Are there more endothermic or exothermic reactions? I'm wondering whether there are more endothermic or exothermic reactions.

Of course the amount of mass will stay the same, but I thought that not every exothermic reaction has an endothermic counter-reaction. Is this stream of thought correct and does that mean that there are more exothermic reactions than endothermic reactions?
In other words, does every reaction have a reverse reaction?

Answering "does every reaction have a reverse reaction?" (are there more endothermic or exothermic reactions):

You've actually hit on a topic people have been exploring for nearly a century:

  Lewis. A New Principle of Equilibrium. PNAS, 1925, 11(3),179-183

    The Law of Entire Equilibrium — Thus I am led to propose a law which in its general form is not deducible from thermodynamics, but proves to be compatible with the laws of thermodynamics in all cases where a comparison is possible. It may be called the law of entire equilibrium, and may be stated as follows. Corresponding to every individual process there is a reverse process, and in a state of equilibrium the average rate of every process is equal to the average rate of its reverse process. The rate at which one group or set of groups L goes over into another group or set of groups L' is the same as the rate at which the groups L' go over into the groups L. Moreover if there are various paths by which the first process occurs, there is an equal number of paths by which the second process occurs, and the rate is the same in both directions along every path. This will be true no matter how detailed are the specifications which define the several groups and the several paths.

The implications of this concept (and your question) have been well-elucidated by SE user Curt F. in two posts:

  Is every chemical reaction in equilibrium?

    Commenters suggest that "irreversible" reactions do not have an equilibrium. This is true, but tautological. In the real world, all reactions are reversible, at least to a (perhaps vanishingly small) degree. To say otherwise would violate microscopic reversibility. A reaction that was 100% irreverible would have an equilibrium constant of infinity. But if =∞, then Δ∘=−ln would turn into Δ∘=−∞. So to get infinite energy we would just have to use 100% irreversible reactions! Hopefully the problems with the idea of "irreversible" reactions are becoming apparent.

  Spontaneous/Non-Spontaneous Reactions and Reversible Reactions

    All reactions can be viewed as reversible from a mathematical standpoint, if the reacting system is "big enough". Reactions that are highly spontaneous in the forward direction mean that at equilibrium, the number of "reactant" molecules will be very very small (but not zero!).
    Many times, "big enough" would mean astronomically big, such as bigger than the Earth. In these cases, reactions can be regarded as "irreversible", although there is no strict, universally agreed upon boundary between "reversible" and "irreversible".

This should implicitly answer your question, as a 'reversed' exothermic reaction is an endothermic reaction. If every reaction is fundamentally reversible (on some scale, whether or not it's reasonable, as Curt F. describes), then there are an equal number of exothermic and endothermic reactions.

Answering "which occurs more frequently: exothermic or endothermic reactions?" (are there more endothermic or exothermic reactions?):

This is an interesting way to look at things. Considering the ever-increasing entropy of the universe and the favorable relationship between entropy and heat, I would guess that there are more exothermic reactions. Exothermic reactions represent energetically favorable transformations and are more spontaneous than endothermic reactions, so it stands to reason that there are "more exothermic reactions."

  Roduner &  Radhakrishnan. In command of non-equilibrium. Chem. Soc. Rev., 2016, 45,2768-2784

    2.7 Heat death: the ultimate end of an evolving universe

      From the second law of thermodynamics we know that the entropy of the universe will increase for all spontaneous processes. Since the total possible reversible heat flow is limited by everything having the same temperature, and when no chemical or nuclear reactions are available anymore which can produce further heat, entropy will reach a maximum but will not go to infinity. This extrapolated end of a universe that evolves further in time has been dubbed heat death. But how will this final state look like? This is a question that is complex to answer. Considering an extremely simplified model universe helps to get a rough idea.

      We first consider a closed system consisting of 1 mol of glucose and 6 mol of molecular oxygen at 25°C and 1 bar pressure (see Fig. 18a). The equilibrium state at the same temperature and pressure consists of the system after complete combustion to 6 mol of CO2 and 6 mol of liquid H2O. The heat of reaction of 2805 kJ will be dissipated from the system...

    5 Concluding remarks

      Heat engines produce work that is typically used to drive processes which do not occur spontaneously. This is achieved by letting the machine operate between two temperatures, where energy is provided to the system at the high temperature and partly converted to work. This can be done only when an unavoidable remaining fraction of the energy input is degraded to lower temperature and dissipated as waste heat. It is shown here that this principle is not limited to heat engines but holds more generally for many other processes in which a subsystem of the universe is to be transformed to higher order, as in the separation by distillation or crystallisation, photosynthesis, solar cells, or self-assembly...

      The simple model universe of Fig. 18, the combustion of glucose, predicted that the final state of the universe will be much warmer compared with today and consists of a higher fraction of atmospheric CO2 and H2O at the cost of O2. One might therefore ask whether the global warming that we see on our planet actually reflects that we are noticeably approaching the heat death of the universe as a consequence of combustion of fossil fuels with corresponding heat and entropy production. The answer is straightforward: the direction of the effect is correct, but it does not quantitatively reflect the approach of our planet to heat death. The solar heat incident on our atmosphere is about 5 orders of magnitude more than the total energy consumption by the human population. CO2 and water produced by combustion just increase the mismatch between incident and reflected solar energy by increasing the greenhouse effect of the atmosphere, thereby amplifying the direct effect by a considerable factor.

If we're to believe that heat death is a logical conclusion of the second law of thermodynamics, then there must be more exothermic reactions to result in an ultimately stagnant system where all available energy has been dispersed as heat and atomic components are at their lowest possible energy states — this outcome, by definition, represents an excess of exothermic reactions.


is water a non-metallic oxide? and hence acidic in nature? I just read about non-metallic oxides in my textbook which says that non-metallic oxides are a compound formed of an oxide ion and a non-metal. It also said that they are acidic in nature. It gave examples of carbon dioxide and sulfur dioxide.

But then wouldn't hydrogen oxide (water) also be a non-metallic oxide and hence, acidic in nature. But most textbooks say that water is neutral. So how's that possible?

When textbooks say that water is "neutral," textbooks mean that the water meets this condition:

$\ce{[H_3O^+]} = [\ce{HO^-}]$.

That's all.

Water can be neutral, in that its hydronium ion concentration equals its hydroxide ion concentration. Water can also be acidic if there is carbon dioxide dissolved in the water. Carbon dioxide can dissolve in water simply by letting the water come into contact with the atmosphere. Water can be basic too if you put a basic solute in water.

Water itself can also act as a weak Brønsted acid by donating a proton. This however is not a favorable process.

Water can also act as a Lewis acid because the oxygen in water withdraws electron density from the hydrogens, giving each hydrogen a partial positive charge. This in turns makes the hydrogens electrophilic - fitting the definition of Lewis acidity.

I think what your textbook is getting at is Lewis acidity. Consider carbon dioxide. This doesn't fit the Brønsted definition of acidity because it has no hydrogen protons to donate. So consider the partial charges on the carbon dioxide molecule. There is a positive partial charge on the carbon because the oxygens are more electronegative than carbon and have withdrawn electron density from the carbon. This makes the carbon electrophilic. And hence carbon dioxide as a whole can be considered a Lewis acid (but do understand that it's the carbon specifically in carbon dioxide acting as a Lewis acid).

So in general, your textbook says that non-metallic oxides tend to be acidic (in the Lewis sense) because the oxygens withdraw electron density from the non-metal (remember that oxygen is one of the most electronegative elements). This in turns creates partial positive charge(s) and makes the centers of these partial positive charges electrophilic.

Also, you could say that non-metallic oxides are precursors to Brønsted acids; consider this Lewis acid/base reaction. Remember that the central conceit of Lewis acid/base chemistry (coordination chemistry) is electron flow; nucleophiles attack electrophiles (not the other way around; that would be the fatal conceit).

So consider this reaction:

$\ce{SO_3 + H_2O -> H_2SO_4}$

We have sulfur with an electronegativity (EN) of 2.6. Oxygen has an electronegativity of 3.5. The difference is 0.9. The bond is clearly polar covalent. In addition, we have not one but three oxygens isolating electron density from the sulfur. Therefore, the sulfur most definitely has a partial positive charge and the oxygens partial negative charges. We say the oxygens are nucleophilic. And back to water, we have already concluded the hydrogen atoms in water are electrophilic due to their positive partial chage. The sulfur trioxide will attack the hydrogens and form the sulfuric acid molecule - a strong Brønsted acid (its first ionization is complete in water solution).


Is it possible to have Volume(solution) < Volume(solvent before adding solute)? I understand the answer to the question "Is it possible to have Volume(solution) < Volume(solvent) + Volume(solute)?" is "yes" -- but it is possible that adding a solute causes the solvent to contract so much that the volume of the solution is less than the initial volume of the solvent?

An ideal answer would include either a specific example of such a case, or an explanation of why it is impossible.

If the above is not possible, is it possible for the volume of the solution to equal the initial volume of the solvent?

Gosh, sure. According to various on-line info (of unknown quality) a 2% by wt. solution of NaOH in water has a density of 1.0207 at 20°C. This means 100 g of it has a volume of 97.97 ml. At 20°C water has a density of 0.9982 meaning that 98 grams of water has a volume of 98.18 ml.


The Crisscross method for finding the chemical formula I am reading this wikipedia article that I don't understand. What I don't understand is:

  suppose we have two elements $X$ and $Y$ having oxidation numbers $x$ and $y$ respectively. Can we prove that the compound formed will be $X_yY_x$.

I tried to understand this method as: I think this method comes from the fact of charge neutrality. Suppose the compound to be formed has formula $X_{y'}Y_{x'}$. The charge on $y'$ number of $X$ elements is $xy'$. The charge on $x'$ number of $Y$ elements is $x'y$. Now by charge neutrality we should have : $$|x|y'=x'|y|$$
 this implies $\dfrac{|x|}{|y|}=\dfrac{x'}{y'}$.

  But I could not further understand why the we divide $x$ and $y$ with their greatest common factor to find the required solution.

I have also found some contradiction examples to this Crisscross rule. Example is : Solid-state-semiconductor $GaAs$, in this Gallium has valancy $3$ and $As$ has valancy $5$.
 The main points on which I want answers are:

  What is the logic behind this method, how to prove this method?
  Who, how and when developed this method?
  Is this method only applicable to ionic compounds?
  Why compounds like $GaAs$ do not follow this rule, how charge neutrality is preserved in these compounds. I think these don't follow this rule because these are not ionic compounds, if yes then how the chemical formula of compounds like solid state $GaAs$ is known(or decided) and how it is justified, how one atom of $Ga$ balances one atom of $As$ ,their valencies are different, how they are bonded(covalently I guess) together?

What is the logic behind this method, how to prove this method?

The logic of the method is in fact trivial. Suppose you have to write a compound knowing his oxidation numbers :
$$\ce{xA^{a} + yB^{b} -> A_{x}B_{y}}$$

Where $a$ and $b$ are oxidation numbers  that have opposite sings,  $A$, $B$ chemical species and $x$, $y$ stoichiometric coefficient $ \in \mathbb{N}$.
We can start from this assumption: if a compound is neutral the algebraic sum of the oxidation numbers of his atoms must be zero ( point .4 of current IUPAC definition of the oxidation state) so we have to find when:

$$ ax+by=0 \space [1] $$

But we assumed that $a$ and $b$ have opposite sings so we can write:

$$ ax - by=0$$

This is a single equation with two variables so you can't solve it easily because it can have several solutions. However when $\textbf{x=b}$ and $\textbf{y=a}$ the equation is always solved, because:

$$ ab - ab=0$$
And so you can always write a compound formula that satisfy equation 1 using this formula:
$$\ce{xA^{a} + yB^{b} -> A_{b}B_{a}}$$

This works for $\ce{Al^{+3} , O^{-2}}$:
$$\ce{xAl^{+3} + yO^{-2} -> Al_{2}O_{3}}$$
 but what's about $\ce{Ti^{+4} , O^{-2}}$?

This turns out $\ce{Ti2O4}$ and not $\ce{TiO2}$ we need another step, we need to divide the subscripts by their greatest common factor gcf(a,b) because we have found another solution of the equation 1 with higher coefficient. In this case the gcf(2,4) is 2 and dividing the stoichiometric coefficient by two we get $\ce{TiO2}$.

Who, how and when developed this method?

I've found it the first time in Chemistry and Human Affairs by William Evans Price, George Howard Bruce, World Book Company, 1949. I don't know who is the inventor.
I think you can easily deduce the crisscross method simply looking at many compounds and their oxidation number! Is not so hard to imagine!

Is this method only applicable to ionic compounds?

No, it works for every compounds that can be describe with oxidation numbers. Oxidation numbers are the real charge in ionic compounds and a formal (or imaginary) charge in compounds with covalent bonds, look at $TiO_2$ example.

Why compounds like $GaAs$ do not follow this rule?

Is not right if we assume that we are dealing with $Ga(+3)$ and $As(-3)$ the equation is this:
$$\ce{xGa^{+3} + yAs^{-3} -> Ga_{3}As_{3}}$$
if we divide the subscripts by the gcf(3,3)  that is 3, we get $GaAs$. So this method will be always valid until you can describe the compound effectively with oxidation numbers, unfortunately not all the oxidation numbers are so easy to determine.

In fact I don't find the criss cross method useful I think is better to calculate the least common multiple of $a$ and $b$ and then divide it by $a$ for $x$ and $b$ for $y$.
I think this photo, from J. Chem. Educ., 2012, 89 (11), pp 1436–1438, summarize the best way to do it:

Edit:

The general solution of |a|x=|b|y  s.t.  x,y∈ℕ is x=n|a| and y=n|b| where n∈ℕ, that is we can use x=2|a|,3|a|...n|a| and similarly y=2|b|,3|b|...n|b| then why use x=|a| and y=|b| only?

(rearranged after the advices in the comment of mannaia)
In fact you can write $Na_{2}Cl_{2}$ without breaking any charge neutrality or valence rule. NaCl is a crystal with a lattice and you  can't consider $NaCl$ like a stand alone unit, what matter is the charge balance, the proportion between positive and negative ions. Maybe if you are dealing with a crystal made of 23456 atoms of $Na^{+}$ and 23456 atoms of $Cl^{-}$ you could  write $Na_{23456}Cl_{23456}$ this is in fact a solution of the equation and is not theoretically wrong however if you are talking about a generic crystal of NaCl you should write the ratio between $Na^{+}$ and $Cl^{-}$ ions it has no sense to choose randomly one of the solutions because has stated mannaia: "there's no need to choose a larger unit, whereas with a smaller one you can fully explain how the crystal is formed". Mathematically speaking we write the ratio to Lowest Terms.

Why always take the "gcf"?

In fact we don't take the "gcf" we divide for the "gcf" because this is the best way to get the ratio in the lowest terms.

Furthermore as explained in the $TiO_{2}$ example.
$$
A=\{(b,a),(2b,2a),(3b,3a),\ldots\}.
$$
Is a set of solutions of the equation but not all the solutions. If you have a=6 b=-4 you have the solution (2,3) too, here is the plot of the solutions in $\mathbb{N}^{2}$.

For compounds with covalent bond is different, because you are dealing with different stand alone units called molecule. For example if you have $C^{-4}$ and $H^{+1}$ the solutions not divide by gcf of the equation have no sense. For example $C_{2}H_{8}$ doesn't exist due the valence of the atoms involved.


Why is AlMg3 called aluminum alloy? In AlMg3, magnesium atoms outnumber aluminum atoms by a factor of three. In terms of mass, the ratio of aluminium to magnesium is a mere $27.0 / (3\times24.3) = 0.37$.

Shouldn't AlMg3 be better called magnesium alloy?

AlMg3 is not the chemical formula of the alloy, it's the product name for a wrought alloy. Its technical sheet specifies the following composition:

Product name
AlMg3
Class of product
Al-Mg alloy for MIG/TIG welding.
Corresponding standards
DIN 1732, SG-AlMg3, AWS A5.10, ER 5754
Nominal composition (weight %)
Al: Bal. Si: 0.4 Mg: 3 Mn: 0.5

Thus, the 3 in the product name refers to 3 % (in weight) of magnesium.


Why does an electron lose energy so that it falls back to a lower energy state than just stay at the higher energy state? Suppose an electron in the 2nd shell gets excited and jumps to the 4th shell. Shouldn't the electron then exist in an unstable equilibrium in the 4th shell, than lose energy and drop down to the 2nd shell? What causes it to come out of the equilibrium position and lose energy? The teacher who gave me this problem to think about said that quantum mechanics plays a role in this. Can anyone please give an explanation?

Actually, this question is not as bad as the comments may make you think. It's actually quite a sensible question, because when we solve the usual, Schrödinger's equation-based, QM problem of energy states of hydrogen atom (or any other atom), we get eigenstates of the Hamiltonian we inserted into the equation. These solutions should thus be stable for infinite time.

In real world, however, an atom isn't isolated. It interacts with the electromagnetic field, and this interaction results in spontaneous emission, which is the reason why an atom can't be in excited state indefinitely.


How can a bond be more stable and more polar at the same time? When comparing the acidic strength of the halogen acids, we see that the acidic strength decreases as $\ce{HI>HBr>HCl>HF}$. It is said that the reason for this is the bond strength of $\ce{HI}$ is the least, while that of $\ce{HF}$ is the greatest. But, the polarity of the bond in $\ce{HF}$ is the greatest, how then is the strongest bond?
I understand that fluorine has smaller valence atomic orbital (2p) than, for example, iodine (5p), and so has more effective overlap with the s-orbital of hydrogen. But how do we reconcile this fact with the fact that the fluorine hydrogen bond is more polar too? The electrons remain more with fluorine in $\ce{HF}$ than with iodine in $\ce{HI}$, so intuitively it seems that the hydrogen fluorine bond may be more easily broken by a base than the hydrogen iodine bond. So what is going on?
In short, how can a bond be stable when it is highly polar?

Edit
My question is regarding bond strength, and not the acidic order. For example, in organic compounds, the leaving group ability of the halide ions decreases as$^1$:
$$\ce{I^{-}>Br^{-}>Cl^{-}>>F^-}$$
This means that the fluorine-carbon bond is the hardest to break. But fluorine has very high electronegativity, and so the bond must be highly polar. But inspite of this, fluorine has the highest bond strength!
So is this just a misconception I have? If it isn't a misconception, then why does the highly polar bond have the maximum bond strength?
Thank you!
$^1$: https://chem.ucr.edu/documents/curricularmaterials/neumantextbook/Chapter7.pdf

The important thing here to keep in mind is that the bond strength is not a spherical chicken in vacuum, meaning that we need to keep the entire process in mind when analyzing all the factors that can influence it. Although the fluorine has a higher electronegativity, it is significantly smaller than Cl, Br, or I (same goes to their corresponding ions). And without going too deep into intricacies, $\ce{F-}$ is $way$ less stable than $\ce{I-}$ and thus, using the OChem language, a much better nucleophile. Due to it's inability to stabilize the charge, it holds onto the hydrogen with all it's might (pardon me for anthropomorphizing the atoms).

SO, when you're thinking about breaking $\ce{H-F}$ bond, you gotta take into consideration what happens with the products. In the example with organic chemistry and leaving groups that you have pointed out, fluorine is the worst leaving group for the precisely the same reason--it cannot stabilize the negative charge well, so it doesn't "want" to leave. However, if you try to do a, say, $S_N2$ reaction in a protic solution (not the best way to go around, but doable), then $\ce{F-}$ will be a much better leaving group than $\ce{I-}$, b/c $\ce{F-}$ will instantly "catch" a hydrogen from the solution and stop being nucleophilic, which according to Le'Chatelie'r principle shifts the equilibrium to where $\ce{F-}$ is a leaving group.


Why stepwise dissociation for acids but not hydroxide bases? My son asked this over the weekend and I didn't have a good answer: how come with polyprotic acids we assume that the release of each $\ce{H+}$ is progressively harder, but with bases like $\ce{Ca(OH)2}$ we write $\ce{Ca^2+}$ and $\ce{2OH-}$ as the products?

What I know about, say, sulfuric acid, seems to suggest that when we put calcium hydroxide in water we should get $\ce{CaOH+}$ and a single hydroxide, and it should be much harder to release that second hydroxide.  What am I missing?

This is not a general rule. Much like with acids for some bases we write progressive capture of one proton at a time. It all depends on the nature of the acid/base.

In the case of $\ce{Ca(OH)2}$ you don’t have a molecule like $\ce{HO-Ca-OH}$ or something like you can assume for $\ce{HO-SO2-OH}$ (sulfuric acid). Rather, you have an ionic compound composed of two $\ce{OH-}$ ions per $\ce{Ca^2+}$ ion. Once we dissolve this, all the ions are released at the same time resulting in $2~\mathrm{mol}$ of $\ce{OH-}$ per mole of $\ce{Ca(OH)2}$ dissolved.

As a different example take sodium oxide $\ce{Na2O}$. If you slowly add water to that compound the oxide ion will grab one proton at a time resulting in $\ce{2 NaOH}$ first before a second set of protons will generate water. This is much the same mechanism as for diprotic acids. (Note, however, that the oxide ion is such a strong base that it is immediately protonated with a single equivalent of water — much like sulfuric acid is such a strong acid that it will immediately release one proton on contact with water.)


What prevents the anode from corroding in electrolytic organic reactions? An electrode can be plated by having a solution full of electrolyte, using the the metal that will receive the plating as the cathode, and the metal doing the plating as the anode. Current is put through the cell and the anode dissolves oxidized metal ions into the solution, which gravitate over to the cathode, and then "plate" onto the cathode.

My question is, what prevents this process from happening when doing an electrolytic organic reduction or oxidation? Imagine you plate an electrode with a specific metal to be used as an anode in a different reaction, what prevents that anode from immediately corroding and having the metal ions fall off into the solution when current is applied? Does the presence of an organic species to be reduced/oxidized take priority over the anode corrosion, thus preventing it?

One would control the composition of the electrodes (mercury, platinum, Ebonex, boron-doped diamond coating) and the potential of the cell so that the organic reaction went forward while the electrodes remained inert.

Metal oxidation is facilitated by Lewis base ligands that drive reaction by stabilizing the ionic products.  Don't have them in there.


Why is H2S more polar than HCl? From EN alone we might expect $\ce{HCl}$ to be more polar but a book says that the bent structure of $\ce{H2S}$ gives rise to its polarity - which is also apparently more than $\ce{HCl}$. Why? Does this have to do with the lone pairs? Lone pairs supply electron density, right?

I have some additional comments. As mentioned previously, by the most straightforward measure of polarity for small molecules (net molecular dipole moment), $\ce{HCl}$ is undoubtedly more polar than $\ce{H2S}$. If some other measure of polarity is being considered, then it needs to be stated for clarity.

I suspect the authors of the book and some of its readers are taking the higher boiling point of $\ce{H2S}$ compared to $\ce{HCl}$ to mean the former is a more polar molecule. In fact, this behaviour holds for all of the group 15/16/17 molecular hydrides. However, this is only an indirect measure of polarity, and there are other confounding factors. In this particular case, I believe the reason $\ce{H2S}$ has a higher boiling point does not have to do with the strength of each intermolecular interaction, but the average number of intermolecular interactions which can be held at once, compared to its neighbouring hydrides ($\ce{PH3}$ and $\ce{HCl}$).

A hand-waving explanation is that the group 16 hydrides have a better "balance" of partially charged atoms compared to the group 15 and 17 hydrides in the same period. For example, looking at the hydrogen bonded molecules of the second period and considering an optimal hydrogen bonded network, $\ce{NH3}$ is capable of being the donor atom for three H-bonds and being the acceptor atom for only one H-bond. In $\ce{HF}$, the situation is inverted; now there are three H-bond acceptor sites, and only one H-bond donor. $\ce{H2O}$, however, is a convenient match of being a H-bond double donor and double acceptor. Thus, water performs more hydrogen bonds per molecule on average than either ammonia or hydrogen fluoride, and has a significantly higher boiling point, even though the actual hydrogen bonds are arguably strongest for $\ce{HF}$. The logic would be similar for the group 15/16/17 hydrides of the third period.

Lastly, I'd like to clarify the role of non-bonding pairs in the dipole moment of a molecule. Electrons in pure atomic orbitals are symmetric with respect to the nucleus and produce no net dipole. In a hybridization picture, one might say that non-hybridized $s$ and $p$ orbitals possess no dipole moment, while hybrid orbitals are not symmetric and do possess a dipole moment. In the case of the second period element hydrides, the lone pairs in hybridized $sp^3$ orbitals would provide a net dipole moment. However, for the hydrides in the third period and below, the molecular geometries are consistent with no hybridization (bond angles close to 90°), and therefore the lone pairs would be present in effectively $s$ and $p$ atomic orbitals, with no effect on a molecule's polarity.

In a more accurate picture without invoking hybridization, the molecular orbitals which contain the lone pairs are almost the same as the central atom's atomic orbitals in the heavier group 15/16/17 hydrides (and hence produce no net dipole), while for the second period hydrides the molecular orbitals are not as symmetric and hence actually produce a dipole.


Why are triple covalent carbon to carbon bonds drawn linearly in skeletal structure? In the link see that cyclic C-C structure with a triple covalent bond is drawn with an angle, probably the only way to draw it but why is it drawn linearly in non-cyclic and not like single or double bond? I am wondering about the reason for this not that its a convention.

Alkyne Wikipedia

Triple bond in carbon means that the carbon atom is sp hybridized. sp hybridized carbon has the two bonded groups at 180 to each other. Therefore it is displayed linearly which is closest to the actual structure. Single and double bonds are at an angle of nearly 109.5 and 120 degree therefore are not drawn linearly.


How does an atom's electronegativity change as it becomes more negatively charged? I had a question about the carbon atom specifically. Carbon has 4 valence electrons, with electronegativity 2.55 (approx). If I was to fire electrons at the carbon atom, each carbon will occupy up to 4 electrons (8 in total in outer shell) to complete the octet. Now, let's assume that this happens slowly such that we have time to observe each carbon atom as it changes its charge, from $\ce{C-, C^2-, C^3-, C^4-}$

My question is: how would the electronegativity of the carbon ions will vary from each other? Will it increase as electrons are added and becomes zero after obtaining full octet? What would happen upon taking away electrons from carbons through help of highly electronegative elements?

Electronegativity is a measure of the ability of an atom to attract electrons to itself (usually in a bonding framework; outside of a bonding framework the ability to attract electrons is electron affinity).

So from simple Coulombic considerations, we can surmise that as your carbon becomes more and more negatively charged, the less it will be able to attract electrons to itself.


what is the white fuzz left behind on basement floor after puddle evaporated My basement is quite dry (knock on wood!).  Recently I brought into the basement for the winter a table from outside.  The table had some ice stuck to it which I couldn't remove, so I simply let it melt on the floor of the basement.  After the puddle that formed evaporated, an interesting artifact was left behind along the the perimeter of the puddle.  It's a raised white line of fuzzy material.

Here's the puddle's perimeter

And here's a close-up of the line that's left behind with a clothespin for scale

Is this efflorescence?  I did some googling, and this is what seems most likely, but I don't know much about chemistry and thought I would ask those who would actually know.  Is it something else?

Thanks.

Three pieces of information from the comments:

It fizzed and disappeared in vinegar.
It had a yellow-orange color in blue flame.
It's salty (note that it's usually a bad idea to taste random stuff you find on your floor).

The only white crystalline material that is a plausible candidate for being there and fizzes in acetic acid (ie vinegar) is calcite - $\ce{CaCO3}$. It's also a mineral that has 'retrograde solubility' meaning it dissolves more readily in cold water. This is probably how it got to the water in the first place. This is also why it precipitates on your electric kettle. It is the same stuff.

The salty stuff that burns yellow-orange is obviously $\ce{NaCl}$, aka table salt. It also probably dissolved with the vinegar as well. Why is the salty taste not quite like table salt? Well, first of all it is mixed with calcite. But, calcite has no taste. It's possible that there is some $\ce{KCl}$ or $\ce{MgCl2}$ in there as well. They are slightly more bitter than $\ce{NaCl}$. This is the stuff they use in low sodium salt.

There's another possibility - it could be any of the epsom salt family, $\ce{MgSO4.$n$\,H2O}$.

As to would this be dissolved in the first place, you can only speculate. You say that it came from a table that was outside. Did people eat there? Could people had spilled salt on it? Is the airborne dust in your area contain a lot of calcite? What about your basement? Could someone have spilled some salts on the floor, maybe even years ago?


Finding CML files for small molecules I'm trying to generate a figure for a physics textbook to show examples of the vector addition of dipole moments in small molecules. For this purpose, I would like to render pictures of a bunch of small molecules, such as CS2, formaldehyde, benzene, SF6, and chloroform. I've got avogadro running on my linux box, but I'm finding it very time-consuming and confusing to construct the molecules using the GUI. Is there anywhere that I could find pre-made CML files for small molecules like these?

DSVA's comment pointed me to a database that contains the information I need. The wikipedia articles contain links to an app called jsmol, but that app appears to be getting its data from a database called ChEBI. You can search the database using search terms such as "formaldehyde" or "sf6." The entry for each molecule contains a link to a .mol file, which avogadro can read.


Can ethanol alone be used as Fuel? Ethanol is a clean fuel and added as an additive to petrol.

Why can't ethanol itself be used as fuel?

Ethanol clearly can be used as a fuel.

But there are several reasons why it isn't.

First is simple economics. Ethanol costs more to produce than gasoline: oversimplifying only slightly, fermentation is more complicated than simple distillation. Those countries that do use mixtures of ethanol in gasoline do so for "environmental" reasons (which are not always well thought through as being made from plants doesn't make something good if associated production costs are high or alternative uses are better). And "environmental" causes are a good excuse to subsidise favoured groups, like US or Brazilian farmers. In the absence of subsidy, there would be very little use of ethanol as a fuel.

But those are not the only reasons. Ethanol carries less energy than the same volume of gasoline. In other words your mileage will be lower; your fuel efficiency will be lower.

Worse, though, ethanol doesn't have exactly the same characteristics as gasoline. Engines may need to be adjusted so the alternative fuels burn equally well, which is annoying but fixable easily in electronically controlled engines. In addition, Ethanol may be more corrosive to some of the components used in fuel pipes (nothing like as bad as methanol, but clearly worse than gasoline). Ethanol is also much more hygroscopic than gasoline, absorbing significant amounts of water from air. This is another potential cause of contamination in fuel systems and may make corrosion worse.

In short, there are no fundamental reasons why ethanol can't be used. But there are a range of practical issues that make it less economical or inconvenient.


Inorganic chemistry reaction mechanisms Are there well developed reaction mechanisms in inorganic chemistry like those in organic chemistry? If yes, please suggest an easy to follow reference.

If this is in regard to mechanisms with organometallic complexes or coordination compounds, there are a set of specific mechanisms that are well defined. These are:

Ligand dissociation / association
Reductive elimination / oxidative addition
Intramolecular insertion (e.g. CO migratory insertion) / elimination (e.g. $\beta$–hydride elimination)
Metallacycle processes (e.g. $\sigma$-bond metathesis)
Nucleophilic / electrophilic addition and abstraction are also possible, similar to organic chemistry.

These mechanisms should be in inorganic textbooks but can also be viewed on the first page here. The later slides in this pdf cover more of the details of each specific mechanism.

In many reaction mechanisms with organometallic complexes, it is a process of a number of these smaller mechanisms occurring, leading to the overall chemical transformation that is stated in a reaction equation. These overall mechanisms can be complicated and need to be evaluated for each case. Knowing the basic underlying mechanisms will certainly help clarify processes that seem quite complex at first glance.

There are subtleties in all of these mechanisms to consider beyond the basic types above. Significantly more detail can be found in these textbooks that I have previously utilized which you may also find informative:

Collman, J.; Hegedus, L.; Norton, J.; Finke, R. Principles and Applications of Organotransition Metal Chemistry. 1987. University Science Books. Mill Valley, California.

Spessard, G.; Miessler, G. Organometallic Chemistry. 1997. Prentice-Hall. Upper Saddle River, New Jersey.


Is there a name, or any specific jargon for a precipitation equilibrium involving a solid reactant? Let's say you have some ions in solution, for example Na+(aq) and OH-(aq), and some partially dissolved solid, MgCO3(s), Mg+2(aq) and CO3-2(aq), which precipitates Mg(OH)2(s) until equilibrium.

So you have two solubility equilibriums interacting with each other:

$$\ce{ MgCO_3(s) <=> Mg^{+2}(aq) + CO_3^{-2}(aq) }$$

and

$$\ce{ Mg^{+2}(aq) + 2OH^-(aq) <=> Mg(OH)_2(s) }$$

overall (with all spectator ions removed):

$$\ce{ MgCO_3(s) + 2OH^-(aq) <=> Mg(OH)_2(s) + CO_3^{-2}(aq) }$$

Is there a name for this type of reaction, and what method would I use to calculate the equilibrium point, with what physical properties?

I would call it something like a "binary solubility equilibrium". And I guess if I know the starting OH concentration and the solubility constants of the two solids, I could work out the equilibrium?

Any clues and key words appreciated. Thanks.

AFAIK, such a reaction has no name. However, if one were so inclined, one could call it a replacement reaction, because the $\ce{MgCO3}$ is being converted into $\ce{Mg(OH)2}$.

To calculate the "equilibrium point," which I take to mean the concentration of all the ions at equilibrium, we would need to know the $k_\text{sp}$ of both $\ce{MgCO3}$ and $\ce{Mg(OH)2}$, as well as the $\ce{pH}$ of the solution (which gives us $[\ce{OH^-}]$).

The solubility products of $\ce{MgCO3}$ and $\ce{Mg(OH)2}$ are $10^{-7.8}$, which is approximately $1.58 \cdot 10^{-8}$, and $5.61\cdot 10^{-12}$, respectively.

To calculate how much magnesium hydroxide precipitates, we would first need to find the concentrations of $[\ce{Mg^{2+}}]$ and $[\ce{CO3^{2-}}]$ using the $k_\text{sp}$ of magnesium carbonate. This gives us the solubility of $\ce{MgCO3}$, which is $1.26 \cdot 10^{-4} \,\, \ce{mol lit^{-1}}$. That means that we have $[\ce{Mg^{2+}}] = 1.26 \cdot 10^{-4} \,\, \ce{mol lit^{-1}}$ in the beaker.

Now, to find out how much $\ce{Mg(OH)2}$ will precipitate, we need to find the reaction quotient $Q_c$ of the reaction.
$$Q_c = [\ce{Mg^{2+}}][\ce{OH^-}]^2$$
(I cannot calculate $Q_c$ here because it depends on $[\ce{OH^-}]$, which will have to be given.)

If $Q_c > k_\text{sp}$, then the reaction
$$\ce{Mg(OH)_2(s) <=> Mg^{+2}(aq) + 2OH^-(aq)}$$
will proceed in the backward direction and magnesium hydroxide will precipitate out. This will continue until the concentration of magnesium ions decreases to such a value that $Q_c = k_\text{sp}$, at which point the reaction will be at equilibrium. Here, I've assumed that the concentration is too large to be appreciably changed by it precipitating out of the solution. The exact value of $[\ce{Mg^{2+}}]$ as well as how many moles of $\ce{Mg(OH)2}$ will precipitate can be calculated with this assumption. (However, if we want, we can do it without this assumption too. The calculations would become more messy and involve solving a quadratic.)


Conducting current in electrolytes I keep trying to figure out how current is conducted through an electrolyte but all I can find are BS half answers. They say the ions conduct, but the specifics are poorly explained or absent. I understand that if you, for example, put sodium chloride in water it will disassociate into positive sodium ions and negative chlorine ions, and that an electric field would cause the ions to move, allowing a current to flow. But what happens once the charges get to the electrodes? Obviously, they can't just stay there in elemental form stably. I would expect the sodium, once given an electron, to be looking for a way to give that electron back. Would it cause the water to split into hydrogen gas and hydroxide ions? On the chlorine side, would the chlorine atoms simply turn into chlorine gas. If this is all correct, it seems one would end up with sodium and hydroxide ions. Does it still conduct then? If so, how so? And don't you have a solution of sodium hydroxide? It seems this process necessitates that the water be split. Is there any salt or other compound which would act similarly but not result in electrolysis (just conducting)?

An approach using a classical model of electron flow

You can think of an electron flowing through a medium as analogous to a pinball bouncing around a pinball machine. The electric (potential) energy difference driving the displacement of the electron is analogous to the gravitational energy difference in the pinball machine. Now consider two pinball machines with differing numbers of obstacles to a falling pinball. The time it takes for a pinball to reach the bottom in each machine would be different by virtue of these obstacles (on average, taking more time with more obstacles). Similarly, the flow of electrons, I, in the presence of an electric potential, V, is higher for media with less resistance, R , to electron flow. We can use this model (Drude see below) to understand Ohm's law: $$I = \frac{V}{R}$$ The observation that electrons flow more readily through electrolyte solutions than non-electrolyte solutions indicates that solutions with charge carriers reduce the resistance of the media. A better understanding of why this is true at a molecular level will require a more in depth look. Notice, though, that we now have answers to your good questions!

  "What happens once the charges get to the electrodes?"

The sodium stays in solution as an ion, but the hydrogen ion (from the hydrolysis of water) can be reduced to form hydrogen gas at the cathode, and the chlorine ion can be oxidized to form chlorine gas at the anode. Electrons can be passed through the solution as solvated electrons under some circumstances, but this is unlikely due to the larger energy barrier to this process (around 3 eV). With sufficient energy, these electrons (called solvated electrons) flow through metal and solutions. The resistance to electron flow in metals is much less than in electrolyte solutions, though.

  "Would it cause the water to split into hydrogen gas and hydroxide ions?" and "I would expect the sodium, once given an electron, to be looking for a way to give that electron back."

Any additional redox reaction would require electric potential. You mentioned the splitting of water or the formation of sodium metal. This would not occur according to the standard electrode potential unless the electric potential provided is sufficiently high to overcome the "uphill" energy barrier and as you mention electrolysis would be involved.

The Drude Model

In 1900, Paul Drude used the following schematic of a lattice of metal ions sitting in a sea of electrons to derive a relationship between electron flow and electric potential applied.

  Drude Model electrons (shown here in blue) constantly bounce between heavier, stationary crystal ions (shown in red). -From wikipedia entry on Drude model.

A metal's electrical resistance, R, (as seen in Ohm's law above) should take the form:

$$R = \left(\frac{m}{nq^2\tau}\right)$$

where,  $m$ is mass of electron; $n$ is number density of electrons; $q$ is charge of electron and $\tau$ is the mean free path. While this classical model does not explain the resistances for all substances perfectly and quantum mechanics can be used to do a better job, it does suggest that if the mean free path for electron flow is increased and the assumptions of the model are valid, the electrical resistance of the media should be reduced.

One way to understand why there is lower electrical resistance (higher electrical conductance) in aqueous solutions of electrolytes is that there is less water in the way of the Drude ion lattice! In other words, in the limit that the solution is a molten salt, we approach a Drudian model. As you add more and more molecules of water to this model, you are reducing the mean free path of the electron and increasing the resistance of the solution.

Notes

The fact that an electrolyte solution is comprised of positive and negative ions does not diminish the utility of Drude's model: electrons are equally attracted (and speed up) moving toward a (metal) cation as they are attracted (and slow down) moving away from a metal cation. The situation is reversed but the effect is the same for a nonmetal anion.
For more evidence of the utility of note 1., read about the additive nature of the molar electrical conductance (inverse of resistance) of ions (cations and anions): Kohlrausch's Law of independent migration of ions. The additive nature suggests that both cations and anions improve the conductance of an electrolyte solution as the note above suggests.


Nomenclature of an organic compound with benzene and double bonds How would the following compound be named?

Attemp of solution

I believe is 4-phenyl-1,2-pentadiene

Since the compound that is given in the question doesn’t have a principal characteristic group (that would be expressed as a suffix and that would determine the senior parent structure), the ring has seniority over the chain when selecting the preferred IUPAC name. In general nomenclature and depending on the context, however, the chain may be favoured to recognize the unsaturated acyclic structure.

The corresponding subsection in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-44.1.2.2 Systems composed of rings and chains (exclusive of linear phanes)

  Two methods are recognized to name systems composed of rings and chains
  (exclusive of linear phanes).

  (1) Within the same class, a ring or ring system has seniority over a chain. When a ring and a chain contain the same senior element, the ring is chosen as parent. Rings and chains are chosen regardless of their degree of hydrogenation. As a consequence, this approach prefers the choice of a ring over a chain in systems composed of cyclic and acyclic hydrocarbons.

  (2) The context may favor the ring or the chain, so that, for example, substituents may be treated alike or an unsaturated acyclic structure may be recognized, or the one chosen has the greater number of skeletal atoms in the ring or in the principal chain of the acyclic structure.

  (…) For selection of a preferred IUPAC name, see P-52.2.8.

  P-52.2.8 Selection between a ring and a chain as parent hydride

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

Method (1): (penta-3,4-dien-2-yl)benzene (preferred IUPAC name; ring preferred to chain)

Method (2): 4-phenylpenta-1,2-diene (unsaturated acyclic structure may be recognized)

Therefore, the preferred IUPAC name for the compound that is given in the question is (penta-3,4-dien-2-yl)benzene.

Note that parentheses are used around substituent prefixes (here: “penta-3,4-dien-2-yl”) to separate locants referring to different structural elements.

Also note that locants are placed immediately before that part of the name to which they relate (i.e. “penta-1,2-diene” and not “1,2-pentadiene”).


Why enzyme-catalyzed reactions mostly go one way? I listened to the lecture about enzymes. Professor downplayed the question in the subject. He said that enzymes are able to work both ways, but practically speaking reactions with overwhelming probability flow in only one direction. He said that a variety of factors cause this, and there is nothing interesting there beyond this.

I am not clear what the reasons are. There is the reaction rate constant. Does this mean that rate constants are very different in value depending on direction? Or the reason is that reaction product molecules are always quickly consumed by the next step, and as a result the concentration of the product is always low?

Some enzyme-catalyzed reactions always go in the same direction in a live cell (e.g. reactions of citric acid cycle, RNA polymerase) and others go in either direction depending on circumstances (for example, some enzymes of glycolysis are also involved in the reverse pathway that makes glucose).

As for reasons why a reaction might go in one direction only, you are on the right track. You mentioned two reasons you could imaging for a preference in one direction:

  Does this mean that rate constants are very different in value depending on direction?

If a reaction is far from equilibrium (i.e. proceeds mainly in one direction), the forward rate and the reverse rate will be very different from each other. This could be because the rate constants a very different from each other (with similar concentrations of reactants and products) or because the concentrations of reactants and products are very different from each other, or both. So the safe thing to say is that the rates differ by a lot (rather than the rate constants).

  Or the reason is that reaction product molecules are always quickly consumed by the next step, and as a result the concentration of the product is always low?

This case also occurs. In RNA synthesis, for example, the reaction of a NTP (i.e. a triphosphate) with the growing RNA occurs near equilibrium, yielding longer RNA and diphosphate (pyrophosphate). In a second step, pyrophosphate hydrolyzes quickly, removing one of the products and making the first step virtually irreversible once the second step has occurred.

In the end, whether the reaction goes in both directions or just in one is not dependent on the enzyme: the enzyme is not capable of changing the position of an equilibrium. Whether the reaction goes forward or backward depends on the equilibrium constant of the net reaction (which the enzyme is not a part of) and the concentrations of the reactants and products.

You could envision a scenario where all enzyme-catalyzed reactions happen to be far from equilibrium under physiological conditions, but this is not the case, as the counterexamples mentioned above show.

Enzymes that catalyze reactions that are far from equilibrium are often regulated (turned on and off in various ways) to regulate entire pathways. If a reaction is near equilibrium and is used in different directions by distinct pathways, regulating the enzyme catalyzing the reaction would affect both pathways, so usually these enzymes are not regulated.


How can normalities be interchangeably used? So I was doing the following question on stoichiometry . The part $\pu{1.80N}\ \ce{HCl}$ can also be written as $\pu{0.9 N}\ \ce{   H_2SO_4}$  but I don't understand why .

$\pu{10 g}$ sample of 'gas liquor' ($\ce{NH}^+_4$ salt) is boiled with $\ce{NaOH}$ and the resulting $\ce{NH3}$ is passed into $\pu{60 ml}$        of $\pu{1.8 N}\ \ce{HCl}$ . Excess $\ce{H_2SO4}$ required $\pu{10cm^3}$ of $\pu{0.40 N}\ \ce{NaOH}$. What is the % of $\ce{NH3}$ in gas liquor?

I got the answer but just can't understand why the two can be used interchangeably.

The fundamental idea behind gram equivalents and normality is that one gram equivalent of an acid reacts with one gram equivalent of a base. This idea eliminates the need of using molarity and thinking about mole ratios for any acid base titration.

Equal volume 1 N HCl will react with 1 N NaOH. Similarly, 1 liter of 1 N $\ce{H2SO4}$  or 1 N $\ce{H3PO4}$ will consume 1 L of 1 N NaOH.

The key calculations is that you should know how to calculate normality from a given molecular weight and a balanced equation:

For example:
$\ce{HCl}$ -> $\ce{H+}$ + $\ce{Cl-}$
gram equivalent weight= formula weight/ (No. of acidic protons) = 36/1 =36

$\ce{H2SO4}$ -> $\ce{2H+}$ + $\ce{SO4^2-}$
gram equivalent weight= formula weight/ (No. of acidic protons) = 98/2 = 49

Can you now make the connection that why "The part 1.80 N HCl can also be written as 0.9 N H2SO4" is quite wrong? because

1 N HCl = 1 N $\ce{H2SO4}$

and both solutions contain 1 gram equivalents of protons.

Yes if your teacher were comparing normality and molarity, then it is different.

Molarity calculations are different and far more easier. By definition, M of X= moles of A/ Total volume in which X is present.

A 1 M HCl just means there is one mole HCl (gas) dissolved in 1 L of solution
Similarly,
1 M $\ce{H2SO4}$ is 1 mole $\ce{H2SO4}$ dissolved in 1 L of solution.

If you were to titrate  1 L of 1 M NaOH with 1 M HCl you would require 1 L of HCl. However, since  sulfuric acid furnishes two moles of proton $\ce{H+}$ for each mole of sulfuric acid, only 500 mL would be required.

In short, 1 M HCl is equivalent to 0.5 M $\ce{H2SO4}$ in terms of titer for $\ce{OH-}$ ion.

Contrast this molarity unit with normality which is almost obsolete except in some South Asian colleges.


test to differentiate between NaCl and zinc hydroxide I had conducted a Salt spray corrosion test on Zinc coated bolts. The test result showed white deposits over the bolts. I want to confirm whether its zinc hydroxide or simply the sodium chloride salt that got deposited. So any  test to differentiate between NaCl and zinc hydroxide?

You can use the large difference in water solubility to determine whether the residue is sodium chloride or zinc hydroxide. Sodium chloride is very soluble in water and zinc hydroxide is insoluble in water. You can do the test as follows:

1) Brush or scrape a portion of the residue onto a piece of paper or card stock.
2) Transfer the residue onto a dark colored plate or bowl.
3) Add enough water to cover up the residue and swirl gently.

Any white solid that remains should be zinc hydroxide as the sodium chloride will dissolve away.


Gattermann-Koch Reaction with Phenol Substrates The Gattermann-Koch Reaction involves the formation of an aryl aldehyde using carbon monoxide and hydrochloric acid as reactants. The reaction is catalyzed by aluminium trichloride in the presence of trace amounts of cuprous chloride.

What I don't understand about this reaction is that according to Wikipedia along with other online texts I've used, the Gattermann-Koch reaction is inapplicable to phenol substrates. The drawback has been left unexplained by all my resources (online and textbooks) and even my professor was oblivious to it.
I am unable to understand why the reaction doesn't apply to phenol and what exact side reactions might occur.

Additionally, why is the explanation so obscure?

[EDIT] I found the explanation. The Gattermann-Koch Reaction is not applicable to phenol and phenol ethers because they "could not be successfully formylated at atmospheric pressure in benzene as a solvent."
"This failure to react was attributed to the insolubility of the cuprous chloride in the reaction mixture."

Can someone please explain this in simple terms?

You have outlined that the formyl chloride is generated as the reactive, electrophilic intermediate under these conditions.

You are planning to perform an electrophilic aromatic substitution, but in a phenol, there's high electron density on the $\ce{O}$ atom. Isn't it conceivable that your main reaction product is phenylformate here, resulting from $\ce{O}$-acylation?


Can calcium hydroxide + sulphuric acid be a way of preparing insoluble calcium sulphate? To prepare a soluble salt, we can react an acid with an insoluble base (hydroxide or oxide).
To prepare an insoluble salt, we can use precipitation method which is by mixing two solutions of soluble compound.

However, can I prepare an insoluble salt by reacting an acid with an insoluble base (hydroxide or oxide)?

For example,
 1. Calcium hydroxide + Sulphuric acid
 2. Lead (II) hydroxide + Sulphuric acid
 3. Lead (II) hydroxide + Hydrochloric acid

No.

The outside layer of the calcium hydroxide reacts with the sulfuric acid to form the calcium sulfate that you want. All good, until you realise that the newly-formed calcium sulfate is now coating the rest of the calcium hydroxide that didn't get to react. In the end you just get a big mess.

From a practical standpoint, even if your method worked, it's just much easier to just mix two solutions together and filter. No need to stir stuff and wait for insoluble powders to dissolve. On top of that, you're dissolving a white powder in order to make another white powder - how would you know when your reaction has finished?

Conclusion: Use calcium chloride and sodium sulfate.


Activity Coefficients from Electrode Potentials Lets say we want the activity coefficient of $\ce{Cu^{2+}}$ ions in solution. Would we simply setup one half cell of a $\ce{Cu}$ electrode in a known concentration of $\ce{CuSO4}$ and then connect this to a Standard Hydrogen Electrode.

Using the standard electrode potential of the reaction $\ce{Cu^{2+} + 2e-  ->  Cu}$  and the Nernst equation could we determine the activity coefficient of $\ce{Cu^{2+}}$ ions ?

Is this all I would need to do ?

Sort of 'yes', sort of 'no'.  Consider the following points:

Activity coefficients are notorious for changing with $\mathrm{p}\ce{H}$, ionic strength, and concentration (among other things).  You might measure the the activity coefficient of $\ce{Cu^{2+}}$ in a $0.7~\mathrm{M}$ $\ce{CuSO4}$ solution and get one value, but try to measure the same thing in a low $\mathrm{p}\ce{H}$ and $1~\mathrm{\mu M}$ $\ce{CuCl2}$ solution and you may get a completely different value.  In practice, people usually end up measuring activity coefficients across a range of conditions that are relevant to their specific use-case.
The Standard Hydrogen Electrode is a theoretical device; and actually constructing something similar is difficult.  In practice, you would just want to use a commercially available double-junction silver-silver chloride or saturated calomel electrode. The potentials of those two are well-known and fairly stable.

See Chapter 2 of "Electrochemical Methods" 2nd Ed, by Bard and Faulkner for more information.


Why is allyl chloride more reactive towards substitution than alkyl chloride? I can understand that the reactivity of alkyl halide is more than vinyl and aryl halides because the halogen atom in alkyl halide is connected to the sp3 carbon.

Whereas in vinyl or aryl the halogen atom is connected to the sp2 carbon.

But how allyl halide is more reactive than alkyl halide?

But how allyl halide is more reactive than alkyl halide?

Reactivity towards $\ce{S_N1}$
The key to reactivity towards $\ce{S_N1}$ is the stability of the formed carbocation. Allyl system stabilises the carbocation through overlap with the vacant p orbital (@gsurfer999 has shown the resonance structures in his answer below). However, note that any allyl halide wouldn't be better at $\ce{S_N1}$ than any alkyl chloride.
In fact, the tertiary alkyl chloride's rate of $\ce{S_N1}$ is faster than that of allylic chloride which is secondary at one end. (due to $\ce{+I}$ effect of 3 methyl groups as well as hyperconjugation)
Visual aid for stabilisation through hyperconugation (from Clayden):

For such comparison, here's a useful table again from Clayden:
Reactivity towards $\ce{S_N2}$:
In $\ce{S_N2}$ the most important factor is the substrate. The reaction proceeds through a single transition state with trigonal bipyramidal geometry and $\ce{\sim sp^2}$ hybridisation of the electrophilic carbon. The p-orbital makes two partial bonds, one with the nucleophile and another with the leaving group. Thus, it's electron deficient.  In such a case, the allyl system provides the additional electron density through conjugation.
However, again note that primary alkyl chloride is better at $\ce{S_N2}$ than allylic system due to least steric hindrance!


Calculating pH of an unknown base, and the pKb The percent dissociation of a base $\ce{X}$ in a $0.002\ \mathrm{M}$ solution is $4.7 \cdot 10^{-3}\ \%$. What is the pH of the solution and what is the $\mathrm{p}K_\text{b}$ of base $\ce{X}$? The autoionization of water may not be neglected.

Attempt at solution: We have the following reactions: $$ \ce{X + H2O(l) <=> XH+ + OH-} \\ \ce{H2O(l) <=> H+ + OH-}$$

Since the percent dissociation is $4.7 \times 10^{-3}\ \%$, this enables us to find the concentration dissociated: $$ \frac{x}{0.002\ \mathrm{M}} \cdot 100 = 4.7 \times 10^{-3},$$ so $x = 9.4 \times 10^{-8}$. Now I did: $$ K_\text{a} = \frac{x^2}{0.002 - x}$$ and I let $(0.002 - x) \approx 0.002$ since $x$ is so small. But this gives me the wrong answer, and it seems I haven't taken the autoionization of water into account yet. How should I solve this problem then?

In such an solution, since the autoionization of water may not be neglected, you should expect $x=9.4 \times 10^{-8}$ do not mean $[\ce{OH-}]=9.4 \times 10^{-8}$. So you can only leave it unknown.

Then, $K_\text{b} = \frac {[\ce{XH+}][\ce{OH-}]}{[\ce{X}]} \approx \frac {9.4 \times 10^{-8} [\ce{OH-}]}{0.002} = 4.7 \times 10^{-5}[\ce{OH-}]$.

However, we still do not know what $[\ce{OH-}]$ is.

If only to consider the autoionization of water balanced, you will get

$[\ce{OH-}][\ce{H+}] = K_\text{w} = 1 \times 10^{-14}$

and this is still not enough.

The answer can only consider by the fact that the solution will be electrically neutral,

$[\ce{OH-}]=[\ce{H+}] + [\ce{XH+}] = \frac {1 \times 10^{-14}}{[\ce{OH-}]} + 9.4 \times 10^{-8}$

$[\ce{OH-}]^2 - 9.4 \times 10^{-8} [\ce{OH-}] - 1 \times 10^{-14}=0$

Since $\Delta = (9.4 \times 10^{-8})^2 + 4 \times 10^{-14} \approx 4 \times 10^{-14}$,

$[\ce{OH-}] \approx \frac {9.4 \times 10^{-8} + 2 \times 10^{-7}}{2} = 2.47 \times 10^{-7}$.

Thus, $\mathrm{pH} = -\log \left(\frac {1 \times 10^{-14}}{2.47 \times 10^{-7}}\right) = 7.39$

$\mathrm{p}K_\text{b} = -\log (4.7 \times 10^{-5} \times 2.47 \times 10^{-7}) = 10.94$.

Sometimes the “electrically neutral” trait will be very useful, this is a good example.


Regenerating the imaginary data of a 2D NMR spectrum with a hilbert transform 2D spectrum take up a large amount of hard disk space. At my last lab it was standard practice to delete the imaginary portions of the spectrum with xfb n (We used TopSpin) and if needed later, to regenerate it from the Real part of the data with a Hilbert transform (Related, the TopSpin Commands and References, ~pg P-233 in version 2.1.2)

My first question: If the raw data is still stored in the ser file (TopSpin Commands and References v2.1.2, pg P-8) why can't you just reprocess the raw data? Is all the data not there? (I suspect this somewhat, as those files are much, much smaller then the processed files, and TopSpin processes 2D spectra as it acquires them, unlike 1D spectra)

My second (highly related) question: Is there a downside to regenerating the imaginary data from the real data? Is it a perfect replication of the original data, or is it an approximation? Why do we ever bother to store this data if we can recreate it at any time?

If you have the raw data (the FIDs in the ser file) still available, there is no reason to use a Hilbert transformation, you can just process with xfb instead of xfb n again. xfb n also doesn't really discard the imaginary part, it processes the spectrum from the raw data again, but doesn't save the imaginary processed data.

The ser file contains all the recorded FIDs, that data is sufficient (together with the parameter files) to recreate the full spectrum. That your Topspin processes 2D spectra automatically is just some specific configuration, it doesn't do that by default.

As far as I understand it, if you zero-fill your data at least once, all the information content of the imaginary part is then present in the real part of the processed spectrum. So I'd guess that under those conditions, the Hilbert transformation shouldn't have any downside, but I'm not totally sure about that.


Hartree product and the Slater determinant I am very new to theoretical chemistry so my question may sound very basic to most. It is known that the Hartree product does not satisfy the antisymmetry principle for electrons while the Slater determinant is able to do so. Thus, the electronic wavefunction is often written as a Slater determinant, rather than a Hartree product. However, I am just wondering if the Hartree product can still be used to approximate the electronic wavefunction or if it is still being use to do so in some theoretical methods? And if the Hartree product is used, what other measures are implemented to ensure that the antisymmetry principle is not violated? Also, in Density Functional Theory, which is used to approximate the wavefunction, the Hartree product or the Slater determinant?

The Hartree product can definitely be used for approximate calculations; as you might know Hartree developed his method for atoms in 1927 (for reference, the Schrödinger equation was discovered in 1926) and he implicitely used a Hartree product for the wave function, although the connection to the exact solution of the many-body Schrödinger equation via the variational principle wasn't know. It was Slater in 1930 (and, independently, Fock) to introduce the determinant form which bears his name. When using the Hartree product one has to account for the Pauli principle "by hand", by allowing only two electrons to occupy the same spatial orbital.
By starting with a Hartree product, enforcing the Pauli principle and applying the variational principle one obtains the Hartree equations, which are similar to the Hartree-Fock ones but are missing the exchange term, usually indicated by $K$; this term is, however, smaller than the  electrostatic term $J$ so Hartree calculations are meaningful and give useful results, although of course very approximate ones.

Kohn-Sham DFT calculations use in principle the Slater product form, but in practice, depending on the exchange-correlation functional used, they lead to equations which may be more similar to the Hartree ones (i.e., without the exchange term) than to the Hartree-Fock ones. In the original LDA (local density approximation) functional the exchange term is dropped and replaced by a functional of the electron density which is much simpler to compute. As a result, DFT-LDA is similar in complexity to the Hartree method, not to the Hartree-Fock (a rather significant simplification). However, most modern functionals for molecules (as opposed to solid state calculations), e.g. B3LYP, re-introduce the exchange term (with a scaling coefficient).

The Hartree form of the wave function can be also used to compute expectation values (e.g., to compute dipole moments); you can check by looking a the Slater-Condon rules  that the results you obtain with Hartree functions are, for one-body operators, the same you'd get using the proper Slater-determinant form.


How can one determine an atomic mass with high accuracy using gravimetric analysis? In the Wikipedia article on gravimetric analysis it is said that gravimetric analysis was used to determine the atomic masses of many elements in the periodic table to six figure accuracy. There are, however, no references to that claim.

Could someone please tell me, the molar masses of which elements can be determined using gravimetric analysis and what the exact procedure is?

Well, interesting question, I would re-word your question as "How can one determine an atomic mass with high accuracy using gravimetric analysis? Precision simply implies that you repeated a measurement and it agrees very well with itself even though it may be wrong. What we desire is high precision and high accuracy.

Almost all the naturally occurring elements had their atomic masses (used synonymously as atomic weight in older books) determined by gravimetry. For that you need to read Richard Theodore's Nobel Prize winning work who meticulously measured atomic weights for many many elements using gravimetry at Harvard in the early 1900s.

Do you know what was the experimental definition of atomic mass which made gravimetry possible? It is weight of an element which combined with 1 part of hydrogen or 16 parts of oxygen (we have to choose one of these but not both). Very sadly, nobody teaches this today so most chemistry students are oblivious to it and yet they can regurgitate the 1/12 of C-12 definition. This definition was invented for some other reasons rather than experimentally measuring atomic weights in the 1960s to resolve disputes among physicists and chemists.

So if I assign a weight of 16 units to oxygen and 65.4 parts of zinc combine with one oxygen in a 1:1 ratio (ZnO), we would assign an atomic weight of Zn as 65.4. Arbitrary as it may sound, this was the experimental way. What Richard did is that he used precipitation reactions using silver chloride and correlated that with oxygen 16.

I provide you with an exercise from Pauling's General Chemistry (forget about moles, Avogadro's, g/mol, molar mass etc.) to solve this problem. Let us go to 1940s, we have the above definition that atomic weight is weight of an element which combines with 16 parts of oxygen.

  In a determination of the atomic weight of iron, 7.59712 g of
  carefully purified ferric oxide, $\ce{Fe2O3}$ , was reduced by heating
  in a stream of hydrogen, and found to yield 5.31364 g of metallic
  iron. Given the atomic weight of oxygen as 15.9994, to what value of
  the atomic weight of iron does this result lead?

Hint: Try to find what weight of iron would combine with 15.9994 parts of oxygen and note that there are two iron atoms and three oxygen atoms.


"Estrontium" on poster I was looking at a poster for an international chemistry olympiad. Its background contains some random elements as periodic table tiles.

One of them said "Estrontium." A google search returned nothing but clip art.
Is this an accepted spelling internationally (similar to the way Spanish speakers prefix "e" on words starting with "s" followed by a consonant) or a careless typo on their part? My chemistry teacher was equally confused.

"Estrontium" is not used as an element name in any language. It appears that the error traces to a single user by the name of Alejo Miranda (listed as from Ecuador) who has posted a large collection of periodic tables and individual element clip art to shuttersock, iphoto, etc. in which "estrontium" is incorrectly used as the English name for strontium.  The Spanish periodic table correctly uses estroncio.  The link to the gllery at shutterstock is

https://www.shutterstock.com/g/Alejo+Miranda?searchterm=estrontium&search_source=base_gallery&language=en&sort=popular&measurement=px&safe=true


How to convert micromolar concentrations into molar quantities, when the volume is also known \begin{array}{c|cccccc}
\text{Sample} & \text{Ammonia} & \text{Phosphate} & \text{Silicate} &\text{Nitrate + nitrite} & \text{Nitrite} & \text{Mn} \\
 & (\pu{\mu M}) & (\pu{\mu M}) & (\pu{\mu M}) & (\pu{\mu M}) & (\pu{\mu M}) & (\pu{\mu g L-1}) \\
\hline
\text{Water} & 0.02 & 0.51 & 9.92 & 9.93 & 0.00 & 0.2 \\
\text{Core 1} & 189.74 & 6.13 & 24.56 & 0.73 & 0.73 & 320.2
\end{array}

The samples are from sediment cores. Different cores have different volume; I need to work in quantities rather than concentrations. I need to calculate the mole (actually micromole) quantities of the $\ce{Mn}$, $\ce{NH3}$, $\ce{NO3-}$ and $\ce{NO2-}$ from their concentrations. The volume for core 1 that is provided here is $\pu{6.86 L}$.

Perhaps I'm being silly, but I am not sure how to do this and how to present this data that is to plotted as follows: molar quantities of the nutrients and metals ($y$-axis) against the oxygen molar concentration ($\pu{\mu M}$) of the water ($x$-axis).

I am also supposed to explain the results in terms of redox chemistry and try to determine approximately at what oxygen concentration in the overlying water these redox reactions in the underlying sediment are initiated.

Molar concentration, or molarity, is moles per unit volume. If you know concentration and sample volume, you can calculate the moles of a particular component in the sample. Since your concentrations are already in micro-molar, when you multiply by the sample volume (in liters), you will have the micro-moles of each component.

For the Mn result, you will get the mass of Mn (in micrograms) when you multiply the test result by the sample volume, so you can convert that to moles using the molar mass of Mn.


Phase abbreviations for non-aqueous solutions When writing a chemical equation, how do you indicate that a chemical is dissolved in a non-aqueous solvent?

For example, it would be inappropriate to write

X (aq)

when X is dissolved in benzene, since aq stands for "aqueous," implying water soluble. A previous answer provided the potential abbreviation (sol), but I wonder if that is actually correct/commonly used.

How would you indicate that X is dissolved in a non-aqueous solvent?

According to IUPAC, general notation for any solution would be $\ce{X(sln)}$ [1, p. 54]:

(vi) States of aggregation
The following one-, two- or three-letter symbols are used to represent the states of aggregation of chemical species [...] The letters are appended to the formula symbol in parentheses, and should be printed in Roman (upright) type without a full stop (period).
$$
\begin{array}{ll}
&\ldots\\
&\text{s} &\text{solid}\\
&\text{sln} &\text{solution}\\
&\text{vit} &\text{vitreous substance}\\
&\ldots\\
\end{array}
$$

Similarly, when denoting relation of a given quantity/symbol, subscript $\mathrm{sln}$ notation is used, e.g. $Δ_\mathrm{sln}G^\circ(T,p^\circ)$ [2, pp. 241, 276]:

4.2 Subscripts and superscripts
4.2.1 Subscripts
$$
\begin{array}{ll}
&\ldots\\
&\text{r} &\text{reference, reduced}\\
&\text{sln} &\text{solution}\\
&\text{s} &\text{saline solution}\\
&\ldots\\
\end{array}
$$

References

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7. (PDF)
Gamsjäger, H.; Lorimer, J. W.; Scharlin, P.; Shaw, D. G. Glossary of Terms Related to Solubility (IUPAC Recommendations 2008). Pure and Applied Chemistry 2008, 80 (2), 233–276. DOI: 10.1351/pac200880020233. (Free Access)


How to remove salt from powdered soup mix? Sometimes I carry powdered soup mix (Knorr/Maggi) when I travel but they are way too salty. Is there an efficient and simple chemical process to lower the salt content?

I know sodium is super soluble and is hard to remove by precipitation, but is there a work around to perhaps only dissolving the salt and take them out? Thanks!

Peel and thickly slice several potatoes and cook these in your soup. Remove them when they have softened. They will absorb the salt while cooking.

You can use this method if you accidently add too much salt to stews or sauces (normally this becomes apparent when reducing.)

The removed potatoes can be discarded or used in another meal, maybe mashed.


Inactivation of enzyme: Why inactivation constants do not follow the Arrhenius equation? The Arrhenius equation may not account for the temperature effect on thermal inactivation rate constants. Why can the thermal inactivation kinetics not be expected to follow Arrhenius? The reason that the inactivation constants do not follow Arrhenius is it because of other chemical processes occuring like diffusion?

Without attempting to elaborate (this is not my area of expertise) I rely on the abstract to an article [1] reviewing the subject for an answer:

  The Arrhenius equation has been widely used as a model of the
  temperature effect on the rate of chemical reactions and biological
  processes in foods. Since the model requires that the rate increase
  monotonically with temperature, its applicability to enzymatic
  reactions and microbial growth, which have optimal temperature, is
  obviously limited. This is also true for microbial inactivation and
  chemical reactions that only start at an elevated temperature, and for
  complex processes and reactions that do not follow fixed order
  kinetics, that is, where the isothermal rate constant, however
  defined, is a function of both temperature and time.

Reference

[1] Peleg et al Critical reviews in food science and nutrition 52(9):830-51 (2012)


Is the Haber Process here proceeding at positive Gibbs free energy change? $\ce{3H2 + N2 -> 2NH3}$ is the forward reaction used in the Haber process, the industrial production of ammonia. After studying Gibbs energy and how it is just another way of saying that the 2nd law of thermodynamics is obeyed if $\Delta G < 0$, I decided to apply it to some common reactions that I know, just to confirm that my understanding of the topic is correct.

When applying it to the Haber process forward reaction, I worked out change in entropy to be $\pu{-199
J K^{-1} mol^{-1}}$. The enthalpy change for the forward reaction is exothermic and has a value of $\pu{-92 kJ mol^{-1}}$. Looking up common reaction conditions for the Haber process, I found that a compromise temperature of $\pu{650K}$ is used.

Applying this to the Gibbs energy equation $-92 - 650(-199/1000)$, I calculated a positive change in Gibbs energy for the reaction of which had a value of $\pu{37.35 kJ mol^{-1}}$.

As a result of this, I am now quite confused as to how the forward reaction can be allowed to proceed by the 2nd law of thermodynamics at a temperature of $\pu{650K}$, and I was wondering where my understanding of the Gibbs energy and thermodynamics falls short. I need help to explain the pitfalls in my calculations, understanding or even whether there is another factor that I am not considering.

It's all about the pressure! The Haber process requires a pressure of about 200 atm. This is how the process becomes favorable, since LeChatelie's principle says that increasing pressure shifts toward the side with less moles of gas.

As long as Q is smaller than K, the reaction will go forward. For example, if nitrogen and hydrogen are supplied at stoichiometric ratio and 1% of a given reaction mixture is product, we would have the following partial pressures:

$$ P_\ce{N2} = 0.2475 \times \pu{200 atm}$$

$$ P_\ce{H2} = 0.7425 \times \pu{200 atm}$$

$$ P_\ce{NH3} = 0.01 \times \pu{200 atm}$$

Accordingly, Q would be

$$ Q = \frac{2^2}{50 \times 75^3} = \pu{1.9e-7} $$

In the Haber process, the concentration of product is kept small by continuously removing product (liquifying ammonia at low temperature) and feeding the remaining reactants back into the reaction mixture.


Compatibility of Grignard reagents with aromatic nitro groups Is it possible to form a Grignard reagent from 1-chloro-3-nitrobenzene, as shown below?

Specifically, will the nitro group react with the Grignard reagent formed?

Aromatic nitro groups are not compatible with Grignards as they react with the nitro group. An example of this being useful is the Bartoli synthesis of indoles from vinyl Grignards (Tet. Letts. 1989, 30 2129, Curr. Org. Chem.. 2005, 9, 163). The first step of the Bartoli is addition of the vinyl Grignard to the O of the N=O of the nitro group so I presume this would be true for other Grignards and is the reason for the incompatibility.


Better functionals to work with transition metals My Masters in inorganic chemistry is focused in magnetic properties of transition metals; great part of the work is done with DFT methods using softwares like ORCA and NWChem, but since my University has a lack of resources, we are performing computations with not much computational power. That is why we are focusing in results and using functionals like PBE and B3LYP (since many papers are showing good results).
Nevertheless, as said by Martin in this thread, today we have "better and more efficient functionals" available, because PBE and B3LYP are very popular, but kind of old. That is why I’d like to ask: what are good, modern, functionals to work with transition metals (mostly Ti…Zn)?

Aditional information:

We are working with systems  that have no more than 200 atoms, most of the time, usually dimers.

This question is like the snake that eats its own tail. There is no functional good enough for metals. To my limited knowledge, fitting too many parameters -which happens with metals- feels like DFT is not first principle anymore and could be called "empirical" (which is also not correct). You allow a number of parameters to to be fit(1,2,more!!) to a specific system. Those dozens of parameters -at the end- may give you desirable results but you "trained" the set of energies according to your will which is not that realistic. You train your dragon not in the jungle but in a cage. The result is never the same. This "fixing" not only contradicts the spirit of DFT but is also a weak form of empirism. It is like adding more "dark force" to the community. Imagine in what extend many of those "good", modern functionals affect research daily. To end, maybe other comments will come up with accurate functionals about Ti or Zn with papers in good journals etc but without being against DFT, I would just keep a small basket.


Estimation of ground state energy from wave function Given the wave functions of a particle in a 1D box below determine (for each pair) if the left or right function gives a better estimation of the ground state energy if the variational method is used. No calculaltions are needed.

I am having some problem understand how to solve this question. We are not allowed to use any graphing calculators and the tip says "try to plot some points and compare it with the exact wave function", which I don't find helpful.
All help is truly appreciated!

The variational method uses a linear combination of functions as an estimate of the true wavefunction. It works by adjusting coefficients in the linear combination to minimize the energy of the trial wavefunction. This is the best estimate of the true wavefunction using the restricted basis set. The variational principle states that adding more (new) basis functions to the linear combination always improves the estimate.

In your problem (or according to your instructions), you don't really optimize anything (except for the last comparison), because the normalization condition fixes the value of the $c_i$ coefficients, but you can estimate which of the two functions appears most like the true wavefunction, as this should return a minimum energy.

The ground state solution has the form $$\psi \propto sin(\pi x)$$
when x is in units of the box length, with bounds at $x=0$ and $x=1$ (and null wavefunction outside these bounds). This function has nodes at $x=0$ and $x=1$ and is concave with a single maximum at $x=\frac12$. By evaluating the functions at the bounds, in the midpoint and at one or more other points you should get an idea about the shape of each function and thereby which are better. It would of course be easier to plot them. Consider to the extent possible symmetry and whether bounds are satisfied (which they must for the solution to be at all physically reasonable - remember that the energy blows up if the wavefunction samples points out of bounds.

The last comparison is a "trick question". There you have to take into account the variational principle which is mentioned in the first paragraph of this answer.


Ensuring constant ionisation of transition metals using plasma ionisation for accurate mass spec Mass spectrometry depends on separation of the analyte based on mass to charge (m/z). ICP-MS is used to detect metals, but under the plasma conditions many of these metals could be ionised multiple times, giving them a variety of (m/z) ratios making a difficult to identify peak for any particular metal. Additionally some metals would have m/z ratios that overlap with others. Then it is conceivable that depending on your solution the quantitation of some elements is impracticable. For example if you want to quanitate aluminium (mass ~27) then a solution high in manganese (mass ~54) would make this impracticable, if manganese were twice ionised to give the same mass to charge ratio as a single ionised aluminium ion. How does the ICP-MS ensure atoms are ionised only once, or that only singly ionised species are detected?

This is further complicated if you consider that the elements rarely exist in solution at oxidation states of 0, but rather aqueous ions with varying oxidation states.

Interesting question. Some of the stuff can be justified. As an analytical chemist, we should keep in mind that no technique is perfect or universal, even if the instrument costs thousands or millions of dollars. Mass spectrometer cannot distinguish isobaric species (or ions), for instance, if two ions have identical mass due to a different location of the isotope within the same molecule, the mass spec won't be able to you. However, if there is a small difference in mass at fourth, or even fifth decimal place, modern day mass spectrometers can distinguish them (see e.g., Fourier transform ion cyclotron resonance mass spectrometers). The more money you add, the fancier the mass spec gets, with it comes better and better mass resolution.

Now come to the ion generating source - the plasma, which is at say 10,000 K. Although the name plasma sounds very fancy as the fourth state of matter, and given that its temperature is higher than the surface of the Sun- this temperature is still too low for atoms to ionize like crazy. It is pretty hard to strip off one, then the second and then the third electron from an atom as you are thinking of Al. Therefore in a typical ICP torch, multiple ionization on a single atom is not a major concern. To satisfy yourself, do a Boltzmann distribution calculation for sodium atom at 10,000 K. You will notice that a very small fraction of sodium atoms will be in the excited state.

  This is further complicated if you consider that the elements rarely
  exist in solution at oxidation states of 0, but rather aqueous ions
  with varying oxidation states.

Once the ions from solution reach the plasma they get reduced to "zero" oxidation state. Consider the plasma as a rich source of free electrons, which are circulating like crazy following the radiofrequency field of the coil. The emission in ICP mainly comes from neutral atoms.


Reaction mechanism of rearrangement I am currently studying for my organic chemistry exam, but there is one problem I do not understand. Unfortunately, I do not have any solutions.

See image for the problem.

My first thought was the Dienone-phenol-rearrangement. But in order to do so, I need to add the alkyl substituent and I need a dienone. So can I deprotonate the phenol using potassium carbonate? (Or is it too weak as a base? The pKa of bicarbonate is around 10, so maybe it is too weak) But if it is possible, then the free electron pair of the oxygen can form a ketone... I could then use the alkene with a chloride substituent, so that the aromatic ring can attack it. Then I would have the dienone. But with the Dienone-phenol-rearrangement, the alkyl substituent can only move by one carbon-atom in the ring...

Is my thought process wrong? Can you please help me?

I did not have any lectures on rearrangement reactions, I had to learn them by myself, but by now, I have not found any similar reactions.

Thank you!

The first reaction is O-alkylation of p-cresol to give a 4-methylphenyl allyl ether derivative 3. The reagent in the first box should be 1-bromo-3-methylbut-2-ene (1; see the top box in the picture), which would undergo $\mathrm{S_N2}$ reaction with phenolic anion (2) in refluxing acetone. Note that potassium carbonate is a strong enough base to complete this reaction (this is surely a strong base than potassium bicarbonate).

The product 3 from alkylation would undergo Claisen rearrangement upon heating to give final alkylated cresol 4 in solvent-free condition. Remember, Claisen found his famous rearrangement first in solvent-free conditions, while he was trying to find the melting point of newly synthesized naphthyl allyl ether! (Picture is from https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3873100/ and modified accordingly)


Why are potentials of cells with magnesium electrode always lower than expected? I have prepared the 1M water solution of magnesium acetate. Then I placed $\ce{Mg}$ and $\ce{Cu}$ electrodes in it. I predict there such reactions take place: $$\ce{Mg-2e^- \rightarrow Mg^{2+}}$$ $$\ce{H_3^+O + e^- \rightarrow\frac12 H_2 + H_2O}$$
The gas bubbles emitting on both electrodes probably confirm my prediction (I did not check is it hydrogen or not).
So I expect to obtain the potential difference about 2.3 V. But really I observe something like 1.62 V.

Then I changed the acetate solution to a water (simple water which we drink) and observed almost same value of 1.60 V. Adding there many crystals of $\ce{Mg}$ acetate (I added them till they could not dissolve more) almost did not change the voltage!

When I added much acid to the solution the potential difference became larger - up to 1.9 V.

So, I have two questions: 1) why the real voltage is so different from the expected value? 2) Why changing the $\ce{Mg^{2+}}$ concentration almost does not change the voltage?

You may say I have dirty magnesium electrode etc. But if you search e.g. experiments with potatoes (when one puts $\ce{Mg}$ and $\ce{Cu}$ strips in potato) you can observe that everywhere it is mentioned that such $\ce{Mg-Cu}$ cell gives 1.7 V.

I have several ideas why can it happen: a)some story with overpotentials, b)oxide film on magnesium surface.
What is the real reason?

At least one significant reason is that the voltage you "expect" is probably for a standard state of one mole per liter of $\ce{H+}$, i.e. at pH 0.  Neither the magnesium acetate nor the plain (tap?) water you used for the experiments come close to this pH. Also, the standard state for $\ce{H2}$ would be one bar of atmospheric pressure.  Probably this condition was not achieved either.

You can calculate the expected potential at any pH using the Nernst Equation.

The half-cell reactions in your system are

$\ce{Mg^0 -> +2e- +Mg^{+2}}$ (this has a potential of 2.372 V not 2.7 V as a mistakenly wrote in a prev. version of this answer)
$\ce{2H^+ +2e- -> H2}$

The total reaction is thus $\ce{Mg^0 +2H+ -> Mg^{+2} + H2}$

So applying the Nernst equation to a one molar magnesium solution at pH 7 (i.e. $\ce{H+}$ concentrations of $10^{-7}$ mol/L) would give:

$E_\text{cell} = E^{\ominus}_\text{cell} - \frac{RT}{zF} \ln Q$
$E_\text{cell} = 2.4 \text{ V} - \frac{RT}{zF} \ln \frac{1}{(10^{-7})^2 }$
$E_\text{cell} = 2.4 \text{ V} - 0.41 \text{ V} = 1.9 \text{ V}$

Obviously, this correction gives you a predicted voltage that is still higher than what you measure closer what you measured.  But you reported using acid and getting a higher potential, so it's clear that the pH effect matters for your system.  But something else must be going on too.  Some other possibilities for other minor effects that tweak the voltage a bit more:

Hydrogen partial pressure is unaccounted for
overpotential
your cell is not open but is conducting current
dirty electrodes etc.
other unaccounted for electrochemical reactions (copper oxidation, etc.)


Bond angle and bond lengths in Arginine Based on the Lewis structure of arginine, the terminal primary amine attached to the guanidine group seems to be $\mathrm{sp^3}$ hybridized and trigonal pyramidal. Also, in quantum mechanical computations, it is shown  that all carbon-nitrogen bonds in the guanidine group are different lengths.

However, in the 3d structure below, the entire guanidine group is planar, indicating an sp2 hybridization. Why is $\ce{N}$ $\mathrm{sp^2}$ and not $\mathrm{sp^3}$ hybridized? Why is the guanidine group planar? Why are all the bond lengths different if all the $\ce{N}$ atoms are $\mathrm{sp^2}$?

L-arginine is the most basic amino acid since it carries guanidyl group in addition to the $\alpha$-amino group. The arginine molecule is a zwitterion with the guanidyl group, rather than the primary $\alpha$-amino group, accepting extra proton from carboxylic acid group (Ref.1). Protonated or not, when consider guanidine as a molecule, there shouldn't be any $\mathrm{sp^3}$ nitrogen in the molecule because it is resonance-stabilized molecule. As a guanidyl group, this principle stays the same and it stays flat. There are few excellent answers to similar question you could find in here. In one answer it states that:

  Guanidine is isolobal to urea, where the carbonyl oxygen has been replaced by an imine $\ce{NH}$. However, in principle it is still the same flat, resonance-stabilised molecule. The main difference is that there is no ‘preferred’ site for the double bond — it could point towards any of the three nitrogens in theory; you could say the diamide resonance is enhanced (Author: Jan).

And, accepted answer showed following diagram to depict the concept of resonance stabilization (Author: Ron):

Further, I'll attached recent crystal structure of L-arginine, which shows clear equivalent three nitrogen atoms with hydrohen bonding network (Ref.2):

  Two planes characterize the arginine molecule: one through the acid group and the other through the extended side chain, which contains the guanidyl group. The dihedral angle between these two planes is $74^\circ$ (Ref.1).

References:

I. L. Karle, J. Karle, “An application of the symbolic addition method to the structure of L-arginine dehydrate,” Acta Cryst. 1964, 17, 835–841 (https://doi.org/10.1107/S0365110X64002250).
Emilie Courvoisier, P. Andrew Williams, Gin K. Lim, Colan E. Hughes,  Kenneth D. M. Harris, “The crystal structure of L-arginine,” Chem. Commun. 2012, 48, 2761–2763 (DOI: 10.1039/C2CC17203H).


Balancing the reaction equation for the oxidation of lead(II) sulfide with ozone While I was reading about the chemical reactions of ozone, I came across this reaction.
$$\ce{PbS + 4O3 -> PbSO4 + 4O2}$$

Though the reaction seems well balanced, but when I attempted to balance it myself, I did this as $$\ce{PbS + 2O3 -> PbSO4 + O2}$$

I understand that there must be some reason that the 1st reaction is given in the book. I'm just not able to figure out why.

Can somebody give an explanation as to why is it so?

This is a semi-standard example of why 'naive' balancing does not work.

Ozone is a source of atomic oxygen, producing free oxygen molecules. So, in 'mild' conditions only one oxygen per ozone molecule would react as a strong oxidizer, and the remaining molecular oxygen would require elevated temperatures to react. Thus the equation would be

$\ce{PbS + 4O3 -> PbSO4 + 4 O2}$

Similarly, in excess of reducer at elevated temperature (though why would you want it with ozone is another matter) all atoms in ozone would react. Thus, the reaction should be

$\ce{3PbS + 4O3 -> 3PbSO4 }$


Comparing equilibrium constants between acetone and benzaldehyde in nucleophilic addition reaction My teacher told me that the $K_\mathrm{eq}$ of acetone is greater than that of benzaldehyde in nucleophilic addition reaction.
But I am not able to understand how is this possible. Acetone has a +I group attached to it, so it will decrease the reactivity of carbonyl. If we talk about benzaldehyde, it has a −M group. Can someone tell what is the correct reason behind it?

The information given by your teacher is incorrect. The $K_\mathrm{eq}$ value of acetone for cyanohydrin formation (a nucleophilic addition reaction) is less than the $K_\mathrm{eq}$ shown by benzaldehyde
According to Organic Chemistry: Structure, Mechanism, Synthesis [1 , p. 595], the values of $K_\mathrm{eq}$ for cyanohydrin formation are as follows:
$$
\begin{array}{lr} \hline
\mathrm{Compound} & K_\mathrm{eq} \\ \hline
\text{Acetone} & 20 \\
\text{Ethanal} & 10^5 \\
\text{Benzaldehyde} & 210 \\
\textit{p}\text{-Methoxybenzaldehyde} & 30 \\
\text{Acetophenone} & 0.8 \\ \hline
\end{array}
$$
Hence the statement that the $K_\mathrm{eq}$ is greater for acetone is incorrect. This seems accurate considering that acetone is more sterically hindered than benzaldehyde, which is a dominating factor for nucleophilic addition reaction since they are kinetically stable reactions.
This document has a much larger cases of compounds that are compared for multiple reactions, in all of which benzaldehyde is more reactive than acetone.
However, there is one minor issue with your reasoning given in the question.

Benzaldehyde does not have a −M group in it. However if you add another functional group into the ring, then we say that the $\ce{-CHO}$ group acts as a −M group. Actually, $\ce{-CHO}$ has a +R effect from the phenyl ring attached to it which makes the ring in benzaldehyde deactivated. So, the carbon in benzaldehyde is less electrophilic than formaldehyde which has no such effects.
Reference

Ouellette, R. J.; Rawn, J. D. Organic Chemistry: Structure, Mechanism, and Synthesis, 1st ed.; Elsevier, 2014. ISBN: 978-0-12-800780-8 (Google Books Preview)


Grignard reagent with sulfur dioxide I have studied that the Grignard reagent $\ce{RMgX}$ reacts with $\ce{CO2}$ to form $\ce{RCOOH}$ after acid hydrolysis. But does the same apply for sulfur dioxide?

Grignard reagents are good bases and also good nucleophiles.
So the alkyl anion will attack the electron deficient carbon atom in $\ce{CO2}.$

But in the case of $\ce{SO2}$ the following conditions aren't present.
First of all, sulfur is more electronegative than carbon. Secondly, sulfur has a lone pair of electrons which the carbon lacks. So this should further discourage an attack by the Grignard's alkyl anion.

So, what actually will happen when Grignards are treated with $\ce{SO2}?$

You get sulfinic acids $\ce{RSO2H}.$

This paper by Dowson et al. [1] states that Grignard plus $\ce{SO2}$ gives sulfinic acids and as a reference quotes an older article by Allen Jr. et al. [2] which states

  The reaction between sulfur dioxide and Grignard reagent has been widely used for the preparation of aromatic and aliphatic sulfinic acids.

I would suggest you read [2] and the references it quotes.

References

Dowson, G. R. M.; Dimitriou, I.; Owen, R. E.; Reed, D. G.; Allen, R. W. K.; Styring, P. Kinetic and Economic Analysis of Reactive Capture of Dilute Carbon Dioxide with Grignard Reagents. Faraday Discuss. 2015, 183, 47–65. https://doi.org/10/gf9hvn. (Research Gate — PDF)
Allen, Jr., Paul.; Rehl, W. Richard.; Fuchs, P. E. Sulfone Formation during Sulfination of the Alkyl Grignard Reagent. J. Org. Chem. 1955, 20 (9), 1237–1239. https://doi.org/10/d42z3x.