Pressure and LeChatelier's Principle We know that if a system's equilibrium is exposed to a stress, the system shifts to relieve that stress. According to my teacher, certain kinds of pressures are stresses and others are not.

Normally, if we just say that the total pressure of a container with a reaction occuring inside is increased, the reaction shifts toward the side with a small sum of the mole ratios (i.e. $\rm 2A +3B \longleftrightarrow C + D$, if increasing pressure reaction shifts to the right to decrease pressure). I understood this formerly, but another case erases this understanding.

My teacher also said that if we add a noble gas to the container in which this reaction is occurring, although the pressure increases because of a new gas, the equilibrium does not shift to counteract this pressure increase. She says it has to do with the partial pressures of the reactants and products of only the particles In the reaction, but not in the total balloon. But this makes no sense to me because by increasing the total pressure of the system as described before, how does this cause an equilibrial stress and this case doesn't? How does increasing total pressure even cause an equilibrium shift? By increasing total pressure we have a proportional increase in the collision rates of all particles involved, do we not? If this is true, then why wouldn't the reaction rates both be doubled, and hence cause the equilibrium to not shift?

If I'm missing some fundamental principle, please let me know.

Thanks!

Le Chatelier's principle is not about pressure, it is about concentrations and temperature (Please also see the good article on Wikipedia for Le Chatelier's principle).

Le Chatelier's principle is also called "The Equilibrium Law", that can be used to predict the effect of a change in conditions on a chemical equilibrium. Chemical equilibrium means that reaction is reversible: it can go as forward, so backwards.

$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$

Lets imagine that $\rm A,\,B,\,C,\,D$ are gases. So you wrote correctly that a pressure increase will cause the reaction equilibrium to shift to the "right" - more products $\rm C,\,D$ will be formed. When we increase the pressure in the system, it means that we reduce volume. We squeeze our system, and there is less volume for molecules to be in. They hit each other more frequently, and the probability of product formation increases. When you reduce the volume by increasing the pressure, you basically increase the concentration of reagents. The same amount of atoms or molecules occupy lesser volume.

I'll make a silly comparison. Let's take this reaction:
$$\rm Boy+Girl\longleftrightarrow Pair$$
If there is too little space on the dance-floor to dance on your own it could be a good idea to find a partner to reduce the volume you both occupied to feel much more comfortable. But when the other people start to leave the club you can unpair with your partner...

Lets take 2 mol of $\rm A$, 3 mol of $\rm B$, 1 mol of $\rm C$, 1 mol of $\rm D$ at room temperature. We will put it in the jar of $\rm 156.8\,L$ ($\rm (2+3+1+1)\,mol \cdot 22.4\, L\cdot mol^{-1}=156.8\,L$). Pressure in the jar is $\rm 1\,atm$.

$P(A)=X(A)\cdot P(total)=2/7\cdot1=0.286\,atm$
$P(B)=X(B)\cdot P(total)=3/7\cdot 1=0.439\,atm$
$P(C)=X(C)\cdot P(total)=1/7\cdot 1=0.143\,atm$
$P(D)=X(D)\cdot P(total)=1/7\cdot 1=0.143\,atm$

Our system is in equilibrium, so we can calculate Equilibrium constant: $$K_P=\dfrac{P(C)^1P(D)^1}{P(A)^2P(B)^3}=\dfrac{0.143\cdot0.143}{0.286^2\cdot0.439^3}=3.17\,atm^{-3}$$
$$K_P=\dfrac{(X(C)\cdot P(total))^1(X(D)\cdot P(total))^1}{(X(A)\cdot P(total))^2(X(B)\cdot P(total))^3}=\dfrac{X(C)^1X(D)^1P^{\Delta n}}{X(A)^2X(B)^3};\,\Delta n=n(C)+n(D)-n(A)-n(B)=1+1-2-3=-3$$

Now lets double the pressure in the jar, by reducing the volume, the temperature did not change so $K_P$ stays the same. Lets calculate the molar concentrations from:

$$K_P=3.17\,atm^{-3}=\dfrac{X'(C)^1X'(D)^1\cdot (2\,atm)^{-3}}{X'(A)^2X'(B)^3},$$
 where

$\upsilon'(A)=2\,mol-2x\,mol$
$\upsilon'(B)=3\,mol-3x\,mol$
$\upsilon'(C)=1\,mol+x\,mol$
$\upsilon'(D)=1\,mol+x\,mol$

Total mols: $\sum=(1+x)+(1+x)+(2-2x)+(3-3x)=(7-3x)\,mol$

$X'(A)=(2-2x)/(7-3x)$
$X'(B)=(3-3x)/(7-3x)$
$X'(C)=(1+x)/(7-3x)$
$X'(D)=(1+x)/(7-3x)$

Solving the eqution gives $x=0.325\,mol$, so now:

$\upsilon'(A)=1.35\,mol$
$\upsilon'(B)=2.03\,mol$
$\upsilon'(C)=1.33\,mol$
$\upsilon'(D)=1.33\,mol$

But it the begining we has $A$=2 mol, $B$=3 mol, $C$=1 mol, $D$=1 mol. The reactions shifted to the right.

$P'(A)=0.45\,atm$
$P'(B)=0.67\,atm$
$P'(C)=0.44\,atm$
$P'(D)=0.44\,atm$

Ok, but when we add inert gas, what happens:

$$\rm 2A+3B+inert\,gas \longleftrightarrow C+D+inert\,gas $$

It does not do anything because even though the pressure increased, the probabilty for 2 atoms $\rm A$ to hit 3 atoms $\rm B$ does not change. The partial pressures stay the same. Why is it so? It can be easily shown. For example we have this reaction:
$$\rm A_g+B_g\longleftrightarrow C_g$$

Lets say we put this reaction in a jar of 134.4 L volume at room temperature, we have 3 mol of $\rm A$, 2 mol $\rm B$, and one 1 mol $\rm C$. 6 mol of gases being put in 134.4 L at RT will give us 1 atm pressure.
$$6\,{\rm mol} \cdot 22.4\,{\rm L\cdot mol^{-1}}=134.4\,{\rm L}$$

The partial pressures are:

$ P(A)=3/6\cdot P_{total}=0.5\cdot1\,atm$
$ P(B)=2/6\cdot P_{total}=0.33\cdot1\,atm$
$P(C)=1/6\cdot P_{total}=0.17\cdot1\,atm$

Now we will add 1 mol of inert gas:
$$\rm A_g+B_g+inert\,gas\longleftrightarrow C_g+inert\,gas$$

Now, the partial pressures are (pressure in the jar increased, because quantatiy of molecules increased, but volume stays the same):

$P(A)=3/7\cdot P_{total}=3/7\cdot 1.17\,atm=0.5\,atm$
$P(B)=2/7\cdot P_{total}=2/7\cdot 1.17\,atm=0.33\,atm$
$P(C)=1/7\cdot P_{total}=1/7\cdot 1.17\,atm=0.17\,atm$

As you can see, the partial pressures did not change (concentration did not change). That is why there is no effect.

Let's take this reaction:
$$\rm Boy+Girl+Cat\longleftrightarrow Pair+Cat$$
You cannot dance with a cat. Adding extra cats on the dance-floor will increase the pressure in the crowd, but will not increase the probability of formation of a dancing pair.

Let's imagine that $\rm A$ and $\rm B$ are liquids, and see what a pressure increase will cause. Usually it does not cause a lot, because it is very difficult to squeeze liquids. No matter how you try, the volume which molecules of liquid occupy stays the same. So increasing the pressure in reaction:

$$\rm A_\ell+B_\ell\longleftrightarrow C_\ell$$

will not do practically anything, because concentrations of reagents will not increase under the pressure.

When I wrote $\Delta H=-100\,kJ\cdot mol^{-1}$ (the enthalpy is negative), it means that in reaction the heat is released.

$$\rm 2A+3B \longleftrightarrow C+D;\space \Delta H=-100\,kJ\cdot mol^{-1} (for\space example) $$

So if we decrease temperature in the jar, reaction will move to the right. It can be understood using this primitive logic. Assume that heat is some substance, that could be treated as "product". If you remove this so called "product" (heat, by temperature reduction) you constantly force reactin to happen from left to right. In reality the equilibrium is just shifted from left to right.

If we increase temperature, it is like we added more of this so called "product" (heat) in the jar, so reaction is moved to the left.

If $\Delta H>0$, you need to think about heat not as "product", but as "reagent".

For better understanding try to "dive" into Equilibrium thermodynamics.


In the Haworth projections of D and L glucose, is the stereochemistry at every carbon except the anomeric carbon just inverted? I'm having a bit of trouble grasping sugar stereochemistry.

Using glucose as an example, if you cyclize D and L glucose, is the stereochemistry at every carbon just inverted from the other form?

Like for example, on carbon 2 for D glucose, the hydroxyl is down. On carbon 2 for L glucose, the hydroxyl would be up.

Is this trend true for every hydroxyl or $\ce{CH2OH}$ group except the one on the anomeric carbon?

Glucose is the hexose that projects its hydroxyl groups at positions C2, C3 and C4 as well as its CH2OH off C6 in the equatorial direction (in its low energy conformer that is). In the β and α configurations, its anomeric carbon C1 projects the hydroxyl in the equatorial and axial direction, respectively.
The D and L forms are mirror images in which all anomeric carbons have the opposite configuration between the two forms as:

Here are the equivalent structures using the Haworth projection.


Mechanism with hypohalite in haloform reaction The haloform reaction is usually done in a solution of NaOH and $\ce{I2}$ (or indeed, any halogen). Now the first part of the reaction mechanism in the wikipedia page and in the most organic books describes the reaction between the enolate and a halogen molecule ($\ce{X2}$).

[Picture source: Wikipedia, URL:https://upload.wikimedia.org/wikipedia/commons/8/8d/Haloform_Schritt_1.svg ]

Now, one of the things that I read in the wikipedia page is that the haloform reaction can also be done with sodium hypochlorite($\ce{NaOCl}$). We also know that hypohalite ions are produced in a solution of base and halogen, which is what we use in case of haloform reaction. So, I was wondering, is there any way to draw the mechanism with hypohalite only, without bringing in the halogen molecule? What would that mechanism look like?

The mechanism of haloform reaction will also be almost similar in the case if we use hypohalites (here, it is $\ce{NaOCl}$).

In the first step, instead of $\ce{OH-}$, the acidic $\alpha$ proton will be taken up by $\ce{OCl-}$ (Remember, $\ce{HOCl}$ is a weak acid (but not as weak as water), so it's conjugate base i.e $\ce{OCl-}$ is a moderately strong base to take up a proton), and $\ce{HOCl }$ will be formed, which will be attacked by the enolate formed in the previous step, and $\ce{OH-}$ will leave to form the $\alpha$-haloketone. The reaction will proceed further two steps to form the tri-halo substituted ketone, and correspondingly $\ce{OH-}$ will be generated. These generated $\ce{OH-}$ can then attack the electrophilic carbonyl carbon give the haloform and the salt of the organic acid .

The mechanism is shown below,


Is it possible to have only liquid in a closed container? I recently read an answer by Aman Rusia for this question: Is a liquid in a container always in equilibrium with its vapour?

In it, they say that

  You can't have a container filled with only liquid. Think about that
  for a moment. Similarly you can't have a container filled with only
  ice. Some water, or water vapour, or even combination of the two will
  appear alongside.

However, my textbook says that

  Imagine that you have a sample of water in a cylinder fitted with a piston at low pressure. Suppose that the temperature is held constant at 50 °C, and that weights are placed on a piston to exert a pressure of 1.0 atm. Only liquid water is present. The piston presses on the surface of the liquid. Now gradually reduce the pressure by removing some of the weights. At first, nothing seems to happen. The high pressure is keeping all the water molecules in the liquid state, and the volume of a liquid changes very little with pressure. However, when so many weights have been removed that the pressure has fallen to 0.12 atm (the vapor pressure of water at 50 °C), vapor begins to appear. The sample is now at the vapor-liquid boundary on the phase diagram. The pressure remains constant so long as the liquid and vapor phases are both present at equilibrium and the temperature remains constant. You are free to pull up the piston by an arbitrary extent, but enough water will evaporate to maintain the pressure at 0.12 atm. When you pull the piston out far enough, the liquid phase disappears; you are now free to modify the pressure of the vapor at will.

Atkins, Jones, and Laverman, Chemical Principles

So, is it true that you can have a container filled with only liquid?

In a tight, solid vessel, you can of course have a liquid (or any condensed phase) completely fill it.

There are basically three possibilities

the inside is at (potentially very) high pressure. no problem, water is compressible, just not very much, and also the vessel has a finite $E$ modulus
the inside is just at the vapour pressure of its content, but (because of the different expansion coefficients of vessel and content) the temperature must be kept perfectly constant at exactly the level where this is the case
the inside is below vapour pressure, but no gas bubble has evolved yet, that's principally an unstable condition, but if the deviation isn't large, it can stay practically forever.

Because of surface tension, there is even a small pressure range where the last case is actually stable.

I say your statement is basically right: You can have a completely filled vessel, but you can't do anything useful in it before it either bursts or bubbles.


Why does chlorine have a higher boiling point than hydrogen chloride? Chlorine has a boiling point of $238~\mathrm{K}$ while hydrogen chloride has a boiling point of $188~\mathrm{K}$.

Hydrogen chloride has dipole-dipole forces so I would expect it to have greater inter-molecular forces and thus a higher boiling point. However, since this is not the case, this must mean that chlorine has a greater intermolecular forces due to London dispersion forces.

But why?  I thought London dispersion forces were miniscule and only had a notable effect in large molecules. Shouldn't the dipole-dipole forces be way stronger than the London dispersion forces?

This is because $\ce{Cl2}$ has close to double the mass and is also a larger molecule compared to $\ce{HCl}$. You were correct in saying that London dispersion forces are weaker but because of $\ce{Cl2}$s size they overcome the dipole-dipole forces in $\ce{HCl}$.


Single electrode potential measurement My textbook says that in a galvanic cell, it's not possible to measure single electrode potential independently. Instead, a Standard Hydrogen Electrode is used as the system is under equilibrium. Can someone explain this concept more clearly?

Electrodes transport electrons produced from one half-cell to another, thus producing an electric charge. You need to have a reference value to compare it to in order to obtain the relative potential. Relative to the Standard Hydrogen Electrode (SHE), that is.

The SHE has a potential that is set arbitrarily to zero giving you a basis with which to compare the potentials from the electrode you're testing.

For further reading, I feel this article on the Standard Hydrogen electrode is pretty descriptive.


Periodic table- quantum numbers I have come across many questions like: "if electron had 3 spins (-1/2,0,+12) then what change will be there in the periodic table?", also sometimes " if the capacity of each orbital becomes 5 then in what period or group will the particular element be?" So what is the relation between quantum no. and the periodic table?

If electrons could have $m_s = -1/2, 0,$ and $1/2$, then the entire structure of the Periodic Table would be different. The s-block would have 3 elements instead of the 2 that we see now, because each s orbital can accommodate 3 electrons, and similarly the p-block would have 9 elements instead of 6.

It is not possible to say, for example, "sodium would be in so-and-so position in the new Periodic Table" because in such a universe, the element sodium would not even exist the way it does in our universe. The new Periodic Table would not have an element with the electronic configuration $\mathrm{(1s)^2(2s)^2(2p)^6(3s)^1}$. What I am basically saying is, there is no one-to-one correspondence between the elements in our universe and the elements in such a hypothetical universe (mathematically speaking, you can, but it would not make any chemical sense). Nevertheless, you could say that sodium is defined to be the element with 11 protons and 11 electrons. In that case, it would have a configuration of $\mathrm{(1s)^3(2s)^3(2p)^5}$ and it would be the fifth element in the Period 2 p-block.

Actually, there is no evidence that the aufbau principle should still hold true in such a universe: the large repulsion that arises from putting 3 electrons into the same orbital might well make filling the 2s orbital before the 1s orbital more favourable. So, our hypothetical "sodium" might have a weird electronic configuration of $\mathrm{(1s)^2(2s)^2(2p)^7}$. The Periodic Table is useful partly because it groups elements with similar electron configurations - which implies similar chemical properties - together. However, if these elements do not obey the aufbau principle, then there is no guarantee that the elements in the same group should have similar chemical properties, and therefore no guarantee that a Periodic Table in such a hypothetical universe would even be useful.

I should end off by saying that the laws of quantum mechanics stipulate that the different values of $m_s$ must increase in steps of $1$. Therefore, it is not actually possible that $m_s$ could ever take on values of $-1/2, 0$, and $1/2$, since these increase in steps of $1/2$. You might say: "Ah! But we could say, what happens if $m_s = -1, 0, 1$." Well, in that case, you have much more to worry about than just the orbital occupancy. Without going into too much detail, the Pauli exclusion principle would no longer apply to electrons, and you would have more problems than just "one orbital can fit 3 electrons".


What phase of matter is shaving cream? Our teacher asked us if shaving cream is a solid, liquid, or gas. She didn't quite know herself.

Shaving cream is a special type of mixture called a colloid, consisting of tiny air bubbles dispersed in liquid. This particular type of colloid can further classified as a foam. Since the mixture itself consists of two distinct phases, it doesn't really make sense to classify the entire mixture as one phase. If you had to pick one though, it is most nearly a liquid since that is the dispersion medium.


Dehydration of alcohol I have seen $\ce{P4O_{10}}$ and $\ce{HSO3Cl}$ being used to dehydrate carboxylic acids giving an anhydride and a ketene respectively.

Can  we use them to dehydrate $\ce{-OH}$ (of aliphatic alcohol) group from an organic compound? I have never seen anyone do that. If we can, then please specify the appropriate conditions for the reaction  to be carried out.

As you've noted, $\ce{P4O10}$ (or $\ce{P2O5}$ ) is a powerful dehydrating agent. According to the Chemical Book, it can actually dehydrate concentrated sulfuric acid to $\ce{SO3}$. So, it's not surprising that it can indeed dehydrate aliphatic alcohols. Using ethanol as an example, ethene can be produced by refluxing ethanol over $\ce{P4O10}$:

$$\ce{C2H5OH ->[P4O10] C2H4 + H2O}$$

Note that the reaction will proceed with gentle or no additional heating (the reaction itself is exothermic), it will just do so more slowly.


Do 2-hexyne and hexa-2-yne mean exactly the same thing? Do 2-hexyne and hexa-2-yne mean exactly the same thing? If so, what is the purpose of having two naming conventions?

For 2-hexyne, the extra naming possibility isn't really a help.

However, consider a more complex molecule, say, a hexynol.  It has both an alkyne and an alcohol.  A name like 2,4-hexynol would be very confusing.  Which group is at carbon 2 and which is at carbon 4?  That is why IUPAC doesn't provide for that possibility.  Instead, Writing it as hex-4-yn-2-ol is actually clearer and more understandable.


How does calcium carbide artificially ripen fruits? Lately, use of chemicals to artificially ripen fruits is in full swing. Firstly, how do they do it? Next, Are there any ways to distinguish between those that are naturally ripened and those that are artificially ripened?

Technically, there is no difference between naturally ripened fruit and artificially ripened fruit... they're both ripened fruit. I suppose you were looking for some sort of simple chemical test to distinguish between them; but like I said, there's no point in it since in both cases you end up with ripe fruit.

Fruits normally emit ethene (ethylene) which is essential to the ripening process. Which is why fruits picked before they're ripe are sometimes wrapped in a bag that seals in whatever little ethene they emit, thereby accelerating the ripening process.

As you would know, calcium carbide does not ripen fruit per se. Rather, it is the reaction of calcium carbide with water/moisture to release ethyne (acetylene) that's important here. Though I do not know of their mode(s) of action, apparently ethyne is almost a effective as ethylene in ripening fruits (I still need to find a reliable reference for this; though it is supposed to be a fairly well known fact).

If you let a fruit ripen naturally, or if you use ethyne instead... there's really no chemical difference at the end (apart from the usual arsenic contaminants that's associated with commercial-grade calcium-carbide).

The actual mechanism by which ethene (ethylene) and ethyne (acetylene) initiate, and thereafter, accelerate fruit-ripening is not really understood. Pectinases play a major role in the ripening process (they break down a structural component of plant cell walls called "pectin", which is why ripe fruit tend to be soft), however the scientific community has not yet arrived at a consensus as to how ethene and ethyne are linked to the activation/production of these enzymes.

However, there are a couple of subtle (non-chemical) difference between naturally and artificially ripened fruit:

Fruit that has ripened prior to plucking tend to be "fuller" than artificially ripened ones. This is because the fruit still gets an influx of nutrients while it's still attached to the tree...in other words; in the time it takes to ripen, the fruit grows as well. Artificially ripened fruit however, tend to be smaller, as they pretty much stop growing as they ripen (off the tree). ["Small", "big", "fuller" are crudely defined terms...you need some "experience" with fruits if you want to put this tip to good use].

Additionally, if you find the fruit being sold off-season (i.e- At a time of the year when it doesn't usually grow/ripen),  for a normal or somewhat cheaper price...then in all likelihood you have an artificially ripened fruit in your hand.

I reiterate:

It doesn't matter how it was ripened. It's ripe, that's all there is to it (from a consumer's PoV). Don't forget to wash and peel it before taking a bite :-)


What is the number of unpaired electrons in cobalt(II) tetrathiocyanate? Find the number of unpaired electrons present in the d orbital (whose lobes are present along the axis) for the complex $\ce{[Co(SCN)4]^{2-}}$.

Since $\ce{{SCN}^{-}}$ is a weak ligand I did not pair the electrons, but still I got the wrong answer. The correct answer is 0. But how?

According to the tetrahedral splitting in crystal field theory, the filling of the orbitals is just opposite to the filling in case of octahedral splitting. Thus filling up the elecrons according to it and obeying the hunds rule (since it is a weak ligand) the number of unpaired electrons are 0 in $d_{x^2−y^2}$ and $d_{z^2}$ whose lobes are directed along the axis.


Evolution of gas during chemical reaction Why is it that whenever hydrogen gas is produced in a reaction, it comes out in the form of tiny bubbles, but whenever Carbon dioxide is produced it comes out in the form of a brisk effervescence like lather?

Likely, any difference you've observed is due to the high solubility of $\ce{CO2}$ in water... it allows accumulation of enough dissolved gas to nucleate and form large bubbles, as in seltzer. BTW, $\ce{Na}$ or $\ce{K}$ etc. do not make tiny bubbles when put in water.


Shapes of ionic compound Can anyone explain the shapes of ionic compounds?  What does it mean? What does the coordinate number mean? I know that in $\ce{NaCl}$, one $\ce{Na}$ atom is surrounded by 6 $\ce{Cl}$ atoms, and vice versa, but can it be explained with the help of coordinate numbers?

Only for solids that have regular and repeating pattern of constituent particles we define shape and geometry.

So, on the basis of order of arrangement of constituent particles solids are classified into crystalline and amorphous.The former has regular pattern of arrangement of particles which repeat itself periodically over the entire crystal (as shown below is the two dimensional structure of quartz)

Amorphous solids consist of particles of irregular shape(as shown below is 2d structure of quarts glass.

If the three dimensional arrangement of constituent particles of crystalline solid is represented diagramatically in which each particle is depicted as a point, the arrangement is called crystal lattice

Unit cell is the smallest portion of crystal lattice which when repeated in all directions generate the entire crystal (see below). There are only 14 possible crystal lattices called Bravais Lattices( after a French mathematician who first described them.

On the basis of unit cell parameters (axial distances and axial angles)
we get the following seven primitive unit cell which defines the shape of solid.

COORDINATION NUMBER
this term is used in crystallography in a somewhat different sense than it is in coordination compounds.

The number of nearest neighbors (or touching particles) that a particle has in a crystal is called its coordination number. All atoms in bravais lattice have the same coordination number.

In a simple or primitive unit cell, around any given atom there would be six equally spaced nearest neighbours.Thus the coordination no=6.

Any other particle in crystal lattice will be at a greater distance from the particle under consideration. This also apply to your NaCl crystal.


Does changing the stoichiometric coefficients affect the Nernst equation? If I take the electrodes as nickel and silver, I've according to the nernst equation that

$\ce{Ni(s) + 2Ag+(aq) -> Ni^2+(aq) + Ag(s)}$

Now $E=E_{0} - \frac{RT}{2F} \ln\left(\frac{[\ce{Ni^{2+}}]}{[\ce{Ag+}]^2} \right)$

But if I double the stoichiometric coefficients, the value inside the logarithm would square up. So if I write the same reaction but with different coefficients, I'll get different values of potential of the cell?

No, you won't get a different value. The number of electrons transfered in that case is 4, so it will be $E=E_{0} - \frac{RT}{4F} \ln\left(\frac{[\ce{Ni^{2+}}]^{2}}{[\ce{Ag+}]^{4}} \right)$.


Meaning of "MS1 and MS2 resolution" and "MS1 and MS2: Unit" in a tandem mass-spectrometry parameter list I'm translating a Certificate of Suitability that very briefly describes procedures used to measure several impurities in a drug substance. There is a short description of the parameters of a Gas Chromatography-MS/MS procedure to detect an impurity called NDBA.
I don't understand the final line here:

What is the meaning of Resolution and especially of Unit? I can get by with "resolution" by simply translating it into Russian without knowing what it means, but there is no suitable Russian word to translate "Unit" - I must use either the Russian words for "Unit operation" or for "Equipment unit" or for "Measurement unit", and I have no idea what this Unit means in the original text.
I tried reading up on MS/MS but I haven't understood it well enough to guess the meaning of this line.

The term "unit" probably means the size of the isolation window used in the QQQ.  The term "unit" means the nominal isolation window is 1 Da (i.e. one mass "unit") wide.
The "resolution" is a standard term in mass spectrometry, and here they are saying it is about 1 Da.


Why is magnesium sulfate acidic? Why are my calculations showing that it is basic? Let's say I was dissolving $\pu{1 M}~\ce{MgSO4}$ into water and I wanted to find its $\mathrm{pH}$. I would go right to the reaction of the dissociation of $\ce{Mg(OH)2}$:
$$\ce{Mg(OH)2 <=> Mg^2+ + 2OH-}$$
The $K_\mathrm{sp}$ expression for this reaction is:
$$K_\mathrm{sp}=[\ce{Mg^2+}][\ce{OH-}]^2$$
According to wikipedia (see sources at bottom), the $K_\mathrm{sp}$ for $\ce{Mg(OH)2}$ is $\pu{5.61*10^{-12}}$
$$\pu{5.61*10^{-12}}=[\ce{Mg^2+}][\ce{OH-}]^2$$
My molarity of $\ce{MgSO4}$ will be the same as that of $\ce{Mg^2+}$, which is $\pu{1 M}$. Substitute that in.
$$\pu{5.61*10^{-12}}=[\ce{OH-}]^2$$
And then solving for $[\ce{OH-}]$ we get:
$$[\ce{OH-}]=\pu{2.37*10^{-6} M}$$
Therefore:
$$\mathrm{pOH}=5.63$$
and
$$\mathrm{pH}=8.37$$
So my final solution of $\ce{MgSO4}$ will be basic. However, on stack exchange it says that the $\mathrm{pH}$ of $\ce{MgSO4}$ is between 5.5 and 6.5. The post does not explain why, but that the general consensus is that that is the range of the $\mathrm{pH}$.
What did I do wrong?
Sources:
https://en.wikipedia.org/wiki/Magnesium_hydroxide
Is magnesium sulfate basic, neutral or acidic?

Your answer lies not in the solubility product but in the acidity constants of sulfuric acid and the basicity constants of $\ce{Mg(OH)2}$, though the solubility product may well come into play. You would need to solve a system of 6 equations, one for each of the:

2 $\mathrm{p}K_\mathrm{a}$ values for the acid
Two $\mathrm{p}K_\mathrm{b}$ values for the base
Proton condition or charge neutrality
And as you will be groping in $\mathrm{pH}$, one to be sure you don't exceed the solubility product.

As sulfuric acid is 'stronger' as an acid than $\ce{Mg(OH)2}$ is as a base, you would expect the answer to be slightly on the "acid side" of neutral $\mathrm{pH}$.


Chemical Biology - synthesizing/designing siRNA I was reading an article on siRNA, and from what I read it sounded like we can edit genes using siRNA. First, did I even interpret this correctly? Second, I read that siRNA was being used to create drugs to help correct mutations. Does this mean that they could also give positive mutations to people? Finally, how does the structure of siRNA differ from that of normal RNA?

At this time, it it possible to completely chemically synthesize oligonucleotides (DNA/RNA) between 30-50 base pairs. siRNAs are a class of RNAs that, as the name suggests, bind to other RNAs in a sequence specific manner. Specifically, siRNAs preferentially target mRNAs, which blocks translation of those mRNAs into proteins. This regulation of gene expression has absolutely nothing to do with "creating genes for a human being".


Why are pi bonds only formed when sigma bonds are formed? While studying about bonding there was one statement that "pi bonds can only be formed only with sigma bonds" as we know that in double bond there is 1 sigma bond and 1 pi bond but then one question arises: Why is a pi bond formed only when a sigma bond is formed? Is it possible to form a pi bond without any formation of a sigma bond?

A $\pi$ bond has a plane of symmetry along the bond axis. It cannot be formed by s-orbitals; it needs at least p-orbitals to be created. $90\,\%$ of all bonds described some time or another are somehow involving carbon, nitrogen or oxygen. (In fact, I probably underestimated). But these elements can only use p-orbitals to create $\pi$ bonds. To do that, one needs a p-orbital that is ortohogonal to the bond axis. So you run into the problem that you have an orbital pointing in one direction, but want to bond into another direction — hardly optimal, especially since there likely is already another orbital pointing in the direction you need to give a $\sigma$ bond.

Transition metals can use d-orbitals for $\pi$ bonding. They can actually point towards the atom they want to bond with so there is a greater chance of using them due to higher overlap. However, there will usually also be a different orbital pointing directly in the bonding direction which again will bond earlier and would give a $\sigma$ bond.


Why does raising the temperature affect the mass of dissolved solute? Consider a solution of $\ce{KOH}$ which is in equilibrium with undissolved $\ce{KOH}$ at $25\ \mathrm{^\circ C}$.
After raising the temperature to $50\ \mathrm{^\circ C}$, the mass of undissolved $\ce{KOH}$ decreases. Why is this?

For many (though not all) solids dissolved in water, solubility increases with temperature. The increase in kinetic energy that comes with higher temperatures allows the solvent molecules to more effectively break apart the intermolecular forces of the solute molecules.

For many salts and some gasses there are visual representations of solubility variation with temperature called a solubility curve, for which the units are $\frac{\mathrm{g~solute}}{\mathrm{100~g~solvent}}$. I could not find one for $\ce{KOH}$, but the Wikipedia page for $\ce{KOH}$ lists the solubility of $\ce{KOH}$ at various temperatures, and it is clear the trend is that it is more soluble at higher temperatures.

As a result of Le Châtelier's principle, substances that release heat upon being dissolved tend to dissolve better in colder water. This accounts for gasses and some salts.

You should note, however, that although $\ce{KOH}$ does evolve a large amount of heat upon being dissolved, as the concentration of $\ce{KOH}$ increases, the process becomes less exothermic, and eventually near the saturation point it becomes endothermic (it becomes less favorable to dissolve more $\ce{KOH}$), which explains why it is more soluble at higher temperatures.


Finding cubic ice CIF files in crystal database Could anyone please tell me where I could find the the structure of cubic ice (ice Ic) as a CIF file?

I have been searching the cambridge structural database using "H2O", "water", "dihyrogen oxide" and "ice" as search key words, but cannot find anything!

One resource you can use is the Crystallography Open Database. Search for structures with two elements O and H, and named “ice”. You will then find several experimental structures for cubic ice:

with downloadable structure files in CIF format: 1, 2. These were done from X-ray diffraction studies and only feature heavy atoms (oxygen) positions.

Although I could not find in any open database a CIF file for cubic ice with hydrogen positions, it is easy to make one. From the paper “High-resolution neutron powder diffraction study of ice Ic”, Le Journal de Physique Colloques, 48.C1 (1987), C1-631:

you can create the CIF file yourself:

data_Ice_Ic
_audit_creation_method         'generated by CrystalMaker 9.2.7'
_publ_section_comment
;
Chemical formula (sum): H2 O

Systematic name: H2 O

Crystal structure data from: W.F. Kuhs, D.V. Bliss, J.L. Finney,  High-resolution neutron powder diffraction study of ice Ic, Journal de Physique Colloques, 1987, 48, C1-631

;
_cell_length_a                   6.358(3)
_cell_length_b                   6.358(3)
_cell_length_c                   6.358(3)
_cell_angle_alpha               90.000(0)
_cell_angle_beta                90.000(0)
_cell_angle_gamma               90.000(0)

_symmetry_space_group_name_H-M     'F 41/d -3 2/m'
_symmetry_Int_Tables_number         227
_symmetry_cell_setting             cubic
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    H1    H  0.5000      0.4670(19) 0.4670(19)  0.4670(19)
    O1    O  1.0000      0.3750     0.3750      0.3750


Mixing Oxalic acid and glycerin I am doing an experiment for a miticide for honey bees. I will be dissolving 600 g of 99.6 % oxalic acid in one liter vegetable glycerin at 150 °F. I have been told it will give off formic acid fumes while mixing and heating. I don't think it will until it get to the melting temperature of oxalic acid. Is there any safety concern? Also what would the shelf life of the mixture be?

You might want to raise the temperature a little bit as the reaction proceeds at 100-110°C to form formic acid. If the temperature is further raised to 260-280°C, then the reaction will proceed to form allyl alcohol.

Formic acid has almost unlimited shelf life if unopened and is protected from heat and light sources. Allyl alcohol is stable in normal storing conditions but it may polymerize into a thick syrup if left untouched for several years.

Discussing about safety concerns is pretty much off-topic here but you might want to check the safety data-sheets: formic acid and allyl alcohol.

Here is a research paper for further reading.

Reaction sources: Google Images


Chromatography in geology I just started reading about the technique of chromatography and learning how important it is.

Now I'm trying to focus on whether is possible to use this technique in the geological fields (fossils and dating for example.)

I'm just brainstorming so any thoughts will help here.

A typical example for the application of models and experimental methods of chromatography for geologic materials (e.g. soil, sediment) is the $K_\mathrm d$ model, which can be used for understanding the long-term behaviour of contaminants in the subsurface environment (e.g. the rooting zone of crops).

The parameter $K_\mathrm d$ is the partition coefficient, i.e. the ratio of the concentration of a contaminant sorbed on a specified solid phase (e.g. soil) to the concentration in a specified liquid phase (e.g. groundwater):

$$K_\mathrm d=\frac{\text{concentration in solid phase (per kg)}}{\text{concentration in liquid phase (per l)}}$$

The partition coefficient $K_\mathrm d$ can be used to calculate the retention factor $k$:

$$k=\frac\rho\varepsilon K_\mathrm d$$

where
$\rho$ is the dry bulk density of the solid phase, and
$\varepsilon$ is the total porosity of the solid phase.
Typical values for soil are $\rho=1.4\ \mathrm{kg\ l^{-1}}$ and $\varepsilon=0.3$.

This parameter is related to the retardation factor $R_\mathrm F$:

$$R_\mathrm F=\frac1{k+1}$$

The retardation factor $R_\mathrm F$ corresponds to the ratio of the distance travelled by the contaminant to the distance simultaneously travelled by the mobile liquid phase:

$$R_\mathrm F=\frac{v_\text{contaminant}}{v_\text{mobile phase}}$$

For example, if the annual precipitation exceeds the evaporation by $300\ \mathrm{mm}$ and the porosity of the soil is about $\varepsilon=0.3$, the drainage water travels with an average velocity of $v_\text{mobile phase}=100\ \mathrm{cm\ a^{-1}}$ downward through the soil.

Assuming a mean value for the partition coefficient of plutonium in all soils of

$$K_\mathrm d=7.4\times10^2\ \mathrm{l\ kg^{-1}}$$

the corresponding retention factor is

$$\begin{align}
k&=\frac\rho\varepsilon K_\mathrm d\\[6pt]
&=\frac{1.4\ \mathrm{kg\ l^{-1}}}{0.3}7.4\times10^2\ \mathrm{l\ kg^{-1}}\\[6pt]
&=3.5\times10^3
\end{align}$$

and the retardation factor is

$$\begin{align}
R_\mathrm F&=\frac1{k+1}\\[6pt]
&=\frac1{\frac\rho\varepsilon K_\mathrm d+1}\\[6pt]
&=\frac1{\frac{1.4\ \mathrm{kg\ l^{-1}}}{0.3}7.4\times10^2\ \mathrm{l\ kg^{-1}}+1}\\[6pt]
&=2.9\times10^{-4}
\end{align}$$

Therefore, the migration rate is

$$\begin{align}
v_\text{contaminant}&=R_\mathrm F\cdot v_\text{mobile phase}\\[6pt]
&=2.9\times10^{-4}\times100\ \mathrm{cm\ a^{-1}}\\[6pt]
&=0.029\ \mathrm{cm\ a^{-1}}
\end{align}$$

i.e. a downward migration of $1\ \mathrm{cm}$ takes about 35 years.


How many grams of CH3OH can be made from 30.0g of CO if it all reacts? Once made by heating wood in the absence of air, methanol ($\ce{CH3OH}$) is now made by reacting carbon monoxide and hydrogen at high pressure:
$\ce{CO_{(g)} + 2 H2_{(g)} -> CH3OH_{(l)}}$

How many grams of $\ce{CH3OH}$ can be made from 30.0g of $\ce{CO}$ if it all reacts?

Moles of CO in equals moles of methanol out (at 100% yield), typically over copper-doped zinc oxide catalyst.  Do the arithmetic.

One gram-mole of a chemical formula is [the magnitude of its atomic mass units] grams.  A nitrogen atom is 14.0067 amu.  A mole of nitrogen gas, $\ce{N2}$, is then 28.0134 grams.


Converting sulfate ion to sulfur trioxide? I have a solution of only sulfate ions (no other positive ions belonging to the sulfate ions, because the positive ions are attached to the intermediate in a catalytic reaction).

I want to convert the sulfate ions to sulfur trioxide by using $\ce{V^5+}$ ions...

So what if $\ce{V^5+}$ ions could be present and could gain 2 electrons from one sulfate ion to produce $\ce{V^3+}$ and $\ce{SO3}$ (however I'm not even sure if $\ce{SO3}$ would be formed)? Here the two negatively charged oxygen atoms will each lose 1 electron to the $\ce{V^5+}$ ion, so the oxidation state of the central sulfur atom is still $+6$, but the oxidation states of the 2 oxygen atoms are $0$.

Could this be possible? And if so, would it be $\ce{SO3}$ that is formed, or other sulfur oxides?

A good analogy to your proposed reaction is the formation of carbon dioxide above a solution of carbonate salts.  Like sulfur trioxide, carbon dioxide is an acid anhydride.  The acid for sulfur trioxide is sulfuric acid, $\ce{H2SO4}$.  The acid for carbon dioxide is carbonic acid, $\ce{H2CO3}$.

Dry carbonate salts like sodium carbonate can be made to release $\ce{CO2}$ by heating to high temperatures.  For example, calcium carbonate decomposes to calcium oxide and $\ce{CO2}$ at around 800 °C to 900 °C.  Since such high temperatures are required, this technique does not work with aqueous solutions at ambient temperatures.   (The water would boil off way before gas is released.)

Thus, sulfur trioxide evaporation from sulfate salts would be even more disfavored at room temperature than carbon dioxide evaporation from carbonate salt solutions, because of sulfur trioxide's affinity for water is much higher than $\ce{CO2}$'s.

Also, the reaction mechanism you have drawn is not a dehydration, but a reduction of the sulfate.  That would not form sulfur trioxide, but a reduced sulfur compound, such as sulfite.  Sulfur dioxide gas evaporation from above sulfite salts is probably more favored than sulfur trioxide evaporation, but that isn't saying much. Either one is still highly disfavored under neutral or alkaline pH.


Peak at m/z = 39 in mass spectrum of 3-chloropropene I've noticed that in the mass spectrum of 3-chloropropene, there is a peak at m/z = 39, which to me, indicates the formation of the C3H3 ion.

However, given that 3 bonds would have to be broken to form this ion, how can the frequency of this cleavage be explained?

It would indeed be $\ce{C3H3^+}$ assuming predominant isotopic species.  Although there can be other isomers, the cyclic isomer, the cyclopropenyl cation is strongly stabilized by aromaticity.

As PLD comments, there are really only two steps, not three, to forming this cation.  The allyl cation ($\ce{C3H5^+},m/z=41)$ is first obtained by loss of the chlorine atom plus an electron.  The aromatic cyclopropenyl cation noted above is then obtained by loss of $\ce{H2}$.


How to use salt spray data in real environments? I want to convert resultant hours of salt spray corrosion test to estimated values in years for real life environments such as marine. In other words, How could I predict the result of real life tests using salt spray data?

I have to agree with Klaus.  ASTM B117 is, at best, a QC procedure.  In my experience, most corrosion and coatings scientist who use B117 do so because it is very well established, but it does not correlate to a real-world measure of performance.  ASTM D5894 is a better cabinet test and there is also an SAE procedure developed for automobiles.


Why does sodium react with water to produce a hydroxide, while zinc produces an oxide? Sodium belongs to the so-called active metals and produces a hydroxide on contact with water. Zinc produces an oxide. Why is it so?

I was given a hint: to read up about electrochemical series. I've read about it, and naturally zinc is way below sodium in the series. Sodium is more enthusiastic about giving away its electrons and forming positive ions.

Zinc is slightly below this bolded line in the series, but I'm afraid I do not fully understand what that means. "Hydrogen gas bubbled through an alkaline solution, shedding two electrons to produce water and neutralize the solution"?

The overall potential of an electrochemical cell is equal to the reduction potential for the substance being reduced plus the oxidation potential for the substance being oxidized. In other words, $ΔE_{rxn} = E°_{red} + E°_{ox}$. Now, all that we need to do is find a list of reduction potentials. From here, I found the following reduction potentials:

$$\ce{Na+(aq) + e- -> Na(s)}: E°_{red}=-2.71 V $$

$$\ce{2H2O(l) + 2e- -> H2(g) + 2OH^{-}(aq)}: E°_{red}=-0.83 V $$

$$\ce{Zn^2+(aq) + 2e- -> Zn(s)}: E°_{red}=-0.76 V $$

A redox reaction occurs between one species being oxidized (losing electrons) and another being reduced (gaining electrons). Since all of these values are reduction potentials, we can flip reactants and products to get oxidation potentials. The oxidation potential for a reaction is negative one times the reduction potential (due to conservation of energy) and since we possess sodium and zinc metal rather than ions, we want for these reagents to be oxidized. This gives us the following equations with oxidation potentials:

$$\ce{Na(s) -> Na+(aq) + e-}: E°_{ox}=2.71 V $$

$$\ce{Zn(s)->Zn^2+(aq) + 2e-}: E°_{ox}=0.76 V $$

Now, you might be wondering what any of this has to do with your original question. It turns out that a reaction is spontaneous (will occur) when the $ΔE_{rxn}>0$. Calculating $ΔE_{rxn}$ for the oxidation of sodium gives us $ΔE_{rxn} = E°_{red} + E°_{ox}=-0.83V + 2.71V=1.88V$, which is greater than zero, so this reaction will be spontaneous. On the other hand, if we calculate the cell potential for zinc being oxidized and water being reduced, we get a negative number, so this reaction is not spontaneous. In other words, sodium will readily reduce water, but zinc will not. Since zinc cannot directly reduce water, it will react with water to produce zinc hydroxide and hydrogen gas. The zinc will then decompose into zinc oxide and water. The reaction mechanism for this is given here:

$$\ce{Zn + 2 H2O → Zn(OH)2 + H2}$$
$$\ce{Zn(OH)2 → ZnO + H2O }$$

You might be wondering why this reaction occurs. It turns out that the standard electrode is:

$$\ce{2H+(aq) + 2e- -> H2(g) + e-}: E°_{red}=0 V $$

which means that while zinc cannot reduce $\ce{H2O}$, it can reduce $\ce{H+}$, reacting with the protons present due to the natural decomposition of water. After losing two electrons, the positively charged zinc will react with the hydroxide found in solution. This creates zinc hydroxide, which is an insoluble compound. Sodium will remain dissociated from the hydroxide, as sodium hydroxide is water soluble. From here, the decomposition of zinc hydroxide is spontaneous, producing water and zinc oxide.


Is there a way to check reactions? I'm very new to organic chemistry.
I would like to practise simple reactions (HX + Alkenes, X2 with Alkenes and so on) with examples i create myself. Is there a way to check if my products are correct?

I have access to Chemdraw and ChemSketch, however I don't know how and if these tools can be used to do that, i.e. draw a reaction , or two educts and let them show the product.

Is there any software available which can do that?

In August 2018 IBM launched a free cloud service called IBM RXN. This neural network is quite young, but it is constantly being trained, and the prediction should get better over time. I haven't used it all that extensively, but it seems it works quite adequately for simple reactions, including the OP's case. For example:

They also published a great paper [1] which describes on which base their AI operates.

References

Schwaller, P.; Gaudin, T.; Lányi, D.; Bekas, C.; Laino, T. “Found in Translation”: Predicting Outcomes of Complex Organic Chemistry Reactions Using Neural Sequence-to-Sequence Models. Chemical Science 2018, 9 (28), 6091–6098. https://doi.org/10.1039/C8SC02339E.
(Open Access)


Sodium benzoate in soft drinks reacting to produce benzene? Regarding chemical synergies, I read: if someone ingests various products containing sodium benzoate (for example soft drinks that exist) AND if one has had some quantity of vitamin C there might be 'reactions' that produce benzene, a carcinogen.

Some people have told me you'd have to drink a lot of pop for this to happen (assuming you have had some vitamin C during the day), but how much pop is a lot?

Given one has had a glass of orange juice say, how much pop would one have to drink before a benzene molecule is formed?

The reaction taking place is described in article Benzene production from decarboxylation of benzoic acid in the presence of ascorbic acid and a transition-metal catalyst and its importance for food safety recently reviewed in Evaluation of Accelerated UV and Thermal Testing for Benzene Formation in Beverages Containing Benzoate and Ascorbic Acid.

To summarize, the reaction CAN occur, namely at elevated temperatures (>50 °C) and UV-light and in presence of transition metal. Even if the reaction takes place, the amount of benzene formed is either below detection limit, or below limit for drinking water.

For the end user it means, that in extremely unlucky situation, the benzene can form in quantity, which is not considered safe to drink in unlimited amount for the whole life (1 ppb). As such, I would consider the benzene formation as very weak argument against such beverages (mainly if compared with other problems, such as sugar content and associated obesity).


change in state (sublimation and deposition) I am slightly confused as to how I should think of changes in state (especially sublimation and deposition). Should I think of sublimation and deposition as skipping the liquid phase (fig 1) or should I think of it as a circular pattern (fig 2).

fig 1

fig 2

Why during sublimation and deposition is the liquid phase skipped and how is it skipped, would not all matter have to be a liquid in between solid and gas?

Is it due to the fact that some substances are unable to be liquids? if so what prevents it from being a liquid?

Firstly, in regards to the diagrams, as mentioned in the comments, both diagrams are correct. One thing that both diagrams imply is that sublimation/deposition is equivalent to the combination of melting/freezing and evaporation/condensation - another way to look at this is the following diagram from the UC Davis ChemWiki page Heat of Sublimation:

An explanation from the webpage is that:

  Though in sublimation a solid does not pass through the liquid phase on its way to the gas phase, it takes the same amount of energy that it would to first melt (fuse) and then vaporize.

In regards to your related questions:

  Is it due to the fact that some substances are unable to be liquids? if so what prevents it from being a liquid?

Not quite, liquids can occur in the right conditions, the explanation is below.

  Why during sublimation and deposition is the liquid phase skipped and how is it skipped, would not all matter have to be a liquid in between solid and gas?

To understand why sublimation occurs requires an understanding of phase diagrams. A generalised phase diagram from UC Davis Chemwiki page Phase Diagrams is shown below:

The sublimation/deposition curve is the line from the origin to the triple point, represented as the red line in the diagram above. A generalisation that can be made is that a substance undergoes sublimation/deposition when either (or both) the temperature and pressure are lower than at the triple point. Consider the following examples comparing the phase diagrams of carbon dioxide and water (from the UC Davis Chemwiki page linked before):

First, carbon dioxide - at 'normal' atmospheric pressure (such as in the typical laboratory - 1 atm), it is lower than the 5.2 atm at the triple point (labelled 'D'). Assuming minimal pressure changes from standard laboratory conditions, as the $\ce{CO2}$ is cooled, it will eventually undergo deposition to the solid phase.

Now, for comparison, consider the phase diagram for water, the triple state (also labelled 'D') occurs at a very low pressure 0.006 atm, pressures lower than that would result in water subliming. Note for the standard laboratory conditions, water maintains the familiar solid, liquid and gas characteristics (depending on temperature).

Note, both do have a liquid phase.


Is the S–O bond strength in sulfur trioxide or sulfite anion larger? Can you please explain the difference between the bond enthalpies in sulfurtrioxide $\ce{SO3}$ and the sulfite anion $\ce{SO3^{2-}}$?

Due to symmetry constraints ($D_\mathrm{3h}$) in $\ce{SO3}$ there 6 electrons in $\pi$ type orbitals. In a wider sense of the term this molecule is Y-aromatic, but the HOMO actually represents two electrons in in-plane lone pair orbitals of oxygen. The LUMO is an antibonding $\pi$ orbital with respect to all $\ce{S-O}$ bonds.

The following valence molecular orbitals were derived at BP86/cc-pVTZ, Energy level decreases from top to bottom (enlarge).

The bonding picture is usually trivialised as each $\ce{S-O}$ bond being a double bond. But this is actually far away from the truth, as it does not respect the charge of $q=+2$ at the sulfur atom and the charges at the oxygens with $q=-\frac23$. This is due to the fact, that the sulfur atom actually only contributes to one of the three $\pi$ bonding orbitals, resulting in electron density deficiency.

However, the $\ce{S-O}$ bond in sulfurtrioxide is still stronger and shorter, than in the sulfite anion. (BO means bond order)

\begin{array}{lrr}
              & \mathbf{d}(\ce{S-O}) & \text{BO}(\ce{S-O})\\\hline
\ce{SO3}      & 1.458                & 1.4                \\
\ce{SO3^{2-}} & 1.578                & 1.1                \\\hline
\end{array}

The main reason for this is, that upon reduction of sulfurtrioxide to sulfite, $$\ce{SO3 + 2e- <=> SO3^{2-}},$$ you will fill the LUMO. As already stated, this orbital is antibonding with respect to the $\ce{S-O}$ bonds. This is energetically unfavourable for the molecule, therefore the molecule distorts to get rid of the symmetry constraints, i.e. it adapts the lower $C_\mathrm{3v}$ point group. This results in pyramidalization and a different bonding pattern.

You will now no longer find any $\pi$ type bonding orbitals. More accurately this molecule is best described consisting of three equal $\ce{S-O}$ covalent single bonds, and three lone pairs at each oxygen. This results in a charge at each oxygen of $q=-1$ and at the sulfur of $q=1$ (which does not fit the traditional Lewis type bonding description.)

This is also reflected in the molecular orbitals (BP86/cc-pVTZ, energy decreases from top to bottom, enlarge).

It is also very important to understand that the latter molecule would not be stable in gas phase, since it is too highly charged. Without compensation of solvent and/ or counter ions, this molecule will most certainly fragmentate.


How to find the rate of effusion given rates? Question

At $25~^\circ\mathrm{C}$, a sample of $\ce{NH}$ effuses at the rate of $0.050$ moles per minute. Under the same conditions, what is the molar mass of a gas that effuses at approximately one half of that rate?

My steps and thoughts

I converted $25~^{\circ}\mathrm{C}$ to $298~\mathrm{K}$ and then I was thinking of using $pV=nRT$ and I got stuck.

I am thinking of using molar mass, but I don't know how to incorporate that.

Please, only give hints not the answer.

HINT:

This question seems to be a simple application of Graham's Law (assuming you know Graham's Law), i.e the rate of effusion of a gas is inversely proportional to the square root of the mass of their particles. This empirical law can be formulated as follows:

$$\frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{\text{M}_2}{\text{M}_1}}$$

Solution:

   I don't know what the OP meant by $\ce{NH}$, I assume the gas in question is $\ce{NH3}$ with rate of effusion ($R_1$) = 0.050 $\mathrm{mol \min^{-1}}$.

 $R_2 = \frac{R_1}{2}$ and $M_1 = 17.031$, plugging all of this in

 $M_2 = M_1\cdot \left (\frac{R_1}{R_2} \right )^2 = 4\cdot M_1 = 68.124$


Is the adsorption of dissolved iodine onto activated carbon physical adsorption or chemical adsorption? For an experiment, I am looking to see the effect of temperatures from 0–100 °C on the efficacy of activated carbon as an adsorbent. My adsorbate is iodine and I am confused as to whether the adsorption taking place below 100 °C is physical or chemical adsorption.

I have also read that physical adsorption decreases with an increase in temperature while chemical adsorption increases and then decreases with an increase in temperature. Additionally, I was wondering whether chemical adsorption takes place at all temperatures.

The mechanism(s) of iodine adsorption into carbons are complex and strongly dependent on each carbon. That is, iodine can be physisorbed in the carbon micro/mesopores, but in carbons with a rich surface chamistry, it can also be chemically adsorbed through charge transfer interactions.
There are many papers about this, but this one shows clear evidence of both mechanisms. (Written by the carbon "connoisseur" Harry Marsh).
They clearly mention:

  Thus, both classical physical adsorption theory and more specific chemical interactions have been proposed to model the adsorption of iodine by carbonaceous materials.

Strong evidence is provided in the paper itself.


How to determine the strongest oxyacid of chlorine? Which is the strongest acid here?

$\ce{HClO}$
$\ce{HClO2}$
$\ce{HClO3}$
$\ce{HClO4}$

I found this question, didn't really get the first answer, but this one in particular seemed interesting.
Apparently, you calculate a $\mathrm{p}K_\mathrm{a}$ for each substance.
According to this website.
$$\mathrm{p}K_\mathrm{a} = -\log_{10}K_\mathrm{a}$$
I need to find out $K_\mathrm{a}$ then. From what I understand $K_\mathrm{a}$ is obtained from dividing the concentrations of products by the concentrations of reagents. However, how am I supposed to do so if I am not given any equilibrium reactions?

In such questions you're supposed to know certain rules of acidity. You can find some of them on this page. In your case fourth rule, concerning oxoacids, is relevant and as you can see there, it's described on this very case. More precise formulation is so called Pauling's Rule described here. In general the more oxo groups connected to central atom, the more acidic compound.

Related: How are Pauling's rules applied to Nitric acid?


Does cold bluing steel prevent rust? I've recently purchased a Damascus knife and am patiently awaiting it's arrival in the mail. I have been researching the potential care and maintenance habits for this craftsman blade, hoping to keep it in pristine condition.

I'm wondering what measure(s) could be taken to prevent a Damascus knife from rusting. I will be crafting the handle from my own scales and pins upon receiving the item, so any manipulation done to the steel can be preformed upon the blade in its entirety. The alloys used are a mix of 1095 and 15n20 steels respectively.

One of the potential means that I've entertained the thought of is bluing, AKA a black oxide finish. Bluing is a method commonly used on gun metal in which various gun parts are submerged in caustic hot salt baths for a period of time in-order to attain the coat. This isn't the only method to obtain the oxide layer, it's but one of many. Some sources have stated that bluing provides rust resistance while others are skeptical since most oxide layers as a result of the process are merely 2.5 micrometres (0.0001 inches) in thickness.

The method I'm most interested in potentially trying is cold bluing. Cold bluing can be achieved either by coating the high-carbon steel in mustard, painting it in a commercially sold "perma-blue formula", or submerging it in an acidic at-home substance such as vinegar or lemon juice (boiling vinegar to enhance the effect).

Here is my question Ultimately TL:DR Can a true rust resistant layer be obtained via cold-bluing methods on high-carbon Damascus steel (1095 & 15n20)? Will the acidic substance such as vinegar or "perma-blue formula" simply eat through the steel with time (while I'm letting the bluing process take place before I've neutralized the blade with something such as baking soda), or will it develop a thicker layer of patina oxide layer? Some sources say the bluing layer is rust, therefore rust can't develop on top of rust. Others say bluing is purely cosmetic and can not prevent rust.

It provides limited corrosion protection. I would suggest just putting oil on it; more corrosion protection and it will not obscure the surface pattern which is  most all that separates Damascus steel from ordinary steels. As I remember regular "bluing" is a complex of ferocyanides ; Oxides of iron are red and black , brown rusts are hydrated oxides. I have now looked at Wiki and see some of the problem.  Molten cyanide salts are use to carbonitride steels;I believe these leave a blue surface- you can't do that. Another source of blue ( and straw and purple) color is different thicknesses of very thin oxide layers. I have a chart showing one hour at 540 F (in air) , or 10 minutes at 590 F, or 20 hours at 485 F will produce a blue. Higher temperature and longer time give thicker oxide. The color of the oxide is determined by the thickness. Assuming the blade is hard like Rockwell C 60, these temperatures will cause some softening. The other methods sound like they will etch the steel . Same conclusion, oil it.


How are N95 masks different from dust masks? There is a fire raging through my hometown right now. People in the area are advised to be wearing N95 masks to filter out smoke and particulates; and there is misinformation being spread that these masks actually don't work for this purpose. What's the science behind the N95 filter? The Wikipedia page didn't really elaborate.

The short answer to your question "What's the science behind the N95 filter?" is that it is what's called a tortuous path filter, removing particulates from a flow of contaminated air by impact and absorption of the particulate onto the filter material or by trapping particles of sufficient size between the fibers constituting the filter.

The term N95 comes from the NIOSH (National Institute for Occupational Safety and Health, a body of the US government) rating where the 95 means that the material is certified to block 95% of particles 0.3 μm or larger in diameter. The limiting factor toward achieving the stated level of performance during usage is how well the mask seals to the face of the user. Whiskers, dirt and improper fit can all allow a larger portion of particles to circumvent the filter material. The image in the OP shows a mask having a valve to facilitate breathing and maintain a seal by opening during the exhale portion of a breath.


Dispersion of Lithium Ions in Nature (Ratio Similar to Isotopes?) I'm working on a project that identifies average characteristics of different elements. I'm stuck on lithium, I can't find a way to ratio lithium I and lithium II that symbolizes how it is in nature. I tried looking around on the internet but I haven't found any mention of ratios to ions only ratios to isotopes. Does anyone know of anything that might help?

I believe I resolved my problem, the attributes I was looking at was spectra and combining the lines of both ions. The problem was I was comparing data given by two authors that used different units, so what I did, which I think solves my problem, was divide the sums of the given intensity by the emission probabilities for which I received the ratio of Li I to Li II as roughly 33 to 6 or 18% Lithium II. More than I or apparently anyone else thought so I may have still made a mistake, either way thanks to everyone who tried and helped.


Calculating the ionic strength for an aqueous solution containing potassium ferricyanide and sodium sulfate I have the $100$ microM of K$_3$Fe(CN)$_6$ is dissolved in water and the following data:
$$\begin{array}{c|c}
\text{Experiment} & \ce{[Na_2SO_4]}/\pu{M} \\ \hline
\mathrm{A} & 0 \\
\mathrm{B} & 1\cdot10^{-3} \\
\mathrm{C} & 3\cdot10^{-3} \\
\mathrm{D} & 8\cdot10^{-3} \\
\mathrm{E} & 1\cdot10^{-2} \\
\hline
\end{array}$$
I want to calculate the ionic strength for each experiment and I know that I should use the equation:
$$I = \frac {1}{2} Σ c_i z_i^2$$
and that it for this case will look like:
$$I = \frac {1}{2} ((1)^2*[\ce{K+}]+(-3)^2\cdot[\ce{Fe(CN)6^3-}] + (1)^2*[\ce{Na+}]+(-2)^2\cdot[\ce{SO4^2-}] )$$
However, when I calculate this for the different experiments, I don't get the number which is in our answer key. There is no calculations in the answers and my answers differ quite much. This is the way I calculate the different ionic strengths:
$$I_{\mathrm{A}} = \frac {1}{2} ((1)^2\cdot(1\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot0+(-2)^2\cdot0 ) = 5\cdot 10^{-4}$$
$$I_{\mathrm{B}} = \frac {1}{2} ((1)^2\cdot(1\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot(1\cdot10^{-3})+(-2)^2\cdot(1\cdot10^{-3}) ) = 3\cdot10^{-3}$$
$$I_{\mathrm{C}} = \frac {1}{2} ((1)^2\cdot(1\cdot10^{-4})+(-3)^2\cdot(1\cdot10^{-4}) + (1)^2\cdot(3\cdot10^{-3})+(-2)^2\cdot(3\cdot10^{-3}) ) = 8\cdot10^{-3}$$
and so on.
In the answer key $I_{\mathrm{A}}=0.6\cdot10^{-3}$, $I_{\mathrm{B}}=3.6\cdot10^{-3}$ and $I_{\mathrm{C}}=9.6\cdot10^{-3}$. As you can see the difference increases and I don't understand what I am doing wrong. I would truly appreciate if someone could point out my mistakes. Thank you!
EDIT
I am also given this but I don't think they matter for the ionic strength:


Finding the mole fraction of nitrogen in a mixture of nitrogen and oxygen Question

The density of a mixture of Nitrogen and Oxygen is $\pu{4.7 g L-1}$.
Its given that $P = \pu{4 atm}$, and that $T = \pu{298 K}$. Assuming both gasses are ideal gasses, what is the molar fraction of nitrogen in this mixture?

Answer

 Mole fraction of nitrogen is 0.82.

My Attempt
I tried using the $PV=nRT$ equation, having $V = 1$ and $n$ being the number of moles of both gasses together.
Then I used a second equation for the density being $D=\dfrac{16n_1+14n_2}{V}$.
However I got a negative answer that is also larger than $1$.

OP has given a good effort to solve the problem using correct path. Only loose point was not considering the diatomic nature of the gases as Safdar Faisal pointed out in a comment. Suppose the amount of $\ce{O2}$ in the gas mixture is $n_1 \ \pu{mol}$ and the amount of $\ce{N2}$ in the gas mixture is $n_2 \ \pu{mol}$ in volume of $\pu{1.0 L}$ container. Thus, two parallel equations can be setup as follows:
$$PV = nRT \ \Rightarrow \ n_1 + n_2 = \frac{PV}{RT} = \frac{4 \times 1}{0.082 \times 298} = 0.164 \tag1$$
$$\frac{32n_1 + 28n_2}{1} = 4.7 \ \Rightarrow \ 32n_1 + 28n_2 = 4.7 \tag2$$
To resolve two parallel equations, take $(1) \times 32 - (2)$, which gives you,
$$(32-28)n_2 = 0.164 \times 32 - 4.7 = 0.538$$
$$\therefore \ n_2 = \frac{0.538}{4} = 0.135$$
Thus mole fraction of $\ce{N2}$ is $\chi_\ce{N2}$, then
$$\chi_\ce{N2} = \frac{n_2}{n_1 + n_2} = \frac{0.135}{0.164} = 0.820$$
Therefore, the mole fraction of $\ce{N2}$ in the mixture is $0.82$ (hence, the mole fraction of $\ce{O2}$ in the mixture is $0.18$).


Still struggling with Huckel rule and aromaticity I'm still learning huckel rule and aromaticity in different compound.
For the 1-methoxy-1,3,5,7-octatetraene, is it correct that it has 8 electrons inside the cycle, but since they will be 8, then it is antiaromatic?
And for the Imidazole, is it 6 electron inside because the N: doesn't contribute to the pi, or is it 8?

Advice
Take a closer look at the lone pairs and double bonds that participate only and only in the delocalization of electrons.
1-methoxy-1,3,5,7-octatetraene
As you can see, 1-methoxy-1,3,5,7-octatetraene has four double bonds in an octagonal cycle. The lone pairs of the adjacent oxygen atoms, even though they directly influence the delocalization by means of the resonance effect, aren't actually part of the cycle and cannot be included amongst the pi electrons of the double bonds. Therefore, your molecule is antiaromatic. Its structure is not planar, in fact it looks something like this (the picture below includes only the cyclic backbone):

For aromatic molecules, planarity is absolutely necessary. But Huckel's rule which I'm sure you're familiar with, provides the reasoning behind the planarity of the molecule.
The imidazole
As for the imidazole, your assumption is absolutely correct. Here we get to know a scenario similar to pyridine:

The lone pair does not participate in the aromatic cycle. This can be easily checked by taking a look at the hybridization of the nitrogen atom. If the lone pair were to be part of the aromatic system, it should have been perpendicular to the other orbitals of the cycle. However, this is not the case. The 2p orbitals of the nitrogen are actually involved in the aromatic cycle, while the aforementioned paired electrons get to be embedded in an sp2 orbital, placed perpendicular to the plane of the cycle. This actually has significant consequences on the basicity of pyridine.
On the other hand, in order not to interrupt the aromatic cycle, the lone pairs of the other nitrogen atos must come into play when counting pi electrons. Thus, the molecule is aromatic.


How to control the optimization step size in Gaussian? I would like to perform a geometry optimization for a weakly bound methane dimer structure with Gaussian 16. During this process, the Newton-Raphson step should be controlled such that its length should not exceed 0.01 bohr. For this purpose, I created the following input (strictly for testing purposes):
%nprocs=120
%chk=test
%mem=100GB
#hf/sto-3g opt=(trustupdate,nrscale,maxstep=1,maxcycles=100,tight)

test

0 1
C        0.000000000      0.000000000     -1.821797542
H        0.000000000      0.000000000     -2.931818538
H       -0.906328348      0.523268916     -1.451790544
H        0.906328348      0.523268916     -1.451790544
H        0.000000000     -1.046537831     -1.451790544
C        0.000000000      0.000000000      1.821797542
H        0.000000000      0.000000000      2.931818538
H       -0.906328348     -0.523268916      1.451790544
H        0.906328348     -0.523268916      1.451790544
H        0.000000000      1.046537831      1.451790544

Here maxstep=1 should restrict the NR step below 0.01 bohr. Nevertheless, in the 9th step, I received the following report on convergence:
         Item               Value     Threshold  Converged?
 Maximum Force            0.001795     0.000015     NO
 RMS     Force            0.000287     0.000010     NO
 Maximum Displacement     0.079485     0.000060     NO
 RMS     Displacement     0.024807     0.000040     NO

It seems to me that the 0.01 bohr criterion is not fulfilled for the optimization displacement. Is there any way to really constrain the step length during optimization?

Gaussian by default uses a dynamic trust radius for optimizations to a minimum (you also ensured this by including the trustupdate option). When you specify a maxstep this is only guaranteed to be the initial trust radius. In fact, the second step will ensure the trust radius is at least 0.05.
You can see the trust radius for a given step (if you have #p in the route section) by searching for "Trust test". In that same line, it will say "DXMaxT set to 5.00D-2", with the value printed being the trust radius used for this step.
To prevent these updates, simply replace TrustUpdate with NoTrustUpdate. Keep in mind that they likely force the radius up to 0.05 for a reason and your optimizations may fail or take much longer without these updates. You should only use a step this small if you think you have a geometry very close to the minimum or a very messy potential energy surface that you want to step through cautiously.


Iron nomenclature differences between Iron(III) and Ferric As I was doing my Ap Chem homework, sometimes the question would use Iron(III) and sometimes it would use Ferric.  What's the difference in usage?  Also, what's the difference between Iron(II) and Ferrous?

There is no difference, in both the examples you cite.  The forms Iron(II) and Iron(III) are preferred today, versus the older forms ferric and ferrous, respectively, because they more explicitly give the oxidation states of Fe, +2 and +3 in this example.


What determines the electronegativity of an element? Is it possible to work out without post-Bohrian mathematics?

There are various slightly different electronegativity scales.  The two most common are the Pauling scale and the Mulliken scale.  Of the two the Mulliken is the simplest.

The Mulliken scale is based on two quantities, both measurable.  One is the ionization energy $(E_i)$, the energy required to remove an electron from a neutral atom.  The other term is the electron affinity $(E_a)$.  That is the energy released when the neutral atom gains an electron.

The Mulliken electronegativity $\chi$ is then given by

$$
\chi = (E_a + E_i)/2
$$

It is that simple!

For more details and explanation see

http://en.wikipedia.org/wiki/Electronegativity


What makes a compound a dehydrating agent? When a carboxylic acid with low molecular weight is heated with $\ce{P2O5}$ we get the corresponding anhydride. My book says that this reaction happens with dehydrating agents such as $\ce{P2O5}$. I have also learned from my teacher that $\ce{H2SO4}$ is also a good dehydrating agent.

My question is what makes a molecule a good dehydrating agent? How can I identify one (or maybe what are the common dehydrating agents?) while solving questions?

Also, I have seen the use of $\ce{P4O10}$, but do $\ce{P2O5}$ and $\ce{P4O10}$ play the same role as a reagent?

To answer your question, what makes a compound a dehydrating agent is, in my opinion, stability, a good affinity for water and possibly inertness (or in special cases, selective activity but in reforming a designated salt, see example below).

Now, $\ce{P2O5}$ is a white powder that loves water, in fact, it reacts violently and exothermically with water forming $\ce{H3PO4}$.  Phosphorus pentoxide is the anhydride of orthophosphoric acid, $\ce{H3PO4}$ , as dehydration of the latter yields $\ce{P2O5}$.

A safe, effective and inexpensive dehydrating agent is dry Calcium chloride. Again a strong affinity for water and a stable salt.

Per one preparation, one can dehydrate aluminum chloride hydrate by gentle heating in a stream of DRY hydrogen chloride gas (which is the gaseous anhydride of Hydrochloric acid), see here. Just heating the hydrate in air can result in basic aluminum chloride formation, but, I suspect, the dry hydrogen chloride gas can convert any created basic salt back to just $\ce{AlCl3}$ and water (removed by heating and contact with the dry gas). Note, gaseous $\ce{HCl}$ is inert with respect to chloride and stable.


About the nomenclature: manganate or manganite? I am a bit confused about the correct name of $\mathrm{LaMnO}_3$. Is it lanthanium manganate or lanthanum manganite?

I was assuming that since $\mathrm{SrTiO}_3$ is called strontium titanate, $\mathrm{LaMnO}_3$ would then be called lanthanium manganate, but I got confused when I heard a collegue referring to it as manganite and after reading a wikiepdia article on $\mathrm{La}_x\mathrm{Sr}_{1-x}\mathrm{MnO}_3$ where they also used the term manganite.

Which is the correct form, or are they simply interchangeable?

Let's go down the ladder.  $\ce{KMnO4}$ is permanganate, Mn(VII).  $\ce{K2MnO4}$ is manganate, Mn(VI).  $\ce{K4MnO4}$ is manganite, Mn(IV).  $\ce{LaMnO3}$ is called both manganate and manganite, but Mn(III). In a proper world it should be lower than that manganite.  They are pedagogically both wrong if it is La(III).

http://dictionary.sensagent.com/potassium%20manganite/en-en/

Manganese oxidation state nomenclature is a real world mess. Drawing nomenclature parallels to Cl(VII) and downward is not valid.  Drawing nomenclature parallels to titanium is poor, for Ti(IV) is the highest common oxidation state.  $\ce{SrO + TiO2}$ gives $\ce{SrTiO3}$, properly called titanate, for it is Ti(IV) on both sides.

Organic had the same problem. The highest valence carbocation was tricoordinate, hence a carbonium ion.  Magic Acid gave $\ce{CH5^+}$.  Carbonium ions were renamed carbenium ions to general confusion.  Then, $\ce{CH6^{2+}}$.  Organic dropped the ball with direction of optical rotation versus geometric left and right.  It was flat out stupid designating E ("trans") and Z ("cis") olefins.  The designation should have been abstract rather than language-originated.


Is melting/boiling point of ionically bonded substance higher than of covalently bound? Is the melting and boiling point of ionic bond usually higher than covalent bond?

I know that compounds with ionic bonds are usually solid at room temperature, so I want other answers than this. (and this question can also be about why compounds with ionic bonds are usually solid in room temperature.)

The answer relates to the strength of the interactions between the component units that make up a crystal or a solid.

The reason why anything is a solid at a given temperature is, crudely, that the interactions between the units that make up the solid (atoms, ions or molecules) are stronger than the amount of thermal energy available at that temperature.

Diamond, for example, is a solid because each unit carbon is bonded to its neighbours by a strong carbon-carbon bond that takes a great deal of energy to break. Non-polar molecular compounds, like candle wax (in reality a mixture but imagine a pure long chain hydrocarbon to keep it simple) are held together by the intermolecular forces between the molecules (often called dispersion or Van der Waals forces). These are much weaker than chemical bonds and depend on the surface area of the molecules. So small non-polar or not-very polar molecules tend to have low melting points. Hexane is a volatile liquid. The shape and symmetry of the molecule matters a bit to these interactions so toluene is a liquid down to -95°C but benzene, which is very similar to toluene but far more symmetric and slightly flatter only melts at about 6°C as the molecules pack together better encouraging stronger interactions. Polar molecules have stronger interactions because the molecular dipoles interact and this produces stronger forces than the non-polar dispersion forces. So phenol (very similar in size to toluene but far more polar) melts at about 41°C.

Simple ionic solid are made from charged ions. Table salt crystals consist of Na+ and Cl- ions. The key interaction between these is electrostatic and this is a strong force that takes a lot of energy to break. The melting point is about 800°C. Permanent electrostatic interactions are strong especially if the components can get close (remember, it's an inverse square law). This explains why not every ionic compound is a solid. There are ionic compounds that are liquid at room temperature. These tend to consist of molecular ions that are large and sometime irregular so forcing structures where the ionic force has to act over much larger distances and, therefore, be much weaker. An example is [BMIM]PF6 which has a melting point well below room temperature:

These ionic liquids are an interesting area of research in chemistry as they sometimes enable much less wasteful chemical reactions in industry (see wikipedia).


Electron cloud distribution Can a please explain me what is electron cloud distribution in simple words?
Electron cloud is the area around which electrons are most probably found. What is meant by electron cloud distribution?

The electron cloud distribution describes the probability of finding an electron in some region of space around the nucleus at some instant of time.

Imagine you have a camera that can take close-up shots of a hydrogen atom showing the electron at an instant in time. Also suppose that camera can keep snapping pictures, say, a thousand times a second, each time on the same image so that you get a multiple exposure. As the camera runs, the multiple images of the electron will begin to form a cloud-like picture which is dense in some places and thin in others. For the single electron of hydrogen, the densest part of this cloud will form a sphere with the nucleus at the center. As you travel inward toward the nucleus, or outward away from the nucleus, the cloud will get progressively thinner.

The thickest regions of this cloud are thick because the electron was found there more often than in the thin regions. So, if we were to take just one snapshot, this is where the electron would be most likely be found. That is, the probability of finding the electron at a given time is greatest where the cloud is densest.

Other elecrton clouds in other element's atoms have different shapes. Do understand that this is a model that explains some chemical behavior but it is not the current "final" model of how electrons behave in atoms.


Using Hydrogen Peroxide to clean toilet I want to use lysol hydrogen peroxide multipurpose cleaner for cleaning my toilet. Is this a dangerous course of action?

Lysol Multipurpose Cleaner with hydrogen peroxide contains only 0.5% hydrogen peroxide. This site (using-hydrogen-peroxide.com) suggests using up to 35% hydrogen peroxide for cleaning toilets. This hydrogen peroxide website (and who knew it had one) may be biased, but that's quite a margin between the product you want to use and what the website says is safe. I would feel pretty comfortable using this Lysol product for cleaning toilets (except that it may actually not be strong enough!).


Name the following coordination complex What are the IUPAC rules when naming a coordination complex like:

  $$\ce{[Pt(NH3)4][Pt(NH3)Cl3]}$$

Now according to my textbook, it is named as tetraammineplatinum(II) amminetrichloridoplatinate(II). Both the $\ce{Pt}$ atoms have the oxidation state +2 .There are no charges assigned to the two complex ions in the formula, so, my assumption is that the entire complex is neutral and thus the oxidation states for $\ce{Pt}$ should be assigned accordingly. However, on assigning the oxidation states as mentioned in the answer the assumption is proven wrong. I'm confused about the mentioned oxidation numbers.

How do you calculate the oxidation numbers of the metal atoms in such a complex?
Shouldn't the charge on each complex ion be mentioned in the formula if the oxidation states given in the answer are correct?

Your textbook contains an error. Either the formula is written incorrectly or the name is. Since according to comments the name is correct, it must be an incorrect formula. Let’s break it down:

If the formula is correct

We have two complex ions that together create a doubly-complex salt. According to the rules of (pseudo-) binary ionic compounds, one ion must be positively charged and the other negatively. By convention, the cation is written first so it is safe to assume that $\ce{[Pt(NH3)4]^{?}}$ be the cation — which fits in well with chemical intuition because $\ce{NH3}$ is a neutral ligand and platinum is a metal.

Therefore, $\ce{[Pt(NH3)Cl3]^{?}}$ must be the anion — very nice, because whe have chlorido ligands that are commonly negative. Since every chlorido ligand adds $-1$ charge to the complex, the anion can be anything from $-1$ to $-3$ with decreasing probability.

The complex needs to be a $1\,:\,1$ binary complex, so the negative charge needs to be equivalent to the positive charge. From here onwards it gets difficult. We have the following possibilities:

$\ce{[Pt^{I}(NH3)4]^{+} [Pt^{II}(NH3)Cl3]^-}$
$\ce{[Pt^{II}(NH3)4]^2+ [Pt^{I}(NH3)Cl3]^2-}$
$\ce{[Pt^{III}(NH3)4]^3+ [Pt^{0}(NH3)Cl3]^3-}$

They all differ in the assumed oxidative states of platinum. We cannot decide a priori which one of these three is correct, although the last one is by far less probable. Technically, we can even assume further diverging oxidation states but they get proportionally less probable. Only if the charge were clearly stated could we distinguish between the cases. For the sake of consistency, here are the names:

tetraammineplatinum(I) amminetrichloridoplatinate(II)
tetraammineplatinum(II) amminetrichloridoplatinate(I)
tetraammineplatinum(III) amminetrichloridoplatinate(0)

If the name is correct

Then we can logically build up the resulting salt from the name. Let’s start with the cation:

It is platinum(II), so we get $\ce{Pt^2+}$
adding four ammine ligands gives us $\ce{[Pt(NH3)4]^2+}$

For the anion:

It is still platinum(II) (although rendered as platinate, so we need to end up with an anion): $\ce{Pt^2+}$
We have one ammine ligand: $\ce{[Pt(NH3)]^2+}$
and three chlorido ($\ce{Cl-}$) ligands: $\ce{[Pt(NH3)Cl3]-}$

Since we have a cation with $+2$ charge and an anion with $-1$ charge, we need two anions per cation, giving us the final formula:

$$\ce{[Pt(NH3)4][Pt(NH3)Cl3]2}$$

So what is correct?

Going by chemical intuition (platinum(II) being a well-known $\mathrm{d}^8$ system that forms square-planar complexes), I would say that the name is correct and the formula isn’t. This also has the implication that they messed up the question by missing a subscripted 2 (rather likely) rather than messing up the answer by getting a Roman numeral wrong (not quite as likely).

Hence there is nothing miraculous or confusing about assigning oxidation states to metals in complexes — and the IUPAC even formally defines oxidation states for promiscuous ligands such as nitrosyl (which is always considered to be a radical neutral ligand, never $\ce{NO^+}$ or $\ce{NO-}$) allowing for a clean-cut naming of complex compounds without knowing the chemistry involved.

The charge does not have to be mentioned in doubly-complex salts for any component. However, it always helps students to give them and in non-trivial cases (like the assumed correct formula) it might be vital to know the overall charges. If there are no charges written, though, the final product has to be assumed neutral.


Calculation of Degree of Dissociation from Ostwald's Dilution Law A "weak" electrolyte, $\ce{A+B-}$, ionizes in solution as:

$$\ce{AB <=> A+ + B-}\tag{1}$$

$$  K_d =\dfrac{\ce{[A+][B-]}}{\ce{[AB]}}=\dfrac{(\alpha c_{0})(\alpha c_{0})}{(1-\alpha )c_{0}}=\dfrac{\alpha ^{2}}{1-\alpha }\cdot c_{0}\tag{2}$$

For a weak electrolyte the Degree of Dissociation, $\alpha$, is less than 1 and can be calculated using two different equations.

Equation (2) can be rearranged to a quadratic equation which could be solved exactly with the standard quadratic formula.

$$ c_0 \cdot \alpha^2  + K_d\cdot \alpha -K_d = 0\tag{3}$$

$$\alpha = \dfrac{-K_d + \sqrt{K_d^2 - 4\cdot c_0 \cdot (-K_d) }}{2\cdot c_0} = \dfrac{-K_d + \sqrt{K_d^2 + 4\cdot c_0 \cdot K_d }}{2\cdot c_0}\tag{4}$$

If we assume that there is little dissociation and that $\ce{[AB] \approx c_0}$ then $1- \alpha \approx 1$ and we obtain an approximate answer that is easier to calcualte:

$$  \alpha' \approx \sqrt{\dfrac{K_d}{c_{0}}}\tag{5}$$

If equation (5) does not give the true value of the Degree of Dissociation. How do you know when it is good enough?

To make it clearer, say the approximate value from formula (5) is 0.15. That is not the true value of the degree of dissociation, but would the approximate solution be good enough?

The OP asked a simple question to which I gave a complex answer. My daughter called such answers "Dad answers".

The OP essentially asked "If the dilute solution approximation for $\alpha$ is used when calculating, how do you know how good the approximation is?"

In his answer user porphyrin focused on the pertinent idea.

Dilute Solution Approximation

Now if we assume that $(1-\alpha) \approx 1$ meaning that very little of the compound dissociated, then we get the standard approximation, $\alpha'$:

$$  \alpha^{'} \approx \sqrt{\dfrac{K_d}{c_{0}}}\tag{1}$$

Exact Solution

The equation below is an exact solution even though it doesn't allow $\alpha$ to be calculated directly. You would have to guess at $\alpha$ values and iterate until you found a solution.

$$\dfrac{\alpha}{\sqrt{1 - \alpha}} = \sqrt{\dfrac{K_d}{c_{0}}}\tag{2}$$

Estimated % Error

Now let's calculate an estimated % error as:

$$\pu{Est \% Error} = 100 \times \dfrac{\dfrac{\alpha'}{\sqrt{1 - \alpha'}} - \alpha'}{\alpha'}\tag{3}$$

Quadratic Solution

The exact value of $\alpha$ can be calculated directly using a quadratic equation. The solution is given below.

$$\alpha = - \dfrac{K_d}{2\cdot c_0} + \sqrt{\left(\dfrac{K_d}{2\cdot c_0}\right)^2 + \dfrac{K_d}{c_0}}\tag{4}$$

Data Table

\begin{array}{|c|c|c|c|c|} \hline
(\alpha^{'})^2=K_d/c_0 & \alpha (quadratic) & approx & true\ \% \ error & est.\ \% \ error \\ \hline
1.00000000 & 0.61803399 & 1.00000000 & 61.80339887 &  \\ \hline
0.81000000 & 0.58192705 & 0.90000000 & 54.65856100 & 216.22776602 \\ \hline
0.64000000 & 0.54162637 & 0.80000000 & 47.70329614 & 123.60679775 \\ \hline
0.49000000 & 0.49663670 & 0.70000000 & 40.94810050 & 82.57418584 \\ \hline
0.36000000 & 0.44641839 & 0.60000000 & 34.40306509 & 58.11388301 \\ \hline
0.25000000 & 0.39038820 & 0.50000000 & 28.07764064 & 41.42135624 \\ \hline
0.16000000 & 0.32792156 & 0.40000000 & 21.98039027 & 29.09944487 \\ \hline
0.09000000 & 0.25835623 & 0.30000000 & 16.11874208 & 19.52286093 \\ \hline
0.04000000 & 0.18099751 & 0.20000000 & 10.49875621 & 11.80339887 \\ \hline
0.01000000 & 0.09512492 & 0.10000000 & 5.12492197 & 5.40925534 \\ \hline
0.00810000 & 0.08604108 & 0.09000000 & 4.60119879 & 4.82848367 \\ \hline
0.00640000 & 0.07686397 & 0.08000000 & 4.07996803 & 4.25720703 \\ \hline
0.00490000 & 0.06759286 & 0.07000000 & 3.56123125 & 3.69516947 \\ \hline
0.00360000 & 0.05822699 & 0.06000000 & 3.04498988 & 3.14212463 \\ \hline
0.00250000 & 0.04876562 & 0.05000000 & 2.53124512 & 2.59783521 \\ \hline
0.00160000 & 0.03920800 & 0.04000000 & 2.01999800 & 2.06207262 \\ \hline
0.00090000 & 0.02955337 & 0.03000000 & 1.51124937 & 1.53461651 \\ \hline
0.00040000 & 0.01980100 & 0.02000000 & 1.00499988 & 1.01525446 \\ \hline
0.00010000 & 0.00995012 & 0.01000000 & 0.50124999 & 0.50378153 \\ \hline
0.000001 & 0.0009995 & 0.00100000 & 0.05001250 & 0.05003753 \\ \hline
0.00000001 & 9.9995E-05 & 0.00010000 & 0.00500012 & 0.00500038 \\ \hline
1E-10 & 9.99995E-06 & 0.00001000 & 0.00050000 & 0.00050000 \\ \hline
\end{array}

So the Est % Error from equation (3) works quite well.

When the Est % Error is 5% or less, $\alpha' \le 0.1$, the Est % Error is quite good.
So for the approximate equation when $\alpha^{'} = 0.1$ you get 1 significant figure, 0.01 gets two significant figures and so on. The improvement doesn't continue forever of course - this is chemistry not math. To get three significant figures or more you'd have to be a fastidious experimenter.
For the approximate equation when $\alpha' = 0.7$ the estimated error is 82.6% which is much too large, but an 82.6% error still tells you that the  approximation $\alpha^{'}$ is no good.
To make user M. Farooq's point in another way, by the time you calculate the approximation from equation (1) and the Est % Error from equation (3), you've pretty much done enough work to calculate the exact answer using equation (4). So why bother with the approximation?

EDIT A comment by user Buck Thorn "I never thought about the usefulness of the estimate to bracket the true value" made me think.

It turns out that we can bracket the true value $\alpha$ using a simple calculation of the approximate value $\alpha'$:

$$\alpha'\cdot(1-\alpha') \lt \alpha < \alpha'\tag{5}$$

\begin{array}{|c|c|c|c|} \hline
(\alpha^{'})^2=K_d/c_0 & \alpha (quadratic) & \alpha'\cdot(1-\alpha') & \alpha' \\ \hline
1.0000000000 & 0.6180339887 & 0.0000000000 & 1.0000000000 \\ \hline
0.8100000000 & 0.5819270490 & 0.0900000000 & 0.9000000000 \\ \hline
0.6400000000 & 0.5416263691 & 0.1600000000 & 0.8000000000 \\ \hline
0.4900000000 & 0.4966367035 & 0.2100000000 & 0.7000000000 \\ \hline
0.3600000000 & 0.4464183905 & 0.2400000000 & 0.6000000000 \\ \hline
0.2500000000 & 0.3903882032 & 0.2500000000 & 0.5000000000 \\ \hline
0.1600000000 & 0.3279215611 & 0.2400000000 & 0.4000000000 \\ \hline
0.0900000000 & 0.2583562262 & 0.2100000000 & 0.3000000000 \\ \hline
0.0400000000 & 0.1809975124 & 0.1600000000 & 0.2000000000 \\ \hline
0.0100000000 & 0.0951249220 & 0.0900000000 & 0.1000000000 \\ \hline
0.0081000000 & 0.0860410789 & 0.0819000000 & 0.0900000000 \\ \hline
0.0064000000 & 0.0768639744 & 0.0736000000 & 0.0800000000 \\ \hline
0.0049000000 & 0.0675928619 & 0.0651000000 & 0.0700000000 \\ \hline
0.0036000000 & 0.0582269939 & 0.0564000000 & 0.0600000000 \\ \hline
0.0025000000 & 0.0487656226 & 0.0475000000 & 0.0500000000 \\ \hline
0.0016000000 & 0.0392079992 & 0.0384000000 & 0.0400000000 \\ \hline
0.0009000000 & 0.0295533748 & 0.0291000000 & 0.0300000000 \\ \hline
0.0004000000 & 0.0198010000 & 0.0196000000 & 0.0200000000 \\ \hline
1.000000E-04 & 0.0099501250 & 0.0099000000 & 0.0100000000 \\ \hline
1.000000E-06 & 0.0009995001 & 0.0009990000 & 0.0010000000 \\ \hline
1.000000E-08 & 0.0000999950 & 0.0000999900 & 0.0001000000 \\ \hline
1.000000E-10 & 9.999950E-06 & 9.999900E-06 & 0.0000100000 \\ \hline
\end{array}


Comparing the dipole moment of water and ethanol Why is the dipole moment of water (1.85D) larger than ethanol (1.66D)?

I thought the inductive effect from ethyl would develop more delta - charge on oxygen.. hence increasing the charge and the dipole moment

It's probably due to the slight difference in electronegativity between H (2.2) and C (2.55)

Here's some data on gas-phase dipole moments from the CRC Handbook:

Water - 1.85 D
Methanol - 1.7 D
Ethanol - 1.66 D
n-Propanol - 1.55 D
n-Butanol - 1.66 D

Using some incorrect data, I pointed to a possible odd-even effect. Using the correct gas-phase dipole moments, we see that these are all pretty similar (i.e., +/- ~0.2D).

Certainly water has the highest dipole moment, but H is slightly more electropositive than C, suggesting that as a likely cause.


Technecium and Technetium Was technecium "recently" renamed technetium in English? I know when I formally studied chemistry (in the USA) that earth, air, water and fire were on the periodic table, but I really remember the element as being technecium.

After German chemists claimed to have discovered element 43 in 1925, the name masurium was used in the first publications. However, the German experiments could not be reproduced.

The official discovery of element 43 was accomplished by C. Perrier and E. Segrè in 1936. They proposed the name technetium (see Perrier, C.; Segrè, E. Technetium : The Element of Atomic Number 43, Nature, 1947, 159, 24.).

In 1949, the 15th IUPAC Conference adopted this name.

The name technetium was also used in the early American literature, for example in Radioactivity Applied to Chemistry; Wahl, A. C., Bonner, N. A., Eds.; John Wiley & Sons: New York, 1951.


What are counterion effects? I was reading the following article on wikipedia about Hydroponics

Unlike most nutrients, plants can tolerate a high concentration of the
$\ce{SO4^2−}$, selectively absorbing the nutrient as needed.
Undesirable counterion effects still apply however.

Can somebody explain me what is meant by counterion effects? Is it just the presence of the counterion, i.e. the concentration of the counterion in solution?

Here is a passage from Marschner P. (2012). Marschner’s mineral nutrition of higher plants. 3rd Edn. London: Academic Press:

In the context of the wikipedia table, it means that while you can have a high concentration of sulfate, you should consider the cation of the substance (e.g. calcium sulfate, potassium sulfate, ammonium sulfate) as well when trying to figure out the effects on a plant.