text:
Does gravity affect the trajectory of gas particles?
We're studying the kinetic theory of gases in school, and one of the points that was brought up was that: "Gases consist of particles in constant, random motion." How is it possible for gas particles to move in straight lines, entirely randomly? Doesn't gravity affect them, or is their mass so small that its effect on their movement is negligible?

It's not due to their small mass since everything at this height experiences about the same gravitational acceleration.

Yes, they do, and when you consider it, you get the barometric formula back.

Curved motion need not be any less random. In fact, this begs the question, what do you understand by "random"? It is the orientation (direction) of the trajectory that might be regarded as random. In addition, that orientational probability is nearly uniform (isotropic).

Yes gravity pulls on gas molecules. That is why the atmosphere doesn't just float off into space.
The gist is that the time between collisions is very short in the lower atmosphere, and the distances very short. The mean free path at atmospheric pressure is only about 70 nanometers. So the assumption is that gas particles travel in a straight line between collisions.

In contrast, imagine gas molecules on the moon, bouncing around like isolated balls as they are likely to hit the ground again before another gas molecule; or the situation up where the space station orbits: a molecule tossed upward will make a suborbital arc befome coming back down to where the air is thicker and hitting something.

Suppose an air molecule travels by $500~\mathrm{m}/\mathrm{s}$ (or $500$ nanometers per nanosecond, if you prefer). The rough time scale for gravity to interfere with that, would be found by dividing by $10~\mathrm{m}/\mathrm{s}^2$, so it is of order tens of seconds before gravity will disturb a body of that velocity much. However, with the number from the above answer, we see that the molecule hits another one every $0.14$ nanoseconds. So the deviation from straight line paths will be negligible.

To add to previous answers, all molecules and atoms are affected by gravity and so the density of the atmosphere is greater at the surface of the earth compared to higher up, which is why climbing on Everest most climbers take extra oxygen (although, remarkably, it has been done without this aid).
At room temperature gas molecules have an average thermal energy of $3RT/2$ ($R$ is the gas constant $8.314 \pu{ J mol^{-1} K^{-1}}$) and thus they have some kinetic energy and can jostle around in a random manner.
As there is nothing but empty space between one collision of a gas molecule and another, by Newton's law they would travel in straight lines at constant velocity  if no force is applied. Now there is a force and this is gravity acting downwards but over the very small distance between collisions in a gas in a room, the effect is negligible. Thus we say that they travel in 'straight lines' but remember that this is only an approximation, but a very good one.
(Notes:  The collisions are called 'elastic' which means that no energy is retained within either molecule after the collision compared to that before it. There is also a distribution of velocities, the Maxwell-Boltzmann distribution, or speed the  Maxwell distribution).


text:
Steric factors greater than 1 or less than 0?
From my understanding, the steric factor accounts for the relative orientation required by reactants in order to react in collision theory. This implies that steric factors should probably always be below 1 and above 0, since the relative orientation logically should always be a limiting factor, but below 0 or equal to 0 would mean that the reaction cannot take place.
However, my textbook, as a thinking point, asked if there could be a case where the steric factor is more than 1. Specifically, the example it gave was the harpoon mechanism between potassium and the iodine molecule.
My question is, can the steric factor be above 1 or less than 0, and how does that apply to the harpoon mechanism?

Can you actually have a steric factor less than 0? That would imply a negative rate.

I dont know, thats what Im asking.

Harpoon reactions have a steric factor greater than one.  https://en.wikipedia.org/wiki/Harpoon_reaction

A negative steric factor would imply a negative rate constant which is not possible. In a harpoon reaction a factor greater than 1 simply means that bigger cross-sections are found to be necessary, compared to those normally used,  to fit experiment to theory. Usually there are so many approximations in transition state theory ( geometry of transitions state, activation energy, bond frequencies, friction due to solvent) that steric factors are added as a sort of loose way of fitting calculation to experiment when one has nothing more concrete.

I have also read (on Wikipedia https://en.wikipedia.org/wiki/Collision_theory#Steric_factor, but without reference indicated ) that a greater than 1 steric factor can happen due to the solvent, making so that "several collisions can take place in a single encounter". I don't know how this phenomenon should be balanced with the underestimated cross-section mentioned by @porphyrin.

@The Quark , yes this is the sort of thing that is often stated to rationalise results, but the evidence for such a thing occurring is weak if non-existent. How could one actually measure this for example? Especially as in any reaction with a reasonable activation barrier many thousands to millions of collisions occur before reaction actually occurs.

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text:
What exactly is the use of photographic films in cameras? How are the pictures generated?
By definition:

Photographic film is a strip or sheet of transparent plastic film
  base coated on one side with a gelatin emulsion containing microscopically small light-sensitive silver halide crystals. The
  sizes and other characteristics of the crystals determine the
  sensitivity, contrast and resolution of the film.

But how exactly is a photographic film utilized in a camera?
I know that it has a sort of a silver coating. But how does it work? I read the Wikipedia article but found it a bit difficult to understand.

While it is common to find claims on the Internet that cameras once used a thin layer of matter (or 'film') nanoengineered to capture and record images completely passively and without the use of electronics, the technology to do this is lost, like Atlantis, in the mists of prehistory, if it ever existed. A camera, like the one in your cellphone, uses an array of photosensors to capture images.

A photographic film consists of a light sensitive emulsion of fine particles (grains) of silver salts in gelatine
spread on a clear celluloid strip or a glass plate.
The grain size is very important to photographers, as this
affects the quality of the pictures produced.
AgBr is mainly used as the light sensitive material.
Some AgI is used in "fast" emulsions. The film is placed in a camera. When the photograph is exposed, light from the subject enters the camera
and is focussed by the lens to give a sharp image on the film.
The light starts a photochemical reaction by
exciting a halide ion, which loses an electron.
The electron moves in a conduction band to the surface of the
grain, where it reduces a $\ce{Ag+}$
ion to metallic silver.

$\ce{2AgBr + h\nu \rightarrow 2Ag + Br2}$

In modern photography only a short exposure of perhaps 1/100th of a second is used.
In this short time, only a few atoms of silver (perhaps 10-50) are produced in each grain exposed to light.
Parts of the film which
have been exposed to the bright parts of the subject contain a lot of grains with some silver.
Parts exposed to paler parts of the subject contains a few grains with some silver, whilst parts not exposed contain none.
Thus the film contains a latent image of the subject.
However, the number of silver atoms produced is so small that the image is not visible to the eye.
Next the film is placed a developer solution. This is a mild reducing agent, usually containing quinol.
Its purpose is to reduce more silver halide to $\ce{Ag}$ metal.$\ce{Ag}$ is deposited mainly where there are already some $\ce{Ag}$ atoms. Thus the developing process intensifies the latent image on the film so it becomes visible. The
correct conditions for processing must be used to obtain an image of the required blackness.

If the film was brought out into daylight at this stage, the unexposed parts of the emulsion would turn black
and thus destroy the picture.
To prevent this happening any unchanged silver halides are removed by placing
the film in a fixer solution.
A solution of sodium thiosulphate is used as fixer. It forms a soluble complex with silver halides.

$\ce{AgBr + 2Na2S2O3 \rightarrow Na3 [Ag(S2O3)2] + NaBr}$

After fixing, the film can safely be brought out into daylight. Parts blackened by silver represent the light parts of the original picture. This is therefore a negative.
To obtain an image with light and dark the right way round, a print must be made.
Light is passed through the negative onto a piece of paper coated with $\ce{AgBr}$ emulsion. This is then developed and fixed in the same way
as before.
EDIT: Wikipedia writes:

In black-and-white photographic film there is usually one layer of
  silver halide crystals. When the exposed silver halide grains are
  developed, the silver halide crystals are converted to metallic
  silver, which blocks light and appears as the black part of the film
  negative. Color film has at least three sensitive layers,
  incorporating different combinations of sensitizing dyes. Typically
  the blue-sensitive layer is on top, followed by a yellow filter layer
  to stop any remaining blue light from affecting the layers below. Next
  come a green-and-blue sensitive layer, and a red-and-blue sensitive
  layer, which record the green and red images respectively. During
  development, the exposed silver halide crystals are converted to
  metallic silver, just as with black-and-white film. But in a color
  film, the by-products of the development reaction simultaneously
  combine with chemicals known as color couplers that are included
  either in the film itself or in the developer solution to form colored
  dyes. Because the by-products are created in direct proportion to the
  amount of exposure and development, the dye clouds formed are also in
  proportion to the exposure and development. Following development, the
  silver is converted back to silver halide crystals in the bleach step.
  It is removed from the film during the process of fixing the image on
  the film with a solution of ammonium thiosulfate or sodium thiosulfate
  (hypo or fixer).Fixing leaves behind only the formed color dyes, which
  combine to make up the colored visible image. Later color films, like
  Kodacolor II, have as many as 12 emulsion layers,with upwards of 20
  different chemicals in each layer.

Well you describe B/W film well. Color film is the same sort of thing but has different layers that respond to red/green/blue light like our eyes.

@MaxW Well the answer is about B/W films only...BTW I added about color films too..

@orthocresol Google Images :-P...I lost the links.The third image is from my texbook.The last image is from Wikipedia...

First image from https://tutsplus-media.s3.amazonaws.com/photo.tutsplus.com/uploads/2013/02/iso_05.jpg

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text:
Anomalous oxidation states of Transition Metals

I have noted down the available positive oxidation states of the first
  row of transition elements (on the Periodic Table) from the respective Wikipedia articles of
  the elements.
$\ce{Sc} - 3, 2,1$
$\ce{Ti} - 4, 3, 2, 1 $
$\ce{V} - 5, 4, 3, 2, 1$
$\ce{Cr} - 6, 5, 4, 3, 2, 1$
$\ce{Mn} - 7, 6, 5, 4, 3, 2, 1$
$\ce{Fe} - 6,5,4,3,2,1 $
$\ce{Co} - 5,4,3,2,1 $
$\ce{Ni}-  4,3,2,1$
$\ce{Cu} - 4,3,2,1$
$\ce{Zn} - 2,1$

Why does $\ce{Fe}$ show not show $+7$ or $+8$ oxidation state?
After $\ce{Fe}$ from $\ce{Co}$ why is there a decrease in number of possible    oxidation states?
Why don't $\ce{Fe,Co,Ni,Cu,Zn}$ show higher oxidation states though they have valence electrons left in $3d$ subshell?

For a full picture, you need to add the possible negative oxidation states.

Regarding the maximum oxidation state achievable, there are two competing trends. Both have to do with ionisation energies.
In going to a higher oxidation state, the additional IEs are always balanced against the release in energy in the formation of stronger ionic/covalent bonds. However, if the additional IE is too large, then the favourable bonding will not be sufficient to bring out the higher oxidation state. A simple example is sodium, which cannot adopt the +2 oxidation state because its second IE is simply too large, even though the lattice energy of a hypothetical $\ce{NaCl2}$ crystal is larger than that of $\ce{NaCl}$.
Firstly, the maximum oxidation state is limited by the number of valence electrons available. This is exactly analogous to the case of sodium; therefore, manganese does not exhibit the +8 oxidation state, because its eighth IE involves ionisation from the 3p subshell, much lower in energy than either the 3d or 4s. This therefore accounts for the increase in the maximum oxidation state going from Sc to Mn.
Secondly, the ionisation energies generally increase going across the 3d block, because of an increase in effective nuclear charge. This is the factor which leads to the decrease in the maximum oxidation state after Mn. For example, the seventh IE of Fe is larger than that of Mn; therefore, Fe does not exhibit the +7 oxidation state.
You can look up the IEs in any relevant textbook. However, note that since the +2 oxidation state is very common in the later half of the 3d block, it is usually the third and higher IEs that are more important in this discussion. So, looking at the trend of first/second IEs won't get you anywhere.

But that's not a good enough reason.You are justifying using IE.But what if you did'nt have access to I.E. values.Then how would you justify the sudden decrease in available oxidation states after $\ce{Fe}$ just by using the electronic configurations ?

You don't need to look up the IE values to know that there is an increasing trend.

You did not understand my question.Your answer does not explain why is there a decrease in available oxidation states.Say,why doesnt Fe show +7 oxidation state,why doesnt Co show +6 oxidation state though electrons are left in 3d? I know you can just look up I.E. values but that is not a reason.

Because to get cobalt to the +6 oxidation state, you need to remove six electrons. Six electrons are too hard to remove in cobalt, because there is a large effective nuclear charge, and the ionisation energies are large, despite the electrons still being in the "valence" 3d orbital. In fact, that's exactly what I wrote in my answer.

@ZOZ please use correct spacing before and after punctuation marks. No space before one, but one space when you're beginning a new sentence.

@Rubisco I never knew of that.I thought that was MS-Word style :-P.Anyway I will follow it in future if you say so...

"Iron does not exhibit the +7 oxidation state". Lu et al., https://doi.org/10.1039%2FC6CP06753K. But yes, we would not see iron(VII) in normally encountered laboratory settings.

In some cases what's anomalous is not only the number but the environment in which the oxidation state is realized. We might expect high oxidation states to be realized by combining the element with oxide or fluoride ligands, but for instance one of the most widely used lead(IV) compounds for many years was $\ce{Pb(C2H5)4}$.
Palladium(V) offers an even more extreme example. Silicon as such has low electronegativity, but the molecular orbital structure of silyl ligands makes them more electronegative and metals bonded to such ligands are considered to be at the positive end of such bonds. Palladium is combined with five silyl ligands to give the $\ce{Pd(V)}$ species identified in Ref. 1.
Reference

Shimada, Shigeru; Li, Yong-Hua; Choe, Yoong-Kee; Tanaka, Masato; Bao, Ming; Uchimaru, Tadafumi (2007). "Multinuclear palladium compounds containing palladium centers ligated by five silicon atoms". Proceedings of the National Academy of Sciences. 104 (19): 7758–7763. https://doi.org/10.1073/pnas.0700450104.

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text:
How would I measure the dissolved carbon dioxide content of a liquid (water to start with)
I would like to measure the dissolved carbon dioxide content of a sample of water.  I was thinking that conductivity might change with carbon dioxide level, and also pH.  I would like to have the measurement sample electronically, so current, voltage, conductivity / impedance (resistance, capacitance, inductance) would all be fine.  My chemistry goes up to A-level; I am an engineer.

This paper looks helpful: http://www.sciencedirect.com/science/article/pii/S0144860910000683
Quoting from the abstract:

Many methods are available for the measurement of dissolved carbon dioxide in an aqueous environment. Standard titration is the typical field method for measuring dissolved CO2 in aquaculture systems. However, titrimetric determination of dissolved CO2 in marine water aquaculture systems is unsuitable because of the high dissolved solids, silicates, and other dissolved minerals that interfere with the determination. Other methods used to measure dissolved carbon dioxide in an aquaculture water included use of a wetted CO2 probe analyzer, standard nomographic methods, and calculation by direct measurements of the water's pH, temperature, and alkalinity. The determination of dissolved CO2 in saltwater based on partial pressure measurements and non-dispersive infra-red (NDIR) techniques with a CO2 gas analyzer are widely employed for oceanic surveys of surface ocean CO2 flux and are similar to the techniques employed with the head space unit (HSU) in this study. (...)


text:
Water and Ice - density
We know that ice has a lower density than water despite both having the same [molecular] mass. I know that as water turns to ice, it expands. As far as I was taught, I know that it has something to do with hydrogen bonding. I would like to know in terms of hydrogen bonding, why ice actually expands?

Note that water is far from being the only compound that has a liquid with higher density than the solid at the melting point. Elements such as Si, Ge, Sb, and Bi do as well, and they certainly have nothing to do with hydrogen bonding. The only real certainty is that a first order phase transition must have a volume change.

I'll point out that chemists use "water" in two different ways. First it just means the chemical with the formula $\ce{H2O}$. Second it can mean more specifically the liquid state of $\ce{H2O}$. Also planetary scientists refer to a number of compounds as various kinds of "ice" that would coat the surface of a planet.

The most straightforward answer is that water in the form of ice I (or $\ce{I_h}$) has an open hexagonal structure (see figure) whereas liquid water does not. In ice the oxygens are precisely tetrahedrally positioned where each oxygen is hydrogen bonded by four neighbouring oxygens with an O..O distance of approx 0.28 nm. The H atoms lie very closely along the O-O axes. Two of the H bonds are short, as in liquid water, and two are long. This is expected from residual entropy calculations and also from thermal and neutron diffraction experiments.
As this ice structure has a rather open lattice this is the reason its density is less at its freezing point than that of liquid water, rather than being due to the high density of water. One could argue that the reason that water density is a maximum at 4 $\ce{^0C}$ (277 K) is that the structure of  liquid water is changing from that of ice $\ce{I_h}$ to the short range structure of the liquid.
Liquid water is, of course, also highly hydrogen bonded but the extra energy it contains means that its structure is far less regular and so becomes slightly denser than ice. As the water molecules can move about more in the liquid than in ice, the hydrogen bonds will bend a little or break as molecules rotate. This will lead to a shorter O - O distance and hence slightly higher density in liquid than in the solid.
see also answers to this post Why really is ice less dense than water?

This figure shows the hexagonal ice structure; The O atoms are shown as dots and the lines are the (almost linear) O-H-O bonds.(Image from Murrell & Jenkins 'Properties of liquids & solutions')
(Ice also exhibits eight different phases depending on temperature  and pressure of which $\ce{I_h}$ is the one found at normal atmospheric pressure. In these other phases, angles are distorted from being tetrahedral)

Could you explain the structure of water a bit more? How irregular is it?

Straight from Wikipedia...

The density of water is about 1 gram per cubic centimetre (62 lb/cu ft): this relationship was originally used to define the gram. The density varies with temperature, but not linearly: as the temperature increases, the density rises to a peak at 3.98 °C (39.16 °F) and then decreases. This unusual negative thermal expansion below 4 °C (39 °F) is also observed in molten silica. Regular, hexagonal ice is also less dense than liquid water—upon freezing, the density of water decreases by about 9%.
These effects are due to the reduction of thermal motion with cooling, which allows water molecules to form more hydrogen bonds that prevent the molecules from coming close to each other. While below 4 °C the breakage of hydrogen bonds due to heating allows water molecules to pack closer despite the increase in the thermal motion (which tends to expand a liquid), above 4 °C water expands as the temperature increases. Water near the boiling point is about 4% less dense than water at 4 °C (39 °F).
Other substances that expand on freezing are acetic acid, silicon, gallium, germanium, antimony, bismuth, plutonium and also chemical compounds that form spacious crystal lattices with tetrahedral coordination.

First, this is basically a plagiarising answer as it builds entirely off its source. Second, this does not answer the question why water is more dense than ice.

@Jan - the why - "These effects are due to the reduction of thermal motion with cooling, which allows water molecules to form more hydrogen bonds that prevent the molecules from coming close to each other." // I said explicitly that I lifted the answer from Wikipedia. The whole point was that I thought the OP should do some simple research.

@MaxW It's ultimately preferable to use the external text/graphics to illuminate your own answer. I am confident that you have more you can add to the answer than just the raw explanation.


text:
Energy balances for a heated reactor
I have a heated mixing reactor, as seen in the figure.
How to set up the energy balance of the heating jacket and the reactor?

The flow rate of the heating fluid $F_\mathrm j$ probably refers to the mass flow rate (e.g. expressed in $\mathrm{kg/s}$) since the indicated flow rates into the heat exchanger and out of the heat exchanger are identical. (The corresponding values for the volume flow rate (e.g. expressed in $\mathrm{m^3/s}$ would be different since the change in temperature causes a difference in density.)
$$F_\mathrm j=\frac mt\tag{1}$$
The change in temperature of the heating fluid $\Delta T$ is given by the different temperatures of heating fluid before and after the heat exchanger:
$$\Delta T=T_\mathrm j-T_\mathrm{ji}\tag{2}$$
The ratio of the amount of heat energy $Q$ (e.g. expressed in $\mathrm{J}$) that is transferred to a system at constant pressure and the resulting increase in temperature $\Delta T$ is given by the heat capacity at constant pressure $C_p$ (e.g. expressed in $\mathrm{J/K}$):
$$C_p=\frac Q{\Delta T}\tag{3}$$
You may know the value of the specific heat capacity at constant pressure $c_p$ (e.g. expressed in $\mathrm{J/(kg\cdot K)}$) for the heating fluid at the average temperature of the heat exchanger, which is defined as heat capacity at constant pressure divided by mass $m$:
$$c_p=\frac{C_p}m\tag{4}$$
Now you can rearrange the definition of specific heat capacity $\text{(3)}$ to solve for the heat $Q$:
$$Q=C_p\cdot\Delta T\tag{5}$$
insert the result for $C_p$ from equation $\text{(4)}$
$$Q=c_p\cdot m\cdot\Delta T\tag{6}$$
the result for $\Delta T$ from equation $\text{(2)}$
$$Q=c_p\cdot m\cdot\left(T_\mathrm j-T_\mathrm{ji}\right)\tag{7}$$
and the result for $m$ from equation $\text{(1)}$
$$Q=c_p\cdot F_\mathrm j\cdot t\cdot\left(T_\mathrm j-T_\mathrm{ji}\right)\tag{8}$$
Therefore, the thermal power $P$ of the heat exchanger (e.g. expressed in $\mathrm{W=J/s}$) is
$$\begin{align}P&=\frac Qt\\[6pt]
&=c_p\cdot F_\mathrm j\cdot\left(T_\mathrm j-T_\mathrm{ji}\right)\tag{9}\end{align}$$
The sign of the values for $Q$ and $P$ depends on whether heat is transferred from the fluid to the reactor or from the reactor to the fluid.
Note that the specific heat capacity $c_p$ generally depends on temperature; therefore, the specific heat capacity of the fluid changes during the passage of the heat exchanger. That is why you should use an appropriate average value. A precise engineering calculation would better be based on the specific enthalpy $h$ of the fluid at the given temperature and pressure before and after the heat exchanger.

Example
The characteristic data of a typical nuclear power plant with a pressurized water reactor are

Coolant mass flow rate $F=18\,800\ \mathrm{kg\ s^{-1}}$
Coolant pressure $p=158\ \mathrm{bar}=15.8\ \mathrm{MPa}$
Reactor pressure vessel inlet temperature $T_\text{in}=291\ \mathrm{^\circ C}$
Reactor pressure vessel outlet temperature $T_\text{out}=326\ \mathrm{^\circ C}$

The specific heat capacity of water at the inlet temperature of $T_\text{in}=291\ \mathrm{^\circ C}$ and a pressure of $p=15.8\ \mathrm{MPa}$ is $c_{p,\text{in}}=5.2553\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$. The corresponding specific heat capacity at the outlet temperature of $T_\text{out}=326\ \mathrm{^\circ C}$ is $c_{p,\text{out}}=6.4388\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$. For simplicity’s sake, we may want to use the specific heat capacity for the average temperature $\tfrac12\left(T_\text{in}+T_\text{out}\right)=308.5\ \mathrm{^\circ C}$, which is $c_p=5.6787\ \mathrm{kJ\ kg^{-1}\ K^{-1}}$.
Using equation $\text(9)$, the thermal power of the reactor can be calculated as
$$\begin{align}P&=c_p\cdot F\cdot\left(T_\text{out}-T_\text{in}\right)\\[6pt]
&=5.6787\ \mathrm{kJ\ kg^{-1}\ K^{-1}}\times18\,800\ \mathrm{kg\ s^{-1}}\times\left(326\ \mathrm{^\circ C}-291\ \mathrm{^\circ C}\right)\\[6pt]
&=3736584.6\ \mathrm{kJ\ s^{-1}}\\[6pt]
&\approx3737\ \mathrm{MW}
\end{align}$$
By way of comparison, the literature value for the reactor core thermal power is $3765\ \mathrm{MW}$.
Note that, in this example, heat is transferred from the reactor to the fluid.


text:
Definition of Nucleophilicity
I have a rather quick question on the definition of Nucleophilicity. To get straight to the point, am I right in saying that the most appropriate definition of Nucleophilicity, regardless of whether the reaction takes place in solvent or in the gas phase, is the following; Nucleophilicity is a measure of how close in energy the HOMO of a Lewis Base is to the LUMO of a Lewis Acid?

As far as I remember from my lectures nucleophilicity has the same definition problems as electronegativity.
So unlike e.g. ionisation energy you cannot define an absolute scale of ordering which you can verify experimentally
and just as in the case of electronegativity there are several possible definitions (compare https://en.wikipedia.org/wiki/Nucleophile).
Similar to electronegativity the definitions arrive on quite similar results. ($\ce{Br^-}$ is a good nucleophile in polar, protic solvents...)
Other formulations should be "qualitatively consistent" with previous ones.
Your alternative formulation has two big problems:

There is no solvent dependency.
You completely omit steric effects and nucleophilicity is also a kinetic phenomenon.

The ability  for an atom or ion or molecule to give away extra electron and form bond with nucleus or some other is known as nucleophilicity


text:
Reduction of carbonyl groups
In such reactions as in the pic, how can we know which carbonyl group will undergo a reaction to give an alcohol. Also, can both of the sites undergo the reaction?

R-O-R' is an ether And R-OO-R' is an ester. Am I right?

Sodium borohydride is a weaker reducing agent than, say, lithium aluminium hydride, so, although it will react with ketones and aldehydes, it is not strong enough to react with carboxylic acids or esters. This generalization is quite broadly applicable, so you already know your borohydride will reduce the ketone carbonyl on that molecule. $\ce{NaBH4}$ is a weaker reducing agent than $\ce{LiAlH4}$ because boron is less electropositive than aluminium. Boron has a Pauling electronegativity of 2.04 while aluminium's pauling electronegativity is only 1.61, therefore $\ce{NaBH4}$'s hydrides are significantly less nucleophilic and they won't be able to react with a weaker electrophile.
To answer your question directly, you have to think about which carbonyl is more electrophilic. Generally, resonance effects are more important than inductive effects when considering something's nucleophilicity/electrophilicity.

The ketone has two resonance forms, while the ester has three, therefore the partial positive charge on the carbon is more delocalized on the ester, making the ketone more electrophilic. The weak $\ce{H-}$ nucleophile provided by $\ce{NaBH4}$ will only be able to react with the stronger electrophile, the ketone.

What if it were two ketones?

No reason it couldn't reduce them both.

The key difference between the carbonyl groups is that one is a ketone and the other is an ester.  Esters are not generally able to be reduced by sodium borohydride, while ketones can be.
The key to understanding the different reactivity is being able to understand the different functional groups.


text:
What is meaning of unit (mg-Pt/cm^2) in relation to fuel cells?
I do not understand what the meaning is of the unit
mg-$\ce{Pt}$/cm$^2$
which I have come across in reading about the theory of fuel cells.
In simple terms, what does this unit mean?

Milligrams of platinum per square centimeter.

This isn't really homework, is it close voters? It's just a misguided asker.

The unit in question means milligrams of platinum per square centimeter.
In your application - fuel cells, and specifically the membranes used to fabricate them - it describes how much of the metal is present per square centimeter of membrane.


text:
Why is the yield of substitution reactions decreased at higher temperatures?
This is according to my text :

Increasing temperature favours elimination over substitution. This is because energy of activation for elimination is higher than that of substitution since higher degree o charges in bonding is involved in elimination reaction. [sic!]

Well I understand why more elimination occurs at high temperatures (I guess this is due to $\Delta G$ being more negative due to multiple product formation as well), but why do substitution reactions have their yields decreased as temperature goes up?
This is according to the master of organic chemistry; (Even though it says that there is a relative increase, the graph shows a decrease in yield of the substituted products):

Aren't they supposed to have their output increased as well, as the activation energy barrier has already passed and rate increases with increase in temperature, or am I missing something fundamental here?

Related: https://chemistry.stackexchange.com/questions/76343/why-substitution-and-elimination-reactions-are-favored-respectively-at-lower-and

In short, this is because both elimination and substitution occur parallely and are competing; amount of elimination product + substitution product = constant. If the elimination rate increases more than the increase in substitution rate, then the ratio of elimination product to substitution product will increase. This is what happens on heating, though the rate of both elimination and substitution increase with temperature.
Broadly, there are two types of reactions: those that are kinetically controlled, and those that are thermodynamically controlled. Most elimination-substitution reactions are exothermic and are kinetically controlled. That means that the proportion of products (elimination or substitution) depends only on the rate of elimination and substitution, and not on the thermodynamic favour for the reaction.
Given that the reaction proceeds to completion, the ratio of elimination product to substitution product is equal to the ratio of elimination rate to substitution rate. Suppose elimination and substitution rates are equal, then the products form in a 50:50 proportion.
The ratio of elimination rate to substitution rate is
$$
\frac{e^{\frac{-\Delta G_{e}^{*}}{RT}}}{e^{\frac{-\Delta G_{s}^{*}}{RT}}} = e^{\frac{\Delta G_{s}^{*}-\Delta G_{e}^{*}}{RT}}
$$
where $ \frac{-\Delta G_{e}^{*}}{RT}$  and  $\frac{-\Delta G_{s}^{*}}{RT}$ are the energy barriers for elimination and substitution respectively.
Since the barrier for elimination is more than the barrier for substitution, the above ratio increases with increase in temperature, and hence more elimination product is formed at higher temperatures.

What do you mean by 'constant'? Are you saying that the net yield of products don't change irrespective of the temperature? I do not understand.

@user73157 , The yield does change with the temperature, but only slightly.
Suppose you start with 1 mole of reactant & run the reaction at 80 degrees C, suppose the yield is 90% & elim:subst ratio is 50:50. We'll get 0.45 mol elim & 0.45 mole subst product.
Suppose we run the reaction at 100 degrees C, the yield may be 85%, but elim:subst ratio may be 80:20. We'll get 0.68 mol elim & 0.17 mol subst product.
The yield does change with temperature, but this effect is (usually) much smaller than the difference in amount of products due to difference in their activation barriers.

Thank you for the explanation! I have now understood it.


text:
Addition of hydrogen bromide to 1,3-butadiene: thermodynamic and kinetic control
The reaction of one equivalent of hydrogen bromide with 1,3-butadiene gives different products at under different conditions:1

The addition of hydrogen chloride also gives rise to similar products.2 I can see that the low-temperature product is formed via an electrophilic addition, with Markovnikov regioselectivity. But how is the other product formed, and why does the major product differ with temperature?

References
1 Kharasch, M. S.; Margolis, E. T.; Mayo, F. R. The peroxide effect in the addition of reagents to unsaturated compounds. XIII. The addition of hydrogen bromide to butadiene. J. Org. Chem. 1936, 01 (4), 393–404. DOI: 10.1021/jo01233a008.
2 Kharasch, M. S.; Kritchevsky, J.; Mayo, F. R. The addition of hydrogen chloride to butadiene. J. Org. Chem. 1937, 02 (5), 489–496. DOI: 10.1021/jo01228a010.

Thermodynamic and kinetic control
This is a classic example of the concept of thermodynamic versus kinetic control of a reaction. Take a look at this energy profile diagram.1 The horizontal axis is a reaction coordinate, and the vertical axis represents Gibbs free energy.

In this scenario, the starting material $\ce{A}$ can react to form either $\ce{B}$ or $\ce{C}$. The formation of the product $\ce{C}$ has a lower activation energy, which means that it will form faster:
$$E_{\mathrm{a},\ce{AB}} > E_{\mathrm{a},\ce{AC}} \qquad \Longrightarrow \qquad k_\ce{AB} < k_\ce{AC}$$
If we keep the temperature sufficiently low, the molecules of $\ce{C}$ - which are inevitably formed faster - will probably not have enough energy to overcome the reverse activation barrier. The forward reactions $\ce{A->B}$ and $\ce{A->C}$ are, under such conditions, effectively irreversible. Since the formation of $\ce{C}$ is faster, it will predominate, and the major product formed will be $\ce{C}$. This is known as kinetic control; $\ce{C}$ is the kinetic product.
Now, what happens if we heat it up? At higher temperatures, $\ce{C}$ is still going to be the product that is formed faster. However, it also means that all the reactions will be reversible. This means that molecules of $\ce{C}$ can revert back to $\ce{A}$ - and since the system is no longer limited by temperature, it will try its best to minimise its Gibbs free energy, which is the thermodynamic criterion for chemical equilibrium. This means that, as the most thermodynamically stable molecule, $\ce{B}$ will be predominantly formed.2 The reaction is said to be under thermodynamic control, and $\ce{B}$ is the thermodynamic product.
A simple definition is that the kinetic product is the product that is formed faster, and the thermodynamic product is the product that is more stable. This is precisely what is happening here. The kinetic product is 3-bromobut-1-ene, and the thermodynamic product is 1-bromobut-2-ene (specifically, the trans isomer).
A disclaimer
Note that not every reaction has an energy profile diagram like this, and not every reaction has different thermodynamic and kinetic products! If the transition state leading to the formation of $\ce{C}$ were to be higher in energy than that leading to $\ce{B}$, then $\ce{B}$ would simultaneously be both the thermodynamic and kinetic product. There are plenty of reactions in which the more stable product (thermodynamic) is also formed faster (kinetic), so do not assume that every reaction fits this mould!
The reaction mechanism
The first step is the protonation of one of the $\ce{C=C}$ double bonds. In butadiene (1), both double bonds are the same, so it does not matter which one you protonate. The protonation occurs regioselectively to give the more stable carbocation:

The more stable cation is not only secondary, but also allylic, and therefore enjoys stabilisation via resonance (or conjugation). This is depicted in the resonance forms 2a and 2b above.
This allylic carbocation, more properly denoted as the resonance hybrid 2, has two carbons which have significant positive charge, and the bromide ion (here denoted as $\ce{X-}$) can attack either carbon. Attacking the central carbon, adjacent to the site of protonation, leads to the kinetic product 4; attacking the terminal carbon, distant from the site of protonation, leads to the thermodynamic product 4.

There are some people who write that 3 results from attack of $\ce{X-}$ on resonance form 2a, and 4 from attack of $\ce{X-}$ on resonance form 2b. This is not correct! Resonance forms do not separately exist, and they are not distinct species that rapidly interconvert. As such, one cannot speak of one single resonance form undergoing a reaction.
With that out of the way, let's examine why 4 is the thermodynamic product, and why 3 is the kinetic product.
The thermodynamic product: trans-1-bromobut-2-ene
It is perhaps simple enough to see why 4 is more stable than 3. It has an internal, disubstituted double bond, and we know that as a general rule of thumb, the thermodynamic stability of an alkene increases with increasing substitution. So, compared to the terminal, monosubstituted alkene 3, 4 is more stable.

Both the trans isomer 4 as well as the cis isomer 5 can be formed via attack of the nucleophile at the terminal carbon, and both are disubstituted alkenes. However, the trans isomer 4 is more stable than the cis isomer 5, because there is less steric repulsion between the two substituents on the double bond. As such, 4 is the thermodynamic product.
The kinetic product: 3-bromobut-1-ene
There are quite a few explanations on the Internet,3–6 and not all of them are correct.
The worst possible argument, which I have thankfully not seen yet, goes something along the lines of: resonance form 2a, being an allylic secondary carbocation, is more stable than resonance form 2b, which is an allylic primary carbocation. Therefore, resonance form 2a exists in greater proportion, and the nucleophile preferentially reacts with it, leading to the formation of 3.
However, this is patently incorrect, as discussed earlier! Individual resonance forms do not exist, and on top of that, such an argument suggests that we are looking for the more stable intermediate. In fact, we should be looking for the more stable transition state. The carbocation is an intermediate, and not a transition state.
The most common argument that I have seen is actually: since resonance form 2a is more stable than 2b, it contributes more towards the resonance hybrid 2. As such, the positive charge on the internal carbon is greater than the positive charge on the terminal carbon. The nucleophile, being negatively charged, is more strongly attracted to the more positively charged or more electrophilic carbon, and therefore attack there occurs faster (the transition state being stabilised by greater electrostatic interactions).
That's actually a very sensible explanation; with only the data that has been presented so far, we would not be able to disprove it, and it was indeed the accepted answer for quite a while.
The catch
In 1979, Nordlander et al. carried out a similar investigation on the addition of $\ce{DCl}$ to a different substrate, 1,3-pentadiene.7
This experiment was very ingenious, because it was designed to proceed via an almost symmetrical intermediate:

Resonance forms 7a and 7b are both allylic and secondary. There is a very minor difference in their weightages, arising from the different hyperconjugative ability of $\ce{C-D}$ vs $\ce{C-H}$ bonds,8 but in any case, it is not very large. Therefore, if we adopt the explanation in the previous section, one would expect there not to be any major kinetic pathway, and both 1,2- and 1,4-addition products (8 and 9) would theoretically be formed roughly equally.

Instead, it was found that the 1,2-addition product was favoured over the 1,4-addition product. For example, at $-78\ ^\circ\mathrm{C}$ in the absence of solvent, there was a roughly $75:25$ ratio of 1,2- to 1,4-addition products. Clearly, there is a factor that favours 1,2-addition that does not depend on the electrophilicity of the carbon being attacked!
The authors attributed this effect to an ion pair mechanism. This means that, after the double bond is protonated (deuterated in this case), the chloride counterion remains in close proximity to the carbocation generated. Immediately following dissociation of $\ce{DCl}$, the chloride ion is going to be much closer to $\ce{C-2}$ than it is to $\ce{C-4}$, and therefore attack at $\ce{C-2}$ is much faster.
In fact, normal electrophilic addition of $\ce{HX}$ to conjugated alkenes in polar solvents can also proceed via similar ion pair mechanisms.9,10 This is reflected by the greater proportion of syn addition products to such substrates.11

Notes and references
1 Smith, M. B. March's Advanced Organic Chemistry, 7th ed., p 272
2 This does not mean that all of $\ce{A}$ will be converted to $\ce{B}$; the reaction is still an equilibrium, and equilibria always go forward and backward. In general, the minimum system Gibbs free energy ($G_\mathrm{syst}$) will occur at a certain proportion of $\ce{A}$, $\ce{B}$, and $\ce{C}$. However, since $\ce{B}$ has the lowest Gibbs free energy, it will be formed in a greater proportion than $\ce{C}$. See, for example, the colourful graph in this answer of mine.
3 Thermodynamic and Kinetic Control. http://www.masterorganicchemistry.com/2012/02/09/can-opener-economics/ (accessed August 8, 2017).
4 Addition of Hydrogen Halides to Dienes. http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch10/ch10-4-1.html (accessed August 8, 2017).
5 Electrophilic Attack on Conjugated Dienes-Kinetic and Thermodynamic Control. https://chem.libretexts.org/Core/Organic_Chemistry/Conjugation/Electrophilic_Attack_on_Conjugated_Dienes/Electrophilic_Attack_on_Conjugated_Dienes-Kinetic_and_Thermodynamic_Control (accessed August 8, 2017).
6 1,4-Addition. http://www.ochempal.org/index.php/alphabetical/a-b/14-addition/ (accessed August 8, 2017).
7 Nordlander, J. E.; Owour, P. O.; Haky, J. E. Regiochemistry of the addition of hydrochloric acid-d to trans-1,3-pentadiene. J. Am. Chem. Soc. 1979, 101 (5), 1288–1289. DOI: 10.1021/ja00499a045.
8 Because of the larger reduced mass and lower zero-point energy, a $\ce{C-D}$ bond is stronger and therefore less willing to donate electron density into an adjacent empty $\mathrm{p}$ orbital. This is the origin of some secondary kinetic isotope effects; in our case, it means that 7a is marginally less stable than 7b.
9 Fahey, R. C.; McPherson, C. A. Mechanism of the hydrochlorination of tert-butylethylene and styrene in acetic acid. J. Am. Chem. Soc. 1969, 91 (14), 3865–3869. DOI: 10.1021/ja01042a030.
10 Addition of $\ce{HX}$ to butadiene in the gas phase gives approximately a $1:1$ ratio of 1,2- to 1,4-addition product, suggesting that an ion pair mechanism (which would favour the 1,2-addition product) does not operate. See: Mascavage, L. M.; Chi, H.; La, S.; Dalton, D. R. Surface-catalyzed hydrochlorination of alkenes. The reaction of the gases hydrogen chloride and 1,3-butadiene. J. Org. Chem., 1991, 56 (2), 595–601. DOI: 10.1021/jo00002a021.
11 For more details and references to primary literature, see Carey, F. A., Sundberg, R. J. Advanced Organic Chemistry - Part A: Structure & Mechanisms, 5th ed., pp 478–482.

Welcome! Can you please link to a resource to understand the ion-pair mechanism in a little bit more detail? Not PhD level detail but some basic concept for a high school student is sufficient. Thank you! (I don't have access to the research papers you linked in the footnote :( and I'm sure I would not be able to comprehend them even if I had access :P

@GaurangTandon there's some discussion in Carey/Sundberg 5th ed., Part A, p 479 onwards (dienes are discussed on p 481). Apart from that I'm not sure. Most undergrad texts won't go into the subtleties of the mechanisms (the exact details depend heavily on exact substrate, temperature, solvent, etc. and it's almost impossible to explain everything).

Roytman and Singleton (J. Am. Chem. Soc. 2020, 142, 29, 12865–12877. https://doi.org/10.1021/jacs.0c06295) assert that a more nuanced consideration of the ion pair explanation is in order.


text:
Rule of mutual exclusion: is the converse true?
There is a "famous" rule in spectroscopy,1 that goes this way:

If a compound is centrosymmetric, then its normal vibrational modes cannot be simultaneously IR and Raman active.

and this is simple enough to prove: for a vibrational mode to be IR active, it has to transform under the same irrep as (one of) the cartesian axes $x$, $y$ or $z$, which are all ungerade. For a vibrational mode to be Raman active, it has to transform under the same irrep as one of $(x^2, y^2, z^2, xy, yz, xz)$, and since $u \otimes u = g$, these irreps cannot be the same as the irreps of $(x,y,z)$. Hence, a vibrational mode, which transforms under a single irrep, cannot simultaneously satisfy both conditions at once.
However, I am curious about the converse statement.

If a compound does not have any vibrational modes which are simultaneously IR and Raman active, is it necessarily centrosymmetric?

If the answer is yes, I would like a proof; if it is no, a counterexample, please.
(Just in case anybody is confused, because I am: A counterexample would be a molecule without a centre of inversion, which has no vibrational mode which is simultaneously IR and Raman active.)

1 Famous enough for me to know, at least.

I just went through a number of common molecule point groups and in all of them that were note inversion-symmetric there was at least one irreducible representation that included a simple translational and a product mode. Looks like if there is a counterexmple it will be an extremely rare case.

Have a look at the point groups O (not $\ce{O_h}$) and $\ce{D_{5h}, C_{5h}}$. I will try and find a pic of O point group molecules.

Have a look at the point groups O (not $\ce{O_h}$) I, $\ce{D_{5h} and C_{5h}}$. A pic of an O point group molecule is shown below. The $\ce{c_2}$ axes are not shown. Taken from molecule-viewer.com web site. The molecule is 'vanadium hexaoxo phosphonato' or more properly $\ce{C8H24O30P8V^+Cl^-}$, octakis(μ 3-methylphosphonato)-hexaoxo-penta- vanadium(v)-vanadium(iv) chloride clathrate !!!

Edit: I checked a bit more and found ; $\ce{ S8 }$

One reference says "Number of Vibrational Active IR Bands: Only Rx, Ry, Rz, x, y, and z can be ir active" which would seem to rule out D5h and C5h.  http://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Spectroscopy/Vibrational_Spectroscopy/Vibrational_Modes/Mode_Analysis another reference oppositely says "is Raman active if the symmetry is the same as
x2, y2, z2, or one of the rotational functions Rx, Ry, Rz" http://www.lasalle.edu/~prushan/IC-articles/lecture%204.pdf which would exclude O and I.

D5h, C5h, O and I are all fair counter examples if OP's statement "for a vibrational mode to be IR active, it has to transform under the same irrep as (one of) the cartesian axes x, y or z, which are all ungerade. For a vibrational mode to be Raman active, it has to transform under the same irrep as one of (x2 ,y2 ,z 2 ,xy,yz,xz)" is completely right, but not if one of those references is correct.

hi, Rx, Ry Rz do not refer to vibrations but to 'rotational' operators such as spin-orbit coupling and whole body rotations

If there is a molecule of $O$ (as opposed to $O_h$) or $I$ (as opposed to $I_h$) symmetry, it would be such a counter example (no mode that is both IR and raman active, and not centrosymmetric).
http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=903&option=4
http://symmetry.jacobs-university.de/cgi-bin/group.cgi?group=905&option=4
Apparently a snub cube is an example of the $O$ point group and in Chemistry for the 21st Century it is stated:

9.5.2.5 Snub Cube (6) We have recently demonstrated the ability of six resorcin[4]arenes and eight water molecules to assemble in apolar media to form a spherical molecular assembly which conforms to a snub cube (Fig. 9.3)


text:
Concentration of electrolyte in rust removal
I will be attempting rust removal from a steel gas tank of about 5 liters by means of electrolysis, with steel wire acting as the anode.
For the electrolyte I've been advised to dissolve $\ce{Na2CO3}$ in water. Approximately what should the concentration be? Just looking for a ballpark estimate.

I've seen 4 grams per liter (roughly 1 tablespoon per gallon for my countrymen and the 15 remaining people in the world still not on the metric system).  You might find Rust Removal Using Electrolysis of interest and/or helpful. Good luck!

The stronger the concentration, the higher the current draw.  I prefer to increase the concentration in increments so I keep current draw within the limits of my power supply. It's not a race.


text:
Group of Potassium and It's Oxidation Number
One of the questions I got wrong on a chemistry Kahoot question was
(1) Potassium is in group $1$, what is its oxidation number?
(a) -1
(b) +1
(c) +2
(d) -2

I don't remember what an "oxidation number" is... And what do they mean by group $1$? I don't quite remember learning that.

All about potassium at Wikipedia https://en.wikipedia.org/wiki/Potassium Here is link to Wikipedia article on the periodic table. Notice group numbers along top of first figure. https://en.wikipedia.org/wiki/Periodic_table

Hm... Our teacher said that the number above was the number of valence electrons the elements had. (Paying only attention to the units place)... So a group number just tells you the number of electrons in the outer shell?

Potassium's oxidation number is +1, oxidation numbers are just hypothetical charges that an atom would have if the compound is ionic(This is not always the case),but it's useful for keeping track of electrons. In this case it means that the Potassium atom losses one electron when forming compounds. Chemical elements are arranged in groups where the number with the letter(e.g.,1A,2A,3B,etc.) shows how many valence electrons an atom has, Group 1A(The first column to the left on the periodic table) has one electron on its valence shell so it tends to lose that electron in order to get a noble gas's electronic configuration and that makes sense given Potassium's oxidation number.

Wait, so if Potassium wants to lose an electron, shouldn't it be -1? What does oxidation number mean?

Oxidation Numbers are like charges, when a neutral atom (Same number of protons and electrons) loses one electron there would be one proton in excess so the charge becomes +1; on the other hand if an atom gains one electron there would be one electron in excess so the charge becomes -1.

Oh, so when a question asks for an element's oxidation number, it's really just asking for the element's charge?

Kind of, you can find oxidation numbers of each element following certain rules such as oxygen when forming oxides has an oxidation number of -2 mostly,all alkali elements have an oxidation number of +1,etc. The sum of all oxidation numbers gives you the total charge of the compound(0 if it's neutral)

There you have the tendencies for oxidation numbers of each element http://www.thecatalyst.org/oxnotabl.html


text:
What is reaction between isopropyl alcohol and acetone?
What would be the probable reaction between acetone, $\ce{(CH3)2CO}$, and isopropyl alcohol, $\ce{CH3CHOHCH3}$, at room temperature without any catalyst in different concentrations?

There would be none.

As Ivan already said there would be no reaction. You might as well have any ketone with any alcohol - the only thing that could conceivably happen would be hemiacetal/acetal formation but that certainly needs acid catalysis.

Unless you excite the acetone (UV irradiation), I'd suppose that nothing happens.

If something happened, you might be in real trouble washing out a reaction vessel with isopropanol as solvent at the end of the day...

with higher temperature , i think oxygen atom of hydroxide group give her lone pair to ketonic group of acetone with 1,3 proton shift give isopropyl isopropanol ether

I've used acetone/ipa mix to flush chromatography columns. There is no reaction between them without another reagent present

I mix rubbing alcohol and acetone together to clean flux off of stained glass that I have soldered. Nothing happens when you mix them.

Obviously these liquids do not react with each other.  But they can be interconverted.  Isopropanol can be converted to acetone simply by heating over a copper catalyst (https://en.wikipedia.org/wiki/Isopropyl_alcohol).  And as many readers know, sodium borohydride or lithium aluminum hydride reduces acetone to isopropanol (https://www.chemguide.co.uk/organicprops/carbonyls/reduction.html).


text:
Preparation of Mercurous nitrite
I read that Prafulla Chandra Ray prepared Mercurous nitrite for the first time and published a paper.
Both Mercurous ions and nitrite ions are quite stable. So they should readily form the compound.

Why is it difficult to prepare it, and how to prepare it?
Can it be obtained by simply mixing $\ce {Hg2Cl2}$ and $\ce{AgNO2}$, where $\ce{AgCl}$ will be precipitated?

1.Why is it difficult to prepare it, and how to prepare it?

Prafulla Chandra Ray prepared mercurous nitrite, $\ce{Hg2(NO2)2}$ by reacting metallic mercury with cold, dilute nitrous and nitric acid. According to his paper(1):

Mercurous nitrite is the product of the combined action of nitrous
and nitric acids on mercury :
$$\ce{2Hg + NO2.H + HO.NO2 -> Hg2(NO2)2 + H2O}$$
Some of the nitrite is decomposed by the nitric acid, the quantity
of nitrous acid thus rapidly growing, until mercurous nitrite and nitrate
are accumulated in molecular proportions. After some, the nitrous acid remains constant in the solution. It has become the catalytic agent between the nitric acid and the mercury,which now interact in the following manner:
$$\ce{4Hg + 4HNO3 -> Hg2(NO2)2 + Hg2(NO3)2 + 2HO}$$

It is difficult to prepare the compound because many mixed species are created in the reaction mix like and it becomes difficult to separate the target product i.e. mercurous nitrite. Also to note that both mercurous nitrite and mercurous nitrate can react with elemental mercury creating hemi- species.
$$\ce{2Hg2(NO3)2 + 2Hg + Hg2(NO3)2 + H2O -> 4HO.Hg2NO3 + 2NO}$$

2.Can it be obtained by simply mixing $\ce{Hg2Cl2}$ and $\ce{AgNO2}$, where $\ce{AgCl}$ will be precipitated?

Although P.C.Ray used elemental mercury and dilute nitrous and nitric acid, nobody tried this reaction route because A) Dealing with mercury salts is already dangerous B) P.C.Ray's sythesis is already famous so nobody has bothered trying it.
References

XX.—Theory of the production of mercurous nitrite and of its conversion into various mercury nitrates by Prafulla Chandra Ray, J. Chem. Soc., Trans., 1905,87, 171-177, DOI: 10.1039/CT9058700171
THE CHEMICAL RESEARCHES OF ACHARYA PRAFULLA CHANDRA RAY by ANIMESH CHAKRAVORTY, Indian Journal of History of Science, 49.4 (2014) 361-370 - PDF
Samanta, S. & Goswami, Sudeep & Chakravorty, A.. (2011). On mercurous nitrite and a basic mercurous nitrate derivative. Indian Journal of Chemistry - Section A Inorganic, Physical, Theoretical and Analytical Chemistry. 50. 137-140 - Abstract
Das, A.K. Thermodynamic and Kinetic Aspects of the Stability of Sir P. C. Ray’s Mercurous Nitrite Compound. Reson 25, 787–799 (2020). DOI: 10.1007/s12045-020-0996-9
Molecule of the Month - Mercurous Nitrite by Rajarshi Ghosh
CXCVIII -The double nitrates of mercury and alkali metals by P.C. Ray
LXVI - Dimercuryammonium nitrite and its haloid derivative by P.C.Ray
190. Theory of the production of mercurous nitrite and its conversion to various mercury nitrate by P.C.Ray

$4 Hg + 4 HNO_3 -> ... + 2 H_2O$, maybe?

Any idea why the reaction with $\ce AgNO2$ should not happens? $\ce AgCl$ should simply precipitate, leaving  $\ce Hg2(NO2)2$ in the solution.

@Archisman, theoritically, the reaction between $\ce{Hg2Cl2}$ and$\ce{AgNO2}$ to yield $\ce{Hg2(NO2)2}$ should occur where $\ce{AgCl}$ is precipitated but it has not been yet performed experimentally. Perhaps, working with mercury salts is quite dangerous.


text:
Variation within a chromatography with the same sample
What are some possible explanations to why the exact same sample in SEC HPLC differ in their peak heights, although it's the same chromatographic profile?
The sample contains cell culture medium (undiluted) into a 10mM phosphate NaCl buffer. Is it possible that the sample is overloaded, might this be the cause for such different results?
After each sample there's a 1x column bed volume (24ml) wash with the phosphate buffer.
Any theories based on experience or otherwise?
Edit:
Sample 1:
Peak 1: 166292μV,
Peak 2: 199835μV,
Peak 3: 168024μV,
Peak 4: 78483μV
Sample 2:
Peak 1: 96936μV,
Peak 2: 142119μV,
Peak 3: 148060μV,
Peak 4: 195584μV
Sample 3:
Peak 1: 56769μV,
Peak 2: 88114μV,
Peak 3: 82351μV,
Peak 4: 176517μV
Chromatogam looks something like this (Peak count from left to right)

Is this UV detection? Something else? Also, typically only integrals and never heights are compared, and integrals are only fully comparable when compared to a standard.

280 nm. Alright, I've retained Area (μV*sec), % Area and Height (μV).. which ones do you suggest I use? I suppose area since that's the integral. Does is matter that the peaks doesn't have very good resolution? I.e the area of the peaks will probably overlap to some degree.

We need more details.  As Jan said, what are you actually detecting?  Ultraviolet or visual (UV) light absorbance?  Conductivity?  Light scattering?  Dye?  It's also not clear how the cell culture medium can be "undiluted" if you mixed it with 10 mM phosphate buffer.  What injection volume?  What column size?  What are the column parameters?  What if you inject blanks (just buffer) after the sample?  Do you see peaks?

Odd that peaks 1,2 and 3 go down in samples 1 to 3, but peak 4 is highest in sample 2.

Curt: I'm detecting with UV-light at 280nm as well as fluorescent light. The peak-values added are UV-light at 280nm. The cell culture medium is injected into the HPLC undiluted i.e it's the same concentration as one can grow on. Injection volume is 200μl. The column is Superdex Peptide
10/300 GL SEC. I use 0.2ml/min which is as low as the column allows me - tried higher flow rate but get less resolution. Buffer gives no signal (as it should).

MaxW: yes, it's a mystery to me too.

One theory of mine is that there's an overload. I'll check out the difference with diluted samples, will get back to you later. http://www.chromacademy.com/chromatography-HPLC-Column-Overload.html

Is this the same sample run 3 times or 3 different samples?

Also, if you're worried about overloading, you can test this hypothesis by simply diluting your samples.

Zhe: it's the exact same sample run 3 times. Yeah, I'm running 40 samples: 10x undiluted, 10x 50/50 WFL (water for lab), 10x 10/90 WFL and 10x 1/99 WFL. Will get the result next week to see if it's overloaded once and for all. One other theory is of course contamination which certainly could give different results..


text:
Why do gases under low pressure conduct electricity?
In J. J. Thomson's cathode ray experiment, he used a discharge tube in order to make the gas inside conduct electricity.
Why do gases under low pressure conduct electricity?

Gases under high pressure can as well - see, e.g., lightning. All a question of the electric field, ionization potential, and mean free path of the electrons.

highly related: http://chemistry.stackexchange.com/questions/37004/does-an-electrically-conductive-gas-exist-at-room-conditions

I think @ Jon Custer is correct. I suspect cathode ray tubes are mostly evacuated to allow the electrons a relatively unimpeded path.

In a microwave oven, hot air at atmospheric pressure conducts nicely: https://www.youtube.com/watch?v=5f46zUgYBKE

The cathode ray tube doesn't conduct electricity per se. It liberates free electrons that are simply moving through a vacuum from one end to the other. In practice, you can't create a real vacuum, so a low pressure gas (as low as possible) is used instead.
Air has a high enough dielectric strength that it will dampen the effect of the applied voltage. That is, it'll be harder to liberate free electrons to shoot down the tube.


text:
How to write electron configuration of Mercury?
I am completely confused about how to write the electron configuration anything higher than Barium because there are so many inconsistencies and misconceptions about the 5d and 4f orbitals. I know that Lanthanum and Actinium have an outer orbital of d. Let's say I needed to write the configuration of Mercury. Wouldn't it be 1s2 (blah blah blah) 5d10 4f14? Because they always want to get to the lowest energy level?  However, apparently, the correct way of writing it is 4f14 5d10...

Typically the noble gases are used as jumping off points to avoid a lot of tedium. It also puts more focus on the last few subshells which is nice too.
So Hg is $\ce{[Xe] 4f^{14} 5d^{10} 6s^2}$

the thing that I am confused about is the order in which people write the orbitals. shouldn't 5d10 be before 4f14?

No. For the electronic configuration order by principle quantum number first, then by s, p, d and f.

How do I find the principal quantum number?

:-( For Hg you have 4f, 5d, and 6s. the digits 4, 5, and 6 are the principle quantum numbers. Remember the quantum numbers for an electron are are n, l, m, and s. The principle quantum number is the n value. The l value is the  azimuthal number and gives the orbital type (s, p, d and f). https://en.wikipedia.org/wiki/Quantum_number#Traditional_nomenclatures


text:
Hydrogen Fluoride, Hydrogen Monofluoride?
I had a student ask why this is not referred to as hydrogen monofluoride. Similar, we call H2S hydrogen sulfide, not dihydrogen monosulfide.
Could we just call those common names, but that by IUPAC naming, it is actually hydrogen monofluoride? Or is it really just hydrogen fluoride. I have seen various answers on other websites, but curious if anyone knows the official answer.
The most common reason I've seen is that because both H and F has a single site open for bonding that there is no other combination possible (so no need for numerical prefixes).

Related: Why is CaCl2 called calcium chloride?

http://chemistry.stackexchange.com/a/59396/34388

VTC as duplicate. I don't think it makes sense to not close this. Not closing this would imply that we would allow one question for every single inorganic salt e.g. LiF, LiCl, LiBr, LiI, NaF, NaCl.......... MgF2, MgCl2, ...........

Thank you all, the answer in the first post was more than sufficient. I agree, it is a duplicate

One helpful thing to keep in mind (and you're clearly already keyed into this) is that IUPAC nomenclature is intended to reduce ambiguity and create consistency with rules that are as simple as possible.


text:
Is there any specific formula to find periodic table periods and valence electrons
I am developing a program, that uses a lot of stuff related to the periodic table. As of now, I am making a part of it the determines the period and family. I know that both of those have to do with how many valence electrons each element has. So I want to know if there are formulas to find the amount of valence electrons, family, and periods?

Well, there's no particular rule, unless you want to also code exceptions for about a quarter of the elements. Perhaps you could ask on Math SE whether there is a function that maps atomic number to valence of the neutral atom?

You might be reinventing the wheel. See, for instance, the python library mendeleev. I suggest looking around for what others have done - it'll save you time.

You might just want to store the end of each period as an array... that saves you some time. You can also see a trend of valence electrons with some exceptions...

The number of electrons in each orbital is double an odd number (the Pauli exclusion princilel states electrons can be in the same energy only if of opposite spin, so there can be two otherwise identical electron energy states [which can be separated in an EM field]).

s (sharp spectral lines): 2 (twice 1)
p (principal spectral lines): 6 (twice 3)
d (diffuse spectral line): 10 (twice 5)
f (fine, or "fundamental", spectral lines): 14 (twice 7)

Using that information, and the fact that the first row has only s electrons, second row s & p, usw., you should be able to calculate the first few rows easily. However, rules become much more complex for transition metals and heavier elements, where an outer shell may be added before all inner orbitals are filled.
Primary valence for the first few rows is simply the number of electrons needed to either complete the current shell (e.g. oxygen is two e- short), or revert to the previous complete, noble gas, configuration (e.g. Na minus one e- is neon). Again, there are complications, such as transition metals with multiple valences, fluorine reacting with some noble gases, usw.

Iirc it was fundamental, not fine :)

Both "fine" and "fundamental" are given in Wikipedia etymology. So much for acronyms -- TLA is a three-letter acronym ;)


text:
How do I know what the end products of the reaction of glucose and fructose?
I am unsure about why the end products of the reaction are what they are.
Let's say I was combining glucose and fructose in a dehydration synthesis reaction to create sucrose. The equation is:
$$\ce{C_6H_{12}O_6 + C_{6}H_{12}O_{6} \rightarrow C_{12}H_{22}O_{11} + H_{2}O}$$
How do I know it is not:
$$\ce{C_6H_{12}O_6 + C_{6}H_{12}O_{6} \rightarrow C_{12}H_{24}O_{12}}$$

You don't "derive" reactions. You recite reactions for different classes of chemicals.

What are some common reactions for different classes of chemicals. Like I understand synthesis, single elimination, combustion, double elimination... But what are the other common reactions? What are the set of rules?

Well, for example, carbohydrates polymerize through condensation.

Yes, but why aren't two molecules of water made and a product of C12H21O9 in the first equation?

Because that's how they react...

If illustrations help you: http://biochemreview.weebly.com/uploads/1/0/4/0/10409756/8373954_orig.gif
two monosaccharides are joined by removal of one water molecule.

It is entirely possible for two sugars to react through a double dehydration reaction to form $\ce{C12H20O10}$.

There are three possibilities. The first is knowing the reaction. I.e., you just know because you rote memorised it, that one molecule of water will be lost and then you apply that.
The second is knowing the mechanism. If you take a good look at it, a water molecule must be displaced for the mechanism to work out. Anything else would generate something highly unstable.
And finally, if you know the product structure, you can calculate double bond equivalents. Both reactants have a single double bond equivalent: $(2 \times 6 - 12)/2 + 1 = 1$. This corresponds to the ring in the reactants’ structures. The product has two rings, thus it requires two double bond equivalents. The number of carbons must stay the same, though. Thus: $(2 \times 12 - n)/2 + 1 = 2$; rearranging gives $n = 22$. Two hydrogens must be ejected and that biochemically typically happens with water.

I am unsure about why the end products of the reaction [of glucose and fructose] are what they are.

Glucose and fructose could react to form a nearly infinite number of products.  The identity of the product(s) depends on the conditions (temperature, pressure, solvents, catalysts, enzymes, etc.) used.  Glucose and fructose could react to form sucrose through a condensation reaction, as you note.  However, they could also react to form trehalulose, turanose, leucrose, palatinose, or maltulose.  And that's just considering a few alpha-linked glucosyl fructosides.  All of them have the formula $\ce{C12H22O11}$.  But so do sugars such as kojibiose, made from two glucose moieties.  Those are also potential products, under certain reaction conditions (e.g. using a kojibiose-specific transglucosylase and an glucose/fructose isomerase).  And as I mentioned in a comment, there's no reason that reactions between glucose and fructose are limited to single condensations.  Many products -- double dehydrations to form difructose anhydrides, or even more severe dehydrations to form hydroxymethylfurfural are possible.

Let's say I was combining glucose and fructose in a dehydration synthesis reaction to create sucrose.

Well, now you have specified a single product out of the infinity of products that are possible.  If you've specified sucrose as the product, not only do you know the product has a formula of $\ce{C12H22O11}$, you also know that it is α-glucopyranosyl-(1→2)-β-fructofuranoside.  Because that's what sucrose is.  No other product is "sucrose".

The equation is:
$$\ce{C_6H_{12}O_6 + C_{6}H_{12}O_{6} \rightarrow C_{12}H_{22}O_{11} + H_{2}O}$$

Yes, it must be this, if you are making sucrose.

How do I know it is not:
$$\ce{C_6H_{12}O_6 + C_{6}H_{12}O_{6} \rightarrow C_{12}H_{24}O_{12}}$$

Well, it might be that $\ce{C12H24O12}$ is a product that could be formed from glucose and fructose under some bizarre reaction conditions.  But even if it were, it wouldn't be sucrose.  Because by definition, sucrose has the formula $\ce{C12H22O11}$.
So, by specifying that the product is sucrose, the only possible reaction is
$$\ce{C_6H_{12}O_6 + C_{6}H_{12}O_{6} \rightarrow C_{12}H_{22}O_{11} + H_{2}O}$$


text:
Why is the segment of the IR spectrum below 1500 wavenumbers called the fingerprint region?
Why is the segment of the IR spectrum below 1500 wavenumbers (cm$^{-1}$) considered as the fingerprint region? It is said that "although the entire IR spectrum can be used as a fingerprint for the purposes of comparing molecules, the 600 - 1400 wavenumber range is called the fingerprint region. This is normally a complex area showing many bands, frequently overlapping each other". However, I don't understand why the range from 600 - 1400 wavenumbers can be particularly complex and is unique to the compounds, for instance, why is the fingerprint region of propan-1-ol and propan-2-ol so different from each other given roughly the same appearance for rest of the spectrum?

Spectra taken from http://www.chemguide.co.uk/analysis/ir/fingerprint.html

I guess the naming "fingerprint region" results from the fact that while I can immediately identify amine, alcohol, carbonyl groups, double and triple bonds and so on in the "upper" and "middle" part of the IR spectrum, the peaks observed in the fingerprint region are often arising from vibrations that are very specific to the very compound you are looking at, thus a "fingerprint" of your compound.


text:
Is there a better equation than the Nernst equation for low concentrations?
The Nernst equation can be applied to the whole cell reaction or a half-cell reaction. Consider the half cell reaction:
$$\ce{Cu^{2+} + 2e^- -> Cu(s) \quad \quad E^0 = +0.337~\rm{V}}$$
The Nernst equation expressed in base 10 log at 25 °C is:
$$E = E^0 - \frac{0.1596}{2}\log_{10} \frac{a_{\ce{Cu^2+}}}{a_{\ce{Cu}}}$$
But $a_{\ce{Cu}} = 1$ by definition, and for low concentrations of $\ce{Cu^2+}$ where $a_{\ce{Cu^{2+}}} \ll 1$ then: $\ce{[Cu^2+]} \approx a_{\ce{Cu^2+}}$.
But when $\ce{[Cu^2+]} < 1$ then the log will be negative so:
$$\lim_{\ce{Cu^2+}\to 0} (0.337 - 0.0798 \times \log_{10} [\ce{Cu^2+}]) = \infty $$
This doesn't seem realistic. For instance $\ce{[Cu^2+] = 1.0\times 10^{-1,000,000}}$ is fine mathematically, but that sort of concentration doesn't make any sense chemically.
Thus what is the limit of the half cell potential as the $\ce{[Cu^2+] -> 0}$?

Edit-1 - My gut reaction to this is that you can't really measure the activity of a single ion. If the ion of interest is low in concentration compared to the ionic strength in general, then it would seem that the activity of that ion should go to zero as its effect on the total overall ionic strength goes to zero.

A low concentration say $\ce{10^{-10} }$ $\pu { mol dm^{-3}}$ still contains a staggeringly huge number of molecules  The formula is ok as its not possible to have negative concentrations and only $6.02x 10^{23}$ molecules $ \pu {mol^{-1}}$

The Nernst equation does not need a replacement for low concentrations. Yes the limit is $\infty$, which essentially means you can't realistically have zero concentration of anything. We may think of absolutely pure water as an idealized situation. Well, as soon as it gets in touch with copper, some $\ce{Cu^2+}$ ions instantly jump into the solution, and from that point on, you have a certain non-zero concentration of them (though probably pretty low), and the equation works just fine.

Note also that the formula actually works best low concentrations. You actually use activities instead of concentrations, and activity is approximately equal to concentration at low concentrations. Not so at higher concentrations.


text:
Can I add citric acid to KCl for iron removal in a water softener?
I recently installed a water softener and having a fair amount of clear iron in our water, I've been using Morton Rust Remover pellets which are basically NaCl with citric acid added as an additive.  I'm guessing the iron removal function has something to do with the citric acid since it's the only notable difference between this salt and regular water softener salt.
At some point I'd like to stop using NaCl and switch the KCl since it seems to be more environmentally friendly.  However, I really notice a difference using the iron removal pellets and I've been unable to find an equivalent product for sale which uses potassium.
Can I simply mix some citric acid in with the potassium in the water softener to achieve the same result?  Or do the citric acid and the sodium have some special relationship which helps with the iron removal?

Your ion-exchange column removes $\ce{Fe^{3+}}$ ions from your water by exchanging them for $\ce{Na^+}$.
When you back-flush you use concentrated solutions of $\ce{Na^+}$ ions and citric acid to remove the iron out of your ion-exchange column. The flush water is waste of course which must be disposed.
The exact chemistry mechanism of all of this depends on a number of factors:

NaCl has a solubility of 36 g/L at 20 C which is 0.62 molar but
KCl has a solubility of 34.2 g/L at 20 C which is 0.46 molar.
I'd expect potassium citrate to be less soluble than sodium citrate but
I didn't find any data to support this notion in a quick search.
I'd expect that you column has a better attraction to $\ce{Na^+}$ ions
than to $\ce{K^+}$ ions.
If you have some sort of automatic monitor then it may not
function properly with $\ce{K^+}$ ions. I'd guess that the monitor
just looks to determine if the water is yellow, so I'd guess that
$\ce{K^+}$ ions would work fine in this regard.
Not sure what pH you'd want. I'd guess maybe half trialkali citrate with
half dialkali hydrogen citrate to keep the back-flush solution
slightly acidic.

All in all my guess would be that potassium/potassium citrate flush solution would work. I'd expect the column to need flushing more often. Maybe doing a flush once in a while with the sodium/citric acid mixture would give the ion-exchange column a good cleaning.
The other consideration here is cost. Since the sodium/citric acid mixture is the common back-flushing salt, I'd expect a potassium/citric mixture to be more expensive.


text:
Why is the formation of slacked lime accompanied by a hissing sound and the formation of bubbles?
My teacher made me write that that the formation of calcium hydroxide (slacked lime) from calcium oxide (qucklime) is described as

$\ce{CaO + H2O -> Ca(OH)2}$

and that the process

forms bubbles
is an exothermic reactions
makes a hissing sound

Why does it make a hissing sound and form bubbles?
There is no gas evolved in the reaction, then why does formation of bubbls take place?
And what is the reason of the hissing sound being produced?

Because your teacher is wrong? The observations sound like the observations of $\ce{Ca + 2H2O -> Ca(OH)2 + H2}$.

Because the reaction is so exothermic that the water starts boiling?

No $\ce{H2}$ gas evolves.

?? everyones confused

This reaction is exothermic enough to be used for self-heating food packaging. That's going to cause some sizzling in the water.

why does sizzling happen when something is really hot though?

Good eye for the inconsistency.  Cultivate that, it will serve you well.  But this is an apparent, rather than a real inconsistency.  The reaction is VERY exothermic.  Check out how they make quicklime to find out how much heat it takes.
Where the water touches the quicklime, enough heat is evolved to boil a little bit of the water.  Hence the bubbles and hiss.

well..i don't really get why the hiss is formed though...like the science behind the production of a sound


text:
How to make a buffered 10% formalin solution
How can you prepare a solution of 50 mL buffered 10% formalin using the following reagents?
100% formalin (100 mL)
Phosphate buffer (100 mL)
I know I can mix 5 mL of formalin and 45 mL of distilled water together to make 50 mL of 10% formalin, but if I add phosphate buffer later the volume of formalin would exceed 50 mL.
Can I mix like 1 mL of formalin and 9 mL of distilled water together and then make up the remaining volume with 40 mL of phosphate buffer? Is the concentration of formalin remaining the same as long as the proportion of distilled water to formalin kept constant no matter how much phosphate solution I add?

As I do not know how accurate that shall be, in all practicability you can just substitute water with buffer solution. You should also be aware, that your way of doing it suggests Volume fraction as concentration, and that might not account for volume contraction. A better way would be to use mass fractions or amount concentration.


text:
How full is a gas cylinder?
There are 7 different gas cylinders filled with:

nitrogen
oxygen
air
methane
hydrogen sulfide
carbon dioxide
sulfur dioxide

How do I determine how much of the respective gases is left?
Nitrogen and oxygen are in a gaseous state, so I reckon a glance at a pressure indicator will do. Air? The same. How about the last four cylinders which are filled with liquids? They have their vapor pressure, but I am not sure that those pressures can be used to accurately calculate the volume of respective liquids. Should the weight of the cylinder be measured?

What is their normal pressures? Pressurise nearly any gas enough and it will liquify.  Most gas cylinders of those gases don't have liquid contents- it would be most inconvenient for users to have the liquid interfere with valves.

@user2617804 Try opening the valve of a CO2 cylinder standing upside down.

Correct.
For the non-liquifying gases (room temperature is far above their critical temperature) knowing the pressure is enough, if you know the volume of your cylinders. (You need a better formula than $pV=nRT$ to calculate the amount of gas at those high pressures. https://en.wikipedia.org/wiki/Van_der_Waals_equation, virial expansion, ...)
Actually the van der Waals equation is not useful, because, although it gives a good  idea why the ideal gas law breaks, it gives very wrong numbers esp. for high pressures. BWR or BWRS are supposed to work well, at the cost of using eight or eleven constants.
For those gases which are partially liquified ($\ce{CO2}$ at $57\ \mathrm{bar}$, $\ce{H2S}$ at $12\ \mathrm{bar}$, $\ce{SO2}$ at $3.5\ \mathrm{bar}$), you need to know the weight (and tare weight) of the cylinders, of course.

I suggest that if a gas cylinder doesn't have a tare weight on it, be wary of buying from that supplier again. Maybe the discovery of PTFE is connected.

To expand on Karl's answer a bit, not only do you need a better equation of state than $PV=nRT$ to accurately measure the contents, depending on the cylinder pressure several of the listed gases may require a correction for effects due to the supercritical phase change.
\begin{array}{c|cc}
\mathrm{Species} & T_c \left(^\circ\mathrm C\right) & P_c \left(\mathrm{psi}\right) \\
\hline
\ce{N2} & -147 & 493 \\
\ce{O2} & -118.6 & 731 \\
\ce{CH4} & -82.6 & 668.5 \\
\ce{H2S} & 100 & 1300 \\
\ce{CO2} & 31 & 1070 \\
\ce{SO2} & 157.2 & 1143 \\
\hline
\end{array}
Data links — $\ce{N2}$ – $\ce{O2}$ – $\ce{CH4}$ – $\ce{H2S}$ – $\ce{CO2}$ – $\ce{SO2}$ $T_c$ $P_c$.
So, even for the gases well above their boiling points at room temperature ($\ce{N2}$, $\ce{O2}$, air, $\ce{CH4}$), if the cylinder pressure exceeds the respective critical pressure for that gas it will still be necessary to weigh the cylinders to get a highly accurate measure of the remaining contents, since supercritical phases generally have liquid-like densities.

This is a misconception. Unless you are really close to the liquid state, density of a gas (supercritical or not) increases largely linearly with pressure. There also is no phase transition between normal gas and supercritical.

@Karl Interesting! So, is it that this far above $T_c$ the density will be substantially that of an ideal gas?

No, but it will not be very far off. The density will always approximately double if you double the pressure... until the gas becomes incompressible. That would be the opposite of what you say.

@Karl But, regardless, the IGL will not be valid for accurate absolute calculations of the mass of the cylinder contents, correct?

Even vdW becomes rather inaccurate at 200 bar, i believe. For any actual technical application, this one would be right: https://en.wikipedia.org/wiki/Benedict%E2%80%93Webb%E2%80%93Rubin_equation

By way of comparison, at the typical state for a full gas cylinder ($p=200\ \mathrm{bar}$, $T=25\ \mathrm{^\circ C}$), the values for the molar volume are $V_\mathrm m=0.1239\ \mathrm{l\ mol^{-1}}$ (ideal gas), $V_\mathrm m=0.1308\ \mathrm{l\ mol^{-1}}$ (nitrogen), $V_\mathrm m=0.1173\ \mathrm{l\ mol^{-1}}$ (oxygen), $V_\mathrm m=0.1277\ \mathrm{l\ mol^{-1}}$ (air), and $V_\mathrm m=0.1021\ \mathrm{l\ mol^{-1}}$ (methane).

@Loong Not that far from ideal behaviour. Interesting numbers, never looked at that. So oxygen has 280g /l compared to 1100 for the liquid, 156 compared to 420 for methane.


text:
Why can't sodium salts be produced through a reaction between a metal and an acid?
I am writing a lab report for the preparation of sodium sulfate crystals. This question came up: Why can the reaction between a metal and an acid not be used to make sodium salts?
In this case, could one not use sodium metal and sulfuric acid to make sodium sulfate, as in the following:
$$\ce{2Na + H2SO4 -> H2 + Na2SO4}?$$

It’s not that you can’t; it’s that you shouldn’t.

Related; if something reacts with water acting as an acid, it will just react even stronger when presented with a proper acid..

Normally, reacting a metal and an aqueous acid will produce a salt and hydrogen gas, as in the case of iron:
$$\ce{Fe (s) + 2HCl(aq) -> FeCl2 (aq) + H2 (g)}$$
However, you should not use the same method to produce, for example, sodium chloride:
$$\ce{2Na (s) + 2HCl(aq) -> 2NaCl (aq) + H2 (g)}$$
because sodium reacts violently with water.

Sodium reacts vigorously with water to form sodium hydroxide and hydrogen gas:
$$\ce{Na(s) + H2O (l) -> NaOH (aq) + 1/2 H2(g)}$$
So, it is advised not to put it into water as explosions can occur. It's also highly unlikely to obtain pure salt solution. So, these salts are mostly made by mixing an acid and an alkali.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Given an unbalanced chemical equation, is it possible to determine the largest possible coefficient that could be used in the balanced equation?
I am writing a program to balance chemical equations, and would love to know if there is an algebraic way to determine the greatest possible coefficient that I would need to check.

I think you might just want to use matrices...

In short
Just use matrices.
In long
Consider this equation to be balanced:
$$\ce{CuSCN + KIO3 + HCl -> CuSO4 + KCl + HCN + ICl + H2O}$$
(source with spoiler)
We are trying to solve this equation:
$$\ce{aCuSCN + bKIO3 + cHCl -> dCuSO4 + eKCl + fHCN + gICl + hH2O}$$
Now, we move every term to the left hand side (warning: handwavey chemistry):
$$\ce{aCuSCN + bKIO3 + cHCl - dCuSO4 - eKCl - fHCN - gICl - hH2O = 0}$$
Transform into matrix form (the element names are just for reference):
$$
\begin{matrix}
\ce{Cu}\\
\ce{S}\\
\ce{C}\\
\ce{N}\\
\ce{K}\\
\ce{I}\\
\ce{O}\\
\ce{H}\\
\ce{Cl}
\end{matrix}
\left(
\begin{matrix}
1 &0 &0 &-1&0 &0 &0 &0 \\
1 &0 &0 &-1&0 &0 &0 &0 \\
1 &0 &0 &0 &0 &-1&0 &0 \\
1 &0 &0 &0 &0 &-1&0 &0 \\
0 &1 &0 &0 &-1&0 &0 &0 \\
0 &1 &0 &0 &0 &0 &-1&0 \\
0 &3 &0 &-4&0 &0 &0 &-1\\
0 &0 &1 &0 &0 &-1&0 &-2\\
0 &0 &1 &0 &-1&0 &-1&0
\end{matrix}
\right)
\left(
\begin{matrix}
a\\
b\\
c\\
d\\
e\\
f\\
g\\
h
\end{matrix}
\right)
=
\left(
\begin{matrix}
0\\0\\0\\0\\0\\0\\0\\0\\0\\
\end{matrix}
\right)
$$
Notice that the biggest matrix is not a square. You must transform it to a square. If the elements are in excess (just like this case), add variables; otherwise, add elements. Set the extra variables/elements to zero:
$$
\begin{matrix}
\ce{Cu}\\
\ce{S}\\
\ce{C}\\
\ce{N}\\
\ce{K}\\
\ce{I}\\
\ce{O}\\
\ce{H}\\
\ce{Cl}
\end{matrix}
\left(
\begin{matrix}
1 &0 &0 &-1&0 &0 &0 &0 &0 \\
1 &0 &0 &-1&0 &0 &0 &0 &0 \\
1 &0 &0 &0 &0 &-1&0 &0 &0 \\
1 &0 &0 &0 &0 &-1&0 &0 &0 \\
0 &1 &0 &0 &-1&0 &0 &0 &0 \\
0 &1 &0 &0 &0 &0 &-1&0 &0 \\
0 &3 &0 &-4&0 &0 &0 &-1&0 \\
0 &0 &1 &0 &0 &-1&0 &-2&0 \\
0 &0 &1 &0 &-1&0 &-1&0 &0
\end{matrix}
\right)
\left(
\begin{matrix}
a\\
b\\
c\\
d\\
e\\
f\\
g\\
h\\
i
\end{matrix}
\right)
=
\left(
\begin{matrix}
0\\0\\0\\0\\0\\0\\0\\0\\0\\
\end{matrix}
\right)
$$
Note: $i$ must be zero.
Now you can use any algorithm for solving homogeneous square matrix equations to continue.

There is no guarantee that any solver gives exact integer multiples and if the matrix is singular (as in your example) simply scaling the result vector won't necessarily work either. You will actually need to use an integer matrix solver. A Gaussian elimination with pivoting that only uses integer multipliers for instance should do this correctly. The smallest common multiplier of all matrix entries should provide the upper bound.

Also don't pad matrices to make them square. It's unnecessary.

Matrices, as shown in DHMO's answer, provide a way to solve systems of linear equations.  But I think that, for most students, it might be more accessible to just solve the simultaneous linear equations directly.
Specifically:  We know that, in a balanced chemical equation, the number of atoms of each element on the left must equal the number of atoms of each element on the right.  Thus, for each element, we can write an equation expressing that.
Consider, for instance:
$$\ce{a H_2 + b O_2 -> c H_2O}$$
Based on the hydrogens, we know that $a=c$.  And based on the oxygens, we know that $2 b = c$.  [To get the number of oxygens, you need to multiple the coefficient for the molecule, $b$, times the number of oxygens in the molecule, $2$.]
Using simple substitution, we know that $a = 2 b =c$. Hence if we pick the smallest possible integer, $1$, for $b$, we have: $a = c = 2$, and thus:
$$\ce{2H_2 + O_2 -> 2 H_2O}$$
Now let's apply this to a non-trivial example, say that in DHMO's answer:
$$\ce{aCuSCN + bKIO3 + cHCl -> dCuSO4 + eKCl + fHCN + gICl + hH2O}$$
For this reaction, our simultaneous equations for each element would be:
Cu, S: $a = d$
K: $b = e$
I: $b = g$
C, N: $a = f$
Cl: $c = e + g$
H: $c = f + 2h$
O: $3b = 4d + h$
If you solve this set of simultaneous equations, either manually, or using an equation solver that can be restricted to integer solutions, then you will get the balanced chemical equation shown here:

 $$\ce{4CuSCN + 7KIO3 + 14HCl -> 4CuSO4 + 7KCl + 4HCN + 7ICl + 5H2O}$$

I obtained this with Mathematica, using the following code:
Reduce[$a == d$ && $b == e$ && $b == g$ && $a == f$ && $c == e + g$ &&
$c == f + 2 h$ && $3 b == 4 d + h$, Integers]


text:
Painting corroded 9V battery contacts with vinegar
I have an old radio with corroded 9V battery contacts, probably from a battery that corroded though I don't know.
I wanted to clean up the contacts with 5% vinegar since that works with pennies, but this is not something I can dunk in a small container since it is attached (with a very short wire) to the radio).
So I want to thicken vinegar to a gel, and not change the pH.  I would rather do this slowly, I can leave the solution on the contact for as long as necessary.
A high school chemistry teacher suggested vaseline as a thickener.  I just wanted to get a second opinion that vaseline will not somehow cause a problem?  Glycerin was also mentioned.  Could either react in any way with the metal electrode?
If I use either, presumably the acididy will decrease by dilution with the thickener.  Is this linear, or does the thickening of the solution have any other affect on acids?  I can try treating the surface for a few hours, or a day, then how could I get it off?  My guess is a cotton swap with water and detergent to clean off the gel?
I am not 100% sure what the surface of the contact is, it could be a nickel alloy, it certainly is a formerly shiny silvery metal, now somewhat dulled with green raised bumps on it from some former chemical reaction.
If you think I should just apply vinegar, or anything else for that matter, with a cotton swab I could do that, I just want to be able to apply something and clean up the surface with a minimum of effort.

Using vinegar, petroleum jelly, or glycerol to clean the contacts would leave some residue.  Vinegar might dissolve it somewhat more quickly than water, but you have to be sure you get all the vinegar off the contact, or it could cause more corrosion.  Petroleum jelly - I think someone was kidding you.  Vinegar & petroleum jelly don't mix.
Much easier: use a small piece of emery paper or steel wool to grind the deposit off.  Keep sanding until the metal is shiny.  Use a damp rag to remove the powder.  Don't let the ground-off powder fall into the circuitry.
Edit:  added steel wool.

The only problem with this is that a 9V battery connector is not exactly an easy shape to clean.


text:
Why is the degree of dissociation for a weak acid α, but for a weak base, it's 1−α?
From the equation used in this answer: Calculating charge on amino acid from pKa
$$\alpha=\frac{1}{10^{\mathrm pK_\mathrm a-\mathrm{pH}}+1}$$
Degree of dissociation of weak acid is $\alpha$. However, when it comes to weak base, degree of dissociation is $1-\alpha$. What is the explanation behind this?

I think this is because $\alpha$ is defined from $\ce{HA <=> A- + H+}$; which for a base would be exactly reversed. Not sure enought about it to answer, though.

@Jan is almost certainly correct. For consistency, we looking at the equation he wrote. Same reason we generally look at pH instead of pOH.