carbon radical in a process called termination. Typically this is done by combination or inhibition.

Combination is the process where two radicals come together to form a bond leaving no free radicals to continue the reaction. This may occur when two radical chains combine or when a selected radical is added to cease the reaction. This radical could be the other half of the initiator which began the radical polymerization.

Combination of two polymerizing chains.

Combination of polymerizing chain with radical

For inhibition, a radical combines with a molecule that can form a stable radical such as oxygen or butylated hydroxytoluene (BHT).

 Radical chain combining with oxygen to form stable, non-ractive radical.

In these cases the radical will eventually find a stray radical or atom to bond with (like water), the unstable carbon radical will not exist permanently.

  I say this because I don't know if there are any polymers made where the bond inside the original monomer is a triple bond. In all the commonly used plastics (those that have recycling symbols) the original monomers have double bonds between the carbon atoms.

Polyacetylene uses a triply bonded monomer in its production.


What is the physical interpretation of "internal pressure"?

See orthocresol's derivation of an internal pressure $\pi_T$ relation $(1)$ here.

$$\pi_T = T\left(\frac{\partial P}{\partial T}\right)_V - P\tag{1}$$

Internal pressure $\pi_T$ measures what you would expect from its definition:

$$\pi_T \equiv \left(\frac{\partial U}{\partial V}\right)_T.$$

It shows how internal energy changes when volume is changed and temperature is constant. For ideal gases the change is zero, something that also follows from the equipartition theorem.

$$\mathrm{d}U_{\text{ideal gas}}= \frac{\nu}{2}nR\mathrm{d}T \implies \left(\frac{\partial U}{\partial V}\right)_T = 0 \ \  \text{for an ideal gas}$$

In other words, the total energy an ideal gas has is not dependent on volume. There is no repulsion nor attraction amongst ideal gas particles.

As a first approximation to model real gases, the van der Waals equation $(2)$ is used.

$$\left(P+a\frac{n^2}{V^2}\right)\left(V-nb\right)=nRT\tag{2}$$

Hence,

$$P = \frac{nRT}{V-nb} - a\frac{n^2}{V^2}\tag{3}.$$

Using relations $(1)$ and $(3)$

$$\left(\frac{\partial P}{\partial T}\right)_V = \frac{nR}{V-nb} \overset{(3)}{=} \frac{1}{T}\left(P + a\frac{n^2}{V^2}\right).$$

Again via formula $(1)$

$$\pi_T = T \cdot \overbrace{\frac{1}{T}\left(P + a\frac{n^2}{V^2}\right)}^{(\partial P/\partial T)_V} - P = a\frac{n^2}{V^2}\tag{4}.$$

Result $(4)$ $\pi_T = a\frac{n^2}{V^2}$ demonstrates that for real gases internal energy will change when volume is decreased or increased. Real particles have interparticular forces that vary with distance. A hint as to why $\pi_T$ is called internal pressure will be given when $(4)$ is substituted into $(2)$:

$$\left(P+\pi_T\right)\left(V-nb\right)=nRT,$$

or even better, $(3)$:

$$P = \frac{nRT}{V-nb} - \pi_T.$$

If $\pi_T > 0,$ then $P$ will be smaller. Therefore, attractive forces are dominant.
If $\pi_T < 0,$ then $P$ will be bigger. Therefore, repulsive forces are dominant.


How to calculate the rate constant at different temperature for the decomposition of dinitrogen pentoxide?

When using Arrhenius equation: you have to multiply the activation energy by $1000$, because it must be in $\pu{J}$ and not $\pu{kJ}$.
you must also divide by $RT$ and not by $T$ as you did.


Difference between chemical potential and gibbs free energy

I know that the chemical potential is the molar Gibbs free energy and that it is an intensive variable and that the gibbs free energy is extensive.

The chemical potential is related to the Gibbs energy in one of its definitions:

$$\mu_i = \left(\frac{dG}{dn_i}\right)_{T,P,N_{j\neq i}}$$

So it is the change in Gibbs energy when one component changes amount while the other components are of constant amount. The dimensions are energy per amount of substance (with units of e.g. J/mol).

The term "Gibbs energy" is used for a lot of things, and the dimensions could be energy, or energy per amount of substance. The Gibbs energy of reaction $\Delta_r G$, arguably the most important Gibbs energy for a chemist, is defined as:

$$\Delta_r G = \left(\frac{dG}{d\xi}\right)_{T,P}$$

where $\xi$ is the extent of reaction with dimensions amount of substance (i.e. amounts of all reactants and products change as $\xi$ changes, according to the stoichiometry specified in the reaction equation). As a consequence, the dimensions of chemical potential and Gibbs free energy of reaction are the same.

For other uses, the term "Gibbs energy" might refer to the Gibbs energy of a given system, and would have dimensions of energy. $G$, like the enthalpy $H$, does not have a defined reference point that is zero, so there is never a specific value associated with $G$. You can, however, use $\Delta G$ to describe the change in Gibbs energy from a defined state to another defined state. This $\Delta G$ would have dimensions of energy, and would be extensive (if you duplicate the system and look at the total, it would have a $\Delta G$ that is twice as large).

It would be easier if the symbols would indicate which quantities are per amount of substance (intensive) and which are not (extensive). Some textbooks use bold typeface for all molar quantities, but introductory textbooks are usually not rigid in terms of dimensions at all.

  I don't understand why G is extensive if it has $\pu{mol−1}$ in its units.

The Gibbs energy of reaction is not extensive. However, if you multiply all coefficients by a constant factor, you also have to multiply the Gibbs energy of reaction by the same factor. For example,

$$\ce{H2 + 1/2 O2 -> H2O}\ \ \ \ \Delta_r G^\circ = \pu{−237.13 kJ mol−1}$$

$$\ce{2H2 + O2 -> 2H2O}\ \ \ \ \Delta_r G^\circ = \pu{−474.26 kJ mol−1}$$

The first is equal to the Gibbs energy of formation of water, and the second avoids fractional coefficients. The two are different because the extent of reaction is linked differently to changes in the amount of the reactants and products.


HCH bond angle in cyclopropane and cyclobutane

Computational investigations of geometry, aromaticity and bond energies at HF/6-31G** theory level [1] revealed the following deformed  geometrical ($\alpha$) and interpath angles ($\beta$ -- the angle between two bond paths) for $\ce{H-C-H}$ bond, which are slightly deviating from the strain-free $109.5^\circ$ due to varying Baeyer strains:

\begin{array}{lr}
\hline
\text{molecule} & \alpha,~^\circ & \beta,~^\circ \\
\hline
\text{cyclopropane} & 114.1 & 117.0 \\
\text{planar cyclobutane} & 107.9 & 109.7 \\
\text{puckered cyclobutane} & 108.3 & 110.0 \\
\hline
\end{array}

P. S. Unfortunately, this data cannot be taken from single-crystal x-ray experiment as this method doesn't allow precise determination of H-positions, and I didn't find a neutron diffraction data on solid cycloalkanes mentioned.

Reference

Cremer, D.; Gauss, J. J. Am. Chem. Soc. 1986, 108 (24), 7467–7477 DOI: 10.1021/ja00284a004.


Once you understand that the acetate ion is more stable the the ethoxide ion, we understand the fact that after releasing The acidic hydrogen the final product formed is more stable for acetic acid as compared to ethanol.
This would lead to the conclusion that acetic acid is a better acid as compared to ethanol.

While comparing the acidity of two species, only the comparative study of conjugate base is enough, since its simply the comparison between compounds.

Comparing the stability with the original compounds will lead you to the conclusion that whether the compound itself has a tendency to act as an acid or base.
Alternately one could compare the stability of the original compound and its conjugate(Which is the acid dissociation constant) and compare its value with the acid dissociation constant of other compunds to compare its acidity.

In any case comparing the stability of the conjugate ions will tell you which one of them is comparatively more acidic/basic than others. It will not let you conclude whether its the nature of the compound to naturally act as an acid/base or not. For that one should compare the stability with original compund or just check its dissociation constant.


I am afraid whoever told you "sulfuric acid carbon" does not know the basic terminology (must be from non-chemistry background). There is no such thing as "sulfuric acid carbon" to begin with. Any carbon obtained by charring with sulfuric acid would be jet black. Nobody would like to have a black chalk.

It is very easy to parse $\ce{CaSO4.2H2O}$ incorrectly.

I think he/she misread the calcium $\ce{Ca}$ symbol for carbon which is $\ce{C}$.

$\ce{SO_4^2-}$ can be easily misrepresented to be sulfuric acid.


Which compound is more reactive towards acetylation, 4-aminophenol or 4-aminocyclohexanol?

4-aminocyclohexanol should be more reactive towards acetylation because the reactivity depends on the availability of the lone pair on N to be able to attack the C atom during acetylation. Since in 4-aminophenol the lone pair resonates in the benzene ring, its availability to attack will get reduced making it lesser reactive towards acetylation.


Difference between Reaction Rate and Rate Law?

Let's use the following gaseous chemical reaction as an example:
$$\ce{H2 + I2 -> 2HI}$$
When we study reaction kinetics, we oftentimes can't observe the changes in concentration of each reactant and product.  Therefore the rate expression helps us to relate the rates of appearance/disappearance of the species present:
$$-\frac{1}{1}\frac{\Delta \left[ \ce{H2} \right]}{\Delta t}\;
= \; -\frac{1}{1}\frac{\Delta \left[ \ce{I2} \right]}{\Delta t}\;
=\; \frac{1}{2}\frac{\Delta \left[ \ce{HI} \right]}{\Delta t}\tag{rate expression}\label{rateex}$$
Note that each rate is modified by the inverse of the stoichiometric coefficient.  I explicitly put $\frac{1}{1}$ in the expression just to emphasize this point.  We can see that multiplying the coefficients of the reaction will not change the rate expression in the following example: let's multiply the coefficients in the chemical reaction by 2 and then write the corresponding rate expression.
$$\ce{2H2 + 2I2 -> 4HI}$$
$$-\frac{1}{2}\frac{\Delta \left[ \ce{H2} \right]}{\Delta t}\;
= \; -\frac{1}{2}\frac{\Delta \left[ \ce{I2} \right]}{\Delta t}\;
=\; \frac{1}{4}\frac{\Delta \left[ \ce{HI} \right]}{\Delta t}\tag{new expression}\label{newex}$$
The $\ref{rateex}$ and $\ref{newex}$ are numerically equivalent.

Full disclosure here: the term rate expression is ambiguous and is also used as a synonym for the rate law. I've also seen the rate expression referred to as the rate relationship.

The rate law or differential rate law relates the rate of a reaction to the concentration (or pressure) of the reactants.  The rate of a reaction is proportional to the concentration (or pressure) of the reactants modified by some experimentally determined number called the reaction order.
$$\begin{align}
-\frac{\Delta\left[ \ce{H2} \right]}{\Delta t}
&\propto\left[\ce{H2}\right]^m\left[\ce{I2}\right]^n\\
-\frac{\Delta\left[ \ce{H2} \right]}{\Delta t}
&= k\left[\ce{H2}\right]^m\left[\ce{I2}\right]^n\tag{rate law}
\end{align}$$
Here, $m$ and $n$ are the reaction orders for $\ce{H2}$ and $\ce{I2}$, respectively, and the sum, $m + n$ yields the overall order of the reaction.  The proportionality $a\propto b$ can be replaced with an equality $a = k b$.  In kinetics, this constant of proportionality is referred to as the rate constant.

In summary: the rate expression tells us how the appearance/disappearance rates of the products and reactants relate to one another; the rate law tells us how the rate is related to the concentrations (or pressures) of reactants and the rate constant is a constant of proportionality that comes out of the rate law.

This answer is a summary of information found at the boundless website and from Tro's most recent general chemistry textbook.


Why do crown ether look the way they do?

Crown ethers don't look like that except in structure drawings

You confuse two completely different things: the drawings we use to describe chemical structures and what molecules actually look like in real chemical environments. We draw crown ethers like this because the drawing accurately describes the connectivity in the structure. In reality crown ethers are flexible and floppy and certainly not flat. we don't add that extra complexity to 2D structure drawing as they get very messy very quickly and obscure the essential features that describe the connectivity in a simple way.

When they are in solution with appropriate ions (for 18-crown-6, potassium ions would be a good example) they form complexes driven by the attraction between the oxygens and the charged ions. These complexes have 3-dimensional structures (with K+ the complex is approximately octahedral as potassium fits neatly into the cavity created by the oxygens when the molecule folds up). It looks a little like this as a 3D structure:

Wikipedia has some good descriptions of this and further pictures.

Crown ethers are not rigid molecules and the actual shapes they take up in 3D space demand a lot on what else is in solution with them. Other, similar, complexing agents are far more rigid. Calixarenes, for example, have a 3D shape like a chalice which is even more misrepresented by a 2D drawing.


Can you measure an IR spectrum of a sample inside a glass vial?

As EdV said, when you have aqueous samples, Raman spectroscopy is the way to go. The problem is not the container (which should not be glass etc),the problem is water. Its absorption will hide the absorption of everything.
If you ever open an FTIR instruments, you will see that there is nothing which is made of glass or quartz because they absorb too strongly in the useful IR range.


Is 1000ml of 'hot-ice' solution dangerous?

The heat generated increases directly with the amount of sodium acetate trihydrate used, but the maximum temperature, ignoring surface effects, should not change at all. After all, if the heat in 300 ml is used to raise the temperature of 300 ml of substance, or if 300 kg is used to raise the temperature of 300 kg, it's being spread out proportionately. No matter how much you use, it won't exceed ~58 C

However, do not try this in a container that would not allow room for expansion, e.g. a glass bottle, as in crystallizing, sodium acetate trihydrate might take up more room. See warning at http://chemmovies.unl.edu/chemistry/dochem/DoChem058.html. You could safely do this on a polyethylene cutting board, though, which melts at ~100 C


How should the hydrated proton be represented in chemical equations?

In addition to entropid's answer, let's remember why we invoke the hydronium ion $\ce{H3O+}$ in the first place.

We use $\ce{H3O+}$ as a shorthand for $\ce{H+(aq)}$, which looks more like protonated water clusters of the generic formula $\ce{H+.(H2O)_{$n$}}\equiv \ce{H_{$2n+1$}O_{$n$}+}$.

Almost ten years ago, a very interesting paper appeared in Science that examined the structure of these clusters ($n=2,3,4,5,6,7,8$). The linked article has been made freely available by Science, but you need to register for a free account at sciencemag.org.

One of the key points of the article was the determination of the inner structure of these clusters. For $n=1$, you have $\ce{H3O+}$, the hydronium ion, which we are reasonably familiar with. When $n=2$, you have $\ce{H5O2+}$, which has a different structure: the proton is evenly shared between two water molecules: $\ce{[H2O\bond{...}H\bond{...}OH2]+}$. Higher order clusters have these two structures, termed the "Eigen" and "Zundel" ions, respectively, at their cores.

The study used a combination of theory (MP2/aug-cc-pDVZ) and experiment (photodissociation vibrational spectroscopy). They found the for $n=2,6,7,8$, the clusters have a Zundel core, while the Eigen core exists for $n=3,4,5$. Whoa! And it gets more complicated as you increase the number of water molecules in the clusters (might be behind a pay wall) - though it turns out the Eigen ion is slightly more common in larger clusters.

In aqueous solutions, we can have an extensive dynamic network of clusters always in flux. In which case, we have a mix of solvated Eigen $\ce{H3O+}$ ions representing protons closely associated with water molecules and Zundel $\ce{H5O2+}$ ions representing protons in transit.

So, we use $\ce{H3O+}$ to keep things simple, and also to satisfy the needs of Brønsted-Lowry acid-base theory, in which every acid-base reaction needs a proton-donor and a proton-acceptor. Thus, the "dissociation" of acids in water is not really a dissociation, but an ionization reaction - an acid-base reaction in which water is the base and hydronium is the conjugate acid.

$$\ce{HA + H2O <=> A- + H3O+}$$

Of course, it is only appropriate to use $\ce{H3O+}$ in aqueous solution. In other solvents, the "proton" has a different structure. In general, the proton is attached to a basic site on the solvent molecule. For example, in methanol:

$$\ce{HCl + CH3OH <=> Cl- + CH3OH2+}$$


The Sackur-Tetrode equation can be written as $$S^{Trans}=R\ln \left( \frac{e}{C\Lambda^3} \right)$$ where $C$ is the concentration and $\Lambda$ the thermal wavelength. You can rewrite this as $$S^{Trans}=R\ln \left( \frac{e}{\Lambda^3 C^\circ\frac{C}{C^\circ}} \right) = R\ln \left( \frac{e}{C^\circ\Lambda^3} \right)-R\ln\left( \frac{C}{C^\circ} \right) = S^{\circ,Trans}-R\ln\left( \frac{C}{C^\circ} \right)$$ where $C^\circ$ is an arbitrary standard concentration. When you combine it with enthalpy and the other entropy terms you get $$G = G^{\circ}+RT\ln\left( \frac{C}{C^\circ} \right) $$


Using the following generic reaction, where an acid donates $\ce{H+}$, i.e. a proton:

$$\ce{HX + B <=> X- + HB+}$$

If a lone pair of electrons on $\ce{B}$ has a higher energy than than than the resultant lone pair of electrons on $\ce{X-}$, then the $\ce{H+}$ on $\ce{HX}$ will be transferred to $\ce{B}$ to form $\ce{HB+}$ in order to stabilize the high energy lone pair on $\ce{B}$ in the form of a bond.

So, there are two things that influence whether an acid will transfer a proton:

The energy of the lone pair on the base
The energy of the lone pair on the resultant the conjugate base

If the conjugate base of the acid is more able to stablize a lone pair than the basic starting material is, an acid-base reaction will occur, and a majority of the basic reactant will end up being protonated.


How do I find miller indices for a plane whose intercepts are fractions of the lattice constant?

The plane closest to the origin has the intercepts you mentioned. The rest have intercepts that are integer multiples of this. In the figure, the origin is on the bottom left of each cell. As the figure shows, the [1 1 1] planes correspond to every second of the [2 2 2] planes because the latter is a multiple (by 2) of the former.

Higher indeces are associated with a lower distance of the planes and with a higher resolution (i.e. larger Bragg angle) of the corresponding reflection.


When you learn molecular orbital theory, you learn to throw the octet rule out the window in favor of identifying the molecular orbitals that form the actual bonds.  Take a look at the seemingly innocuous case of carbon dioxide discussed in Chemistry LibreTexts (illustration from this reference):

There are really two related ways in which the molecular orbital reality blows the octet rule away even though you thought the familiar Lewis structure $\ce{O=C=O}$ obeyed this rule.
First:  The octet rule assumes that electrons in covalent bonds are shared between pairs of atoms at the ends of the respective bonds.  In the above diagram they're not.  Most of the occupied orbitals share electron density between all three atoms and those that don't are concentrated at the two oxygen atoms, not at the ends of either carbon-oxygen bond.  This three-way sharing is especially notable with the pi bonds; the structure I drew above is really not an accurate way to describe the pi bonds.  You really need a combination of pairwise valence bond structures even to begin to approach the molecular orbitals:
$\ce{\overset{-}{O}-C#\overset{+}{O}}$
$\ce{O=C=O}$
$\ce{\overset{+}{O}#C-\overset{-}{O}}$
In effect, the carbon-oxygen bonds with this multiple-atom sharing are not just double bonds.  They are a mixture of single, double and triple bonds providing more overall bonding than just plain pairwise double bonds could have.
Second:  Given this multi-way electron sharing the number of occupied orbitals that are at least partly shared by any atom can go well beyond four.  In the case of carbon dioxide this is especially true of the oxygen atoms, each of which has a finger in the pie of all sixteen valence electrons.  In the case of carbon dioxide it is true that each oxygen atom cannot have more than half of that electron density close to it, but that is a property of the specific compound carbon dioxide (which has only sixteen valence electrons to work with among all three atoms) and not a universal law of nature.
This should give you some idea of what goes on with a more complex structure such as your orthonitrate ion.  Electrons are not shared pairwise but among multiple atoms in molecular orbitals, and with this multi-way electron sharing the bonds are not really confined to specific pairs of atoms.  What looks like single, sigma-only bonds between the nitrogen and each oxygen atom are really a combination of sigma and pi bonding interactions, delocalized through a complex molecular structure; and each $\ce{N-O}$ interaction is thereby a combination of single bonds, multiple bonds and even purely ionic interactions.  Assigning a meaningful atom-by-atom electron count to such a structure is, as Mithoron suggests, fuzzy math.


Do entropy changes have an equivalent of bond enthalpy?

Please note that you are calculating the $\Delta U$ with your bond breaking/forming method. If there is low volume work done, then you can approximate:
$$
\Delta U \approx \Delta H
$$

For the entropy: Yes and no. I have no idea for liquids but qualitatively I would guess that it is not possible to have a quick and dirty method as you wish for this phase. And even for gases and solids it is quite a long work.

For gases you can calculate the entropy for every component itself and then add/substract.

Translational (Sackur-Tetrode):
$$
S_\text{trans}= R \ln \left( (2\pi m k_B T)^\frac{3}{2} \frac{V_m}{N_lh^3} \exp(5/2) \right)
$$

Rotational:
$$
S_\text{rot}=R\left(\ln\left(\frac T\Theta_\text{rot} \right) +1 \right)
$$

Vibrational:
$$
S_\text{vib}=R \left(\frac{\Theta_\text{vib}}{T} \frac{1}{\exp\left( \frac{\Theta_\text{vib}}{T}\right)-1} - \ln \left(1-\exp\left( \frac{-\Theta_\text{vib}}{T}\right) \right)\right)
$$

For solids you could calculate the heat capacity $C_p(T)$ with the Debye modell and then integrate from 0K.

I am really sorry, but I do not think, that there are easier methods.


Is there a specific term for reactions of the form A + B → C + D?

There are 4 'base-types' of chemical reactions:

Synthesis reactions ($\ce{A + B -> AB}$)
Decomposition reactions ($\ce{AB -> A + B}$)
Substitution reactions ($\ce{A + BC -> B + AC}$)
Metathesis reactions ($\ce{AB + CD -> AD + CB}$)

Given that you talk about polymers that in one case combine to form a new (single) polymer: $\ce{A + B <=> C}$ you could call this a synthesis reaction.

You other example, $\ce{A + B <=> C + D}$, is an exchange of parts of the polymers so that would be a metathesis reaction.

For polymers there are a couple of additional options to name the types of reactions, but these depend a lot on the specific chemistry (e.g. hydrolysis if water is one of the reactants) so that is probably difficult if you have 'abstract' polymers.


R & S Stereochemistry of an unsaturated ring system

I don’t really know how you arrived at your reasoning, but it seems to fail as soon as double bonds are involved. If you encounter a double bond, it must be expanded with ghost atoms to two single bonds in the following manner:

$$\ce{(R2)C=C(R'2) -> (R2)C(C)-C(C)(R'2)}$$

The carbon atoms in bracket are ghost atoms not connected to anything else. These ghost atoms are treated just like normal atoms when determining priorities. You can read more on Wikipedia, which also includes a better graphic example. (But since your case is rather simple, we can do without.)

Let’s first determine the tertiary alcohol’s carbon’s absolute configuration:

$\ce{O}$ from the hydroxy group.
$\ce{C}$ (along the ring to the left)

$\ce{C}$ (the quarternary carbon)

$\ce{C}$
$\ce{C}$ ← This is the first point of difference.
$\ce{(C)}$ — this is the double bond’s ghost atom.

$\ce{H}$
$\ce{H}$

$\ce{C}$ (along the ring upwards)

$\ce{C}$

$\ce{C}$
$\ce{H}$ ← This is the first point of difference.
$\ce{H}$

$\ce{H}$
$\ce{H}$

$\ce{C}$ from the methyl group.

$\ce{H}$
$\ce{H}$
$\ce{H}$

Thus, the priorities are $\ce{OH} > \ce{CH2C_{(q)}} > \ce{CH2CH2R} > \ce{CH3}$, because the quarternary carbon that one of the two $\ce{CH2}$ groups is bonded to beats the secondary carbon at the top of the right-hand ring. That makes that carbon’s absolute configuration (S).

Similarly, for the carbon attached to iso-propyl:

$\ce{C}$ (the quarternary carbon)

$\ce{C}$
$\ce{C}$
$\ce{C}$

$\ce{C}$ (the iso-propyl group)

$\ce{C}$
$\ce{C}$
$\ce{H}$

$\ce{C}$ (the secondary carbon)

$\ce{C}$
$\ce{H}$
$\ce{H}$

$\ce{H}$

As you can see, we only need go down to the second level to find a difference: $\ce{C(C3)} > \ce{C(C2H)} > \ce{C(CH2)} > \ce{H}$ Thus, that carbon’s absolute configuration is (R).


It might be scientifically correct but it is linguistically misleading

The sentence "diamond is an element" can be seen to be misleading when compared to the sentence "diamond is an allotrope of the element carbon". Or even "diamond consists of the element carbon".

The issue is that clear language should distinguish between the form and the composition of an element with more than one allotrope. Saying "diamond is an element" confuses the fact that diamond is made from carbon with the fact that it is made from an infinite tetrahedral array of carbon atoms. It also subtly excludes the fact that there are other allotropes.

For clarity in language a specific form of an element (an allotrope) should not be referred to as "an element" but should be referred to as one form of that element. As in "diamond is one form of the element carbon, buckminsterfullerene is another form."


The high covalent characted is just explained by the fact of $\ce{Pb^{4+}}$ being very polarizing and $\ce{Cl-}$ fairly polarizable.

The inert pair effect would give an explanation about why it is difficult to reach that oxidation state, and why it has high tendency to get reduced, but there is no need to require to this concept to explain the covalent character of the compound.

Regarding your last question, I do not see why the inner pair concept should be rejected. Is just another model that can explain in a simple way properties or behaviors of certain elements. There are better models and orbital calcualtions that can also do the job? Of course, but that is no reason for avoiding the use of the model. Models that are still useful to explain stuff easily, can be employed even if more accurate ones do exist. One good example is ligand field theory, which is still useful even when molecular orbitals theory is more accurate and can still explain the same (and many more) things than the former.
Even though, a theorem or model HAS to be rejected if a new one is found that explains properties better and is incompatible with the old one.


What is the citric acid concentration, [HCit], in the lemon juice?

For any mono-protic acid, when the acid dissociates in water, every H+ created will create one CB-.

  H-Cit  <--> H+ + Cit-

For each 1 H-Cit. 1 H+ and 1 Cit- is produced upon dissociation.
Therefore, the [H+] = [Cit-]

For a Ka expression, set these two [] equal to each other, and the expression for [H+]*[Cit-] = x^2

Use this to determine the amount of [Cit-] created when you create x amount of [H+].


Why was Avogadro's number chosen to be the value that it is?

Your question implies that you already know that it was a choice rather than something derived from first principle. There are some numbers that are derived from first principles. In math, $\pi$ and $e$ are not a choice, but can be derived from their properties. In physics, the fine structure constant is not a choice, but would be measured to get the same value by aliens on a different planet (at least as far as we can tell).

[...] Avogadro's number [...]

To be correct, it is actually Avogadro's constant $N_A$ that we use more often in chemistry (for identities such as $k_B = R / N_A$, where $k_B$ is the Boltzmann constant and R the universal gas constant). Avogadro's constant has the dimension (in the SI unit system) of one divided by amount of substance. So the number in front of the unit will change as you change the unit, just like 30 minutes is the same as 0.5 hours even though 30 is not the same as 0.5.

I also understand that it is not necessary to know the history of the number in order to apply it. But it would seem to me that the number wasn't chosen arbitrarily.

The value of Avogadro's constant is linked to the choices that were made when setting up the metric system. There is one definition, that of the meter, that is linked to our planet (so aliens on another planet would have no reason to choose it). The meter was chosen to be approximately 1/10,000,000th of the quadrant of the Paris meridian (i.e distance from Paris to South Pole to North Pole to Paris divided by 40,000,000). Then, they made an artifact to define the meter exactly. Next, the kilogram was chosen to be approximately the mass of 1/1000th of a cubic meter of water (at some temperature). It is no coincidence that the density of water is very close to 1 kg/L. Again, they made an artifact to define the kilogram exactly.
Once the kilogram was defined, the mole was chosen to contain the same number of particles as (pick your favorite, they are all approximately correct at this point) one gram of protons or neutrons, 16 g of oxygen atoms or 12 g of pure carbon-12 atoms.
How are the definition of kilogram and mole related to Avogadro's constant?
Avogadro's constant tells us how many particles there are per mole of particles. The mole was defined with reference to the gram and fundamental constants such as the mass of a proton or of a single carbon-12 atom, so once these choices are in place, Avogadro's constant is no longer a choice.
Relationship to current definitions of units
The SI units (including meter, kilogram and mole) are defined in a different way today, but their values are still the same up to many significant digits. So you can still trace back the choice of the Avogadro constant to choices made when defining the meter, the kilogram and the mole.


Deriving the entropy of mixing of a non-ideal solution

The calculation of the mixing energy is extremely complicated and only approximate solutions have been obtained as far as I am aware. I shall attempt to outline one of these methods*.

All molecules are assumed to be spherical and of equal size and situated on a lattice with co-ordination number z . Each pair of neighbours AA of type A molecules contributes energy $w_{AA}$, similarly $w_{BB}$ for B molecules and energy $w_{AB}$ for AB pairs. The total energy due to these  interactions
is
$$E_T=N_{AA}w_{AA} + N_{BB}w_{BB}+N_{AB}w_{AB}$$
where $N_{AA}$ is the number of AA pairs and so on. Counting the number of  $A$ interactions and number of B ’s gives
$$zN_A=2N_{AA} + N_{AB}  ~~~\text{and }~~~ zN_B=2N_{BB} + N_{AB} $$

and thus  combining these equations gives
$$E_T=\frac{zN_Aw_{AA}}{2}+\frac{zN_Bw_{BB}}{2}+wN_{AB}=E^T_{AA}+E^T_{BB}+wN_{AB}$$

where $w=w_{AB}-w_{AA}/2-w_{BB}/2$ is the energy gain on mixing per pair of AB neighbours.

In statistical mechanics the partition function is the key to all other thermodynamic functions. In this case partition function is $$
Z= \sum_{N_{AB}}Z_AZ_Bg(N_A,N_B,N_{AB})e^{(-E^T_{AA}-E^T_{BB}-wN_{AB}/KT)}$$

and where $g(..) $ is the number of arrangements of $N_A$  A’s , $N_B$ B ’s and $N_{AB}$  pairs of molecules. Separating out the terms due to pure A and B molecules gives

$$Z=Z’_AZ’_B \sum_{N_{AB}}g(N_A,N_B,N_{AB})e^{-wN_{AB}/KT}$$

and as the Helmholtz energy is
$F=-kT\ln(Z)=F_A+F_B+\Delta F$
and for mixing only the $\Delta F$ term is important the others being the energy of each pure component.
Thus
$$\Delta F = -kT\ln\left(\sum_{N_{AB}}g(N_A,N_B,N_{AB})e^{-wN_{AB}/kT}  \right) $$

The change in energy can be obtained from the Gibbs-Helmholtz equation $(E = F-T(dF/dT))_V$ and is
$$\Delta E= w\frac{\sum_{N_{AB}} N_{AB} g(N_A,N_B,N_{AB})e^{-wN_{AB}/kT} } { \sum_{N_{AB}}g(N_A,N_B,N_{AB})e^{-wN_{AB}/kT} } $$

and this is an equation that seems to have no exact solution and so this is where approximations have to be made.

The difficulty is in the exponential terms, as they contain $N_{AB}$ the aim is therefore to replace these with an average value, and thus the exponential terms become constant wrt $N_{AB}$. It can be shown that
$$\sum_{N_{AB} }g(N_A,N_B,N_{AB}) =\frac{(N_A + N_B)!}{N_A!N_B!}$$
and
$$\bar N_{AB}=w\frac{\sum_{N_{AB}}N_{AB}g(N_A,N_B,N_{AB})}{\sum_{N_{AB}}g(N_A,N_B,N_{AB})} = z\frac{N_AN_B}{N_A + N_B} $$

By replacing $N_{AB}$ by $\bar N_{AB}$ in the partition function produces

$$Z=Z’_AZ’_B \sum_{N_{AB}}g(N_A,N_B,N_{AB})e^{-w\bar N_{AB}/KT)}
 = \frac{(N_A+N_B)!}{N_A!N_B!}\exp\left(-\frac{zw}{kT}\frac{N_AN_B}{N_A+N_B}\right)$$

the change in energy is then

$$\Delta E = zw\frac{N_AN_B}{N_A+N_B}= w\bar N_{AB}$$

and this can be re-written as molec fractions to produce
$$ \Delta E= z(N_A+N_B)x_1x_2w$$

based on text in Rushbrooke ‘Introduction to Statistical Mechanics’


Resonance Forms in Electrophilic Aromatic Substitution

Well I'm too lazy to open up chemdraw to make a nice drawing so my handwriting will have to do.

The way I see this kind of question is that, to interconvert resonance structures, electrons need to move. Of course, you'd represent that with curly arrows (top half of the picture).

Now for the deprotonation step, imagine you were using the first resonance structure. You'd draw the first two of the three arrows that I drew in the bottom half of the picture. (Or you might omit drawing the base, although for simple mechanisms like this I prefer having everything accounted for.)

What's the third arrow for (the one going from a double bond to the adjacent single bond)? It's the reverse of the arrow you used to interconvert the resonance structure. So you could think of the first two arrows being the "actual" reaction, and the third arrow being a "bookkeeping" arrow to make sure that everything ends up nicely aromatic.

Of course, that's kind of a flawed view because the reaction as shown has to have all three arrows or else you end up with a product that doesn't make sense. But using this idea of "the original reaction" + "reverse resonance arrow" works for every single instance of the question that you are asking, i.e. how to draw a mechanism involving an alternative resonance form. As long as you are able to draw arrows to show the interconversion between resonance forms, you should never run into a problem drawing a mechanism, regardless of what resonance form you use.

Now just a note on what you said - "It seems like most of the time the first form is simply chosen to illustrate the return of aromaticity." You're absolutely right about that. Using the first resonance form is just the simplest option since it involves the fewest arrows. Otherwise the writer of the book is just giving him/herself extra work... and nobody likes that.


Nucleophilicity in Polar Aprotic Solvent

It's because of the anion stability. Fluoride is less stable than iodide, therefore more likely to donate electrons.
In protic solvents, the fluoride was stabilised by the partially positive hydrogen end. Therefore, it is less nucleophilic then iodide.


Merge rigid parts and flexible results after docking (Vina)

A Useful Script for AutoDock4 and Vina That Merges Flexible and Rigid PDBQT Output
« Prosciens.
http://prosciens.com/prosciens/a-useful-script-for-autodock4-and-vina-that-
merges-flexible-and-rigid-pdbqt-output/
Link to the script: http://prosciens.com/prosciens/oldproscienssarl/files/flexrigidpdbqt2pdb_template.sh


Oxidizing properties of sulfuric acid: can it oxidize hydrochloric acid to chlorine?

Concentrated sulfuric acid cannot oxidise chloride or fluoride, see here for a detailed explanation here

Concentrated sulfuric acid is not a strong enough oxidizing agent to oxidize fluoride or chloride. In those cases, only the steamy fumes of the hydrogen halide—hydrogen fluoride or hydrogen chloride—are produced. In terms of the halide ions, fluoride and chloride are not strong enough reducing agents to reduce the sulfuric acid. This is not the case for bromides and iodides.


Alkaline etching of glass tends to be less rapid than HF and leaves a smoother surface. This may not be what you do for the first etch on fresh glass, but it may smooth out the etch somewhat after an HF etch. The smoothing may also be caused by silicate ions that form in the alkaline etch fluid, so perhaps a solution of potassium (or sodium) silicate (ortho or meta) would be gentler. A polysilicate might be too slow.


What is the difference between nuclear fission and nuclear fusion?

Nuclear fission involves the splitting of a heavy nucleus into two lighter fragments plus the release of energy.  As an example here is one route by which $\ce{^235U}$, a fissile material, can split apart.

$$\ce{^{235}_{92}U + ^{1}_{0}n -> ^{141}_{56}Ba + ^{92}_{36}Kr + 3^{1}_{0}n + 200 MeV}$$

Note the 3 neutrons generated in the fission process, they are part of the chain reaction; they allow the reaction to propagate at an even faster rate as they go on to react with 3 other $\ce{^{235}_{92}U}$ nuclei.

Nuclear fusion involves the fusing together of nuclei to form a heavier element.  Again, energy is released in the process.  Here is an example of a fusion reaction involving the combination of 2 deuterium nuclei.

$$\ce{^{2}_{1}H + ^{2}_{1}H -> ^{4}_{2}He + ^{1}_{0}n + 17$.$6 MeV}$$


Enthalpy of neutralisation for weak acids and weak bases at standard state

Q1, well you are correct to a certain level. standard state of a solute in a solution of any specified solvent is the state having solution under pressure exactly equal to standard pressure 1 bar,composition of solute equal to 'a standard composition', solution must be an 'ideal solution (perfect solution)', and finally solution must show 'properties of an infinitely diluted solution'. here standard composition refers to either, standard amount concentration (1 mol/dm3), standard molality (1 mol/kg) or standard mole fraction (1). note that temperature is not a part of definition of standard state. this is a hypothetical state for any temperature we use or what standard composition you may chose. here you have chosen standard molality,its ok.(note that we can prove that for an ideal solution showing properties of infinite dilution at a given pressure and temperature molar enthalpy of solute is a constant what ever its composition may be.plug into somewhere that defines you exactly the terms 'perfect solution',infinite dilution and properties of infinite dilution to understand this, its easy). infinite dilution means amount concentration or equivalently any other composition term of solute approaching zero. however acid  composition is molality =1mol/kg as you prefer or any other standard composition, but yet it is not infinitely diluted although it shows properties of infinite dilution. key thing here to understand is that infinite dilution is one thing and showing properties of infinite dilution is another. this is not the only hypothetical state you are dealing with, standard state of a gas existing in pure form , as a solute or in a mixture is the pure form of gas at pressure 1 bar and gas is ideal or equivalently compressibility factor =1. going back to your question, we can talk about strong and weak acids or bases. in fact Ka and Kb values are based on this.  Q2; strong and weak are relative terms. the better word is strength. even strong acids or bases absorbs heat .


In mass spectrometry, organic compounds can be ionized by different methods. The compounds ionized by electron impact ionization display particular fragmentation patterns. When subjected to mass spectroscopy, cleavage of the $\ce{C-C}$ bond next to the oxygen or nitrogen can usually occur, when the compound is ionized by electron impact ionization.

Let’s look at what happens to 3-pentanol:
$$\ce{CH3CH2CH(OH)CH2CH3 ->[e- beam] CH3CH2CH(O^.+H)CH2CH3 }$$

$$\ce{CH3CH2CH(O^.+H)CH2CH3 -> CH3CH2CH=O^+H + ^.CH2CH3}$$

Now, base peak (most probably), would be $\ce{[CH3CH2CH=OH]+}$, which should be at $ 59 \ m/z$. Note that in general, the molecular ion of an alcohol (here it is $ 88 \ m/z$) is either trace or non-existent, probably loosing $\ce{H2O^.}$.

Now, we'll look at what happens to 1-pentanol:
$$\ce{CH3CH2CH2CH2CH2-OH ->[e- beam] CH3CH2CH2CH2CH2-O^.+H }$$

$$\ce{CH3CH2CH2CH2CH2-O^.+H -> CH2=O^+H + ^.CH2CH2CH2CH3}$$

Now, base peak (again, most probably), would be $\ce{[CH2=OH]+}$, which should be at $ 31 \ m/z$ ($88-31=57$ mass loss). Note that here again, the molecular ion is either trace or non-existent, probably loosing $\ce{H2O^.}$, and possibly $[M-18]$ ion should be significant here, since there is equal chance of cleaving $\ce{C_1-O}$ bond (giving peak at $ 70 \ m/z$) or $\ce{C_2-C_3}$ bond (giving peak at $ 31 \ m/z$):

However, according to the mass spectrum given, $\ce{C_5}$-carbon chain has undergone McLafferty type rearrangement (which is common in MS of aldehydes and ketones) to give the base peak at $ 42 \ m/z$ by loosing a water and ethylene simultaneously from parent ion (see the insert in above diagram).


This is how the graph between the rate constant and temperature looks like:

This might not seem intuitive looking at the Arrhenius equation which says:
$$k=Ae^{\frac{-E_a}{RT}}$$
After using some mathematical prowess someone might say that as $T\rightarrow \infty, ~k\rightarrow A$ and so the graph should look something more like:

Which on first glance seems fair but doesn't match the aforementioned curve.
To understand what is causing the problem we must check everything that might change during a change in temperature.
Upon examination we can find that only two things might cause this discrepancy, $E_a$ and the pre-exponential "constant", $A$.
The activation energy of a system is also independent of the temperature and thus the only possible problem point left is the pre-exponential "constant". Turns out that this constant isn't a constant in true sense, which isn't that disturbing of an idea if you know what that constant signifies.
The pre-exponential constant shows how many effective collisions can take place in a reacting mixture. It is defined as,
$$A=pZ$$
Where,
$A$ is the pre-exponential constant.
$p$ is the steric factor, which takes into account the steric effects during an effective collision.
$Z$ is the collision frequency.
Now as it might seem intuitive, the collision frequency is related to temperature and increases with increase in temperature according to the following relation(using equations for collisions between similar molecules to prevent unnecessary complexity):
$$Z=4nr\sqrt{\frac{\pi RT}{M}}$$
Where,
$Z$ is collision frequency.
$n$ is molecules per unit volume.
$r$ is the radius of molecule.
$R$ is the universal gas constant.
$T$ is temperature.
$M$ is the molar mass of the gas.
Thus the collision frequency and corresponding the pre-exponential constant tend to infinity when $T\rightarrow \infty$, which matches with our observation of $k\rightarrow \infty$ when $T\rightarrow \infty$.
Reference:
The Effect of Temperature on Reaction Rates https://chem.libretexts.org/@go/page/24271 (accessed May 14, 2021).


Deducing the shell when transition occurs such that two photons are discharged

You've got yourself all twisted around with 1's and 2's.

The problem states that "the ion emits two photons in succession with wavelengths 108.5 nm and 30.4 nm." So the first photon emitted has a wavelength of 108.5 nm and the second photon has a wavelength of 30.4 nm.

Furthermore the problem stated that "on transition to ground state." So after the second photon was emitted the helium ion is in its ground state which means that the single electron is in the 1s subshell.

The gist here is that you have to work backwards. Since you know for the 30.4 nm photon was from a transition that ends up in the 1s subshell, then did the electron start in the 2s subshell, or the 3s subshell and so on. So once you figure out the two subshells for the 30.4 nm photon, then you repeat the procedure for the 108.5 nm photon.

The relative wavelengths of the two photons isn't the important clue here. In other words you can't always say that the shortest wavelength will always be to the ground state for some electronic transition.


No astatonium ion with hydrogen is known, but it appears that a cation $\ce{S7At^+}$ is formed by replacing a sulfur atom in $\ce{S8}$ with an isoelectronic $\ce{At^+}$ moiety (the actual oxidation state of astatine, with sulfur being more electronegative, is $+3$).  The ion is proposed to explain the incorporation of astatine into the sulfur precipitate when a sodium thiosulfate solution is decomposed by acid in the presence of astatide ion.

Primary reference:

G. M. W. Visser, E. L. Diemer, Radiochimica Acta 33 145 (1983).


Testing for asbestos using readily available chemicals and tools?

Is there a chemical test for $\ce{[Si2O5]^2−}$ that can be done at home?

First $\ce{[Si2O5]^2−}$ isn't a poly atomic ion like sulfate ($\ce{SO4-}$) or nitrate ($\ce{NO3}$) but represents a stoichiometry in the material.

Unfortunately at this time a test without x-ray diffraction is not possible. Asbestos/chrystolite is not just a chemical formula, but a crystal structure. It's not just what it is made of but also how the atoms are structured that makes it potently carcinogenic. If you were to take asbestos and melt it then cool it sufficiently fast to form a glass (amorphous/non-crystalline substance), it would not have similar carcinogenic properties as the crystalline material.

  I am aware that one test involves viewing the sample under a sufficiently high-powered polarizing light microscope, but not everyone has such a tool.

It is worth noting that using polarized can be done at home with polarized filters but is a presumtive test not a direct method. Essentially all chrystolite will rotate light but not all minerals that rotate light are chrystolite.


Which occurs first: nitration or oxidation of aniline by concentrated nitric acid?

Concentrated $\ce{HNO3}$ is ‘only’ around $70~\%$. To add a nitro functionality by electrophilic aromatic substitution, typically the $\ce{NO2+}$ cation is required which is formed only by strong dehydrating agents such as sulfuric acid:

$$\ce{O2N-OH <<=>[\ce{H2SO4}][-\ce{HSO4-}] O2N-OH2+ ->[][-\ce{H2O}] O=\overset{+}{N}=O}\tag{1}$$

Concentrated nitric acid is not strong enough to dehydrate iself. Thus, nitration will typically not be observed.


How to interpret the instruction of the dilution of hydrogen peroxide?

Take $\pu{200 mL}$ of the $30\%$ solution, add water to a final volume of $1$ liter. Then you have a $6\%$ solution. Next, take $\pu{59 mL}$ of the $6\%$ solution and add water until it is $1$ liter.

After the first dilution it's not exactly $6\%$, because the density of $30\%$ $\ce{H2O2}$ is $\pu{1.135 g/cm3}$. It's more like $6.6\%$, but the instructions are referring to it as $6\%$. If you want exactly $6\%$ you should dilute by weight instead of volume.


Can you find [SO₃²⁻] with just the two acid dissociation constants Kₐ₁ and Kₐ₂?

No, it is not possible.

For example, as total concentration of acid approaches zero, concentration of sulfate would also approach zero.