Reduction of molality of methanol How many grams of water would you add to $\pu{1.00 kg}$ of $\pu{1.44m}$ $\ce{CH3OH (aq)}$ to reduce the molality to $\pu{1.00 m}$ $\ce{CH3OH}$?

I have tried $M_1V_1=M_2V_2$. I have tried converting it to kg of $\ce{CH3OH}$ and subtracting it from the $\pu{1.00kg}$ to get kg of $\ce{H2O}$. All of which have worked to no avail.

If the molality is $1.44 \ce{m}$, it means that, there is 1.44 moles of methanol in $1 \ce{kg}$ of water.

To prepare this solution, we add the mass of  $1.44\times 32= 46.08\ce{g}$ of methanol to $1000\ce{g}$ of water.

The mass of the final solution is $1000+46.08= 1046.08\ce{g} $.

Here, you have 1 kg of this solution (water + methanol). So, it contains $\frac {46.08 \times 1000}{1046.08}=44.05\ce{g}$ of methanol.

The mass of water in this one kg-solution is $1000-44.05=955.95\ce{g}$

The objective of the problem is to prepare a solution of molality $1 \ce{m}$,

i.e, it is prepared by adding $32\ce{g}$ of methaonl to $1000\ce{g}$ water.

It can also be prepared by adding $44.05\ce{g}$ of methanol to $\frac {44.05 \times 1000}{32}=1376.56
\ce{g}$ of water.

Our solution already contains $955.95\ce{g}$ of water.

So, we need to add only $1376.56-955.95=420.61\ce{g}$ of water.


Controversy over products of iodate and bisulfite? We know that when we are trying to recognize iodate, we use sodium bisulphite to the solution and with the help of starch paper recognize the evolved $\ce{I2}$.But when I studied the books I found actually Iodide is formed? What's the solution to this controversy?

Not much controversy here, IMO. There is more than one reaction involved, with slightly more complex kinetics.

As so often, Wikipedia already knows, so I will just quote the relevant part of the following article: http://en.wikipedia.org/wiki/Iodine_clock_reaction

  Iodate variation

  An alternative protocol uses a solution of iodate ion (for instance potassium iodate) to which an acidified solution (again with sulfuric acid) of sodium bisulfite is added.

  In this protocol, iodide ion is generated by the following slow reaction between the iodate and bisulfite:

  $$\ce{   IO3^{−} + 3 HSO3^{−} -> I^{−} + 3 HSO4^{−}}$$

  This is the rate determining step. The iodate in excess will oxidize the iodide generated above to form iodine:

  $$\ce{   IO3^{−} + 5 I^{−} + 6 H+ -> 3 I2 + 3 H2O}$$

  However, the iodine is reduced immediately back to iodide by the bisulfite:

  $$\ce{   I2 + HSO3^{−} + H2O → 2 I^{−} + HSO4^{−} + 2 H+}$$

  When the bisulfite is fully consumed, the iodine will survive (i.e., no reduction by the bisulfite) to form the dark blue complex with starch.


Is it possible to speed up radioactive decay? I’m aware that elements like $\ce{^14C}$ have a known half-life, which means that over a span of roughly $5730$ years, half of the $\ce{^14C}$ atoms decay into $\ce{^14N}$.

Are there any substances known to speed up this process, that applying it to $\ce{^14C}$ will catalyze a chemical reaction resulting in $\ce{^14N}$, or sticking them in a particle accelerator and slamming them together will result in $\ce{^14N}$, or otherwise transmute elements?

It is possible to modify nuclear decay rates using chemistry, though it is rare and the effect is usually very small. Here I summarize the information available in this link. You may want to see the references within.

There is a type of nuclear decay called electron capture, where a nuclide directly captures an electron from the innermost electron shells and transforms a proton into a neutron. Therefore, there is coupling between the nucleus and the wavefunctions of the innermost electrons in this form of radioactive decay.

Usually the core electrons are very weakly disturbed by changes in chemical environment, but they are changed slightly. For especially light atoms, where the core electrons are very close to the valence shell, such as $\ce{^7_4 Be}$, this change results in a measurable difference in decay rate, varying by 0.1-1% compared to the isolated atom.

In a more dramatic case, there is a subset of beta decays called bound-state beta decay, where the electron released by the decaying neutron is immediately captured by the nucleus after decay. Apparently, if the parent atom is stripped completely bare of its electrons, and if the energy involved in the nuclear decay is comparatively low, once again there is a meaningful coupling between nuclear and electronic states. For the case of rhenium-187, the neutral atom $\ce{^187_75 Re}$ has a half-life of 42 billion years, but upon full ionization to $\ce{^187_75 Re^75+}$, the half-life is reduced to 32.9 years! For dysprosium-163, the neutral atom $\ce{^163_66 Dy}$ has half-life so high it has not been measured to decay, but when ionized to $\ce{^163_66 Dy^{66+}}$ its half-life is reduced to 47 days!


What makes a radical 'free'? The qualifier 'free' seems to be ubiquitously attached to discussion of radicals as highly reactive species with unpaired spins. What, precisely (or imprecisely, as the case may be) does 'free' really mean?

Free as in free speech? Free beer? Freedom of movement? Kinetic accessibility?

I liked this question because I had never thought much about it. However, it's not such a mystery because the answer is on Wikipedia's “radical” page:

  Historically, the term radical was also used for bound parts of the molecule, especially when they remain unchanged in reactions. These are now called functional groups. For example, methyl alcohol was described as consisting of a methyl "radical" and a hydroxyl "radical". Neither are radicals in the modern chemical sense, as they are permanently bound to each other, and have no unpaired, reactive electrons. However, they can be observed as radicals in mass spectrometry when broken apart by irradiation with energetic electrons.

So, the free in free radical was originally used to distinguish it from a radical bound to a carbon skeleton.

Digging into historical chemistry texts reveals that “free radical” becomes really widely used around 1930:

However, some quotes for earlier works use it, though it does not appear so frequently that it may be considered a common expression. In other words, at that time simply the juxtaposition of the adjective free with radical, became synonymous with group (or today's noun substituent). For example:

  “This hydrocarbon was formerly supposed to be the free radical of the methyl group, but no methyl compounds have in any way been derived from it, whilst by the action, of chlorine upon it, ethyl chloride is obtained as the product” (1870, Lessons in elementary chemistry)

or

  “Although, however, the free radical methylene is unknown, many of its compounds have been prepared, of which some, such as the haloid ethers, &c.” (1884, A Treatise on Chemistry, Henry Enfield Roscoe and Carl Schorlemmer)

Though more quotes could be dug and provided, the ones I've read support this idea that free radical initially meant “the chemical group as a molecule by itself”.


Which has higher priority between unsaturated ring and unsaturated chain in the nomenclature of organic compounds? According to the IUPAC rules of nomenclature, alkenes have higher priority over rings but what if the ring also contains a double or a triple bond?
For eg:

Is this compound going to be named as 1-(2-chlorocycloprop-2-enoxy)propene or 1-chloro-3-propenoxycyclopropene?

The name given by ChemDraw is: (E)-1-chloro-3-(prop-1-en-1-yloxy)cycloprop-1-ene.
The ring is determined to be the parent structure.  This is rationalised in the 2013 edition of the IUPAC Blue Book on organic nomenclature by the following rules (pg 451):

P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)
Two methods are recognized to name systems composed of rings and chains (exclusive of linear phanes).
(1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent.  Rings and chains are chosen regardless of their degree of hydrogenation.  As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons
(2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated a cyclic structure may be recognised, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

So either the chain or ring may be used as the parent for the name, but the ring will define the preferred IUPAC name.
The preference of ring over chain is further clarified in the following rule (pg 596):

P-52.2.8 Selection between a ring and a chain as parent hydride
Within the same heteroatom class and for tha same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a prefered IUPAC name. In general nomenclature, a ring or a chain can be a parent hydride (see P-44.1.2.2).

All emphasis added.


Ordinary differential equation for molecule collision I have no background in chemistry.

But, I have a good background in mathematics and computer programming.

I have planned to simulate the collision of two simple molecules $\ce{Br2}$ and $\ce{H2}$ just for fun and using state space and Ordinary Differential Equation (ODE) to simulate.

I have considered 24 states for simulations of their collision. The state vector is X, and the state space representation is:

X'=f(X,t)

Atom numbering

1: Br
2: Br
3: H
4: H

The 24 states for vector X are:

x1, x2, x3:  position of atom 1 in x, y, z axis
x4, x5, x6:  position of atom 2 in x, y, z axis
x7, x8, x9:  position of atom 3 in x, y, z axis
x10,x11,x12: position of atom 4 in x, y, z axis
x13,x14,x15: velocity of atom 1 in x, y, z axis
x16,x17,x18: velocity of atom 2 in x, y, z axis
x19,x20,x21: velocity of atom 3 in x, y, z axis
x22,x23,x24: velocity of atom 4 in x, y, z axis

To make a simulation, I need to know two things:

Are the mentioned states enough to describe the entire system?  Are there more states to be added?
What are the forces applied to each atom at each moment?

$\frac{d x_1}{d t}=x4$ (derivative of position is velocity)

$\frac{d x_4}{d t}=\frac1{m_{Br}}(???????)$ (derivative of velocity is acceleration)

There are two types of molecular simulation (in a broad sense), namely
1) Monte Carlo simulation
2) Molecular dynamics simulation
And they can be mixed together to give rise to new algorithm. According to your problem statement I am assuming you are trying to use molecular dynamics simulation. There are lots of things you should consider for your simulation system. According to your problem statement your atoms have position and velocity, but the problem is when they will collide each other they will loose energy and will come to a halt after few time steps. Either you can use thermostat to keep the temperature constant (Anderson thermostat and Nose-Hoover thermostat), or you can use constant pressure molecular dynamics simulation. For a beginner I would like Anderson thermostat as this is very simple technique. Then you have to use Newton's second law to determine the force and velocity on each atoms. After that you can use numerical integration technique to find new position based on F=ma formulation. Now you will repeat this task to get the trajectory of individual atoms.

If you want to find the diffusion coefficient of atoms, you can use lattice gas models or Ising models based on random Monte Carlo movement. I wrote a python code for finding diffusion coefficient. Also you can use Markov Chain Monte Carlo (MCMC) method to determice the stochastic evolution of the system.

For your better understanding I would suggest to use Understanding Molecular Simulation by Daan Frenkel & Berend Smit


At what frequency does a non-polar molecule acquire a dipole able to participate in London Dispersion forces? London Dispersion forces are forces between non-polar non-ionic molecules that the random fluctuation of their electrons makes them temporarily dipoles.

What is the frequency that this occurs?
Every how many milliseconds does a molecule like this becomes a dipole?
Are there such numbers for some specific molecules out there?

According to Prof. Loren Williams of Georgia Tech, the fluctuations occur on the femtosecond $\left(10^{-15}\ \mathrm{s}\right)$ time scale:

  Figure 13 shows how fluctuating dipoles of liquid Xenon (or Helium or Neon, etc) are coupled. Darker blue indicates higher electron density. The fluctuations are correlated and are very fast, on the femtosecond ($10^{-15}$ second) timescale. Adjacent Xenon atoms experience electrostatic attraction from the transient dipoles. Two different representations of fluctuating dipoles are shown.

This corresponds to a frequency of $1\ \mathrm{PHz}$ (petahertz), which is about five hundred thousand times faster than most modern computer processors.

Another estimate of the time scale (albeit one devised by me and thus of potentially suspect validity), is the SI value of the atomic unit for time, derived by dividing the reduced Planck constant by the atomic unit for energy, the Hartree (values are rounded a bit):

$$
\left\{atomic\ time\right\} = \frac{\hbar}{E_\mathrm{h}} = \frac{1.05457\times 10^{-34}\ \mathrm{J\cdot s}}{4.3597\times 10^{-18}\ \mathrm{J}} = 2.41888\times 10^{-17}\ \mathrm{s} \approx 0.024\ \mathrm{fs} = 24\ \mathrm{as}
$$

That last value is $24$ attoseconds, which is a well and truly short span of time. For comparison, this corresponds to a frequency of $41\ \mathrm{PHz}$.

Whatever the exact value, I have to believe the fluctuations are fast enough so as to be impossible to measure directly. (I would be fascinated to learn otherwise, though!)  Thus, even if different atoms/molecules do actually have different frequencies of London dipole oscillation, we have no way of finding out these values.


What is the phase of matter that has flexible volume, but constant shape? In this YouTube video it argued that:

Solids have constant volume, and constant shape
Liquids have constant volume, but flexible shape
Gases have flexible volume, and flexible shape (in fact they don't have volume and shape at all)

So, this table came to my mind:

Do we have a phase of matter that fits in the red cell? In other words, do we have a phase of matter that keeps its shape, and only changes its volume? If so, what is the name of that phase?

There is a mathematical argumentative approach called reductio ad absurdum...i.e., bring your own logical reasoning to the point where the result is an absurd statement if you accepted the original statement as true. In this case you can accept that there is a phase which can change its volume but keeps it geometrical shape intacr. You can try disproving your own statement using this method for solids, liquids, and gases.

Do we have a phase of matter that fits in the red cell? In other words, do we have a phase of matter that keeps its shape, and only changes its volume? If so, what is the name of that phase?

a) When solids experience stress, there is strain = deformation, change in volume.
b) Liquids are almost incompressible and they have no shape, but under extreme extreme pressures their volume does change, typically, > 100,000 psi they can become solids. The solid will have the shape of container in which it became a solid.
c) Gases acquire the shape of the container (actually a bad statement, i.e., gases occupy the space provide by the container), and their volume can be easily changed by stress.
d) The fancier state of matter, plasma, it is also an ionized gas, so it does not have a geometrical shape. However, it can be confined by using magnetic fields so that it does not touch the walls of the container, and perhaps its "volume" and "shape" can be manipulated independently. A true physicist can better answer it.


Mathematical justification for Le Chatelier's principle At equilibrium

$$
\begin{align}
K &= \exp\left(\cfrac{TΔS^\circ - ΔH^\circ}{RT}\right)\\
⇒ \frac{\mathrm d \ln K}{\mathrm dT} &= \frac{ΔH^\circ}{RT^2}
\end{align}
$$

If enthalpy change is positive, change in $\ln K$ w.r.t. $T$ is positive. Hence, $\ln K$ and therefore $K$ increase (position of equilibrium shifts to the right) as $T$ increases. And, $\ln K$ and therefore $K$ decrease (position of equilibrium shifts to the left) as $T$ decreases.

If enthalpy change is negative, change in $\ln K$ w.r.t. $T$ is negative. Hence, $\ln K$ and therefore $K$ increase (position of equilibrium shifts to the right) as $T$ decreases. And, $\ln K$ and therefore $K$ decrease (position of equilibrium shifts to the left) as $T$ increases.

Can you give a similarly mathematical reasoning for why a reaction

$$\ce{A(g) <=> 2 B(g)}$$

has its equilibrium position shifted to the left as the pressure of the surroundings increases?

I think that I might have an answer to my own question:

$$K=exp \left( \frac{TΔS^o-ΔH^o}{RT} \right) $$

implies that K is dependent on T, but not on total pressure of system, p (nor concentrations).

For
$$ A_{(g)}⇌2B_{(g)} $$
$$ K= \frac{{p_{B}}^2}{p_{A}} $$
Partial pressure of a gas = mole fraction (mf) x total pressure of system

If p is increased then K initially increases:
$$ K=\frac{(mf_B.p)^2}{mf_A.p} $$
$$ ⇒ K=\frac{{mf_B}^2.p}{mf_A} $$

But p doesn't affect K, yet K has increased. So the mole fraction of A must be increased by shifting the position of equilibrium to the left (thus decreasing the mole fraction of B), in order to re-establish the value of K to before p was increased.


What happens if a polymer's length is equal to its persistence length? So, I've been looking into polymer chemistry to model breaking polymers for a biological model, and I've been a bit stumped by the concept of persistence length.

I know that if a polymer is significantly longer than its persistence length, it needs to be treated as though the ends are completely independent of each other thanks to thermal fluctuations bending it all over (the analogy I have is treating the polymer like a piece of cooked spaghetti). If I model a polymer like this, I need to use the words "3 dimensional random walk" to describe its shape mathematically.

I also know that if a polymer is significantly shorter than its persistence length, it can be treated like a rigid rod that might be a bit bendy, like a steel rod. This is significantly easier to model: I can just treat it as a line/arc that moves through space as a single particle.

But what happens if a polymer's length is about equal to its persistence length? How can the polymer be modeled then? What precisely does it mean for the polymer's movement?

EDIT: I've been debating the merits of just treating it like a flexible rod (like the significantly shorter than persistence length example) and calling it good just because of model simplification. If I do that, how off would I be from reality?

EDIT2: I've been doing further research and found something called a "worm like chain" model. I'm pretty fuzzy on what it means for the fiber, and I'm having trouble grasping how the fiber would move from an intuitive perspective; the visual for how I'd need to treat it isn't working, and the visual would help immensely with the actual modeling for the rest of the equations I need.

The persistence length is a statistically defined property - in other words, it is defined by looking at many lengths of many polymer molecules over a long period of time. It is the length (L) at which a vector drawn tangent to the polymer is no longer correlated in space with another tangent vector at distance (L). In other words, it tells you how well the position of one section of a polymer is correlated to another section. Beyond the persistence length, you would not expect to find any correlation. Below it, you would expect correlations.

In this case "correlations" are measured by the cosine of the angle between the two vectors. When the angle is 90 degrees on average, the cosine is zero, and when it is zero degrees, the cosine is 1. So, for lengths beyond the persistence length, the amount of correlation is zero, and for lengths much less than the persistence length the amount of correlation is one. For lengths close to the persistence length, the correlation is between 0 and 1.

To give you an idea of how this works, take a look at the following illustrations:

The red lines represent tangent vectors along the backbone of a polymer chain, spaced at regular intervals. We can translate all of these vectors and put their origin at the same point, and then find the angle between them:

If we find another polymer chain, or watch this one for a little while and take another "snapshot" and do the same thing, we might end up with something that looks like this instead:

We can even take the same very long chain and find all of the vectors that are a distance 1 apart, 2 apart, and so on. If we do this for all possible distances over hundreds of polymer molecules, and average over many snapshots for a very long time (nano- to micro- seconds for a melt or something in solution), what we will eventually find is something like this:

The correlations drop off quickly with length, then more slowly, until finally they are completely uncorrelated.

It is important to realize here that at any given time, for any given polymer, the tangent vectors could be almost anything. You wouldn't expect two adjacent bonds to have a very small angle between them, and in general you would expect more flexibility over longer lengths of chain compared to shorter lengths, but you would not be able to predict the shape or angles with any degree of certainty, other than to say that on average, beyond the persistence length the tangent vectors would not be correlated, and that the amount of correlation would decay exponentially with length.

The examples you gave of a stiff rod vs. cooked spaghetti are meant to demonstrate this intuitively, but your question is about what happens when the persistence length and the polymer itself are the same length. The answer is that the polymer will be somewhere between a very flexible piece of cooked spaghetti and a stiff rod. Imagine taking a piece of spaghetti and bending it into a loop, but not enough that it breaks. The circumference of 1/4 of the loop would be something like the persistence length, and the response of the spaghetti in the loop would be similar to a polymer that is about the same length as its persistence length. However, keep in mind that polymers are constantly in motion, and so "flexible rods" and "cooked spaghetti" are not great analogies in the first place. It's more realistic to think of them as constantly wriggling worms, or something like that. Statistical descriptions of correlation lengths and times are the best we can get, unfortunately.

Another concept that might help you visualize this is that of the Kuhn length - this is twice the persistence length, and is the length at which sections of polymer are uncorrelated with each other in space. Practically, it means that the motion of the polymer within the Kuhn length is irrelevant to the motion of the overall chain. On a large scale, the motion of the Kuhn "blobs" is what matters. In your case, the polymer is about as long as the persistence length, which means it is one-half of the Kuhn length. This means motions along the polymer are correlated, and so you can't ignore them (if they are important to your model.)

Practically, what this all means is that your model is probably going to be fairly complicated, because you can't average out or ignore terms that can be ignored in either of the extreme cases.

If you post some more details about your model, I could probably give you more specific information.


why is laser printer toner magnetic? I had an old toner cartridge lying around, so I took it apart and emptied the toner into a jar. I had heard that one could make a ferromagnetic fluid using toner powder and oil, so I mixed a little powder with some kerosene that I had on hand. Turns out, kerosine does something wierd to the polymer part of the toner, and instead of becoming a solution, the toner powder coagulated into a rubbery mass that had a consistency similar to chewing gum.

When I put a powerful magnet from a hard drive next to it, the ball was attracted to it, which surprised me a little, as I had seen on wikipedia that toner had no magnetic components (any more) So, the questio is, Why is printer toner powder magnetic? Is it because carbon is paramagnetic?

Also, what happened to the polymer when kerosine was added?

EDIT

A clarification-
I think the polymer is a resin (based on what the kerosene did to it)

One of the major ingredients in modern printer toner cartridges (for laser printers) is magnetite, Fe3O4. It is used a charge control ingredient, known as a triboelectrification agent. 'Tribo' comes from the Greek meaning 'to rub' (carfeul if you google that one), and essentially refers to the creation of static charge through friction. This is required to have the printer particles stick to the paper. You can read about triboelectrification here. Magnetite, of course, is magnetic, and this is why your toner can produce a ferromagnetic fluid.

Other ingredients in toner cartridges include waxes, polymers and pigments. Many formulations are water or alcohol based and the toner is actually prepared as an emulsion which is filtered and dried. In order to get this material dissolved into kerosene, you probably need to add an emulsifying agent. Oleic acid is commonly used to achieve this, as it is a long chain acid that has both polar (hydrophilic) and non-polar (hydrophobic) properties.


Does a covalent bond constitute of 1 or 2 electrons? A covalent bond is pictured as two electrons shared by two different atomic nuclei, which means that each of the two nuclei has two electrons. The problem I see with it is that it implies that if a hydrogen atom is neutral then a hydrogen nucleus in a covalent bond must be negatively charged since it has one more electron than a hydrogen atom. Which implies that a hydrogen molecule is negatively charged which is a contradiction (molecules are neutral).

Alternatively, one can say that the covalent bond is composed out of a single electron, which acts as a negative potential for both hydrogen nuclei hence it exerts negative potential of two in the molecule. But then the question is, where does the second electron go when two hydrogen atoms form a covalent bond?

My question does not relate only to this specific example, it's about covalent bonds in general.

The bond in your picture is made up of two electrons, and in the Lewis formalism so is every single bond. Yet there is no negative charge. We can think of this in two ways:

Remember that the bond electrons are shared. So each atom contributed one electron and, upon dissociation, would usually take back that one electron. Counting only that one electron will give a neutral hydrogen atom, not a hydride ion.

Charges on an atomic or molecular level are due to a mismatch of the numbers of protons and electrons. In the hydrogen molecule, you have two nuclei consisting of one proton each, meaning an overall $2+$ charge. Furthermore, you have two electrons that get shared in the covalent bond giving an overall $2-$ charge. $+2 + -2 = 0$, so the overall molecule is neutral.
If you wanted to place both electrons on one atom to form a hydride anion, that would mean the other atom is left with zero. Instead of a hydrogen molecule $\ce{H-H}$ you would have a separated ion pair $\ce{\overset{-}{H}\bond{...}\overset{+}{H}}$.

Once you leave the very simplified Lewis formalism and turn to theories such as molecular orbital theory which explain molecules much more accurately, this way of thinking will be lost altogether since there is no longer an obvious localisation of two electrons to a single bond.


Reagent notation: meaning of dash(-) and line(|) or slash(\) in reagents I've seen a lot of reagents which have a slash between two compounds, such as Lindlar's catalyst which is $\ce{H2}$ along with $\ce{Pd|BaSO4}$. I know that the sulphur is a poison here but in general, what does the \ or | mean? And is it supposed to be a \ or a |? Are both equivalent? Does it mean "and" or "or" or something else?
$\ce{Cr2O3-Al2O3/Mn2O3}$, Also what does the dash mean here? In?
EDIT: This in the context of the reagents mentioned above arrows in organic chemistry. If I'm not wrong,  $\ce{Cr2O3-Al2O3/Mn2O3}$ is used to convert hexane to benzene by cyclization followed by aromatization.

The '|' in $\ce{Pd|BaSO4}$ implies that the reaction is taking place in the presence of $\ce{Pd}$ as a metal catalyst. This metal provides surface area for reaction to occur.
Also '/' in $\ce{Cr2O3-Al2O3/Mn2O3}$ means that $\ce{Cr2O3}$ can be used in combination with either $\ce{Al2O3}$ or $\ce{Mn2O3}$.


Colligative properties and sizes of particles Why do colligative properties only depend on the number of particles and not the size?

*My argument being that since the root cause of these properties is the occurrence of solute particles on the surface, reducing the vapour pressure of the solvent, shouldn't bigger molecules, such as urea, give a much more appreciable deviation in properties, as compared to smaller ones like NaF?

Colligative properties From Wikipedia

  In chemistry, colligative properties are properties of solutions that depend on the ratio of the number of solute particles to the number of solvent molecules in a solution, and not on the nature of the chemical species present. The number ratio can be related to the various units for concentration of solutions. The assumption that solution properties are independent of the nature of solute particles is only exact for ideal solutions, and is approximate for dilute real solutions. In other words, colligative properties are a set of solution properties that can be reasonably approximated by assuming that the solution is ideal.

Emphasis in above text is mine.

So it the assumption that colligative properties are independent of particle size, not the reality. The assumption is very useful, but not perfect.


Does the property of additivity for the amounts of different substances hold up for ideal gas law? A mixture contains $\pu{8.00 g}$ each of $\ce{O2},$ $\ce{CO2},$ and $\ce{SO2}$ at STP.  Calculate the volume of this mixture.  Which of the gases would exert the greatest pressure and why?

Here is my work:

$$
\begin{align}
n(\ce{O2}) &= \frac{\pu{8.00 g}}{\pu{32.00 g mol-1}} = \pu{0.250 mol}\\
n(\ce{CO2}) &= \frac{\pu{8.00 g}}{\pu{44.01 g mol-1}} = \pu{0.182 mol}\\
n(\ce{SO2}) &= \frac{\pu{8.00 g}}{\pu{64.07 g mol-1}} = \pu{0.125 mol}
\end{align}
$$

$$n_\mathrm{tot} = \pu{0.250 mol} + \pu{0.182 mol} + \pu{0.125 mol} = \pu{0.557 mol}$$

$$pV = nRT \quad\implies\quad V = \frac{nRT}{p}$$

$$V = \frac{\pu{0.557 mol}\times\pu{0.0821 atm L mol^-1 K^-1}\times\pu{273 K}}{\pu{1.00 atm}} = \pu{12.5 L}$$

Considering $p = nRT/V,$ the gas that takes up the least amount of volume with the greatest amount of particles will exert the greatest pressure. This will be $\ce{O2}$ because it has the lowest molar mass/volume.

Is this how handling the amounts of substance work (can different substances be added like this)?

The question would be better phrased as which gas contributes most to the total pressure? The contribution is referred to as the partial pressure. The sum of the partial pressures is the total pressure.

The formula you would use to compute the partial pressure $p_i$ of gas component $i$ is either of the following:

$$p_i=\chi_i p = \frac{n_i}{n}p = \frac{n_i}{V}RT$$

where $n_i$ is the amount of component $i$ in the gas and $\chi_i$ its mole fraction. So the argument "the gas that takes up the least amount of volume with the greatest amount of particles will exert the greatest pressure" is not quite right. It suffices to say that the gas that contributes the greatest amount of particles will contribute (exert) the greatest pressure. If you mix identical masses of each gas, that will be the gas with lowest molecular mass.


What is the exact definition of Salinity? I am a bit unclear on the definition of salinity.  I have always thought of it as simply total dissolved ions.
Wikipedia seems to agree "saltiness or dissolved salt content" but many other sources seem to suggest that only certain ions contribute.
So is there a list of specific anions and cations that contribute to salinity and is there an underlying logic involved in their selection or is it simply arbitrary? Is the definition dependent on the solute in question?

The term 'salinity' is measure of 'saltiness', particularly used in reference to sea water.

The exact scientific definition has evolved over time to incorporate our ever growing understanding of physics and chemistry.

In 1889 the International Council for the Exploration of the Sea (ICES) used a measure of salinity based on the total dissolved solids of salt (g/kg) found in seawater after evaporation and drying at 480$^{o}$C.

By taking samples of ocean water from around the world and at different depths, oceanographers have determined that whilst the concentration of salts varies, the relative ionic composition remains essentially constant. The average salinity of the world's ocean water is usually stated as 35 g/kg, with the relative ionic composition of this "average sea water" made up as follows:

The major cations present in 1 kg of sea water are Na$^{+}$ (10.7653 g), Mg$^{2+}$ (1.2942 g), Ca$^{2+}$ (0.4117 g), K$^{+}$ (0.3991 g) and Sr$^{2+}$ (0.0079 g).

The major anions present in 1kg of seawater are Cl$^{-}$ (19.3534 g), SO$_{4}$$^{2-}$ (2.7124 g), HCO$_{3}$$^{-}$ (0.1178 g), CO$_{3}$$^{2-}$ (0.0122 g), Br$^{-}$ (0.0674 g), F$^{-}$ (0.0013 g), B(OH)$_{4}$$^{-}$ (0.0066 g).

The process of determining salinity from evaporation & drying can be quite difficult, particularly on board ships.

In 1963, the ICES adopted a new definition based on "chlorinity" measurement (using titration). The "average sea water" with ionic composition given above was determined to have chlorinity of 19.374%. By measuring the relative chlorinity of a sample of sea water, the salinity could be calculated using the formula S = 1.80655 Cl.

In 1978, with the advent of modern electronics, the United Nations Educational, Scientific and Cultural Organization (UNESCO), International Council for the Exploration of the Sea (ICES), Scientific Committee on Oceanic Research (SCOR) and International Association for the Physical Sciences of the Oceans (IAPSO) adopted a method based on electrical conductivity measurements compensated for temperature. The Practical Salinity Scale of 1978 (PSS78) defines S (a dimensionless quantity) from conductivity and temperature measurements, in the range 2 $\leq$ S $\leq$ 42, by the following formula:

S = 0.0080 - 0.1692$K_{15}$$^{1/2}$ + 25.3851$K_{15}$ + 14.0941$K_{15}$$^{3/2}$ - 7.0261$K_{15}$$^{2}$ + 2.7081$K_{15}$$^{5/2}$

Where $K_{15}$ is the ratio of conductivity of the sea water sample to that of potassium chloride (KCl) solution of mass fraction 32.4356x10$^{-3}$, at 15$^{o}$C and atmospheric pressure.

A formula is also provided for compensation of the conductivity measurement at different temperatures (see http://unesdoc.unesco.org/images/0004/000479/047932eb.pdf)

In 2010, the Intergovernmental Oceanographic Commission released "the international thermodynamic equation of seawater (TEOS-10)" in which it recommended a return to the use of "Absolute Salinity" to take into account variations in the composition of sea water which can occur for example in regions where rivers run into the ocean or where some other regional variability results in "anomalous" sea water composition.

The rational for this approach is based on understanding the effect each chemical component of seawater has on its thermodynamic properties such as density, pressure, temperature and Gibbs function. Various measures of salinity are defined and their relationships with thermodynamic properties described.

The Added-Mass Salinity, which incorporates the Reference Composition Salinity of "standard sea water" (the Practical Salinity Scale) with an adjustment for the Absolute Salinity Anomaly (mass fraction of material which must be added to standard seawater to achieve the concentration levels in the sample) is sometimes difficult to determine in-situ.

Modern techniques for determining Absolute Salinity involve taking various other measurements such as conductivity, temperature, pressure and density by incorporating measurements from sensitive electronic and optical instruments with data-driven computer algorithms.

The Density Salinity, which is the salinity argument in the thermodynamic expression for density which gives the sample's actual measured density, can be used where accurate measurements of density are available, such as a vibrating-tube densimeter or optical refractometer. When combined with Practical Salinity, temperature and pressure measurements, this can provide an accurate and easily determined measure of the amount of dissolved material in seawater (within a certain range).

Composition anomalies can be estimated by computer algorithms which take into account longitude, latitude and depth data. The accuracy of such algorithms is dependent upon the collection of accurate data from field measurements and laboratories across the world and is therefore an evolving process.

(For more details, see: http://www.teos-10.org/).


What is active mass? I'm aware of the fact that active mass is defined as the molarity of a substance, but my textbook states that

  "Active masses are dimensionless quantities but for our purposes we generally take them with dimensions of molarity, partial pressures etc."

What point is my textbook trying to get across to me? Why do active masses have no dimensions? Why have we defined this quantity in the first place, if we use it interchangeably with the term "molarity"?

Edit: Please keep in mind that I have only just graduated high school and would be very grateful if you could explain this concept to me in simple terms.

The term "active mass" is a historical term.

The concept of an equilibrium constant was developed by Cato Maximilian Guldberg and Peter Waage.  The Law of Mass Action has also been referred to as the Law of Guldberg and Waage, historically.

Guldberg and Waage defined the term "active mass" in the 1867 Études sur les affinitès chimiques.

  l'on peut choisir un volume arbitraire, par exemple un centimètre cube. Les
   quantités d'un corps qui se trouvent dans lcc du volume total se nomment la
  masse active du corps.

or roughly translated:

  One can choose an arbitrary volume, for example a cubic centimeter. The
  amounts of a body located in 1cc of total volume are called the active mass of the body.

And as stated in The Laws of Reaction Rates and
of Equilibrium J. Chem. Ed. 1956, 33 (4), p 178 :

  Guldberg and Waage in their study of chemical
  affinity formulated the law of the rate of chemical reactions
  as proportional to the "active mass" of the reagents.
  This they called the law of mass action and
  very clearly defined "active mass" as amount per unit
  volume. (emphasis added)

So, from a historical point of view, it is false that "active mass" is dimensionless.

However, in relating Gibbs energy to the equilibrium constant, it is necessary for the equilibrium constant to be dimensionless, because only the logarithm of a dimensionless quantity is logical.  For this reason, the text mentioned in the OP redefines "active mass" to be dimensionless.


Can acidified or neutral KMnO4 oxidise toluene to benzoic acid? Why is alkaline $\ce{KMnO4}$ used in the oxidation of toluene to benzoic acid? Can acidified or neutral $\ce{KMnO4}$ be used in this conversion?

Here are the three equations describing the reduction of manganese (and concurrent oxidation of whatever substrate may be present) under basic, neutral and acidic conditions respectively.

$$\ce{Mn^{+7}O4- +e- ->~ Mn^{+6}O4^2-~~~~[basic]}$$
$$\ce{2H2O + Mn^{+7}O4- + 3e- ->~ Mn^{+4}O2 + 4OH-~~~~[neutral]}$$
$$\ce{8H+ + Mn^{+7}O4^{-} + 5e- ->~ Mn^{+2} + 4H2O~~~~[acidic]}$$

Acidic conditions are the most economical as 5 electrons are transferred per mole of manganese, but oxidation can be achieved under all 3 conditions. A problem with neutral conditions is the precipitation of insoluble $\ce{MnO2}$ and the need for its subsequent separation. A problem with all 3 conditions is that oxidation will only occur if the material to be oxidized is somewhat soluble in the reaction medium.

The milder basic conditions are generally preferred for the oxidation of aromatic alkyl side chains because with the milder basic conditions an organic co-solvent that is soluble in water can be employed to facilitate solubility of the organic substrate.


Why is benzyl alcohol a slightly stronger acid than cyclohexanol? Cyclohexanol has a pKa = 16, while benzyl alcohol has a pKa = 15.4

The given explanation is that the "polarity of the benzyl group stabilizes the negative charge on oxygen".

My understanding is that when resonance is not a concern, benzyl groups are primarily electron donating by the inductive effect... which would destabilize charge on oxygen.

I know the inside of the aromatic ring is more negative, and the outside is more positive. Is it this positive charge that stabilizes the negative charge on oxygen?

(This CW answer is adapted from an earlier comment by ron.)

In both cases, the carbon directly attached to oxygen is $\mathrm{sp^3}$ hybridised. However, the next carbon in the two compounds have different hybridisations. In cyclohexanol, it is also $\mathrm{sp^3}$ hybridised; but in benzyl alcohol, it is $\mathrm{sp^2}$ hybridised.

In general, $\mathrm{sp^2}$ hybridised carbons are more electronegative than $\mathrm{sp^3}$ hybridized carbons. This is because the electron density in an $\mathrm{sp^2}$ orbital has more s-character (50%) compared to one in an $\mathrm{sp^3}$ orbital (33%), and is consequently more strongly attracted to the carbon nucleus. Therefore, the carbon attached to oxygen in benzyl alcohol will be more electron deficient (more positive) and will be better able to stabilise the negative charge on oxygen in the alkoxide conjugate base, via the inductive effect.


What is the result when the polarity of the solvent is increased in nucleophilic substitution? I read in the Guidebook to Mechanism in Organic Chemistry that increasing the polarity of the solvent slightly reduces the rate of an $S_N2$ reaction. Why does this happen?

Another question:
We know that in $S_N2$, using a polar protic solvent would reduce the rate as the nucleophile gets solvated by hydrogen bonding, but I found in Solomon's Organic Chemistry that the $S_N1$ reaction rate increases in a polar protic solvent. Why does this occur? The nucleophile still gets solvated right?

The standard textbook explanation is based on a comparison of charge separation in the ground state and the transition state (TS).  If there is more charge separation in the TS than the ground state, then polar solvents stabilize the TS more than the ground state and the reaction will be accelerated. On the other hand, if there is more charge in the ground state than in the TS, then polar solvents will stabilize the ground state more than the TS and consequently the reaction will slow down.

Let's look further at the effect of charge separation and solvent on these two specific reactions and see how they are affected.  Let's start with the $\ce{S_{N}1}$ reaction.

  I found in another book that the SN1 reaction rate increases in polar
  protic solvent. Why does this occur?

In the $\ce{S_{N}1}$ reaction, the rate determining step (rds) is formation of the carbocation intermediate. Let's say we start with a charged nucleophile and a neutral compound.  Moving along the reaction coordinate, by the time we reach the carbocation intermediate, the starting materials (charged nucleophile plus neutral compound) have transformed into a charged carbocation plus a charged leaving group plus the charged nucleophile.  Moving from starting materials to the intermediate has created additional charged species (carbocation plus leaving group).

$$\ce{Nu^- + R-X~~ (the ~starting~ materials) -> [R^+ +X^{-} + Nu^- ]~~ (the~ TS)}$$

Since the transition state leading to these intermediate species will strongly resemble the intermediate, it will also have more charge separation then the starting materials and therefor, will be more stabilized by polar solvents then the starting materials.  This stabilization of the transition state will lower its energy and speed up the rds.  In other words a polar solvent has sped up the rate of the $\ce{S_{N}1}$ because more charge exists in the transition state than in the ground state.

  I read in a (book) that increasing the polarity of the solvent
  slightly reduces the rate of SN2 reaction. Why does this happen?

Now, lets perform the same analysis on the $\ce{S_{N}2}$ reaction.  Again we start out with a charged nucleophile and a neutral compound.  The rds in this case is the formation of a trigonal bipyramid species.

$$\ce{Nu^- + R-X~~ (the ~starting~ materials) -> [\delta^{-} ~~ X...R...Nu ~~\delta^{-} ]~~ (the~ TS)}$$

In the ground state there was one unit of negative charge residing fully on the nucleophile. In the TS that unit of charge is now spread out over both the nuceophile and the leaving group; the charge has become much more diffuse, the charge density per unit volume has decreased in the TS compared to the ground state.  In this case, we would expect a polar solvent to stabilize the ground state (starting materials) slightly more than the still charged - but more diffusely charged - TS.  Stabilizing the starting materials slightly more than the TS would slightly increase the energy required to pass over the TS.  Hence we would expect polar solvents to slightly slow down the rate of an $\ce{S_{N}2}$ reaction.


Why Aren't Chlorides Of The Noble Gases As Prevalent As Their Fluorides? I can't find the answer to this question on this SE website, and I apologize for the repetition if it has been answered before. It is my understanding that compound formation has only been observed for the noble gases argon, xenon, and krypton, while no compounds have ever been observed for neon and helium, and that this might have something to do with both the decreased ionization potential of larger noble gas atoms and the high value of electronegativity of the halogen (usually fluorine) that bonds with the noble gas atoms to form compounds. So, my question is why aren't compounds formed from argon, xenon, and krypton bonding with chlorine, bromine, and iodine as prevalent as those formed from those noble gases and fluorine? Could the smaller size of the fluorine atom be a reason for its apparently higher reactivity?
-John

You know that the bigger the noble gas atom, the easier it is to ionize it. By the same token it is also easier for a Xe atom than for a He atom to share some electron density in a covalent bond.  That is, Xe is more reactive (a very little bit) than He (not at all).

For halogens, the tendency is in some sense the opposite: The smaller the halogen atom the more eager it is to gain electron density -- that is, the more electronegative it is.  Fluorine is so electronegative that it can persuade Xe, Kr and even Ar (with difficulty) to give up some electron density to form a covalent bond.  Cl is not quite so electronegative, and Br and I even less.  Basically you need the least "noble" of the noble gases and the greediest halogen of all to even have a chance at a bond.

One can justify this handwaving with thermodynamic calculations and molecular orbital arguments (though the latter are often handwavy too).


Regular decrease in the atomic radius of 3d series While comparing atomic radius, two factors are important:

A. Decrease in size due to increase in effective nuclear charge

B. Increase in size due to increase in shielding effect

I was surprised to find a regular decrease in the atomic radius of 3d series elements on this site because that would mean that effective nuclear charge is dominating in case of each element more than shielding factor.

http://crystalmaker.com/support/tutorials/crystalmaker/atomic-radii/index.html

I have also attached the screenshot of the values of atomic radii of 3d series elements(given in Angstrom).

On some previous stackexchange posts like this one

Variation in atomic sizes in the transition elements

I have found that from Ni to Zn there is a regular increase in size.

Are the values shown on this site wrong?
If yes, can you please provide the correct values.
 If not, can you please explain why shielding wouldn't be as effective as increase in nuclear charge?

Periodic Table Website confirms these atomic radii.
So why does the atomic radius decrease? The nucleus of the atom gains protons moving from left to right, which increases the positive charge of the nucleus which attracts electrons. Although electrons are also added as the elements move from left to right across a period, these electrons reside in the same energy shell and therefore do not offer increased shielding.

Source: Periodic Trends


By what mechanism do acids deprotect primary silyl ethers? Standard protocols report the selective deprotection of primary silyl ethers in the presence of secondary silyl ethers under acidic conditions, as exemplary shown in scheme 1, a reaction by Hartmann and Kalesse.[1]

Scheme 1: Selective deprotection of a primary silyl ether[1]

Numerous other examples can be found in the literature using numerous acidic reagents such as campher sulphonic acid,[2] para-tolylsulphonic acid[3] and even acetic acid[4] or $\ce{HCl}$.[5] Similar conditions can be used in the case of TIPS ethers.[citations needed]

I was wondering what the mechanism of this transformation looks like, and especially why it seems to be selective to primary silyl ethers. Answers that reference papers actually investigating the mechanism are preferred, but others are welcome, too.

List of abbreviations:

b.o.r.s.m: based on recovered starting material
Me: methyl
PPTS: pyridinium para-tolylsulphonate
TBS: tert-butyldimethylsilyl
TIPS: tetra-iso-propylsilyl

References:

[1] O. Hartmann, M. Kalesse, Org. Lett. 2012, 14, 3064. DOI: 10.1021/ol3011387
[2] Y. Mori, K. Yaegashi, H. Furukawa, J. Am. Chem. Soc. 1997, 119, 4557. DOI: 10.1021/ja9701523
[3] S. Kiyooka, K. A. Shahid, F. Goto, M. Okazaki, Y. Shuto, J. Org. Chem. 2003, 68, 7967. DOI: 10.1021/jo034901c
[4] B. H. Hart, S. K. Verma, H. Rapoport, J. Org. Chem. 2003, 68, 187. DOI: 10.1021/jo026499s
[5] A. B. Smith III, B. S. Freeze, I. Brouard, T. Hirose, Org. Lett. 2003, 5, 4405. DOI: 10.1021/ol035697i
[citations needed] I think, you get the picture, I’m not going to go looking for even more …

Fluoride-mediated deprotection of silyl ethers proceeds through a pentavalent silicon pathway:[1]

This mechanism proceeds the same for both primary and secondary silyl ethers, and there's no reason the mechanism for acid-catalyzed deprotection would be any different. Though I couldn't find a mechanism proposed in literature, it seems likely it would proceed through a similar pentavalent intermediate (Nu = nucleophile):

Steric effects are what drive the chemoselectivity of these reactions. Usually the differences in steric environments around the carbinol carbons is enough, but the use of different silyl groups on the two alcohols can further increase yield of the desired product. These steric effects are particularly pronounced in the following transformation, in which a primary TBS ether is selectively deprotected in the presence of a primary TBDPS ether and a secondary TIPS ether:[2]

Reaction conditions can also have a major effect on the chemoselectivity of desilylation reactions. For example, when HF•pyr is used in a THF-pyridine mixture, selective removal of the primary TBS ether in a protected triol was achieved in 84%. After oxidation of the newly deprotected alcohol to a carboxylic acid, the secondary TBS ether was cleaved using HF•pyr without excess pyridine, leading to lactonisation:[3]

Nelson and Crouch have written a review on the selective deprotection of specific silyl ethers in a compound with multiple silyl protecting groups.[4]

References

Kim, S.-K. Chitin and Chitosan Derivatives: Advances in Drug Discovery and Developments; CRC Press, 2013; p 80.
Larivée, A.; Unger, J. B.; Thomas, M.; Wirtz, C.; Dubost, C.; Handa, S.; Fürstner, A. The Leiodolide B Puzzle. Angew. Chem. Int. Ed. 2011, 50 (1), 304–309 DOI: 10.1002/anie.201005850.
Körner, M.; Hiersemann, M. Enantioselective Synthesis of the C8−C20 Segment of Curvicollide C. Org. Lett. 2007, 9 (24), 4979–4982 DOI: 10.1021/ol702092h.
Nelson, T. D.; Crouch, R. D. Selective Deprotection of Silyl Ethers. Synthesis 1996, 1996 (09), 1031–1069 DOI: 10.1055/s-1996-4350.


explain the reaction mechanism $\ce{ROH + NaBr + H2SO4 -> RBr + NaHSO4 + H2O}$

What is exactly the reaction mechanism.

Is it $\ce{NaBr + H2SO4 -> NaHSO4 + HBr}$ and then

$\ce{HBr}$ reacts with $\ce{ROH}$?

If yes the why in the reaction of $\ce{NaBr}$ and $\ce{H2SO4}$ does $\ce{H2SO4}$ oxidise $\ce{NaBr}$?

HBr reacts with ROH?

Exactly. It's just a nucleophilic substitution.


Equilibrium constant from mole ratio Chapter 4, problem 13 from the Chemical Priciples [1, p. 170]:

Equilibrium concentrations
Experiments have shown that at $\pu{60 °C}$ and $\pu{1 atm}$ total pressure, the equilibrium ratio of $\ce{NO2}$ to $\ce{N2O4}$ in moles in a closed vessel is exactly $2:1$.
a) Calculate the equilibrium constant, $K_\mathrm{c}$, for the dissociation of $1$ mole of $\ce{N2O4}$ into $2$ moles of $\ce{NO2}$.

My answer:
The chemical equation is $\ce{N2O4 <=> 2 NO2}$ and at equilibrium the ratio is $2:1$ so shouldn't
$$K_\mathrm{c} = \frac{[\ce{NO2}]^2}{[\ce{N2O4}]} = \frac{(\pu{2 mol L-1})^2}{\pu{1 mol L-1}} = \pu{4 mol L-1}?$$
The answer in the back of the text book is $K_\mathrm{c}= \pu{0.0488 mol L-1}$.
I do not understand how this answer was calculated from the information provided.
References

Dickerson, R. E.; Gray, H. B.; Haight, G. P. Chemical Principles, 3d ed.; Benjamin/Cummings Pub. Co: Menlo Park, California, 1979. ISBN-13: 978-0-8053-2398-6.

I suggest to pick one gaseous component, say, $\ce{N2O4}$, and, applying ideal gas law, express $K_\mathrm{c}$ as a function of the partial pressure of the chosen gas. Since at the equilibrium the ratio $n(\ce{N2O4}):n(\ce{NO2})=1:2$ is exact (e.g. molar fractions are $1/3$ and $2/3$, respectively), and the volume doesn't change, you can immediately deduce that

$$[\ce{N2O4}] = 2[\ce{NO2}];$$
$$p(\ce{N2H4}) = \frac{p_\mathrm{tot}}{3}$$

where $p_\mathrm{tot}$ is the known total pressure of the system. Using ideal gas law

$$[\ce{N2O4}] = \frac{p(\ce{N2H4})}{RT} = \frac{p_\mathrm{tot}}{3RT}$$

and the law of mass action

$$K_\mathrm{c} = \frac{[\ce{NO2}]^2}{[\ce{N2O4}]} = \frac{(2[\ce{N2O4}])^2}{[\ce{N2O4}]} = 4[\ce{N2O4}] = \frac{4p_\mathrm{tot}}{3RT},$$

one can plug in the parameters and calculate $K_\mathrm{c}$:

$$K_\mathrm{c} = \frac{4\cdot\pu{101325 Pa}}{3\cdot\pu{8.314 J mol-1 K-1}\cdot\pu{333 K}} = \pu{48.8 mol m-3} = \pu{4.88e-2 mol L-1}$$


How to know the formula of different compounds? For example, if I didn't know the compound tin chloride, and I was asked to write out its formula, how would I know how to? Because it seems unreasonable to know the formula of every compound in existence. Many thanks for the help!

In my experience if someone says something like tin chloride they are usually insinuating you know the most common oxidation state that metal is found in when complexed with the chloride atoms. This is a bad practice of them if they haven't explained further but I digress... In this example with the metal center of the complex being tin, the most common oxidation state is +2. This means the formula in its neutral state would be $\ce{Sn^{+2}}$ with 2 chlorides with a (-1) charge making the molecule neutral at $\ce{Sn(II)Cl2}$.

Different effects cause  common oxidation states for specific elements. These are outlined further with Mendeleev's Periodic Law. It breaks down into how shielded the nucleus is by electrons, and how big the radius of the atom is.

A further explanation of tin's common oxidation state calls upon knowledge of the inert pair effect, where elements to the right of the second and third row transition metals such as: In, Tl, Sn, Pb, Sb, Bi form compounds that are less reactive than one would think where the oxidation state is 2 less than the expected group. This is caused by the $\mathrm{ns^2}$ electrons being relatively inert. The cause of that nonreactive state is because of two reasons specifically:

Larger-than-normal effective
nuclear charges in these elements.
Lower bond energies in their compounds.

A good book to use referencing this subject further is Descriptive Inorganic, Coordination, and Solid-State Chemistry by Glen E. Rodgers. This is the book I own and used for when I studied periodic trends, and inorganic chemistry as an undergraduate. It's a great book for this subject, among others.

Also there are commonplace names that are used as somewhat of a slang for certain compounds used in research frequently. These can be learned through experience, or you can just google common names... All formal names are dictated by the rules set forth by the IUPAC. Learning those rules would be good idea. Also keep in mind that in some research projects you will encounter molecules that have informal nicknames too.


Is cobalt(II) in the hexaamminecobalt(II) complex high spin or low spin? I want to determine the  electronic configuration of $\ce{Co}$ in $\ce{[Co(NH3)_6]Cl2}$ with Crystal Field Theory.

I know that since the complex is octahedral, there would be degeneracy breaking and the d-orbitals would undergo splitting to form a $\mathrm{t_{2g}}$ level having $3$ orbitals and an $\mathrm{e_g}$ level above that with $2$ orbitals.

As cobalt is in the $+2$ oxidation state, it would have a $\mathrm{d^7}$ configuration.

Thus, if the complex is high spin, the electronic configuration is $\mathrm{(t_{2g})^5}\ \mathrm{(e_g)^2}$ and $\mathrm{(t_{2g})^6}\ \mathrm{(e_g)^1}$ if it is low spin.

This is where I ran into a problem, what type of complex is this?

Ammonia is generally considered strong field ligand, but it is weaker than say, $\ce{CO}$ as it is a σ only donor.

Wikipedia does not give anything definitive and I get inconclusive/contradictory information elsewhere on the Internet, so can someone please clarify whether the electronic configuration of cobalt in $\ce{[Co(NH3)_6]Cl2}$ is high spin or low spin?

Theoretically, you cannot predict a priori whether a compound is high- or low-spin. For some reason, a lot of people seem to think that it depends only on the ligand and that it is possible to unambiguously use the position of the ligand in the spectrochemical series to figure out whether a complex is high- or low-spin. This is a very narrow viewpoint and leads to lots of mistakes: for example $\ce{[Co(H2O)6]^3+}$ is low-spin although $\ce{H2O}$ is fairly low on the spectrochemical series. In truth it depends on (at least) the ligand, the metal, as well as the oxidation state, and there is no magic formula or rule that allows you to combine all three factors.

However, as a general rule of thumb, most 3d metal complexes are high-spin. If you return to the fundamental criterion for high- vs low-spin, i.e. the relative magnitudes of $\Delta_\mathrm o$ and the pairing energy, you will find that for many 3d metals, $\Delta_\mathrm o$ is small due to the poor overlap of the 3d orbitals with ligand orbitals. $\ce{[Co(NH3)6]^2+}$ falls squarely into this category.

The main exceptions occur when you have ligands that are very high on the spectrochemical series. These are nearly always π-acceptors, such as $\ce{CN-}$, bpy, phen, etc. Even so, it should be noted that there are some 3d π-acceptor complexes that are still high-spin, such as $\ce{[Co(bpy)3]^2+}$, so this shouldn't be taken as a rule but rather a rough generalisation. The other big exception is when you have high oxidation states, mainly +3 or higher. So, for example, Co(III) is nearly always low-spin except in $\ce{[CoF6]^3-}$. But there are plenty of high-spin Mn(III) and Fe(III) complexes, so even there it is not a clear-cut rule.


How can over grinding lead to reduced API extraction in HPLC analysis? I am looking for some advice regarding particle size distribution (PSD) & how this can be affected by grinding a tablet.
During an assay by HPLC, the method requires grinding of a tablet before dissolving in order to extract the active pharmaceutical ingredient (API) into solution. Of the tablet, the API is only 0.5% m/m of the tablet. I have been told that over grinding the tablets can affect extraction of the API and this is supposedly related to PSD as the API is the smallest particle found in the tablet amongst other excipients.
Could someone explain how "over grinding" would cause a detremential effect? Thanks.

I sort of feel this is actually more of a mathematical/statistical/modeling related question than traditional chemistry.
In essence, one argument is that the sample derived effective API concentration is impacted by possible changes in the particle size distribution due to grinding. Now, there are, in effect, known statistical distributions that address particle size distributions (PSDs), and associated methods to assess parameters of said distributions relating to sample means and variability, and changes thereto.  This would seem to be a possible component of the analysis.
As background, I reference Wikipedia on the topic, which importantly notes, to quote:

The PSD of a material can be important in understanding its physical and chemical properties...It affects the reactivity of solids participating in chemical reactions, and needs to be tightly controlled in many industrial products such as the manufacture of printer toner, cosmetics, and pharmaceutical products.

as both surface area (a function of particle size) and concentration influence a reaction.
Also, some detail on PSD distributions:

The log-normal distribution is often used to approximate the particle size distribution of aerosols, aquatic particles and pulverized material.
The Weibull distribution or Rosin–Rammler distribution is a useful distribution for representing particle size distributions generated by grinding, milling and crushing operations.
The log-hyperbolic distribution was proposed by Bagnold and Barndorff-Nielsen[9] to model the particle-size distribution of naturally occurring sediments. This model suffers from having non-unique solutions for a range of probability coefficients.
The skew log-Laplace model was proposed by Fieller, Gilbertson and Olbricht[10] as a simpler alternative to the log-hyperbolic distribution.

Now, for example, in the case of the Log-normal distribution, the best estimate of the parent population mean particle size (which here can be used to estimate total concentration) is obtained by working with the log values of the sample particle sizes. Wikipedia on the Log-normal distribution notes that the sample mean (say 'm') and variance (${s^2}$) of these log-transformed values can be employed to more accurately specify the parent population (aka, the Log-normal) mean, which turns out to be ${Exp(m + 0.5*s^2)}$. Note, simply taking the straight average of the observed Log-normals size values is not recommended as a Log-normal random deviate can be very noisy being the ${Exp()}$ of a Normal random deviate, hence the use of a log-transform. The Weibull distribution is another example of where the parent population mean is a function of two of the distribution's to-be-estimated parameters.
Further, it may indeed be the case that excessive grinding affects particle size distribution and reaction rate. However, this should be verified experimentally to determine the potential impact. Perhaps using a proxy, that produces a more readily measurable product (as in gas evolution, or strongly colored compound) may be of assistance.
Finally, this educational source comments on the inter-connection between reaction rate, mass and concentration, noting an exception, to quote:

The concentration of the reactants determines the speed of the reaction. In simple reactions, an increase in the concentration of reactants accelerates the reaction. The more collisions over time, the faster the reaction can advance. The small particles have less mass and more surface area available for the collisions of other particles. However, in other more complex reaction mechanisms, this may not always hold true. This is often observed in reactions involving huge protein molecules with large masses and convoluted structures with reaction sites buried deep within them that are not easily approached by collision particles.

So, still some experimenting required here in an obviously specialty discipline with a possibly large molecule.


Difference between the terms "autocatalysis" and "(branching) chain reaction" Is there a technical difference between the terms "autocatalysis" and "branching chain reaction"? They both seem to refer to something very similar, namely sequences of reactions in which one of the intermediates is also one of the products, leading to the possibility of exponential growth. Is one of the terms more specific than the other, or are they generally understood to mean the same thing?

There do seem to be some differences in how the terms are used. I've never seen the formose cycle, for example, referred to as a chain reaction, which does seem to imply that the term "chain reaction" means something more specific. If this is the case, what is the defining difference between the two terms?

Here is a diagram of the formose cycle (from the German Wikipedia entry, because it has a much better diagram than the English one) - is there a specific reason why it would not be referred to as a chain reaction?

An autocatalytic reaction is a reaction where the product catalyzes the reaction. An autocatalytic reaction takes the following form (or more complex variations of it).

$\ce{A + B -> 2B}$

An autocatalytic reaction must produce the exact species $\ce{B}$ that cataylzes the reaction. More about autocatalytic reactions, including examples, can be found at this question and this answer.

A chain reaction is one where every step of the mechanism produces another activated intermediate of some kind that can continue to react under the same conditions. Chain reactions continue until they terminate by means specific to the type of chain reaction. Chain reactions may appear to be autocatalytic, but there is a net consumption of both reagents, and the end product does not tend to cataylze the reaction:

$$\ce{A + B}* \ce{ -> C}$$
$$\ce{C + B -> D + B}*$$
Net:
$$\ce{ A + B -> D}$$

In the generic example above, the second step produces more $\ce{B}*$, which can reenter the first step. Overall, however, the reaction is not autocatalytic, since no more $\ce{B}$ is being produced and the concentration of $\ce{B}*$ is likely unchanging (steady state). All chain reactions have this feature. While the reactive species may appear on both the reactant side and the product side, there is no net change in the number of reactive species present in the reaction.

A classic introductory chain reaction is the radical halogenation of an alkane like methane. This reaction's mechanism is divided into three regimes: 1) Initiation (generation of radicals where there were none previously), 2) propagation (the chain part), and 3) termination (in which there is a net reduction in radicals, usually by recombination).

Initiation:
$$\ce{Cl2 ->[hv] 2Cl*}$$

Propagation:
$$1) \ \ \ \ \ \ce{Cl* + CH4 -> HCl + CH3*}$$
$$2) \ \ \ \ \ \ce{CH3* + Cl2 -> CH3Cl + Cl*}$$

Termination:
$$\ce{CH3* + Cl* -> CH3Cl}$$
$$\ce{2CH3*  -> CH3CH3}$$

The above example is a cyclic chain reaction. Chain reactions can be linear, for example chain-growth polymerization. In this kind of chain reaction, two species, where one is the activated end of a polymer chain $\ce{P}*$ and the other is a new monomer $\ce{M}$, react to form a new activated chain end $\ce{P}'*$ that is one monomer unit longer. $\ce{P}'*$ can then reenter the reaction and react with another $\ce{M}$. Again, this reaction is not autocatalytic. The number of activated chain ends is not increasing (or decreasing).

$$\ce{P}* +\ce{M -> P'}*$$
$$\ce{P'}* +\ce{M -> P''}*$$

The activated chain ends, $\ce{P}*$, can be radicals, anions, cations, the products of ring-opening, or organometallic complexes.


Why do some chemical reactions require many steps? I was a college student of physics ten years ago, and recently started to learn biochemistry. I enjoy finding out that some familiar concepts in physics play important roles in biochemistry such as Entropy and Gibbs free energy.

For example, as a (ex-)student of physics, I am happy to know that Gibbs free energy helps to decide the directions of chemnical reactions. I feel this is a good example where a sort of fundamental law of physics determines how phenomenon looks like.

However, I still can not understand why the chemical reactions in a body need to be so complex. Many chemnical systems consists of more than several steps to acheive their purposes. According to wikipedia, glycolysis takes ten steps through its process. Why so many steps are necessary?

I tried to find out a physical law which prohibit that glycolysis process is achieved by one or two steps. But I could not find an answer.

I would like to know (or discuss) whether there is a physical law which makes the chemical system such a complex one (many steps required).

My assumption is that some physical law prohibit the existence of a enzyme which realize a one-step process of glycolysis.

The number of steps required to model a reaction really depends on what time scales you are interested in looking at. Let's take the process of burning natural gas with air. If you only care about the final products of the combustion after a really long time, like let's say hours, then a single step reaction would be a perfectly good model. In fact, in situations where you only care about the global heat release and maybe the speed of the reaction at one particular condition, single-step or few-step reactions are just fine.

But let's say you are really interested in what's happening. When we look at a global reaction, let's say:

$$ O_2 + 2H_2 \rightarrow 2H_2 O $$

We'll assume this is in air, but the nitrogen doesn't really react. Nice and simple, right? Well, for the reactions to occur, you need some $O_2$ molecules to collide with $H_2$ molecules. But it's not enough that they collide. They need to collide with enough energy to break the bonds between them. And it's possible that one of the molecules breaks at lower energy than the the other. When they break, they break into 2 O's or 2 H'2. So our simple reaction really starts out as:

$$ O_2 + M \rightarrow 2 O + M $$
and
$$ H_2 + M \rightarrow 2 H + M $$

where $M$ is any other molecule. Okay, so now we've got a mixture of $H$, $O$, $H_2$, $O_2$, and $N_2$ floating around and bumping into each other. If they bump together with enough energy, we can start forming other things. Things like $HO_2$, $OH$, $H_2O_2$ and of course, $H_2O$.

Because these reactions rely on collisions, you need to form some things before you can form other things. So before an $O$ and an $H$ can collide to form $OH$, the $O_2$ and $H_2$ have to collide with other things and break apart. So there's finite time scales with all of the reactions, and they tend to cascade through a series of reactions before reaching the final product. If you are looking at time scales similar to those of the reactions, then these details really matter.

Hydrogen reactions are super simple. Imagine how complex organic molecules are. Even things like methane or propane have hundreds or thousands of steps to go from the complex molecule through all of the radicals to the final species. And the time scales the reactions occur over are all based on how frequently the collisions between them occur, and how much energy those collisions have.


Does H2CO3 exist in solution? When $\ce{CO2}$ reacts with water, it supposedly forms $\ce{H2CO3}$, which is the aqueous form of $\ce{CO2}$. But it is hard to isolate, but its $K_\mathrm{a}$ value is pretty low (on the order of $10^{-7}$) indicating it should be found in a non dissociated form? So does $\ce{H2CO3}$ exist in solution?

Aqueous carbonate solutions contain four different solute species:

$\ce{CO2(aq)}$
$\ce{H2CO3}$
$\ce{HCO3-}$
$\ce{CO3^2-}$

The equilibrium for the hydration reaction

$$\ce{H2O + CO2(aq) <=> H2CO3}\tag1$$

lies to the left. The corresponding equilibrium constant $K$ at $25\ \mathrm{^\circ C}$ is about

$$K=\frac{\left[\ce{CO2(aq)}\right]}{\left[\ce{H2CO3}\right]}\approx650$$

(literature values of $K$ range from $350$ to $990$.)

The commonly used first acid dissociation constant of carbonic acid $\mathrm pK_{\mathrm a1}=6.35$ (at $25\ \mathrm{^\circ C}$) actually is a composite constant that includes both the hydration reaction $\text{(1)}$ and the protolysis of true $\ce{H2CO3}$ $\text{(2)}$.

$$\ce{H2CO3 <=> H+ + HCO3-}\tag2$$

The constants $K$, $K_{\mathrm a1}$, and the actual acid dissociation constant $K_{\ce{H2CO3}}$ of true $\ce{H2CO3}$ are interrelated according to

$$K_{\mathrm a1}=\frac{K_{\ce{H2CO3}}}{1+K}\approx\frac{K_{\ce{H2CO3}}}{K}$$

Calculated values for $\mathrm pK_{\ce{H2CO3}}$ range from $3.8$ to $3.4$. Thus, true $\ce{H2CO3}$ is stronger than acetic acid and formic acid.

The second acid dissociation constant of carbonic acid is $\mathrm pK_{\mathrm a2}=10.33$ (at $25\ \mathrm{^\circ C}$).

Therefore, the predominant species in aqueous carbonate solutions (at $25\ \mathrm{^\circ C}$) are

$\ce{CO3^2-}$ at $10.33\lt\mathrm{pH}$
$\ce{HCO3-}$ at $6.35\lt\mathrm{pH}\lt10.33$
$\ce{CO2(aq)}$ and $\ce{H2CO3}$ at $\mathrm{pH}\lt6.35$

Thus, true $\ce{H2CO3}$ primarily exists in acidic solutions. However, because of the hydration equilibrium $\text{(1)}$ and the corresponding equilibrium constant $K\approx650$, most of the true $\ce{H2CO3}$ is converted to $\ce{CO2(aq)}$.

(All values taken from: Stumm, W.; Morgan, J. J. Aquatic Chemistry, Third Edition; John Wiley & Sons: New York, NY, 1996; pp 150–152.)


How would I know that a compound and its isomers are enantiomers? How would I know that a compound and its isomers are enantiomers? Considering the chiral carbons

Consider the compound is 2-bromo-2-chlorobutanedioic acid.

I am assuming that your question is this:
"How to check whether two given compounds are enantiomers of each other?"

First, you check for structural or geometrical isomerism.
If there isn't any, you check whether the compounds are diastereomers. (Check whether they are not the mirror images of each other.)
If they are not diastereomers, you have two possibilities:

Either the compounds are enantiomers of each other.
Or they represent the same compound.

You could try this approach:
First note the groups attached to the chirality centre. The Carbon is obviously an sp3 hybridized atom with four different groups attached to it.
Interchanging any two groups attached to the chirality centre produces the chiral compound's enantiomer.
So you could try interchanging the positions of the groups attached to the chirality centre until you get the other compound. If the number of swaps is even, the two compounds are identical. If odd, they are enantiomers of each other.


Why does ozone, being heavier than air, not settle down? Is this because of some equilibrium between diffusion and gravity ? What exactly is responsible?

While gravity does cause some separation of atmospheric constituents at high altitudes, the reason why ozone doesn't migrate to the ground has more to do with its chemistry.

Ozone is pretty unstable and is formed from some chemical reactions and from the interaction of oxygen with UV light. But it is unstable and decays fairly quickly. This means that is only appears anywhere when the reaction forming it keeps happening.  This is true in the upper atmosphere which is why we have an ozone layer. Some hydrocarbons (not just from pollution from cars but also from volatile stuff released by trees) can react to generate ozone at ground-level. But we only see a significant concentration when the reaction forming it is happening.

The effect of gravity on the separation of the components of the atmosphere is small, especially in the lower atmosphere where diffusion and turbulent circulation mix things up thoroughly. In the upper atmosphere the dominant process is diffusion and gravity can have more of an effect. This paper describes the process and has some details of the calculations about the size of the effect. It summarises the situation like this:

  Turbulent mixing keeps the relative concentrations of gases nearly constant in the lowest 100 km. At higher altitudes, molecular diffusion controls the concentrations, with the lighter gases becoming relatively more abundant with increasing altitude.

It is a worthwhile read if you want the detail.


How does the derivative of concentration make sense? First of all, I would like to make clear that I don't know very much about chemistry, and I'm not very sure this is the right place on Stack Exchange for this question, so please let me know if it needs migration.

I was reading a PDF about differential equations that mentioned equilibrium reactions and where the law of mass action was used. I looked it up on Wikipedia then looked up reaction rate because I was unfamiliar with the concept, and went to its formal definition.

How does the derivative of a concentration with respect to time even make sense? I seems to me that the concentration of a substance is  discontinuous in time and increases or decreases in steps.

Imagine a reaction

$$\ce{aA + bB <=> cC}$$

where $A,B,C$ are reactants and products and $a,b,c$ are stoichiometric coefficients. Now, we see that:

$$\ce{\frac{a}{N_A}A + \frac{b}{N_A}B <=> \frac{c}{N_A}C}$$

where $N_A$ is Avogadro's Number. These are the smallest stoichiometric coefficients for this equation. This reduced equation happens $N_A$ times in the above equation, considering it has an efficiency ($\eta$) of $100\%$. And this reduced equation takes some short time $\Delta t$. Before this reaction, the concentration of $A$ is $[A]_0 = \frac{\nu_A}{V}$ and after it, $[A]_{\Delta t} = \frac{\nu_A'}{V} = \frac{\nu_A - a/N_A}{V}$. I see no in-between that could link these two steps. How could we even define the concentration during the reaction? The only way I see it makes sense it to define the concentration on the interval $(0, \Delta t)$ as $[A]_t = [A]_0$. There is clearly a jump at $\Delta t$. And this doesn't happen only once. The graph of concentration vs time looks like a very compressed and scaled down floor function.

What am I missing?

There are two ways to look at this. In the first, we note that all thermodynamic and kinetic analyses rely on the assumption of very large populations of molecules. The curve of concentration vs time is made of discrete points, but the large population ensures that the change in concentration resulting from addition or subtraction of one molecule is so small that it can be treated as infinitesimal. Thus, we can create a smooth curve by interpolating between points, and the result will be continuous and differentiable. From a practical standpoint, any experiment to determine the concentration will result in discrete points separated by much much more than the change of one molecule, so we are implicitly assuming that the behavior between points can be treated as a well-behaved continuous curve.

The second, and perhaps more mathematically rigorous, view is that we never differentiate the actual concentration v time "curve". Instead, we are claiming that the change in concentration over time can be represented by a function which is continuous and differentiable. The only difference between this and the first view is semantic. Does $[A](t)$ refer to the actual concentration of A or to a function that is a proxy for the concentration? It doesn't matter.

Either way, the key point is that traditional kinetic and thermodynamic analyses assume a population large enough that the discrete behavior can reasonably be treated as equivalent to a continuous function.

[On a side note, there are those who would extend your argument to the $dt$ part of the derivative, as there is no proof that time is infinitely divisible, but that's a discussion for another time and place]


Why can solid lubricants operate only at lower temperature in oxidizing environment but higher temperature in reducing environment? The two main dry lubricants are graphite and molybdenum disulfide. They offer lubrication at temperatures higher than liquid and oil-based lubricants operate. Dry lubricants are often used in applications such as locks or dry lubricated bearings. Such materials can operate up to 350 °C (662 °F) in oxidizing environments and even higher in reducing / non-oxidizing environments (molybdenum disulfide up to 1100 °C, 2012 °F).

Source: Wikipedia contributors, "Dry lubricant," Wikipedia, The Free Encyclopedia, https://en.wikipedia.org/w/index.php?title=Dry_lubricant&oldid=870516892 (accessed July 24, 2019).

Why can solid lubricants operate only at lower temperature in oxidizing environment than in reducing environment?

Graphite is elemental carbon, and carbon is unstable in the presence of oxygen because they react to form carbon dioxide. The rate of this reaction: $\ce{C + O2 = CO2}$ is very slow and has a high activation energy, so a piece of graphite can sit in oxygen at room temperature for millions of years without anything happening to it. Once temperature increases, the reaction becomes much more rapid. From my own experience, a fist sized chunk of graphite can survive for several hours when heated to above 1000 °C in atmospheric oxygen.

Molybdenum sulfide is composed of two things that are likewise not stable in Earth's surface atmosphere: molybdenum and sulfur. It reacts by: $\ce{2MoS2 + 7O2 → 2MoO3 + 4SO2}$. As with graphite, this reaction is strongly temperature dependent.

There isn't a single temperature at which the lubricant is "not stable" any more. It is a balance of the required lifetime of the lubricant, the temperature, and how oxidising or reducing the environment is.


What is the purpose of a rate law? Other than determining whether reactants' concentrations can affect the rate, what purpose does it serve? Does it have any real world application in chemistry research? Also what is the point of knowing that a reaction is "second-order"?

Thanks!

Suppose you want to design an industrial scale continuous flow chemical reactor to produce some product stream from a set of feed streams.  You need to determine how big to make the reactor so that enough product is produced, given the reactant flow rates that you are feeding.  To figure this out, it's pretty important to know the rates of the reactions occurring within the reactor.  It's also important to know the amount of heat you have to add or remove from the reactor to keep it operating properly.  This is also determined by the reaction rates.

Another example is in atmospheric science where you need to know the rates of chemical reactions (including photolytic) occurring in the atmosphere so that you can quantitatively evaluate problems such as fluorocarbon/ozone and global warming.


Predict the species formed by the reaction of mercury(II) chloride and iodine $\ce{HgCl2}$ and $\ce{I2}$ both when dissolved in water containing $\ce{I-}$ ions the pair of species formed is

A. $\ce{HgI2, I3-}$
B. $\ce{HgI2, I-}$
C. $\ce{HgI4^2-, I3-}$
D. $\ce{Hg2I2, I-}$

Answer: option (c)
In this question, I was able to predict the formation of $\ce{I3-}$. So, I could eliminate option (b) and (d).  But, between option (a) and (d), I could not distinguish. Is there any specific way to guess answers for these type of questions in exam?

You are right about the formation of $\ce{I3-}$ ion (from the reaction of $\ce{I2}$ and solution containing iodide ion, let's say $\ce{KI}$ solution). Reaction between $\ce{HgCl2}$ and $\ce{KI}$ solution at $\pu{70 ^\circ C}$ produces mercury(II) iodide ($\ce{HgI2}$):
$$\ce{HgCl2 + 2 KI → HgI2 + KCl}$$
But the story doesn't end here. This mercury(II) iodide reacts with $\ce{KI}$ solution to form potassium tetraiodomercurate(II)($\ce{K2HgI4}$):

Crystallizing from a concentrated aqueous solution of mercuric iodide
with potassium iodide is the monohydrate $\ce{KHgI3.H2O}$, which is pale
orange. In aqueous solution, this triodido complex adds iodide to
give the tetrahedral tetraiodo dianion.

So, the answer is  C. but there is a typo. It should be $\ce{[HgI4]^2-}$. The overall reaction is:
$$\ce{HgCl2 + 5KI + I2 -> K2HgI4 + KI3 + 2KCl}$$


Net charge of Peptides at pH1 and pH0 Taking Glycine and aspartic acid. As both the amino acids are fully protonated at pH 0 ad pH 1, how does the peptide bond form? Is it a two step mechanism where the OH from an alpha carboxyl group bonds with a hydrogen from the alpha amino group of the other amino acid to form water then the H+ from the amino group being released to form a peptide bond?

First, note that the protonated amino group of an amino acid like glycine or aspartic acid has a pKa of about 9.6 - the number for glycine. At pH 1 it is indeed "fully protonated", but more correctly about 1 molecule in 10^(8.6) will have a free amino group at any given point in time (always in flux due to the rapidity of proton transfer). This amino group can react with a carboxyl group (from another amino acid) at the carbonyl carbon to give a tetrahedral intermediate, which in turn can eliminate water to give the peptide bond.

Under equilibrium at pH 0 or 1, the energetics typically favor the reverse of this reaction - hydrolysis of the peptide bonds to give the amino acids - this is what goes on in the stomach when you digest a protein - but it is possible that these conditions could produce a small amount of material from the reverse reaction - the amide-bond forming process.


Relationship between Energy Level and electron position In Bohr's Model of the atom there are set energy levels that are used to show the energy of the electrons in the atom. In the quantum model there are orbitals that show a probable location of the electrons within the atom.

How does the changing energy level of an electron affect it's position in the atom? For example, Helium has two electrons that both fit in the first "s" orbital. When the Helium electrons absorb the correct quanta of energy they would jump up to the next energy level, but what happens to their position in their orbital? As I understand, they can't move to the second "s" orbital, right, so would their distance from the nucleus just become greater while remaining in the first "s" orbital, or is its change in energy level a difference in the oscillation of the electron within its orbital?

The term "position in their orbital" is incorrect. The position of the electron is smeared out over the whole orbital.

I guess, you are still thinking about the electron as a particle, which moves around in the space defined by the orbital. That's wrong!

The orbital IS the electron (at least for one-electron systems; for many-electron systems this is just approximately true). The "probability density" is just about the likelihood of a measurement, if you would do one. Then the electron suddenly behaves like a particle, and needs to decide on a certain position.

One might try to make many of such measurements with very small time steps in between them. Any attempt to connect them to some continuous path or trajectory would fail. The electron would just appear to be randomly jumping around, popping up at different locations, no matter how short the time steps between the measurements.

During a transition to a higher state, the electron does not move. It just changes its wavefunction, instantaneously when absorbing the photon. And as the probability density depends on the orbital (aka wavefunction), the probability density changes as well. There is no need to consider the exact position, as in quantum mechanics we cannot grab it. It is smeared out. It is at every position at the same time (just with varying probabilities).


Reference on Pople basis set I need to write reference on 6-311++g(d,p) basis set in my work where I used it on atoms of I-II periods. Does anyone know where it was first published?

You should look on Basis set exchange

After you select your basis set of interest, on the bottom left you click "Get References for selected basis", and in this case you obtain (I did it for elements C,H,O):

  References for the basis set

  H

## VTZ Valence Triple Zeta: 3 Funct.'s/Valence AO for H,He
Krishnan, R., Binkley, J. S., Seeger, R., Pople, J. A.
Self-consistent molecular orbital methods. XX. A basis set for
        correlated wave functions
J. Chem. Phys., 72, 650-654 (1980)
10.1063/1.438955

## 1P Polarization functions associated with 6-311G for H,He
Krishnan, R., Binkley, J. S., Seeger, R., Pople, J. A.
Self-consistent molecular orbital methods. XX. A basis set for
        correlated wave functions
J. Chem. Phys., 72, 650-654 (1980)
10.1063/1.438955

## Diffuse functions asociated with 31G Hydrogen
Clark, Timothy, Chandrasekhar, Jayaraman, Spitznagel, Günther W.,
        Schleyer, Paul Von Ragué
Efficient diffuse function-augmented basis sets for anion
        calculations. III. The 3-21+G basis set for first-row
        elements, Li-F
J. Comput. Chem., 4, 294-301 (1983)
10.1002/jcc.540040303

  C,O

## VTZ Valence Triple Zeta: 3 Funct.'s/Valence AO
Krishnan, R., Binkley, J. S., Seeger, R., Pople, J. A.
Self-consistent molecular orbital methods. XX. A basis set for
        correlated wave functions
J. Chem. Phys., 72, 650-654 (1980)
10.1063/1.438955

## 1P Polarization functions associated with 6-311G
Krishnan, R., Binkley, J. S., Seeger, R., Pople, J. A.
Self-consistent molecular orbital methods. XX. A basis set for
        correlated wave functions
J. Chem. Phys., 72, 650-654 (1980)
10.1063/1.438955

## Diffuse functions asociated with 6-31G
Clark, Timothy, Chandrasekhar, Jayaraman, Spitznagel, Günther W.,
        Schleyer, Paul Von Ragué
Efficient diffuse function-augmented basis sets for anion
        calculations. III. The 3-21+G basis set for first-row
        elements, Li-F
J. Comput. Chem., 4, 294-301 (1983)
10.1002/jcc.540040303


E2 step in dehydration of primary alcohol using concentrated sulfuric acid The familiar two-step $\mathrm{E2}$ mechanism of the dehydration reaction of ethanol using concentrated sulfuric acid is given below (forgive me for not using diagrams):

$\ce{H-CH2-CH2-OH + H2SO4 -> H-CH2-CH2-OH2+ + HSO4-}$
$\ce{HSO4- + H-CH2-CH2-OH2+ -> H2SO4 + CH2=CH2 + OH2}$

I have certain doubts about the viability of the second step, as $\ce{HSO4-}$ is a weak base.

Maybe $\ce{H2O}$ is used as the base in the second reaction instead? If so, is the catalyst regenerated at all?

What actually happens, as always, depends on the exact conditions and concentrations which you did not add to your question. However, you do state whether the catalyst is regenerated; as such, I assume you are talking about the following reaction:

$$\ce{C2H5OH ->[\ce{[H2SO4]}] C2H4 + H2O}\tag{1}$$

I.e. a dehydration of ethanol using catalytic sulfuric acid.

The first thing we have to consider is the $\mathrm{p}K_\mathrm{a}$ values of sulfuric acid and ethanol (and water; which will be present anyway unless you accidentally found dry sulfuric acid). Using a very quick ballpark, we can assume: $$\mathrm{p}K_\mathrm{a}(\ce{H2O}) \approx \mathrm{p}K_\mathrm{a} (\ce{C2H5OH}) > \mathrm{p}K_\mathrm{a}(\ce{H2SO4})\tag{2}$$

Since sulfuric acid is catalytic, we can assume that the first (complete; non-equilibrious) step will be:

$$\ce{C2H5OH + H2SO4 -> C2H5OH2+ + HSO4-}\tag{3}$$

$\mathrm{p}K_\mathrm{a,2} (\ce{H2SO4}) \approx 2$, which is only two units less than water. Therefore, — under the assumption that we added $10~\mathrm{mol}\text{-}\%\ \ce{H2SO4}$ — we might assume that some additional molecules of ethanol (and/or water) are protonated under equilibrium conditions:

$$\ce{C2H5OH + HSO4- <<=> C2H5OH2+ + SO4^2-}\tag{4}$$

Thus, part of the sulfuric acid will be present as sulfate, the majority as hydrogen sulfate. Sulfate is a stronger base (albeit still weak) and thus the final mechanistic step is probably better written using sulfate rather than hydrogen sulfate:

$$\ce{H-C2H4-OH2+ + SO4^2- -> HSO4- + H2O + C2H4 ^}\tag{5}$$

Of all the bases present at equilibrium, sulfate is the strongest as its conjugate acid’s $\mathrm{p}K_\mathrm{a} > 0$. The general order of basicity of all compounds in solution is:

$$\ce{SO4^2-} > \ce{H2O} \approx \ce{C2H5OH} > \ce{HSO4-}\tag{6}$$

Finally, note that the reaction does not have to proceed via $\mathrm{E2}$ — an $\mathrm{E1}$ mechanism is almost equally viable.


How to calculate the temperature from heat of neutralisation? The heat of neutralization of $\ce{HCl(aq)}$ by $\ce{NaOH}$ is $\pu{-55.9 kJ/mol}$ $\ce{H2O}$ produced. If $\pu{50ml}$ of $\pu{1.6M}$ $\ce{NaOH}$ at $\pu{25.15^\circ C}$ is added to $\pu{25ml}$ of $\pu{1.79M}$ $\ce{HCl}$ at $\pu{26.34^\circ C}$ in a plastic foam cup calorimeter, what will be solution temperature be immediately after the neutralization reaction has occurred?

I'm always overwhelmed by the question if it provided me too much information. I don't know what equation and concept to use first. (But I know all related-equations in this unit.)
I need a detailed explanation to understand! I'm not sure which one is the initial temperature? $\pu{25.15^\circ C}$ of $\ce{NaOH}$ or $\pu{26.34^\circ C}~\ce{HCl}$?

First, you need to recognize what the limiting reagent is.  Because NaOH and HCl react in a 1:1 ratio, and there is less HCl ($\pu{25mL}$ of $\pu{1.79M}$ is fewer moles than $\pu{50 mL}$ of $\pu{1.6M}$), HCl is the limiting reagent.
Second, given the amount of limiting reagent, how much heat will be released (or absorbed).  $\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$
Third, you need to approximate that the solution has the heat capacity of water, which is $\pu{4.18 kJ/K L}$.
If you mix two volumes of the same substance at different temperatures, the temperature of combined volumes will be approximately the volume-weighted average.  So here: $$(50 \times 25.15 + 25 \times 26.35)/75 = 25.55$$
So finally calculate the amount the temperature of $\pu{75 ml}$ of "water" will increase when $\pu{0.025L} \times \pu{1.79M} \times \pu{55.9 kJ/mol}$ of heat is added, and add this to $\pu{25.55 ^\circ C}$.


What is the mass of 1 atom of carbon? I know that relative atomic mass of $\ce{^{12}C}$ is $12~\mathrm{u}$.

Therefore mass of $1~\mathrm{mol}~\ce{C} = 12~\mathrm{g}$
\begin{align}
 \text{mass of }6.022\cdot 10^{23} \text{ C atoms} &= 12~\mathrm{g}\\
 \text{mass of }1 \text{ C atom} &= \frac{12~\mathrm{g}}{6.022\cdot10^{23}}\\
&=\boxed{1.99\cdot10^{-23}~\mathrm{g}},\\
\end{align}

but

\begin{align}
 1~\mathrm{u} &= 1.66\cdot10^{-24}~\mathrm{g}\\
\implies \text{Mass of }1~\ce{^{12}C}\text{ atom} &= 1.66\cdot 12 \cdot 10^{-24}~\mathrm{g}\\
&= 19.92\cdot 10^{-24}~\mathrm{g}\\
&= 1.992\cdot 10^{-23}~\mathrm{g}.\\
\end{align}

Which is correct?

Also what is dimension formula of relative atomic mass, molar mass?

Why is unit of molar mass $\dfrac{\text{gram}}{\text{mole}}$ and not just $\text{gram}$ Since when has this $\text{mol}$ become a unit? It's just a number.

Both approaches are correct.

Avogadro's number is $6.02214129\times 10^{23}$ and represents the number of carbon-12 atoms in 12 grams of unbound carbon-12 in the ground electronic state.

$12$grams$/6.02214129\times 10^{23} = 1.9926467\times 10^{-23}$grams

The unified atomic mass unit (u) is $1.660538921 \times 10^{-24}$ grams

$12 \times 1.660538921 \times 10^{-24}$ grams $ = 1.9926467\times 10^{-23}$grams


Difference between chemical energy and potential energy I know that chemical energy is the potential energy stored in the arrangement of atoms within molecules. Forming new chemical bonds releases energy.

what does potential energy mean?
Is it the energy released when new chemical bonds are formed?
Or is it the needed energy to break the bonds of the compost?

There exist many different potential energies. Maybe you would prefer an analogy. For example let's consider an object on a table (a ball, a book or a piece of chalk). It has a potential energy which is not chemical for the moment. It is not visible. It is a hidden energy. It can only be shown if you give the object an opportunity to fall down and to go to a place where the potential energy is lower. During the fall, the potential energy is transformed into kinetics energy (its speed increases). When it touches the ground, the kinetics energy is transformed into heat which is another sort of energy. This gravitational potential energy is equal to mgH, where m is the mass of the falling object, H is the beginning height of the table, and g is a universal constant (9.81 m/s2 on Earth). Before the fall, the object has a sort of hidden energy, which we call gravitational potential energy (or mechanical energy), and this energy could only be expressed if you give the object an opportunity to go to a lower place. The object has a higher gravitational potential energy on the table than on the ground.

It is the same in chemistry. H2 and O2 have a similar hidden energy, which is a chemical potential energy. If you break the bonds H-H and O-O, you need energy to do it. You give energy to the atoms. It is similar to the operation of picking up the object on the ground and put it higher, on the table. If you imagine going back to H2 and O2, you would get the energy you had consumed for breaking the bonds. But now, if you prefer recombining the H and O atoms into H2O molecules, you will get much more energy (much more heat).

The object on the ground is similar to H2 and O2. When you give it some energy, you can lift it up to the table, or better to a state of separated H and O atoms. But now if the table is just at the edge of an abyss, you can get more energy by letting the object fall down the cliff. In this case you obtain H2O and not H2 and O2. The chemical potential has decreased to a very low level if the final state is H2O. The fall has produced much more heat than the recombination H2 + O2.

Chemists often use this analogy between height and chemical potential energy (which is usually called enthalpy). They draw short horizontal lines above one another to simulate  different states, as if one line represents the table (H and O atoms), and another one (lower) the level of the ground (H2 and O2). Of course the level of H2O is another horizontal line : much much lower.

Have you followed me ? Is it what you needed to understand the potential energy ?