What should be the order of specific conductance (κ) of floride, chloride and bromide ions? The equivalent conductance order is $\ce{Br-} > \ce{Cl-} > \ce{F-}$.

$$Λ = \frac{κ}{c},$$

so $Λ$ is proportional to specific conductance?

But in ion chromatography we get peak area where $\ce{F-} > \ce{Cl-} > \ce{Br-}$ (specific conductance, μS/cm) for the same concentration, let us say 1 ppm.

Why?

ppm is ambiguous quantity, it can be v/v, w/v,w/w,n/n.

By other words, all 1 ppm very probably is not the same molar concentration, unless it is n/n ppm, relating itself to the molar amount of the solution.

$\Lambda = \kappa /c$ is correct, but $c$ must be the molar concentration.

$$\Lambda[\pu{Sm^2mol^-1}]=\frac{M[\pu{g/mol}]}{c[\pu{ppm as g/m^3}]}\cdot \kappa[\pu{S/m}]$$


Electrolysis of aluminum in a solution of magnesium sulfate So I recently started looking into electrolysis to produce chemicals and I have only produced a few chemicals from it like copper hydroxide and copper sulfate. However I can’t seem to figure out what this wight precipitate that is produced in a electrolysis between  a aluminum cathode and a aluminum anode in a solution of magnesium sulfate.I am relatively new at chemistry and I am only in high school so I can’t figure out this precipitate. I would also like to know if there is a new dissolved salt being formed. Thanks in advance all ideas are much appreciated.

Aluminum is not a good idea for carrying out electrolysis, because this metal is covered by a thin layer of aluminum oxide (or alumina) that is insulating. The current passing between these electrodes must be really weak, because the resistance of this layer is high. So that you need a high tension to break this layer and make an electrolysis. But, as soon as aluminium is in contact with water, a new aluminum layer is formed. The majority of the electric energy is wasted and lost for heating the solution. This is non-sense. The powder that you get is probably aluminum oxide coming from the broken alumina layer I told you about earlier. And magnesium is not involved in this reaction.


What does "René" mean in René alloys? René alloys are nickel-based super alloys (René 41, René 88, René N5 etc. etc.). But how did it get the name René? Is it an acronym or contraction of other terms? It was created by Earl Ross and the General Electric Materials Science group, so it doesn't seem to be the name of the discoverer.

Similar to inox for inoxydable (inoxidable), it equally is a French coinage.  To tune mechanical properties of iron, mixing additives leads to numerous alloys of iron.  In case of rene or rené as contraction of résistant (durable) and on base of Nickel, and is example of the superalloys.

This source technologie des métaux et alliages - particulièrement en aéronautique (technology of metals and alloys, especially in aviation industry) by Dominique Ottello (2003), section 4.6.1.1 "Les alliages de nickel: / the alloys of nickel" on page 15 states:

  Ils  conviennent pour les trois types de sollicitations : MÉCANIQUES, THERMIQUES et CHIMIQUES (très grande résistance à la fatigue, au fluage et à  la corrosion à chaud). Il existe plusieurs familles d'alliages de NICKEL :

  Les alliages NICKEL + CUIVRE : alliages "MONEL" qui présentent une bonne résistance à la corrosion chimique, utilisables jusqu'à 500 °C.
  Les alliages NICKEL + CHROME + FER : alliages "INCONEL" qui sont très tenaces et utilisables jusqu'à 700 à 800 °C.
  Les alliages NICKEL + CHROME + COBALT : alliages "NIMONIC" et alliages
  "RENE". Ils sont très tenaces et présentent une bonne résistance au FLUAGE, utilisables jusqu'à 950 °C.

  NOTA : Dans les alliages de NICKEL à haute température, on rencontre :

  Des alliages plus spécialement réservés à la FONDERIE.
  Des alliages pour CORROYAGE.
  Des alliages à durcissement structural par traitements thermiques.

For our non-Francophon readers:

  These [Nickel-derived alloys] are suitable for three fields of application / strain: mechanical, thermic, and chemical (very high endurance against fatigue, compression, and corrosion at high temperatures).  There are several groups of nickel-based alloys

  alloys based on nickel and copper:  "monel" alloy, with a good resistance against chemical corrosion, serviceable up to $\pu{500 ^\circ{}C}$ [i.e., $\pu{932 F}$]
  alloys based on nickel, chrome, iron:  "inconel"alloy, tenacious and deployable up to $700$ or $\pu{800 ^\circ{}C}$ [i.e., $1292$ to $\pu{1472 F}$].
  alloys based on nickel, chrome, and cobalt: "nimonic" and "rene" alloys. These are very tenacious and exhibit a good resistance against traction, deployable up to $\pu{950 ^\circ{}C}$ [i.e., $\pu{1742 F}$].

  Note: One finds, among the nickel-based alloys with high melting point:

  alloys foreseen / dedicated for deployment in foundries / casting houses
  alloys for wrought (e.g., forging, rolling, etc.)
  alloys for hardening by thermal treatment

Voilà !


Deviation from ideal gas law curves So I was reading through my textbook on deviations from an ideal gas, and they had plotted a curve of experimental PV/nRT against the pressure.

They talk a lot about why the volume is really smaller than what the ideal gas predicts, and so on, and thats why as the pressure increases, the deviation from the ideal gas law increases. They further accredit the imbalances to IMFs, and how when the molecules are in closer contact, they tend to form stronger and more IMFs. This all makes sense. This would imply that as the pressure increases or as the temperature decreases, the deviation from the ideal gas would be increasingly over 1. However, when looking at the curve they present, it isn't a clear direct path from no deviation to much deviation. For example, when looking at Methane, CH4, at 200 atm it has a certain negative magnitude, yet at 400 atm, it is back to acting like a ideal gas.

What causes this section of the curve[P=0 to P=400] of Methane that doesn't follow the general trend of increasing pressure --> more deviation from ideal gas?

There are 2 deviation terms of the opposite sign.

The negative deviation term relates to the attractive forces between molecules, decreasing pressure,and is directly related to condensation ability of real gases.

The positive deviation term  relates to non-negligible own volume of molecules, compared to the total gas volume, leading to higher than ideal pressure, as with the higher own volume, the frequency of collisions with wall is higher.

These 2 effects sum to the overall deviation and in some circumstances may cancel each other.

For "permanent gasses" like hydrogen, helium, neon, nitrogen, the negative term is small and there is more or less steady increase of positive deviation with pressure.

For gases that are easy to turn into liquid ( see critical temperature and pressure ) like methane or carbon dioxide, the negative term is dominant for lower pressures and temperatures. If they are high enough, the dominance is overtaken by the positive term.

A lot of insight can be gained from van der Waals equation Wikipedia page


Volume increase when air is enriched with water When air is enriched with water, how much volume increase can be expected?

I am aware that the volume increase will not be proportional (i.e. $\pu{1L}$ air + $\pu{10mL}$ gasified water will result in a volume of less than $\pu{1.01L}$).

So how big is the volume increase actually?

The short answer is that the behavior of the system you describe should very well approximate that of an ideal gas at room temperature and 1 atmosphere, and that the addition of $\pu{0.01L}$ of water vapor to $\pu{1.00L}$ of air will give you a total volume very close to $\pu{1.01L}$ (assuming constant temperature and pressure).

The departure from ideal behavior for water vapor strongly depends on relative humidity. While the absolute humidity given in the question is only $1\%$, this would correspond to $100\%$ relative humidity at $\pu{6^oC}$, and greater than $100\%$ relative humidity at lower temperatures. Gas molecule interactions so strong that condensation occurs essentially represents the opposite extreme of the assumptions made for an ideal gas.

So, at the extreme of very low temperatures, essentially all of the water vapor would condense and your calculation would have a high bias of 1% if you were to assume ideal gas behavior.  However, at room temperature (i.e. $\pu{20^oC - 25^oC}$) this will not be the case and the assumption of ideal behavior would introduce negligible error.


Why are alkali salts so soluble? There's a certain solubility rule stating all carbonates, phosphates and hydroxides are insoluble unless mixed with hydroxides or alkali metals.

Why is this? What makes the alkali metals so special?

So the reason for carbonates and phosphates being insoluble in water because in an acidic environment, they are both protonated and lose their negative charges. The symmetry of their binding groups produces identical resonance structures at high pH values (since rotating one double bond will shape it into a superimposable structure of another). This allows any given oxygen at a time to posses the double bond (in reality the pi bond to the phosphorus or carbon is shared throughout all the bonded oxygens and each oxygen maintains an equal distance away from the central atom).

This partial negative charge (created by lack of firm placement of the pi electrons) can be neutralized by bonded hydrogens (this will create a small positive charge, however hydrogen's ionization energy is much less than oxygen's electron affinity). The lack of polar-ness of the phosphates and carbonates makes them insoluble in water.

In a basic environment, (created by hydroxides) the carbonates and phosphates will become deprotonated and charged. The charge on these molecules will allow them to be more miscible with water.

https://en.wikipedia.org/wiki/Carbonate


Would cooling of a battery using cooler affect its life time? My smartphone gets over heated where I use mobile data for long period. Last day I found a cooler of an old cpu, the thing silver in colour to absorb heat(with eventually gives out heat as it has more surface area on the other side). I used to keep my phone above it and it gets cooled. So my question is whether it will cause any problem for the battery if I tries to cool it using the same? I am asking this because there are a lot of thermodynamics involved in batteries and would it cause any problem to my battery life?

Let's see what I know. Not really sure whether this is a terrific answer to your question, but it's what I have.

Through knowledge of using laptops, I know that lithium-ion batteries will decrease in performance when overheated. This probably applies to most redox reaction batteries for the reason that the reaction proceeds more quickly when temperature is increased, and as a consequence, kinetic energy of the molecules are increased.

Increasing the temperature of the battery also increases internal resistivity, which is somewhat more of a physics question. The electrons collide with the material that they are flowing through much more frequently in a higher temperature situation as opposed to a lower temperature situation. This doesn't mean, however, that you can dip your phone in LN2 and have it work forever. There's obvious physical limitations to everything.

What you're doing to your phone (a.k.a. cooling it) shouldn't have a problem at all on your phone's performance, mainly because you're doing to it what others do to their laptops - placing it on a cooling pad.


What measures of the dielectric constant mark whether the solution is ionic, polar and nonpolar? At what numerical values of ε, then what organic compounds if any mark these thresholds.  How is that only for a vacuum has ε=1?  This is absolute lowest limit?

Just compare the dielectric constant of known substances to non-polar, polar, metallic designations. For example, the dielectric constant of water is around 80.0 and water is polar. H2 has a very low constant value near 2 I believe and it is very non-polar. Metals have very large epsilon values moreover.  Rule of thumb, the larger the dielectric constant, the more polar the substance or material. In math terms, the potential felt between 2 charges is V= (q1 q2) / (4pi epsilon r) where q1 and q2 are different charges separated by r. The higher epsilon, the smaller V becomes. How charges interact with each other hence depends on the dielectric constant of the medium or material.


Effect of temperature and Le Chatelier's principle The question
  $$\ce{2NO(g) + H_2(g) <=> N_{2}O(g) + H_{2}O} + \pu{351kJ}$$
  If the temperature is increased in the reaction above, which of the following occurs?

  A. The concentration of $\ce{H_{2}O}$ increases
  B. The concentration of $\ce{N_{2}O}$ decreases

My teacher said the answer is B;
however, I thought that since the reverse reaction is endothermic, adding more heat would produce more $\ce{2NO}$ and $\ce{H_2}$ which will cause an increase in pressure. But in order to maintain the same pressure, the system needs to decrease the number of moles so it favors the forward reaction because there are 3 moles of the reactants and 2 moles of the products.
Did my teacher make a mistake? If he did not can you please explain why he is correct.

Your teacher is right, let's see why. First off, here's our reaction $$\ce{2 NO + H2 -> N2O + H2O}\qquad \Delta H = -351\ \mathrm{kJ/mol}$$
So the reaction to the right is exothermic, it gives off heat.  That tells us the products are more stable than the reactants and at equilibrium $\mathrm{\frac{[Products]}{[Reactants]}~>1}$.  Here we could guess that if we heat the reaction (put heat back into the reaction), we will tend to level out the equilibrium thereby pushing the reaction back to the left.

To examine this guess let's work an example.  You were given an enthalpy of reaction, but for a "back of the envelope" calculation, let's run a comparison and assume that the change in entropy will be roughly the same for the two conditions (temperatures) we're going to compare.  With this assumption, we can let $$\Delta G= -RT\cdot\ln K_\text{eq}= -351\ \mathrm{kJ/mol}$$  We know $\Delta G$, $R=8.31\times10^{-3}$ and let's say the data is given for room temperature, so $T=298\ \mathrm K$.  Solving for $K_\text{eq}$ we find $K_\text{eq}=3.6\times10^{61}$.  Now let's let our reaction equilibrate $100^\circ$ above room temperature (now $T=398\ \mathrm K$).  $\Delta G$ and $R$ are fixed, so lets resolve the equation and we now find $K_\text{eq}=1.2\times10^{46}$, indeed heat has pushed our reaction to the left.  The teacher was right.


Why is double bond stronger than single bond? If a pi-bond is weak due to its sideways overlapping, then how can we say that breaking a double bond is a bit difficult? Is it because of the slight increase in energy as compared to a single bond to break it?

A double bond is composed of a π-bond and a σ-bond. In order to break a bond such as this you must provide the molecule with the energy needed to break the σ-bond alone plus the one needed to break the π-bond alone. In other words, you must provide the sum of those two amounts of energy. Thus, it goes without saying that the sum is a greater number than the two energies alone.


Why does the melting point get lower going down the Alkali Metal Group with increase in atomic number? Why does the melting point get lower  going down the Alkali Metal Group with increase in atomic number?

You have to understand that melting point is related to the bonds in a metal between the metal cation and the 'sea of electrons'. With an increase in atomic number, you have an increase in electron shells.

As the radius of atoms get larger down the group, you could say that the force holding them together is spread over the greater area and hence, the metal cations are more weakly bonded. And coming back to my first point, when the bonding is weaker, the metal's melting point will decrease.


Be and B ionization energies Why does Be have a higher ionization energy than B? I know that it has a filled 2s orbital, but what's the more accurate explanation?

We essentially deal with two different trends here and I think the answer lies within analyzing these trends. First we have that the ionization energy obviously increases with increased nuclear charge/atomic number. Secondly we have that with a higher magnetic quantum number (the s-, p-, d- and f-orbitals) the energy of the orbital increases and therefore that the ionization energy decreases. The decrease in ionization energy that the higher magnetic quantum number introduces is larger than the increase that the single extra positive charge introduces.

We should also take Hund's rule in account, which is favorable for a lower ionization energy of $\ce{Be}$ compared to $\ce{B}$. This factor is however relatively small compared to the above ones.

To verify some of these trends yourself, you can take a look at this table of ionization energies.


Why copper wires submerged in salt (or fresh) water oxidize much faster while they conduct electrical current? On a lot of flooded vehicles (salt/fresh water), I've noticed that wire that carries more current is always the one with most destruction (at the connector/splice), in comparison with lower current carrying wire. Wire turns black on the outer layer, accumulating green/white oxidation(?) on top of it.

How does electricity affect the rate of oxidation?

Electricity in a wire submerged in water effectively turns it into an electrochemical cell, and then it is no surprise that the anode gets oxidized pretty quickly. A modest potential of a few volts would suffice to oxidize any metal, even gold. This works for AC as well (a wire oxidizes during the positive half-wave and then does nothing during the other half-wave).


Is carbon dioxide IR inactive? Wouldn't CO2 be IR inactive because of its non-polar bonds?

I am told that carbon dioxide is IR inactive.

You're right, that's not true.  Since carbon dioxide is linear it has $3n-5 = 4$ vibrations and they are pictured below.

The symmetric stretch does not result in a change (of the initially
zero dipole moment), so it is ir-inactive.
The asymmetric stretch does result in a change in dipole moment so it is ir-active.
The bend also results in a change in dipole moment so it too is ir-active.

We expected 4 vibrations and I've only listed 3. This is because the "bend" (let's start by placing the molecule along the x-axis) can occur in the y direction and the z direction.  But these two motions are the same, just deforming in different directions, the bend is said to be degenerate, accounting for the "fourth" vibration.

To sum up, carbon dioxide has 2 ir-active vibrations.

Edit - response to example added (question d) by OP

Question d is incorrect.  Either the author 1) inadvertently switched the column headings (IR active, IR inactive) or 2) meant to use some molecule other than carbon dioxide.

The first 3 rules you learn for interpreting IR and Raman spectra are

The number of molecular vibrational modes equals 3n-6 (3n-5 for linear molecules), where n is the number of atoms.
An ir active band will be observed if a vibration results in a change of the dipole moment. The initial dipole moment in the molecule's equilibrium geometry can be zero; all you need is a change.  The terms "polar" and "non-polar" can be confusing, they often mean different things to different people.  Leave "polar" out of the criteria for ir activity and stick with dipole moment, it is a much better understood term.
the rule of mutual exclusion, it states that, for centrosymmetric molecules (molecules with a center of symmetry, like carbon dioxide), vibrations that are IR active are Raman inactive, and vice versa.  So for carbon dioxide there is 1 Raman band and two IR bands.

Here's a link to a recent SE Chem question: How can I deduce the linearity of XeF2 from the IR spectrum? where these rules were used to determine the structure of a molecule.


Enthalpy of formation and bond energies of CH2(OH)2? I was wondering if I could have the bond energy and enthalpy of formation in kJ/mol for $\ce{CH2(OH)2}$?

I honestly searched the Internet and my textbook and I couldn't find it.  I'm sure I could have missed it somewhere but nevertheless I need to calculate the energy produced by the reaction of $\ce{CH2O}$ and $\ce{H2O}$.

If someone could give me that info that would be exceedingly helpful.

Firstly, regarding the bond energy, you will have to specify exactly what bond you are talking about, you can't ask for the bond energy of a compound because compounds have many bonds. This doesn't apply for diatomics (for example, you can ask what the bond energy of $\ce{N2}$ is; although that is loose usage of terminology, the meaning is clear). But methanediol is obviously not a diatomic.

Secondly, from what I understand, you need to find the enthalpy change for the hydration of formaldehyde: $\ce{CH2O_{\;(aq)} + H2O_{\;(l)} -> CH2(OH)2_{\;(aq)}}$ and your approach was to find the enthalpies of formation of the individual compounds. That's a good idea and more often than not, it works because  enthalpies of formation are widely tabulated and easy to find. But it's often worthwhile just searching for the enthalpy change of the reaction you are interested in, especially if it is a relatively well-known or simple reaction (like this one). Anyway, this recent paper gives the value of $\Delta _{\text{r}}H$ for the reaction above as $-39.0 \text{ kJ mol}^{-1}$.


How does ammonium nitrate explode on its own? I thought ammonium nitrate was an oxidizer that needed to be mixed with fuel to form a high explosive (e.g., ANFO). But apparently there have been accidental explosions involving just the "fertilizer". Are these explosions also detonations? What is the chemical formula for the process?
$$\ce{NH4NO3 -> ???}$$
Part of my motive for asking is the news today (Aug 4, 2020) of an explosion in Beirut. Initial reports say it was caused by "2750 tons of stored ammonium nitrate".

It is known that ammonium nitrate decompose exothermically when heated to form nitrous oxide and water. This paper1 notes that the irreversible decomposition of ammonium nitrate occurs at the temperature range of $\pu{230-260 ^\circ C}$.
$$\ce{NH4NO3 ->[t >230 ^\circ C] N2O + 2H2O}$$
They also further noted that beyond $\pu{280 ^\circ C}$, $\ce{NH4NO3}$ is  capable of rapid, self-accelerating decomposition (to the point of detonation).
But at the detonation temperature, $\mathrm{t_d}$ (the temperature at which compounds detonate), ammonium nitrate fully decomposes to nitrogen, oxygen and water releasing a tremendous amount of energy.
$$\ce{2NH4NO3 ->[t_d] 2N2 + O2 + 4H2O}$$
In the context of Beirut explosion, the question that raised was "when did ammonium nitrate reached detonation temperature, and why did it suddenly explode?". According to a news report from cnet.com:

When heated to above 170 degrees Fahrenheit, ammonium nitrate begins
to undergo decomposition. But with rapid heating or detonation, a
chemical reaction can occur that converts ammonium nitrate to nitrogen
and oxygen gas and water vapor. The products of the reaction are
harmless -- they're found in our atmosphere -- but the process
releases huge amounts of energy.[...]
Additionally, in the explosion, not all of the ammonium nitrate is
used up and exploded. Some of it decomposes slowly creating toxic
gases like nitrogen oxides. It's these gases that are responsible for
the red-brown plume of smoke seen in the aftermath of the Beirut
explosion, Rae said.

So, my theory is that ammonium nitrate started heating (from the fire) releasing all sorts of nitrogen oxides (the red fumes). This fire further accelerated the reaction, further heating the remaining ammonium nitrate to the point of detonation and that's when ammonium nitrate exploded instantaneously releasing tremendous amount of energy which send shockwaves around the site along with a white mushroom shaped cloud (from @DDuck's comment) which could probably be nitrogen and/or water vapours where the humid air (water vapor laden air) condensed due to the explosion(@StianYttervik) with release of nitrogen. It is a sad and quite devastating incident.
References

On the Thermal Decomposition of Ammonium Nitrate. Steady-state Reaction
Temperatures and Reaction Rate By George Feick and R. M. Hainer, 1954 (PDF)
Reaction Rates of Ammonium Nitrate in Detonation
Melvin A. Cook, Earle B. Mayfield, and William S. Partridge
The Journal of Physical Chemistry 1955 59 (8), 675-680
DOI: 10.1021/j150530a002 (PDF)
https://en.wikipedia.org/wiki/2020_Beirut_explosions


What causes incomplete combustion? During the combustion of hydrocarbons, there is a difference between the amounts of carbon or hydrogen that results in incomplete or complete combustion of the material.

My question is, besides from an insufficient amount of oxygen, what can cause incomplete combustion?

What are the physical or chemical properties of the hydrocarbon responsible for incomplete combustion, such as the hydrogen to carbon ratio (mass, volume, number of atoms, etc.) or the saturation.

Please keep the reply suitable for a student studying high school to early university chemistry.

According to the Iowa State University page Carbon Monoxide Poisoning: Checking for Complete Combustion (AEN-175) suggest the following mechanisms that cause incomplete combustion:

  Insufficient mixing of air and fuel.
  Insufficient air supply to the flame.
  Insufficient time to burn.
  Cooling of the flame temperature before combustion is complete.

The main notion is the lack of air to cause complete combustion (either through lack of supply, mixing or time for combustion).

However, it should be noted that, practically speaking, according to the ChemWiki page Burning Alkanes, the size of the hydrocarbon has an affect in 'normal' conditions:

  Provided the combustion is complete, all the hydrocarbons will burn with a blue flame. However, combustion tends to be less complete as the number of carbon atoms in the molecules rises. That means that the bigger the hydrocarbon, the more likely you are to get a yellow, smoky flame.

The reason being (according to ChemWiki) is:

  If the liquid is not very volatile, only those molecules on the surface can react with the oxygen. Bigger molecules have greater Van der Waals attractions which makes it more difficult for them to break away from their neighbors and turn to a gas.

So, the larger the hydrocarbon, the less likely it would vaporise sufficiently, thus less chance of the optimal mixing of air and fuel, resulting in incomplete combustion.

However, provide the right conditions with sufficient supply of air and optimal hydrocarbon-air mixing ratio, it is still possible to cause complete combustion, in the general formula (for many hydrocarbons):

$$\ce{C_{x}H_{y} +O2 ->H2O + CO2}$$

A quite comprehensive resource explaining each of the factors is the University of Tulsa chapter Fuels and Combustion.


Will gamma radiation cause "thermal" decomposition of sodium bicarbonate? I am conducting research for a new drug that contains powdered sodium bicarbonate, and the drug will need to be sterilized after placement into it's container/closure system. Typically, this is performed with steam, at 122.5 °C for 15 minutes, but in the case of sodium bicarbonate, this would cause decomposition:

$$\ce{2 NaHCO3 → Na2CO3 + H2O + CO2}$$

The alternative is to sterilize with ionizing (gamma) radiation. Will this also cause decomposition? Any help or referral to another authority is much appreciated.

There is an equilibrium between decomposition and recombination, i.e. as $\ce{CO2}$ is given off, it joins back to the  $\ce{Na2CO3}$ to form $\ce{NaHCO3}$.

  $\ce{2NaHCO3 <-->Na2CO3 + CO2}$

So for moderate doses, given at moderate rates, it's unlikely great pressure would develop, but you'd need to test to see if the drug is degraded by decomposition products.


Why does bubbling carbon dioxide through calcium hydroxide result in a precipitate? Why does the addition of carbon dioxide to calcium hydroxide create the precipitate calcium carbonate? What happens chemically for such a molecule to form? How exactly do their interactions come into play?

Now CaOH isn’t terribly soluble in water but the parts that do dissolve, disassociate into their respective OH- and Ca2+ ions. So when we bubble CO2 through the CaOH solution the calcium ion is attracted to the partial negative charge on the oxygen atoms of carbon dioxide. So that forms CaCO2 so my question is where does the other oxygen of calcium carbonate come from? Is it snatched from a hydroxide ion?

Calcium carbonate also has a resonant structure, I don’t know what that is exactly but it’s like electrons are being shared uniformly throughout a given structure so that means carbons original double bonds get disbanded too I guess? If anybody can explain what happens here as well, it would be great.

Here are my drawings, they may be simplified but it’s all I know:

The important thing to consider is the chemistry of carbon dioxide in water, forming carbonic acid. Note also that calcium hydroxide is at edge between being a strong or a weak base.

$$\begin{align}\
\ce{CO2 ^ + H2O &<=> CO2 . H2O} \tag{1} \\
\ce{CO2 . H2O &<=> H2CO3} \tag{2} \\
\ce{H2CO3 &<=> H+ + HCO3^-} \tag{3} \\
\ce{HCO3^- &<=>  H+  + CO3^2- } \tag{4} \\
\ce{ Ca(OH)2 &<=> CaOH+ + OH- } \tag{5} \\
\ce{CaOH+ &<=> Ca^2+ + OH-} \tag{6} \\
\ce{Ca^2+ + CO3^2-  &<=> CaCO3 v} \tag{7}
\end{align}$$

Yes, bonds in $\ce{CO3^2-}$ are equivalent, with strength and length somewhere between a single and a double bond. It is similar as  for $\ce{SO4^2-}$, $\ce{SO3^2-}$, $\ce{NO3^-}$, $\ce{ClO4^-}$, with sharing the "doublebondness" within the former two, and sharing the charge in all of them.


What is the bond energy of H-O? Wikipedia disagrees with itself What is the bond energy of H-O? Wikipedia disagrees with itself:

On English Wikipedia:
467 kJ/mol

On French Wikipedia:
366 kJ/mol

What is the right value?

According to the 97th edition of the CRC Handbook of Chemistry and Physics, table 1 Bond Dissociation Energies in Diatomic Molecules [1, p. 9-76]:
$$D^\circ_{298}(\ce{H-O}) = \pu{(429.74\pm 0.03) kJ mol-1}.$$
The value has been taken from the article by Ruscic et al. [2]. Note that this value is strictly for the diatomic molecule $\ce{OH}$ in gas phase and is determined as the standard enthalpy change for the fission process
$$\ce{O-H → O + H}.$$
References

Haynes, W. M.; Lide, D. R.; Bruno, T. J. CRC Handbook of Chemistry and Physics: A Ready-Reference Book of Chemical and Physical Data, 97th ed.; Taylor & Francis Group (CRC Press): Boca Raton, FL, 2016. ISBN 978-1-4987-5429-3.
Ruscic, B.; Feller, D.; Peterson, K. A. Active Thermochemical Tables: Dissociation Energies of Several Homonuclear First-Row Diatomics and Related Thermochemical Values. Theor Chem Acc 2014, 133 (1), 1415. DOI: 10.1007/s00214-013-1415-z.


How to optimize the molecular geometry under PyMol? I already now about the Optimize.py plugin, which seems to be working well, but I don't see an option to prevent it from changing double bonds to single ones. If Optimize isn't capable of such, how can I achieve a minimized molecule from a hand-built one?

As far as I know Optimize uses Open Babel. I could imagine that the problem is, that it doesn't export the state of bonds at all.

Disclaimer: I don't typically use PyMol, so I only made a quick look at the Optimize.py script.

You're right that the problem is with the bond orders. The Optimize.py script uses a PDB file as an intermediate representation that it sends to Open Babel for force field optimization and other properties.

The problem is that PDB does not support bond orders. So yes, you don't get double-bond information back, and I can't verify if Open Babel is getting bond orders from PyMol and Optimize.py.

I looked through the PyMol commands and it seems as if there's a way to write an MDL Molfile (which would have bond orders) but not read one - PDB is the only option for reading and writing in a PyMol script.

My suggestions:

Either push the PyMol developers to support other formats (e.g., Sybyl .mol2 or MDL Mol) in the scripting commands.
Use another program like Avogadro to draw and optimize the geometry and export to read into PyMol later.

(Disclaimer: I'm an Avogadro and Open Babel developer, but there are many programs that could be used.)


Did Berzelius unify the names of chemical elements using the Latin language? I asked this question because the Chinese Wikipedia community is trying to decide whether it should use Latin names or IUPAC names for the first introductory sentence for its articles on chemical elements.

One of the central argument by those support Latin names is the statement that Swedish chemist Berzelius unified the names of chemical elements using the Latin language common to European countries.

They cite the source in Chinese language here:
http://www.cnki.com.cn/Article/CJFDTotal-ZHHJ198805015.htm

"Swedish chemist Berzelius first proposed to use Latin, a language common to European countries, to unify the naming of chemical elements, thereby changing the chaotic status of naming them." (My translation)

What I have found from the literature is that Berzelius's proposal helped unifying the symbols of chemical elements, but the names remain diverse in language sources (not all are Latin) and some remain disputed (not all unified). Thus, I gather that Berzelius's proposal helps, but it is far from the statement that he unified all the names of chemical elements using Latin. If so, it begs the question why IUPAC names for Gold, Silver, etc. are not Latin.

Thus, your help is needed to answer the question as stated in the title.

What I have found from the literature is that Berzelius's proposal helped unifying the symbols of chemical elements, but the names remain diverse in language sources (not all are Latin) and some remain disputed (not all unified). Thus, I gather that Berzelius's proposal helps, but it is far from the statement that he unified all the names of chemical elements using Latin.

This is essentially true. Berzelius was the first to suggest using symbols abbreviated from the names of the elements and this system, with some modifications, is still used today. Most of Berzelius' symbols were abbreviated from Latin names but a large number of the elements discovered since then have had symbols drawn from their common names and some of Berzelius' original symbols have been dropped in favour of more modern versions. Dalton had earlier used a system of drawn symbols to represent the elements but this was not widely taken up as the symbols were difficult to draw and memorise. (source)

Looking though this wikipedia article on the etymology of chemical elements, the most common language of origin appears to be Greek (argon, chlorine etc.), with Latin coming in second (mercury, rhenium etc.), including quite a large number of names which come originally from another language (often Greek) via Latin.

Additionally there are a large number of elements which are named after places (Swedish - ytterbium, yttrium, etc. ; English - americium, californium, etc. to name just a few) or people (mendelevium, einsteinium, fermium etc.) and have been given the Latinate endings -ium or occasionally -um (see this question for more discussion). Also note that according to IUPAC recommendations:

  For linguistic consistency, the names of all new elements should end in “-ium”

The common, and IUPAC recognised, names of gold and gilver both have old Germanic origins and are recognised as official names because of their widespread usage. I have never seen the Latin names for gold, silver or lead (aurum, argentum and plumbum) used in a serious chemical context, although of course the elements' symbols (Au, Ag and Pb) are derived from them.

As such I would not recommend using the Latin names as the main recognized name in a Wikipedia page, although a side note about them might be justified.

EDIT: Thanks to @KlausWarzecha for pointing this out. There is one exception where the latin names (aurum, etc.) come into play: in anionic complexes, the latin version of the name is officially used. $\ce{[Cu(CN)4]^{3-}}$ is tetracyanocuprate(I), $\ce{[Ag(Cl)2]^-}$ is dichloroargentate(I), $\ce{[Au(Cl)4]^- }$ is tetrachloroaurate(III). However, this does not change the viewpoint that the Latin names should not be used when commonly discussing the element.


Does solute concentration vary significantly with depth? If I dissolve glucose in an ordinary beaker filled with water, are the concentrations of glucose at the top and bottom significantly different? What if we use NaCl as the solute? Is there a molecule large enough to ``sink'' to the bottom, and if so would it still be classified as a solute? Likewise, are there solutes which are lighter than H2O, so they float upwards?

(1)If I dissolve glucose in an ordinary beaker filled with water, are the concentrations of glucose at the top and bottom significantly different?
(2) What if we use NaCl as the solute?

1 & 2 are basically the same question. The difference in gravity between the top of the beaker and the bottom of the beaker is real but insignificant. So due to gravity alone there won't be a concentration gradient.
However there are solutions which can be centrifuged which will create a concentration gradient. Essentially the centrifuge creates a large gravity gradient. Close to center gives lower gravity, far from center gives higher gravity. Once you stop the centrifuge however the concentration gradient is not at equilibrium. So over time the concentration gradient would disappear.

(3) Is there a molecule large enough to ``sink'' to the bottom, and if so would it still be classified as a solute?

The above discussion about a centrifuge confirms the basic notation, but there is an unstable gradient, not a sinking per se.
Obviously if you drop a marble in a glass of water it sinks. But the marble isn't dissolved in the water.

(4) Likewise, are there solutes which are lighter than H2O, so they float upwards?

This is a bit more interesting. Open a plastic bottle of soda and put a balloon over the neck of the bottle.  Carbon dioxide was dissolved in the soda under pressure. When the soda is opened, the pressure is released and the excess carbon dioxide will "float up" out of the soda and blowup the balloon. Since the gas phase is less dense than the liquid phase (soda) it is on top. The gas doesn't really "float" on the soda however. It is also important to realize that once equilibrium is established the carbon dioxide remaining in the soda is homogeneously distributed throughout the soda.
A different sort of twist are soap molecules which will "float" to the surface, not because they are lighter, but because of the surface tension of water. The famous soap/pepper experiment illustrates this.
Another sort of twist here is a bartender who makes fancy cocktails by floating one layer of a liquor over another due to the higher layers being less dense. The segregation of the layers is unstable however. One good swirl with a straw and the layers mix.

For 3 and 4 there is what is called a phase diagram. For example it shows if two liquids will mix, or if there will be two phases where one liquid would float on the other.
Think of gasoline and water. There is a gasoline layer floating on the water so there are two liquid phases because there is an interface. However a very tiny amount of gasoline will be dissolved in the water and a very tiny amount of water will be dissolved in the gasoline. In this case the gasoline that is dissolved in the water is distributed homogeneously throughout the water. Likewise the water dissolved in the gasoline is distributed throughout the gasoline.

I used gasoline as a conceptual example. Gasoline however isn't a pure compound but rather a mixture. So for those that want to be picky substitute n-hexane for gasoline.


Why doesn't the Brady's test (2,4-DNPH) work on carboxylic acids? Why does the Brady's test, the reaction of 2,4-dinitrophenylhydrazine with aldehydes and ketones, fail with carboxylic acids?

2,4-Dinitrophenylhydrazine (2,4-DNP or 2,4-DNPH) reacts readily with aldehydes and ketones via a condensation reaction (the lone pair of electrons on the terminal amino group in 2,4-DNPH makes it a strong nucleophile and the condensation starts by the nucleophilic 2,4-DNPH attacking the electrophilic carbonyl carbon) to produce the corresponding hydrazone.  The hydrazine is usually a brightly colored yellow, orange or red compound, so the reaction is often used to test for the presences of an aldehyde or ketone

2,4-DNPH does not react with amides, esters or carboxylic acids.  As shown below for the case of an ester, an extra resonance structure can be drawn for these 3 types of compounds as compared to a ketone.  This extra resonance structure delocalizes some of the positive charge away from the carbonyl carbon onto the adjacent hetero-atom (oxygen in the case of the ester or carboxylic acid, nitrogen in the case of an amide).  This makes the carbonyl carbon less electrophilic in these compounds, and consequently attack by the nucleophilic 2,4-DNPH is less favored.


Why are alkanes 'non-polar' I am unsure with regards to the specific reason why alkanes are non-polar as I have heard different explanations. My teacher said that it is because, although there is a difference in electronegativity between carbon and hydrogen atoms, the carbon is shielded by the hydrogen atoms from reacting with the external environment and therefore, since only the hydrogen can react, it is non-polar (something along the lines of it is the difference between delta + and delta - ends that actually make a molecule polar). From searching on the internet, however, I have found the explanation that it is because the difference in electronegativity is so small (0.3) that it is non-polar, which makes sense, however it contradicts what my textbook states: That only bonds between elements with equal electronegativity are non-polar. Can someone clear this up for me? Thank you!

It is because of the very small difference in electronegativity between carbon and hydrogen.This is the reason why alkanes are very less acidic,do not show electrophillic substitutions etc.


Why does no antitoxin for botulinum type H exist yet? Antitoxins against the botulinum toxins type A–G produced by Clostridium botulinum are known. Why does none exist against the type H botulinum toxin yet?

The premise of this question is based upon outdated information.

According to Study: Novel botulinum toxin less dangerous than thought

  Researchers from the Centers for Disease Control and Prevention (CDC) and the University of Wisconsin (UW), writing in the Journal of Infectious Diseases (JID), said it turns out that the novel toxin can be blocked by available antitoxins after all. That suggests that it isn't really a novel toxin, the authors said

(Attention grammar police: The above is an exact quote from the linked reference.  If you feel the compulsion to keep editing this again and again, please do it in a way that makes clear it is no longer an exact quote.)


Generating PDB file from electron density without force field minimization Is there a step-by-step tutorial on how to generate a PDB file from an electron density file (*.ccp4)?

For my current project, I am trying to take some statistics of some PDB files. However, I do not think that the empirical energy function (MM) implemented in x-plorer could accurately reproduce experimental or QM data. As a result, I would like to use x-plorer to generate the raw PDB file myself (without minimization), but since I have never done this before, I was wondering if somebody could provide me some guidance on how this could be done.

The process of making structural models of macromolecules (the content of PDB files) is called model building, rather than generating, because it normally involves tedious manual building of the model in a graphical program such as Coot. A few programs can do automatic model building, but then the structure needs to be finished manually. Maybe this will change in the future.
All these programs have rather steep learning curve, so it's probably not what you are looking for.

The next problem is that you plan to start from an electron density map.
Assuming that the experimental method was crystallography, not Cryo-EM, the electron density map is derived from both reflection data and the model (and at the same time the model is built into the map, iteratively). So for the purpose of re-building or validation people generally use reflection data, not the map.


Internal energy for Phase Transitions Can we justify that "For sublimation of a solid at 1 atm  $\Delta U>0$ at low temperature and $\Delta U<0$ at high temperature?"
I read several answers and came to know that $U=f(T)$ is valid only for ideal gases at $\pu{1 atm}$ pressure where $T$ is temperature.
Internal energy is the sum of kinetic energy of molecules and chemical potential energy of molecules.
For phase transition even if the temperature remains constant, its internal energy changes.
I tried  to justify  by considering that while sublimation as solid changes to gases its degrees of freedom increases and we may say internal energy is a function of degrees of freedom so  as temperature increases so $U_2 > U_1$ so $\Delta U >0$ but this contradicts what we have to prove.
Also similarly what can  we say about internal energy for other phase transitions ?

Answer given is: ABCD
source:Physical Chemistry (Vol. II) (Class XI) Chemistry Module For JEE (Main+Advanced) Paperback – 1 January 2018
by Resonance Eduventures Limited

Can we justify that "For sublimation of a solid at 1 atm  $\Delta U>0$ at low temperature and $\Delta U<0$ at high temperature?"

No.
$\Delta U>0$, always, for sublimation, because of the energy needed to separate the atoms or molecules in changing from the solid to the gas phase.
As for the enthalpy,
$$H = U +PV \Rightarrow \Delta H = \Delta U+\Delta (PV)=\Delta U+ P\Delta V+ V \Delta P$$
Thus at constant pressure:
$$\Delta H = \Delta U+ P\Delta V \approx  \Delta U+ nRT$$
[Here I've used the ideal gas law to approximate the volume of the gas, and ignored the volume of the solid, which (at 1 atm, room temperature) is about 3 orders of magnitude smaller.]
I.e., $\Delta H$ will be even more positive than $\Delta U$, because of the pV-work required to make space for the gas.
The reason why sublimation becomes more favorable as the temperature increases is because sublimation has a positive $\Delta S$, and $\Delta G = \Delta H - T \Delta S$.  But, irrespective of temperature, at any temperature and pressure at which the solid can exist,  $\Delta U_{solid->gas} >0 $.
$\Delta U >0$ for liquid->gas and solid->liquid phase transitions as well (with the possible exception of the solid->liquid phase transitions for helium-3 and helium-4 at extremely low temperatures which, at least according to https://en.wikipedia.org/wiki/Enthalpy_of_fusion, have $\Delta H < 0$; but solid helium doesn't exist at the 1 atm pressure specified by the OP).
*Yes, you might be able to come up some extreme hypothetical mechanical system where the solid is under such great pressure that the intermolecular forces are so far into the repulsive part of their potentials that $\Delta U_{solid->gas} <0 $.  But that's clearly not what the OP had in mind, since s/he specified 1 atm.  And, in addition, such an arrangement would require that only the solid, but not the gas, be at that extreme pressure.  So, strictly speaking, that $\Delta U$ would not be for the sublimation alone, it would be for the sublimation plus the pressure change.


Hybridisation of ClO2 Find the hybridization as well identify the pπ-pπ as well as pπ-dπ bonds in $\ce{ClO2}$.

$\ce{ClO2}$ has 2 $\sigma$ bonds, 1 lone pair, 2π bonds and 1 odd electron.

Hybridisation is equal to number of $\sigma$ bonds + lone pairs. Since we consider odd electron a lone pair like in $\ce{NO2}$ therefore hybridisation is coming to be $\ce{sp^3}$. There will be no pπ-pπ bonding as all p orbitals are hybridised and there will be 3pπ-dπ bonds

But this is given wrong according to my textbook. What is my mistake?

I actually take issue with the question (which was asked of you not by you), as I think this oversimplifies a molecule, that still isn't well understood in the first place.
That is why I find this question incredibly difficult to answer and why I find that Nuclear Chemist's answer dangerous, because it propagates myths about bonding that have been disproved already.
But enough of the commentary, let's clear up some issues.

Find the hybridization as well identify the pπ-pπ as well as pπ-dπ bonds in $\ce{ClO2}$.

$\ce{ClO2}$ has two $\sigma$ bonds, one lone pair, two π bonds and one odd electron.

There comes a time, where you should realise that hybridisation is not always a useful concept to understand bonding. This time is now.
Your analysis of the bonding is not wrong, but maybe a bit too simple.
The reason for this is that the molecule has a very high symmetry, but it also has an unpaired electron. The problem here is that orbital symmetry cannot match with spacial symmetry, which means that hybridisation doesn't actually work here.
A bit more precise: The electron density of α-spin electrons is different from that of β-spin electrons. The hybridisation of the orbitals to produce either electron density need to also be different. In the very least you would have to consider two different hybridisation schemes at once.
All possible states (that is independent of any hybridisation model) then will be in superposition to form the actual ground state. This ground state will have a complicated bonding situation.
Of course you can use VSEPR theory to predict the shape of the molecule, that is a very good starting point. You will find out that the molecule is bent, and that matches the crystal structure. A bent molecule of the form $\ce{B-A-B}$ always has $C_\mathrm{2v}$ symmetry, it therefore always has π-orbitals. Whether those orbitals are essentially bonding or not, is an entirely different question.
It also does not tell you which atomic orbitals will be involved.
One obvious fault in the question is that there is no involvement of d-orbitals whatsoever in the molecules. The d-orbitals of chlorine are simply not accessible energetically; at least not in the way that it would form stable bonds. Anyone who is jibber-jabbering about 'octet-expansion' has stopped learning about chemistry about 20 years ago. (There are plentiful discussions about this on chemistry.se alone.)
As long as that last part of the question is not meant as a trap (which would be even more awful), it alone shows that the exercise is not well-thought about.
The π-orbitals will therefore be entirely composed out of p-orbitals of the involved elements.
If you look close enough you will realise, that writing a valid Lewis structure of the molecule is not that hard, and you don't even need π-bonds in the first place.
These are solely dictated by symmetry.
In principle that molecule is of a similar system like allyl-systems (with two extra electrons), or ozone (with one more electron). You will need a better than rudimentary grasp at molecular orbital theory to understand it.
Let's assume that things would be easier, and try tor roll with that for as long as possible. I have written a couple of times that terminal atoms can be (as a rule of thumb) regarded as having sp-orbitals.
Therefore the two oxygen will use one sp-orbital to form a σ-bond towards chlorine, the other one is an in-plane-σ-lone-pair. That leaves for every oxygen one in-plane-p-lone-pair, which we will (in the spirit of the approximation) disregard as having any effect on bonding. There is one remaining out-of-plane-p-orbital per oxygen. These will, due to symmetry, participate in π-bonding.
From VSEPR-theory we know the molecule is bent. The approximate hybridisation of the orbitals of chlorine that match that would be sp². Let's just assume that this is the case. Two of these will be the counterpart for the σ-bonds. The remaining will be either a lone-pair, or singly occupied. Already in this very simplified view, the problems start here. Let's further assume, that because it has higher s-character, the orbital will be lower in energy and therefore be doubly occupied. That leaves the odd electron in an out-of-plane-p-orbital, which will participate in π-bonding.
In this simplified and assumption riddled thought-experiment, the π-bonding will form three molecular orbitals pictured below.

Essentially the stabilisation from the π-bonds itself should be small, and could be described as π-backbonding, or is similar to negative hyperconjugation; maybe negative π-backbonding (!?).
This all was deduced from a very reduced and simplified set of theory; the reality is probably by far more complicated. Some of the greatest minds of (theoretical) chemistry have been puzzled about $\bf\ce{ClO2}$, and we have not even started to discuss the three-electron bond they assumed in between. Let's just say that this is not an undergrad study problem.
I know this doesn't really answer your question, and this is not very satisfactory either, but this is science: It often gives you more questions than answers.


Why are axial bonds are longer than equatorial bond in sp3d hybridized atoms? Why are axial bonds are longer than equatorial bond in case of $\mathrm{sp^3d}$ hybridization? I have done some research but I can't seem to find the answer.

You are asking for $\mathrm{sp^3d}$ hybridisation, but I do not know of a case where $\mathrm{sp^3d}$ hybridisation actually happens. Either it does not make sense to discuss hybridisation at all (the iron in pentacarbonyliron) or the hybridisation is actually not $\mathrm{sp^3d}$ but $\mathrm{sp^2 + p}$ (the phosphorus in $\ce{PCl5}$). I shall discuss the latter, because I believe it could be the one you are on about.

Phosphorus pentachloride is often drawn like in figure 1 below for simplicity. This depiction implies five identical covalent 2-electron-2-centre bonds. However, that does not agree with the octet rule.

Figure 1: Simplified structure of $\ce{PCl5}$.

Instead, one can draw a set of two mesomeric structures, each one conforming with the octet rule — see figure 2.

Figure 2: Mesomeric structures of $\ce{PCl5}$ conforming to the octet rule.

This already hints us towards the answer: Rather than assuming two bonds which are equal to the other three bonds we need to consider three ‘classical’ 2-electron-2-centre bonds to the equatorial chlorines and one 4-electron-3-centre bond to the two axial chlorines. This bond’s order is $0.5$ rather than $1$. Typically, the lower the bond order the weaker a bond and therefore the greater the bond length is — the experiment is in fine agreement with theory.

But no bonding discussion is truly complete without an orbital consideration. Check out the three collinear p-orbitals of phosphorus and the two chlorines, that together form three molecular orbitals labelled $\Psi_1$ to $\Psi_3$ in figure 3.

Figure 3: Representation of the three molecular orbitals that form the 4-electron-3-centre bond.

As you can see, the lowest MO $\Psi_1$ is bonding with respect to both $\ce{P-Cl}$ bonds. The highest MO $\Psi_3$ is antibonding with respect to both. And the middle one is bonding with respect to $\ce{Cl-Cl}$ but nonbonding if we add the phosphorus atom. We need to fill in four electrons into these three orbitals ($\ce{PCl3}$ has one lone pair, and we are effectively using the bond of a $\ce{Cl-Cl}$ molecule as our second electron pair). Thus, the bonding and the nonbonding orbitals are filled. Bond order can now be calculated by:

$$\text{bond order} = \frac{(\text{electrons in bonding orbitals}) - (\text{electrons in antibonding orbitals})}{\text{number of bonds}\times 2}\\= \frac{2-0}{2 \times 2} = 0.5$$

Again, a lower bond order typically correlates with a greater bond length.


Is organic carbon the same thing as organic compounds? Are they the same thing or is there some minute difference? I have looked all over the web and you would think this would be a common question but I can't find any answer. If they are different then what is the difference?

Thanks

Organic compounds typically refers to particular, identified compounds and/or compound classes.

Organic carbon, ususally expressed as total organic carbon (TOC), is a relevant term in waste (water) management.

In this context the total carbon (TC) is expressed as the sum of total inorganic carbon (TIC) and total organic carbon (TOC).

TIC and TOC are often measured as the amount of $\ce{CO2}$ liberated from a sample.

UPDATE

While Ivan Netetin's remark on the broader context is certainly true, terms like TOC and COD (in German CSB), the chemical oxygen demand, play an important legal role in the question whether and under which conditions waste water may be submitted to the public sewer systems. This involves water directly from procdution processes, as as the elution of material from from landfills.

In Germany, this is specified in the Abwasserverordnung - AbwV. Here, §6(3) states

  Ein nach dieser Verordnung einzuhaltender oder in der wasserrechtlichen Zulassung festgesetzter Wert für den Chemischen Sauerstoffbedarf (CSB) gilt unter Beachtung von Absatz 1 auch als eingehalten, wenn der vierfache Wert des gesamten organisch gebundenen Kohlenstoffs (TOC), bestimmt in Milligramm je Liter, diesen Wert nicht überschreitet.

Translated to non-legalese English, this means: Business-specific limits are considered to be met (= it is legal to submit the water to the sewer) if the fourfold TOC does not exceed the COD given by law.


Why is the first ionization enthalpy of zinc higher than expected? The following sentence is stated in my current chemistry book:

  The first ionization enthalpy of zinc is higher because it represents an ionization from the $\rm 4s$ level.

Now zinc does have an ionization enthalpy of $\pu{906 kJ/mol}$, which is higher than copper's $\pu{745 kJ/mol}$ or nickel's $\pu{736 kJ/mol}$ in the same series. But I am not sure that the reason for this increase stated in the book is correct. For example, even copper has an ionization from the $\rm 4s$ level, but it does not have anywhere near as high ionization enthalpy as zinc. Is the book correct, or is there some other reason for this?

It is because the shell is complete.

Zinc's electronic configuration is $\rm [Ar] 3d^{10} 4s^2$, which is a complete configuration for a transition metal. The next element in the same period is $\ce{Ga}$, which starts filling $\rm 4p$ orbitals ($\rm [Ar] 3d^{10} 4s^2 4p^1$).

The zinc's electronic configuration causes a sort of special stability because of the filled electronic states. Also, the effective nuclear charge is greater than $\ce{Cu}$'s, so it's more difficult to ionize $\ce{Zn}$.


Etymology of saturation degrees (-ane, -ene, -yne) in aliphatic compounds I like to know where the names for saturated, unsaturated double, and unsaturated triple bonded aliphatic compounds came from.

What is the reason/etymology behind the suffix -ane, -ene, and -yne?

I suppose it parallels something like the inflection-based umlauts in Germanic languages (which may be seen rudimentary in modern English for example in words like mouse/mice) but that is utter speculation. Yet I did not find anything in the web (I think some good extensive Chemical Etymological Encyclopaedia is really missing).

I found the following information through a website linked to Yale University.

The naming structure seems to have evolved from some of the early nonsystematic names given to hydrocarbons. It started with Dumas who in 1835 came up with the name Methylene for the $\ce{CH2}$ radical. The "-ene" suffix is a Greek feminine patronymic which was used to denote that Methylene was the "daughter of" (derived from) the "spirit of wood" or what we now call methanol (Dumas believed methanol had the formula $\ce{CH2\cdot H2O}$).

The naming structure caught on for ethylene since it was related to ethyl(even though the meth- in Methylene came from a Greek work for wine and not the numerical prefix used in methyl).

When the use of these feminine patronymics (-ene, -ine, -one) became common during the 1860s, Hoffman proposed making a consistent nomenclature that followed the standard progression of vowels: a e i o u. This led to the suffixes -ane, -ene, -ine(-yne), -one, and -une to refer to increasingly unsaturated hydrocarbons.

Since -one was already in use for ketone structures, only the first three suffixes caught on.


Do Ligands electrons fill the d-orbitals of the metal ion? We know that ligands make coordination bonds with d-orbitals of transition metals forming complexes. My question is, if that happened, all d-orbitals will be filled and no electron transition could happen between eg and t2g orbitals to be responsible for magnetism or colour formation.

I need clarification.

You are mistaken. The electron pairs donated in the coordinate bonds by the ligands do not fill the d orbitals of the metal ion.

The basis of crystal field theory is the electrostatic repulsion between the d orbitals of the central metal ion and the electron cloud of the ligands. This repulsion then leads to the breaking of the degeneracy of the orbitals, giving rise to disparity in energy between different sets of d orbitals. This arises due to the different spatial orientations of the 5 d orbitals (i.e. $xy, yz, xz, x^2-y^2, z^2$). Thus, they experience different amounts of repulsion from the ligands when the ligands are oriented in different geometric configurations (i.e. octahedral, square planar, tetrahedral etc.) around the central metal ion.

For more details on crystal field theory and an alternative explanation known as ligand field theory (based on molecular orbital theory), you can refer to the following discussion:

In an octahedral complex, what happens to the electrons donated by the ligand?

Since the d orbitals are not filled up by any ligand electrons, the electronic transitions between the orbitals of different energy levels (such as in the case of $t2g$ and $eg*$) are certainly possible.

Now, you may wonder: Where do the donated electrons go?

Well... based on valence bond theory, they are used in the formation of covalent bonds between the central metal ion and the ligands. These covalent bonding interactions are described as the overlap of the ligands' atomic orbitals with the electron pairs with the hybrid atomic orbitals of the central metal ion.

For example, in the case of a hexaaquairon (III) complex, the iron (III) ion has 5 electrons, each singly occupying one d orbital. As the d orbitals are occupied, the vacant 4s, 4p and 4d orbitals are hybridised to give 6 $sp^3d^2$ orbitals to be used to overlap with the atomic orbitals of the ligands.

Thus, the donated electrons actually are "shared" between the valence atomic orbitals of the ligand and the valence hybrid orbitals of the metal ion.

Hope I have clarified your doubts.


What's the difference between precipitate and turbidity? In many tests for various radicals, a "<color> precipitate" or "<color> turbidity" is mentioned. For example, lead acetate with chromate ions gives a yellow precipitate, but hydrogen sulphide with potassium dichromate gives a yellow(or white) "turbidity".

In the reaction itself, the turbidity is noted as a precipitate with a down arrow, but I'm wondering if there is any difference between the two. It looks like turbidity refers to a colloid/suspension being formed, but I'm not sure.

I'd like to know:

The physical differences/distinguishing features between turbidity and precipitate, if any.
The observable differences/distinguishing features between the two, if any.

My understanding is that turbidity is an adjective describing the presence or degree of particulate matter suspended in a liquid (see turbidimetry), whereas a precipitate is a particulate that can result in turbidity.

The turbidity generated by a precipitate is a function of its ability to form a stable suspension. If the precipitate simply drops to the bottom of a container or readily flocculates, the resultant turbidity will be low. The turbidity of a suspension is also going to be a function of the solvent and the nature and concentration of dissolved ions - a dramatic example of this is the rapid clarification of muddy freshwater at river deltas mixing with salt water.

If the word turbidity is being used as a noun, it is probably referring to a light-occluding suspension.


Why is Hessian Matrix called 'force-constant' matrix? Hessian Matrix is a square matrix containing the elements as the second-order partial derivatives of energy-function of a molecule; the derivative is done with respect to geometric coordinates of the molecule.

However, I couldn't get why Hessian matrix is referred to as force-constant matrix; as is done in the book I'm following viz. Computational Chemistry: Introduction to the Theory and Applications of Molecular and Quantum Mechanics by Errol G. Lewars and in many other sources.

If the energy function is a quadratic function which is mostly the case for a molecule following 1D PES; the function which a Hookean spring follows for small displacement, then the second derivative does indeed give the value of force constant, that is $$\begin{align}E- E_0&= k(q-q_0)^2\\ \implies \frac{\mathrm{d}^2 E}{\mathrm{d}q^2}&= 2k\;.\end{align}$$

However, for molecules following multidimensional PES, most of the time, the energy function is not simple quadratic; how can the second order derivative give the force constant? Second order derivatives vary from point to point unlike the simple case above.

So, can anyone tell why Hessian matrix is called force-constant matrix?

You are right that multidimensional PES is not simple quadratic, but close to the PES well minima it can be approximated as harmonic oscillator. Also if you really want to include the real PES you can include anharmonicity in your calculation. In that case you have to include an anharonic constant $x_e\;.$ Your PES can be written as
$$E=\left(n+\frac{1}{2}\right)h\nu-x_e\left(n+\frac{1}{2}\right)^2h\nu$$

Usually in computational chemistry the energy can be expanded into a Taylor series
$$E=E_0+\frac{\mathrm{d}E}{\mathrm{d}x}(x-x_0)+\frac{1}{2}\frac{\mathrm{d}^2E}{\mathrm{d}x^2}(x-x_0)^2+\textrm{higher order terms}\;.$$
And you can usually neglect those first order term and higher order terms for most of the practical problems. And then your PES will follow the trajectory of a harmonic oscillator.


How do I perform a gravimetric analysis of Portland cement? To determine the silica ($\ce{SiO2}$) content in Portland cement by gravimetric analysis it is digested in $\ce{HCl}$ and the silica residue is eventually measured.

The chief constituents are $\ce{Ca2SiO4}$ (or $\ce{2CaO.SiO2}$), $\ce{4CaO.Al2O3.Fe2O3}$, $\ce{Ca3(AlO3)2}$ (or $\ce{3CaO.Al2O3}$), and $\ce{CaSiO3}$ (or $\ce{CaO.SiO2}$). Gypsum is also present. What species are present in solution after the $\ce{HCl}$ has reacted with the cement? I assume $\ce{CaCl2}$ and $\ce{H2O}$ would be present, but would $\ce{AlCl3}$ also form? How would I approach this problem?

http://books.google.com/books?id=VeXQAAAAMAAJ&pg=PA2&lpg=PA2&dq=Gravimetric+Analysis+of+Portland+Cement+%22aluminum
Copy whole URL. Page 2 for clickable index, then page 18.
DOI: 10.1039/AN9709500051

Digest it in "strong" HCl (tough lab for employment, worse if you service the fume hoods).  Everything except silica comes out as soluble chlorides (hydrated complex ions).  That says something about Portland cement in acid environments and, unless modified, in seawater.


PVC reacting or self-decomposing? I have a PVC inflatable device (a custom made inflatable pillow) which is, according to the spec sheet, made out of PVC. The PVC material itself, however, is not the one you'd find on usual swimming rings or such, but something closer to that of the PVC pipes (matte texture, white color).

The problem I encountered is, after some time of storage (deflated of course) in a dark and dry room, some kind of an oily fluid appears on the surface of the device.

Initially I thought that, as it is a pillow, it might have accumulated some of the body sweat over time, and after some time it came out as the material was returning to its original size after being stretched due to inflation. So I thoroughly washed it, but after a couple of days of storage, oily stains can be seen again on it's surface. I didn't kinda think it through, but it might have been helpful to collect the accumulated fluid before I washed it off to run a pH test on it (can't do much other tests at home).

So, may that be a sign of PVC degradation of some sort, and if so, what is that? This occurs whether it's stored in a polyethylene bag or just plain put on a wooden shelf with nothing in contact.

I was looking in various sources but most only specify that PVC degrades either under higher temperatures (above 60ºC) or in contact with quite agressive fluids.

Picture of plain material texture:

Picture of material with stains that appear (reflective):

What you are seeing is most likely PVC plasticiser.

Pure PVC is a hard, inflexible solid. It doesn't make for good flexible objects. To make it flexible, plasticisers are usually added. These plasticisers are often compounds like diisononyl phthalate (DINP), diisodecyl phthalate (DIDP) and di-2-ethylhexyl phthalate. These are, when pure, oily liquids. They often slowly migrate from the PVC object and can do so faster in certain circumstances (eg they will be pulled out when the PVC is in contact with hydrocarbon solvents).

So, most likely, you are seeing small amounts of plasticiser migrate from the PVC and not any degradation of the polymer itself. Eventually, of course, if enough plasticiser leaves the PVC object it will become brittle and inflexible. But it may be quite some time before that happens.


Why do pregnant women turn toilet seats blue? My partner is pregnant, and thought perhaps she was going mad when one day she found our toilet seat had turned blue. After some investigation, it seems this is a real phenomena, see here for a nice example, with the image from this link shown below.

One explanation suggested is dye transfer from new maternity jeans, however we find this unlikely for a variety of reasons:

This is late pregnancy and the jeans have been worn for some time without this effect.
My partner assures me she takes her trousers off to go to the toilet, and even her underwear, and there is not a white patch where the underwear would have been (and therefore no dye transfer)

The phenomenon is also described as possibly relating to taking certain pharmaceuticals for Crohn's disease. My partner happens to take (Sulfasalazine) which is used for this (although it's not what she's taking it for) and she's never encountered this before, despite taking them for some time. Another suggestion we've seen is that it is from the dye in a batch of pregnancy vitamin tablets she's started taking, which are indeed blue, although, she's also been taking these daily for a month or so.

However, we don't know the answer, and these are just ideas, so why has our toilet seat turned blue?!

The question cannot be answered without a chemical analysis.

However, given that there exist toilet seats with silver based anti-microbial coatings, the bluish grey color may come from metallic silver particles produced by a reduction process similar to the photographic development process.

You could proof the presence of silver by applying Caffenol, an alkaline aqueous solution of instant coffee, washing soda, and vitamin C (ascorbic acid) to the toilet seat. I suggest to use Caffenol-C-M whose preparation is described here. Let the Caffenol react on the toilet seat as long as possible. If grey or black stains are formed that persist after cleaning, your toilet seat contains silver with high probability. Note that the stains cannot be removed anymore.

If the test is positive, the next step would be to investigate why your pregnant partner has reducing powers.


IUPAC nomenclature: "Smallest sum of locants"? While naming any compound, the numbering should be done such that position of substituted groups gives smallest sum.

I'm struggling with the IUPAC naming for cyclic compounds. How exactly does this rule work?

The guideline that you have mentioned (‘While naming any compound, the numbering should be done such that position of substituted groups gives the smallest sum.’) does not exist in the IUPAC recommendations.
The corresponding section in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) actually reads as follows:

P-14.3.5 Lowest set of locants

The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’.
(…)

Therefore, the smallest sum of locants can lead to wrong results.
For example, for 2,7,8-trimethyldecane, the smallest sum of locants would yield the wrong name ‘3,4,9-trimethyldecane’, since $2+7+8=17 > 3+4+9=16$.

Likewise, for 2,2,6,6,7-pentamethyloctane, the smallest sum of locants would yield the wrong name ‘2,3,3,7,7-pentamethyloctane’, since $2 + 2 + 6 + 6 + 7 = 23 > 2 + 3 + 3 + 7 + 7 = 22$.

For 1,6,7-trimethylnaphthalene, the smallest sum of locants would yield the wrong name ‘2,3,5-trimethylnaphthalene’, since $1 + 6 + 7 = 14 > 2 + 3 + 5 = 10$.

For 1,6-dimethylcyclohex-1-ene, the smallest sum of locants would yield the wrong name ‘2,3-dimethylcyclohex-1-ene’, since $1 + 6 = 7 > 2 + 3 = 5$.

Anyway, the IUPAC recommendations have more than 1500 pages. Hence, you should be careful when applying any simplified or abridged version of the nomenclature rules.


Are intermolecular forces a type of chemical bond? My chemistry teacher told me that chemical bonds are of two types: intramolecular and intermolecular. He said that intermolecular forces come under the category of intermolecular chemical bond.

I have never read such statement anywhere. Nor can I find anything on the Internet that would support this statement.

My understanding is that chemical bond is a force that holds atoms together in a chemical species. Since intermolecular forces do not hold atoms together, they should not be termed as chemical bond.

So, are intermolecular forces a type of chemical bond?

The IUPAC definition of "chemical bond" is:

  When forces acting between two atoms or groups of atoms lead to the formation of a stable independent molecular entity, a chemical bond is considered to exist between these atoms or groups. The principal characteristic of a bond in a molecule is the existence of a region between the nuclei of constant potential contours that allows the potential energy to improve substantially by atomic contraction at the expense of only a small increase in kinetic energy. Not only directed covalent bonds characteristic of organic compounds, but also bonds such as those existing between sodium cations and chloride anions in a crystal of sodium chloride or the bonds binding aluminium to six molecules of water in its environment, and even weak bonds that link two molecules of $\ce{O_2}$ into $\ce{O_4}$, are to be attributed to chemical bonds.

So the answer is "yes" in some cases.

See also the IUPAC definition of "molecule":

  ... must correspond to a depression on the potential energy surface that is deep enough to confine at least one vibrational state

So, for example, a water-water dimer, held together by hydrogen bonding, has a monomer-monomer potential energy surface that is deep enough to confine at least one vibrational state, and it would be appropriate to refer to the hydrogen bond as a chemical bond.


Why will a strong acid neutralize as much base as a weak acid? This is a simple concept that I can't seem to understand. Why will a strong acid neutralize as much base as a weak acid, if the acids are of the same volume and concentration? A strong acid will dissociate more in solution and thus have a greater number of $\ce{H^+}$ ions, as far as my understanding goes. The opposite if true for the weak base. So wouldn't the strong acid neutralize more base than the weak acid, because it has more $\ce{H^+}$ ions to neutralize the $\ce{OH^-}$ ions of the base with?

Strong acid is dissociated all but completely: you have a certain amount of $\ce H^+$ and start neutralizing them, one drop at a time. As you do so, their numbers steadily decrease. It feels like sawing a huge log with a hand saw: you see the amount of job, and you do it until it is all done.

Weak acid is dissociated only partially. You see a relatively small amount of $\ce H^+$, but as you start neutralizing them, more molecules of acid get dissociated and their $\ce H^+$ ions stand in place of those you've neutralized. It feels like sawing a thin log, but a peculiar one: it seems to regenerate under the saw.

So at the end of the day, it is the sheer amount of acid that matters, rather than immediate concentration of $\ce H^+$.


Stability of 3d metal fluorides and iodides in different oxidation states I read that the fluorides of 3d metals in lower oxidation states, e.g. $\ce{VF2}$, $\ce{TiF2}$, and $\ce{CuF}$ are thermodynamically unstable. For example, $\ce{CuF}$ disproportionates to $\ce{Cu + CuF2}$, as described on Wikipedia.

On the other hand, their iodides tend to be more stable (so, for example, $\ce{CuI}$ can be easily obtained from a redox reaction between $\ce{Cu^2+}$ and $\ce{I-}$). Why is this the case?

In general, iodides stabilise lower oxidation states and fluorides stabilise higher oxidation states, e.g. $\ce{CuF2}$ versus $\ce{CuI}$. This can be explained with some thermodynamics. Consider

$$\ce{M(s) + $\frac{n}{2}\,$X2 (g/l) -> MX_n (s)}$$

In order to form a halide salt with a higher oxidation state (i.e. larger $n$), you need to pay some energetic costs to get the metal to the higher oxidation state, namely:

more electrons are removed from the metal, so you need more ionisation energies
you need to generate more $\ce{X-}$ anions, so you need to break more $\ce{X-X}$ bonds (bond dissociation energies) and add more electrons to them (electron affinities).

This seems bad, so why would any higher oxidation state ever be formed? The answer is that when you pay these costs, you recoup some of the energy by forming more bonds between $\ce{M^n+}$ and $\ce{X-}$. If this bonding is strong enough, this can outweigh the costs described previously, and that makes the higher oxidation state more stable than the lower one.

For fluorine, the $\ce{M-F}$ bond is strong (regardless of whether you consider it to be ionic or covalent) and hence this energetic gain is large. On the other hand, the formation of fluoride ion is not particularly costly: the $\ce{F-F}$ bond is easy to break and fluorine loves picking up electrons, so the electron affinity works in your favour.

Overall, this means that there is a thermodynamic force for formation of fluorides in high oxidation states. Conversely, if you consider iodides instead, the formation of iodide ions is not as easy as the formation of fluoride. And the $\ce{M-I}$ bonding is weaker, too, so there isn't even much incentive to go to higher oxidation states.


Change in current in an electrolytic cell over time Why does the current increase over time in an electrolytic cell with a $\ce{CuNO3}$ electrolyte. The current increased about $2~\mathrm{A}$, and the temperature increased during the experiment to around $55\ \mathrm{^\circ C}$.

I’m not sure whether the formation of hydrogen and oxygen gas at the electrodes would decrease the internal cell resistance causing the current to increase. Or is there something else causing this?

The electrolyte conductivity is dependent on the temperature of the electrolyte. Higher temperature increases ions mobility, thereby increasing electrolyte conductivity. The conductivity increases around 2% per degree (assuming linear increase of conductivity). More info here.

As Burak Ulgut mentioned, temperature increase is due to ohmic (Joule) heating. This can be estimated accordingly: $Q=V \cdot I \cdot t$


Are there any common (preferably somewhat complex) organic molecules that help combat ocean pollution? I'm participating in an activity where I need to present a topic/molecule in organic chemistry, and I am pretty passionate about "reversing"/helping with cleaning the ocean. I was wondering if there are any organic molecules that help with this or with lessening the effects of pollution. If you could let me know what solutions they are used in that would be a great help also. Thank you very much.

Chemical surfactants or dispersants are useful for cleaning up oil spills. They break up the oil slick and allow the oil to become dispersed in the water, so that naturally occurring bacteria can break down and oxidize the hydrocarbons in the oil. Dispersants are typically designed to be water-soluble at one end and oily at the other end--basically soaps or detergents. See http://masgc.org/oilscience/oil-spill-science-dispersant-bkgrnd.pdf for more details.


Why is carbon monoxide a "weak" lewis base? A lewis base is substance that donates an unshared pair of electron.

In CO, a lone pair is present on both carbon and oxygen each. Moreover, lone pair of electron on carbon is localised so CO will act as lewis base.
What causes the CO to become weak lewis base?

Also tell, what "Lewis acidity of CO" would mean? Actually , I read that lewis acidity of CO is very small? But I am not able to understand it.

$\ce{CO}$ is usually considered a Lewis base, as the lone pair on carbon readily donates it's electrons, for example forming transition metal coordination complexes. Although oxygen is more electronegative than carbon, the carbon is more rich in electron density in the $\ce{CO}$ molecule as it basically has a full formal negative charge. Additionally, products formed by the carbon bonding to Lewis acids are much more favorable that products that would be formed by bonding to the oxygen atom.

It is also possible for CO to act as a Lewis acid however. When acting as a ligand with certain low-valent transition metals, the metal can redistribute electron density back to the $\ce{CO}$ ligand in a process called back-donating. In this case the $\ce{CO}$ molecule accepts electrons via it's antibonding $\mathrm{\pi}$ orbitals, making it act as a Lewis acid.


Van der Waals forces What exactly are Van der Waals forces? How do they arise and how can an instantaneous dipole cause another dipole? Shouldn't this make a chain reaction that causes all matter to eventually become polar?

The post that was linked to you in comments contains  a nice, short description of the 'traditional' categories of van der Waals interactions. I suggest you take a look at that, it might just be enough to clear things up. The reason I am writing this post is that the categories introduced there are defined somewhat arbitrary, so I'll try to introduce some terminology that (as of today) is considered correct in the field.

So given a molecular system, it is quite easy to factorize the forces into intermolecular and intramolecular forces. Intramolecular forces are what hold the molecules together. These are (almost completely) covalent bonds, and these interactions would be same for one single molecule and the bulk phase of the matter. But we know, from the real life, that matter does not only exist as an ideal gas: we got liquids, solids, and even the gases are non-ideal. So evidently, there has to be forces between molecules. These forces are collectively called van der Waals forces for historical reasons.

Now there are three different types of interaction between molecules. There are electrostatic, induction and dispersion interactions.

Electrostatic effects are simple classical interaction between charged particles. Thesy simply come from the Coulomb law, can be either attractive or repulsive, and they are strictly additive (because Coulomb law is additive)

It is tricky to properly understand induction and dispersion. These two effect come from quantum mechanical perturbation theory. Without getting into details, perturbation theory means that the wavefunction of two molecules together is not simply the sum (or average) of the wavefunctions of the separate system, but there is an interaction between them, but this interaction can be expressed in terms of the interactionless wavefunction. The expression for the second-order energy correction is given by the Rayleigh-Schrodinger picture as follows (this is called the polarization approximation):

$$W^{(2)} = - \sum_{mn}' \frac{ <00|H'|mn><mn|H'00> }{W_{mn}-W_{00}}$$

In this expression, $|m>$ represents the mth excited state of the original wavefunction of molecule A and $|n>$ represents the nth excited state of the original wavefunction of molecule B. This expression could be read: the energy difference between the two separate systems and the interacting system depends on the excitations in the separate systems through an interaction operator $H'$.

The sum can be factorized into three components:

$$W^{(2)} = - \sum_{m\neq0}' \frac{ <00|H'|m0><m0|H'00> }{W_{m0}-W_{00}} - \sum_{m\neq0}' \frac{ <00|H'|0n><0n|H'00> }{W_{0n}-W_{00}} - \sum_{m\neq0,n\neq0}' \frac{ <00|H'|mn><mn|H'00> }{W_{mn}-W_{00}}$$

Here, the first term contains the energy contributions of the interactions of the ground state of A with the excited states of B. This is called the induction energy of B. The second term is the induction energy of $A$, which again, expresses the total response of the molecule A to the ground state of B. The third term is the so-called dispersion interaction, which is then a mixing of the excited states of both molecules.

So these expressions are the defining expressions for induction and dispersion. One should know that both of these interactions are always attractive , they are non-additive, and they are always present, between charged and neutral particles too. Also don't get confused that the energies are expressed in terms of excited states: there is no excitation going on in the system, it's just the mathematical tool that we used as definitions.

The total intermolecular force, or van der Waals force is then usually the sum of electrostatic, induction and dispersion terms. For the sake of completeness we might include resonant energy couplings and magnetostatic interactions too, but these are either rare or so small that we can neglect them.


Why the H+ ion remains with the molecule even though there is no bond between it and hyperconjugative structure? In hyperconjugation, it is known that form an $\alpha$ hydrogen, one $\sigma$ electron goes to the corresponding carbon carbon bond, and there is apparently or practically, no bond between the hydrogen and carbon. I read in a textbook,

  although one $\ce{C-H}$ bond of the alkyl is shown to be broken in each hyperconjugative structure, $\ce{H+}$ is never free from the rest of the molecule.

But then, still there is no bond. Is there any specific reason why the $\ce{H+}$ ion remains with the molecule though there is no bond attaching the ion to the rest of molecule?

Resonance structures do not exist on their own! Resonance is not a rapid equilibrium between different structures. It is a description of a single delocalised structure using contributing forms which only involve 2-centre-2-electron bonds. It is therefore meaningless to think about what would happen to an atom in a resonance contributor if it were left around.

In your example the two structures are resonance contributors to the single delocalised structure and so the hydrogen ion cannot move away from the molecule because it never exists on its own as a hydrogen ion. It is always bonded to the rest of the molecule.

Hyperconjugation can be explained through resonance but it is much more clearly explained through molecular orbital theory. In this picture you have donation of electron density from the $\ce{C-H~\sigma}$ orbitals into the empty $p$ orbital of the positively charged carbon. This is possible because the two orbitals can be aligned in a plane due to rotation around the C-C bonds.


How do water insoluble medications get absorbed in the blood stream? If a chemical is ingested (supplement or drug) orally, which is insoluble in water, how can it be absorbed into the body?  Is there something inside a capsule that the compound is mixed with?

Well, firstly the capsule itself is not entirely insoluble. You can see this for yourself - if you drop one into water, it will start to dissolve very slowly and the water will turn cloudy. This process usually goes faster in the stomach, which has a very acidic environment, or the small intestine which is more alkaline. The capsule material is designed to dissolve when it's eaten, hence releasing the drug - if it didn't dissolve, the whole capsule would just pass through your digestive system and come out... at the other end.

On to the issue about the drug itself. In general designing a drug is not very simple at all, because it has to survive the very same enzymes that break down our food - if the enzymes break down the drug then it becomes useless. For example, if you tried to eat insulin it wouldn't do you any good - insulin is a protein and will be immediately digested by protease enzymes. If the drug survives those enzymes, it has to be able to diffuse through the lining of the small intestine in order for it to enter the bloodstream. The cell membrane, being a non-polar lipid bilayer, doesn't easily let polar or charged molecules diffuse across it. So normally drugs have to be somewhat lipophilic, which is a fancy way of saying non-polar. But at the same time, as you rightly note, they also have to be water-soluble. Most drugs try to strike a balance between the two.

One way that drugs are absorbed is by reversible protonation. Many drugs have amine functional groups with a $\text{p}K_{\text{aH}}$ of around 6-8. At the physiological pH of 7.4, an equilibrium will exist in which both the unprotonated and protonated forms exist in significant amounts. The unprotonated form can cross cell membranes more easily, and the protonated form is more soluble in water.

There are other drugs that don't diffuse across cell membranes very well, due to their large polarity. They may be absorbed via different means, such as active transport (in which the cell uses energy in order to let the molecule enter), or facilitated diffusion (which is the same as diffusion, except that a protein helps out). An example is levodopa; it is structurally similar to the amino acid phenylalanine and the transport protein for phenylalanine can't quite tell the difference. Being polar molecules, these drugs are already water-soluble.

In fact, that's only half the story, because different drugs have different targets. Some need to go to the brain and for this to happen, they have to cross the blood-brain barrier. Some need to stay in the intestines to kill off bacteria - so you have to make them really polar so that they can't diffuse into the bloodstream. Absorption is only a small part of a drug's journey in the human body. If you are interested in finding out more, look for a textbook on medicinal chemistry and read the section about pharmacokinetics - it should have all the info you want.


Is there a reliable way to determine if a chemical system is more suited to be calculated using a multireference method instead of DFT? Some time ago I recall hearing a professor say that looking at a DFT result (I don't remember exactly what, a TD-DFT, spin contamination etc...), he could tell that the system was multireferenced.

My question is: do we have anything in a DFT calculation that is indicative of a multireference system?

You should have a look at the so called FOD (Fermi Occupied Density) analysis by Grimme et al. That is accessible by ORCA 4 very easy and straight forward by specifying "FOD" as a keyword. [1-3]

From ORCA Input Library [1]:

  The FOD plot above (plotted with Chemcraft) indicates a significant and delocalized FOD. Multiconfigurational methods should be used to study this molecule and any single-reference result (HF, DFT, MP2, CC should be regarded with suspicion)

Via Multiwfn you can access the "local total/dynamic/nondynamic electron correlation function" and "total/dynamic/nondynamic electron correlation index" [4,5].

Visual abstract from [5]. You can see the dynamic and non-dynamic correlation from which you also can see if single reference methods are acceptable or not.

[1] ORCA Input Library, FOD analysis
[2] S. Grimme, A. Hansen, Angew. Chem. Int. Ed. 2015, 54 (42), 12308-12313
[3] C. A. Bauer, A. Hansen, S. Grimme, Chem. Eur. J. 2017, 23 (25), 6150-6164
[4] E. Ramos-Cordoba, P. Salvadorc, E. Matito, Phys. Chem. Chem. Phys. 2016, 18, 24015-24023
[5] E. Ramos-Cordoba, E. Matito, J. Chem. Theory Comput. 2017, 13 (6), 2705–2711


Can ionic compounds dissolved in organic solvent conduct electricity? Ionic compounds dissolved in water, as an aqueous solution, can conduct electricity.

Can they conduct electricity if they are dissolved in organic solvent?

Sure they can.  Even if most of the ions are in ion pairs or other complexes, they can still move by being transferred from one complex to another.  All you need is to get the ionic compound to dissolve, which in most cases requires a polar solvent with which the ions will not react.
A rather unexpected case involves magnesocene, whose bonding is a mixture of ionic and covalent.  When dissolved in ether or THF, magnesocene forms enough ions not only to impart significant conductivity in solution, but also to be proposed for use in batteries:

Unlike ferrocene, magnesocene displays slight dissociation and subsequent ion association in polar, electron-donating solvents (such as ether and THF).[1]
$\displaystyle {\ce {MgCp2 <=> MgCp+ + Cp-}}$
$\displaystyle {\ce {MgCp2 + MgCp+ <=> Mg2Cp3+}}$
$\displaystyle {\ce {MgCp2 + Cp- <=> MgCp3-}}$

Note the single charge on all ionic species.  This is to be expected in organic solvents with their weaker polarity and ion solvation power relative to water.
Cited reference:
1.
Schwarz, Rainer; Pejic, Marijana; Fischer, Philipp; Marinaro, Mario; Jörissen, Ludwig; Wachtler, Mario, "Magnesocene-Based Electrolytes: A New Class of Electrolytes for Magnesium Batteries," Angewandte Chemie International Edition 2016, 55(48), 14958–14962.  https://onlinelibrary.wiley.com/doi/abs/10.1002/anie.201606448


Propene react with ICl When propene react with ICl , it gives an inoptically active compound ,why.

According to me there is chiral carbon so it should be optically active

The product is optically active compound. However reaction is not stereoselective and therefore both stereoisomers are formed without one being in excess to the other. Product is a racemic mixture of both isomers and therefore should not expose optical activity because effects of both isomers just cancel out.

Reagents or substrate must be optically active in first place in order to get optically active mixture of products.


chemical possibility in oil - 8th graders have questions for a project I have a team of kids (8th graders) that are doing the First Lego League competition this year. As part of that they have to come up with a possible solution to a problem that relates to the human water cycle.  They need to contact experts and share their thoughts and get feedback of their solution.  We would greatly appreciate any feedback.  I've listed what the problem is they are trying to solve and their suggested solution below, and then some questions.

Problem:  Oil spills in water (caused by various things - accidents, hurricanes, etc)

Their proposed solution:  Create an environmentally safe chemical that can be put in the oil, that if there is a leak, as soon as the oil comes in contact with water, the chemical causes the oil to solidify thus sealing the leak or at least slowing it down.  The solidified oil would be buoyant so that it could be easily skimmed out of the water.  The solidified oil could also be reprocessed back into usable oil.

Questions the kids have:
    Does such a chemical currently exist that can do this?
    If so, why is it not being used - any downside to using it?
    If not, is it possible to invent such a chemical?  Now or in the future?
    What kind of components could be used to do create such a chemical?

We really appreciate any response you can give.
Thank you so much!

A chemical that can be added to crude oil so that the oil solidifies upon contact with water to either seal leaks or to make the oil recoverable sounds like a very good idea. Unfortunately, such a chemical is exceedingly unlikely to exist.

When processing crude oil, you typically want the oil itself and not any additives. That means, the chemical in question would have to be added in concentrations as low as possible. Therefore, it’s reaction with water must be very strong. But this in turn means that the traces of water present everywhere (yes, even inside your sample of crude oil) would almost be enough to cause the reaction. Therefore, outside of water-free atmospheres (used in laboratories under inert gas to perform water and air-sensitive reactions) this chemical would hardly be usable.

A second problem is getting the chemical reasonably dissolved. Crude oil is a very viscous mixture unless heated. To use the chemical effectively, it would need to be added almost immediately and distributed well in the sample of crude oil — which would mean immediately heating the crude oil to such temperatures that efficient mixing is possible (likely under pressure to not lose the more valuable but also more volatile petrol components) right after getting it out of the ground. Especially on oil rigs, I see major difficulties with this process; and it is basically on oil rigs that you would want this chemical the most.

Finally, there is the ‘how would it work’ problem. I have extreme difficulty imagining a reaction with water that would lead to solidification. The only idea I can come up with is a compound that would be liquid at standard conditions but adds water to become a solid which crystallises out of the crude oil phase. However, this solid would likely not do you the favour of crystallising where you need it and you would need so much of it that my first argument becomes invalidated. Therefore, I conclude that such a compound cannot exist.

If such a compound exists, I’m sure we would have heard of it or it would be in use (or have been in use if there are significant and severe environmental issues with it that are worse than the issues with crude oil itself.


Ionic equilibrium equation What form is the ionic equalibrium equation? I have problem asking for it, but I'be done so many different equalibrium equations that I'm starting to get mixed up on which one is called what... Which one is it?

An ionic equilibrium equation is simply one that involves ions.

Example : $\ce {HA <=> H^+ + A^-}$