The bond in coordination complexes This is a very basic question and I'm surprised it only just struck me. The nature of the bond in coordination complexes is a coordinate covalent bond. Only the ligand donates electrons for bond formation between the metal and ligand.

Why is it that we never place a positive charge on the ligands? For example, in $\ce{[Pt(NH3)5Cl]Br3}$, $\ce{NH3}$ is attached to three Hydrogens and the lone pair is donated for forming the coordinate bond. So there should be a positive charge on $\ce{NH3}$, right?

In a typical covalent bond, such as is found between carbon and hydrogen in
methane, each atom is considered to contribute one electron to the two-electron,
two-centre bond which is formed. The metal-ligand bond is often envisaged as a second type of covalent bond in which we still have a two-centre, two-electron bond, but where
both of the electrons come from one of the atoms or from a molecule. This type of
bond is known variously as a coordination, a dative covalent or a donor-acceptor
bond.

It was recognized early on that the formality of dative covalency in coordination
compounds presents some difficulties. Many inorganic compounds are conveniently
thought of as ionic salts in which there is an essentially complete charge separation
between cationic and anionic species. Compounds of the group 1 and group 2 metals
tend to be readily categorized as ionic. Dissolution of a salt of a group 1 or a
group 2 metal results in the formation of solutions containing solvated cations and
anions - only in concentrated solutions are there significant cation-anion
interactions beyond simple ion-pairing. However, the ionic model does not appear
to be suitable for the description of the properties of many transition-metal
compounds. For example, the compound $\ce{K4[Fe(CN)6]}$ dissolves in water to give
solutions containing solvated potassium ions and the $\ce{[Fe(CN)6]^{4-}}$ ion, rather than
solvated potassium, iron(II) and cyanide ions. The interactions between the cyanide
and the iron(II) centre appear to result in longer lived species than result from
simple electrostatic interactions of the type observed in sodium chloride. This is, of
course, the sort of argument which led to the development of the description of
coordination compounds in terms of donor-acceptor interactions between the
ligands and the metal centre.
Let's have a look at the consequences of the formation of a donor-acceptor
bond in a little more detail. If the donor-acceptor bond is completely covalent,
then there is a net transfer of one unit of charge from the donor to the acceptor as
a direct consequence of the equal sharing of the electron pair between the two
centres. This result leaves a positive charge on the donor atom and a negative
charge on the acceptor atom. The limiting ionic and covalent descriptions of a
complex cation such as $\ce{[Fe(H2O)6]^{3+}}$ are shown below.

As commented above 'ionic' structures are not in accord with the
chemical properties of coordination compounds. However, there are also a number
of objections to the covalent description. The charge distribution is such that the
iron(III) centre bears a three minus charge, whereas the oxygen atoms of the water
ligands each bear a single positive charge. This would be unrealistic in view of the
electronegativities of these elements ($\ce{Fe}$, 1.8; $\ce{O}$, 3.5), which predict that the $\ce{Fe-O}$ bond should be polarized in the sense $\ce{Fe^{\delta +}-O^{\delta -}}$. The problem was addressed by Pauling, who recognized that, in reality, it was not appropriate to describe most bonds as being purely 'ionic' or purely 'covalent'.
He developed a description of the bonding between a metal and its ligands which included considerable ionic character in the metal-ligand bonds within a basically covalent regime. In order to determine the amount of ionic character within a given metal-ligand bond, Pauling framed his electroneutrality principle. In this, he opined that the actual distribution of charges within a molecule is such that charges on any single atom are within the range -1 to +1.
It can be seen how this works by reconsidering the ion $\ce{[Fe(H2O)6]^{3+}}$. An ionic
description results in a +3 charge on the metal centre, whilst a covalent description
gives the metal a -3 charge. Now the electroneutrality principle suggests that the
ideal charge on the metal centre is zero. That would be achieved if the iron centre
gains a total of three electrons from the six oxygen donor atoms; in other words, if
each oxygen loses one half of an electron rather than the whole electron which the
fully covalent model demands. Pauling describes this situation as 50%
covalent (or 50% ionic).

Pauling's model gives a good indication that the bonding situation in transition metal complexes is rather complex since the bonds are highly polarized (An 'accurate' view of the bonding in transition metal complexes can only be gained by more sophisticated methods such as the molecular orbital method).
Thus, Lewis structures are not particularly well suited to depict the bonding situation in coordination compounds because Lewis structures are intended for covalent compounds only.


How chemists find out the electronic properties of elements (atomic numbers) How did chemists find how many electrons, neutrons, and protons each element had?

I mean I can understand the Mendeleev's table but without really knowing how you make one...

One could do it in the following way.

The atomic number $Z$ can be determined by examining the x-ray spectrum of an atom.
The frequency $f$ of the so-called K-alpha line is related to the atomic number $Z$ by Moseley's law
$$
    \sqrt f = k_1 \cdot \left(Z - k_2\right) \, .
$$
The atom is electrically neutral, so the number of electrons is obviously calculated from the atomic number.
And the number of neutrons is also simply to choose: the only requirement at this stage would be the correct atomic weight for the element in question.


Does diamond have a pungent smell like graphite? I was writing down a chemical equation with my pencil and noticed the pungent smell of the graphite. This made me wonder if other substances made of carbon, such as diamond, have a similar pungent smell?

None of the carbon allotropes have a smell. If you smell something from a pencil lead (graphite), then it will be from the impurities in it.

Graphite is made from impure organic compounds that frequently contain sulphur. It is likely that you are smelling this. Mercaptans and other sulphur compounds can be very smelly. on its own


Why is NO2 formed when N valency is 3 Sorry if the question is silly but this is troubling me more than enough. Here's what I know about chemical formula: Reverse the valencies of the combining elements. But in this case, $\ce{NO2}$ is the formula for nitrogen dioxide while N valency is 3. Please explain it to me if I am wrong somewhere.

The rule of swapping the valencies of the atoms to get their ratios only works for a very specific set of compounds and is more of a convenient trick than a rule based on any sort of hard theory.

In particular, it works for binary 'ionic' compounds such as $\ce{NaCl}$, $\ce{CuCl2}$, $\ce{Al2O3}$ etc. because the valencies of the atoms correspond to the charges on the ions and the net charge has to be zero.

However, it obviously doesn't work for a huge range of other compounds where there is more complicated bonding. For example there are a whole range of covalent nitrogen oxides of varying degrees of stability which do not fit the rule ($\ce{N2O}$, $\ce{NO}$, $\ce{NO2}$, $\ce{N2O3}$, $\ce{N2O4}$ etc).

Ultimately the only way to determine the formula of any arbitrary substance is through experiment, typically with a mass spectrometer.

If you are interested in finding out more about the different types of substances that are possible I suggest that you start reading about different types of bonding, in particular ionic versus covalent, to begin to understand the different types of compounds that exist.


How is the dissolution of acetic acid that makes its aqueous solution a poor electrolyte? I'm having a difficulty understanding the following quote from Wikipedia - Dissociation:

  Acetic acid is extremely soluble in water, but most of the compound dissolves into molecules, rendering it a weak electrolyte.

To what (non-ionic) molecules can acetic acid be dissolved? If the answer is $\ce{CH3COOH}$ then in what way is it extremely soluble, if it dissolved to itself?

There are two questions being asked here:

Why is aqueous acetic acid a weak electrolyte?

When acetic acid is dissolved in water there is an equilibrium reaction:
$$\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$

Since acetic acid is a weak acid, the equilibrium position lies well to the left, with only a small fraction of the acetic acid molecules reacting to form ethanoate and hydronium ions. The presence of this small amount of ions results in aqueous acetic acid being a weak electrolyte.

Why is acetic acid highly soluble in water?

Since the vast majority of acetic acid molecules do not dissociate when a sample is dissolved in water, the solubility has to do with the interactions between acetic acid molecules and water molecules. Water has a network of hydrogen bonds between molecules in its liquid phase and so when a substance dissolves in water this bonding is disrupted. However, acetic acid is able to form many new hydrogen bonds to water molecules and so this results in a highly favourable interaction, leading to the high solubility of acetic acid in water.


Mass percentage of potassium chlorate in decomposition reaction When heated, a $3.40\ \mathrm g$ sample mixture of $\ce{KClO3}$ decomposes and produces $685\ \mathrm{cm^3}$ $\ce{O2}$ gas at $30\ \mathrm{^\circ C}$ and $101\ \mathrm{kPa}$. What is the mass percentage of $\ce{KClO3}$ in the mixture?

What I've done so far is write out a balanced equation which is:

$$\ce{2KClO3 -> 3O2 + 2KCl}$$

Where do I go from here?

Generally, it is preferable to write down the complete quantity equation before plugging in the numbers. If necessary, perform all mathematical operations on quantities symbolically. Do not plug in the numbers until you have only one equation for the desired result. In the following, we use this method to calculate the amount of oxygen $n_{\ce{O2}}$, the mass of potassium chlorate $m_{\ce{KClO3}}$, and the mass fraction of potassium chlorate $w_{\ce{KClO3}}$ directly from the parameter values given in the question (although only the last calculation is actually required to answer the question).

Amount of oxygen

For simplicity’s sake, you may use the ideal gas law to estimate the amount of oxygen $n_{\ce{O2}}$ as follows. The volume $V=685\ \mathrm{cm^3}=0.000\,685\ \mathrm{m^3}$, the temperature $T=30\ \mathrm{^\circ C}=303.15\ \mathrm K$, and the pressure $p=101\ \mathrm{kPa}=101\,000\ \mathrm{Pa}$ are given in the question. The molar gas constant is $R=8.314\,459\,8(48)\ \mathrm{J\ mol^{-1}\ K^{-1}}$.

$$\begin{align}
p\cdot V&=n_{\ce{O2}}\cdot R\cdot T\tag1\\[6pt]
n_{\ce{O2}}&=\frac{p\cdot V}{R\cdot T}\tag2\\[6pt]
&=\frac{101\,000\ \mathrm{Pa}\times0.000\,685\ \mathrm{m^3}}{8.314\,459\,8\ \mathrm{J\ mol^{-1}\ K^{-1}}\times303.15\ \mathrm K}\\[6pt]
&=0.027\,448\,609\ \mathrm{mol}\\[6pt]
&\approx0.027\,4\ \mathrm{mol}
\end{align}$$

By way of comparison, the amount of oxygen calculated using standard reference data for oxygen instead of using the ideal gas law yields $n_{\ce{O2}}=0.027\,47\ \mathrm{mol}$. Considering the estimated uncertainty of the result obtained by using the ideal gas law, the difference is not significant. Hence, the approximation using the ideal gas law is acceptable in this case.

Mass of potassium chlorate

The chemical equation
$$\ce{2KClO3 -> 3O2 + 2KCl}\tag7$$
indicates that $2\ \mathrm{mol}$ of potassium chlorate yield $3\ \mathrm{mol}$ of oxygen. Therefore,
$$n_{\ce{KClO3}}=\frac23\cdot n_{\ce{O2}}\tag8$$

The molar mass of potassium chlorate is $M_{\ce{KClO3}}=122.55\ \mathrm{g\ mol^{-1}}$. Since the molar mass $M_{\ce{KClO3}}$ is defined as
$$M_{\ce{KClO3}}=\frac {m_{\ce{KClO3}}}{n_{\ce{KClO3}}}\tag9$$
the mass of potassium chlorate $m_{\ce{KClO3}}$ is
$$\begin{align}
m_{\ce{KClO3}}&=M_{\ce{KClO3}}\cdot n_{\ce{KClO3}}\tag{10}\\[6pt]
&=M_{\ce{KClO3}}\cdot \frac23\cdot n_{\ce{O2}}\tag{11}\\[6pt]
&=M_{\ce{KClO3}}\cdot \frac23\cdot\frac{p\cdot V}{R\cdot T}\tag{12}\\[6pt]
&=122.55\ \mathrm{g\ mol^{-1}}\times\frac23\times\frac{101\,000\ \mathrm{Pa}\times0.000\,685\ \mathrm{m^3}}{8.314\,459\,8\ \mathrm{J\ mol^{-1}\ K^{-1}}\times303.15\ \mathrm K}\\[6pt]
&=2.242\,551\,324\ \mathrm g\\[6pt]
&\approx2.24\ \mathrm g
\end{align}$$

Mass fraction of potassium chlorate

The mass fraction of potassium chlorate $w_{\ce{KClO3}}$ is defined as
$$w_{\ce{KClO3}}=\frac{m_{\ce{KClO3}}}{m}\tag{13a}$$
where $m=3.40\ \mathrm g$ is the total mass of the sample. Therefore,
$$\begin{align}
w_{\ce{KClO3}}&=m_{\ce{KClO3}}\cdot\frac1m\tag{13b}\\[6pt]
&=M_{\ce{KClO3}}\cdot \frac23\cdot\frac{p\cdot V}{R\cdot T}\cdot\frac1m\tag{14}\\[6pt]
&=122.55\ \mathrm{g\ mol^{-1}}\times\frac23\times\frac{101\,000\ \mathrm{Pa}\times0.000\,685\ \mathrm{m^3}}{8.314\,459\,8\ \mathrm{J\ mol^{-1}\ K^{-1}}\times303.15\ \mathrm K}\times\frac1{3.40\ \mathrm g}\\[6pt]
&=0.659\,573\,919\\[6pt]
&\approx0.660=66.0\ \%
\end{align}$$


Can absolute enthalpy values be used to determine reaction enthalpy? Our teacher added this question to an exam, and there was a huge disagreement with my classmates and teacher about the correct answer. One side favoured answer 2, the other side favoured answer 4 being the false statement.

I personally selected answer 4, as it did contradict with the other 3: option 1 has $\Delta H < 0$, option 2 has $\Delta H = H_\mathrm{f} -H_\mathrm{i} = \pu{-500kJ}$ and option 3 has $\Delta H = \pu{-500 kJ}$,  so option  4 contradicts the above. However I would like to hear your thoughts.

  Which of the following options that refer to the enthalpy of a specific chemical reaction is not correct?

  $H_\mathrm{i} > H_\mathrm{f}$
  The enthalpy of the system has altered from the initial value of $\pu{1600 kJ}$ ($H_\mathrm{i}$) to the final value of $\pu{1100 kJ}$ ($H_\mathrm{f}$)
  $\Delta H = \pu{-500 kJ}$
  $\Delta H = \pu{500 kJ}$

Assuming that 3 are true, the only self-consistent set are 1,2 and 3.

Importantly, statements 3 and 4 contradict one another, so one of them must be the incorrect statement.

Consider a reaction with a change in enthalpy ($\Delta H$) of $\pu{-500 kJ/mol}$, and statement 3 is true:

\begin{align}
  \Delta H &= H_\mathrm{f} - H_\mathrm{i}\\
  \implies H_\mathrm{i} &= H_\mathrm{f} - \Delta H\\
  H_\mathrm{i} &= H_\mathrm{f} + \pu{500 kJ/mol}\\
  \implies H_\mathrm{i} &> H_\mathrm{f}
\end{align}
We've shown that statement 1 is true also.

As wolphram says in the comments, you can't measure an absolute value of enthalpy, but you can measure it relative to an chosen zero enthalpy.  If you chose zero enthalpy such that $H_\mathrm{i} = \pu{1600 kJ/mol}$, then by the first equation above we can say that:
\begin{align}
  H_\mathrm{f} &= \Delta H + H_\mathrm{i}\\
  \implies H_\mathrm{f} &= \pu{-500 kJ/mol} + \pu{1600 kJ/mol}\\
\implies H_\mathrm{f} &= \pu{1100 kJ/mol}
\end{align}
We've shown that statement 2 is true as well.

Statement 4 contradicts statement 3 (as well as the other two), so statement 4 must be incorrect.


Is gravity an obstacle for homogenous solvation due to diffusion? If you have a glas with a spoon of salt ($\ce{NaCl}$) in it and you carefully add the water($15^\circ\mathrm{C}$). A lot of the salt will not be dissolved immediatelly, it will take a while for full dissolution.

When the salt dissolves, it is said that the salt is mixed equally in the water. But $\ce{NaCl}$ is split in two ions $\ce{Na+}$ and $\ce{Cl-}$. $\ce{Cl}$ has a mass of $35$ and Na $23$. But due to gravity (or is really curved space in the glass a better word) you should expect that the $\ce{Cl}$ atom will be more at the bottom then the $\ce{Na}$ ion. Is that right or is there someting I'm missing?

So can you say that after an hour the salt is evenly diffused through the water and that there are probably more more $\ce{Na+}$ ions at the top of the glass, so that the top of the salt water will taste more salty due to presence of sodium, which has more power of giving salty tast?

At normal Earth gravity, for all practical purposes homogeneous solvation is not hindered by gravitational force.
However: For some solutes such as sucrose and $\ce{CsCl}$, though, the "artificial large gravity" of centrifugation is capable of establishing concentration gradients in their solutions. One example in common laboratory use is buoyant density centrifugation:

One of the most common methods of fractionating DNA is to exploit differences in density between different types of DNA in concentrated salt solutions. Caesium chloride is the most commonly used. The method involves prolonged high speed centrifugation of purified DNA in concentrated caesium chloride. The centrifugation produces a gradient of salt concentration in the centrifuge tube and the DNA migrates to occupy a position in the gradient which corresponds to its own buoyant density.


Why is boric acid written as B(OH)3 and not H3BO3? Why is boric acid written as $\ce{B(OH)3}$ and not just $\ce{H3BO3}$, like it should be for an acid?

This is mostly because boric acid is commonly believed to have different acidity mechanism. While common acids generally dissociates

$\ce{HNO3 <=> H+ + NO3-}$

boric acid is commonly believed to associate with water

$\ce{H2O + B(OH)3 <=> H+ + [B(OH)4]-}$

this type of acidity is rather common for metal hydroxides, hence the way of writing.

For that matter, not every hydrogen should be written at the beginning of a formula of an acid, but only actually acidic ones. So, $\ce{H2SO4}$, but $\ce{HPH2O2}$


Can compressed CO2 sprayed onto a hot mass/liquid be an effective means to cool it? I thought the answer was obvious, but endless google searching has revealed nothing but speculation and people who have completely conflicting opinions. Some people say $\ce{CO2}$ absorbs no heat, others say it can cool things extremely rapidly. Anyone who has ever had compressed $\ce{CO2}$ sprayed on themselves or on hands knows it can cause very rapid drop in temperature on what ever it hits very fast.

So, if I have a really hot mass of liquid, would spraying compressed $\ce{CO2}$ on it be an effective means of freezing/solidying the mass? - Assuming the liquid is well above room temperature (100+) and turns to a solid after a 20-30 degree drop in temperature?

I understand why the $\ce{CO2}$ is cold coming out of the tank, as it is decompressed, but is it effective at absorbing heat?

Celebrity chef Alton Brown produces ice cream using a device called the Jet Cream.  He sprays $\ce{CO2}$ from a fire extinguisher at a stream of chocolate milk, instantly producing ice cream. The only difference in your scenario and his is the starting temperature of the liquid.

To say that $\ce{CO2}$ absorbs no heat is absurd.  If it is colder than what it touches, it should absorb heat.


Boiling solutions with two salts If you dissolve two different ionic compounds (that don't make a precipitate) in water and then boil the solution so all the water evaporates, the salts will re-crystallize. Is it possible that you will get two compounds that are different from the ones you orignially dissolved? Is there any way to predict whether you will get the original compounds or new ones?

Mixing and matching different anions and cations can easily produce different compounds.  For instance, sodium and potassium chloride, and sodium and potassium nitrate, are all soluble in water and do not react to form a precipitate. When the aqueous solution dries, all four compounds are possible.  So, mixing sodium chloride and potassium nitrate will result in 2 anions and 2 cations in solution, allowing all four pairs to possible precipitate.

Now, will the precipitate as pure compounds or not is another question.  That requires looking at the relevant phase diagrams and seeing what is thermodynamically favored (and consideration of kinetic factors like how fast you evaporate the liquid).

As an example, the phase diagram for KCl and NaCl exhibits a miscibility gap starting at 505C.  This means that below that temperature they will phase separate into a Na-rich chloride and a K-rich chloride.  At 100C the concentration of the 'impurity' phase is less than 1%.

The nitrate diagram also shows a two phase region below 100C.

So, mixing (Na,K) (Cl,NO3) will result in all four precipitating out.


Why do only ketones exhibit keto-enol tautomerism? Why can't esters or acids or their derivatives exhibit tautomerism? I mean if they did, then there would be a conjugation between lone pair electrons of oxygen and pi electrons adding to their stability. Then why don't they exhibit similar tautomerism?

Esters, amides and acids do exhibit tautomerization to the enol form, but the enol content in these compounds is just much less than that found with aldehydes and ketones. Why?

The keto-enol process is an equilibrium; to understand where an equilibrium lies, we must compare the relative stabilities of the products (enol) and reactants (carbonyl compound).

Let's compare some resonance structures for an ester and a ketone and their corresponding enols.

The ester is stabilized by resonance structures such as A, whereas in the case of the ketone, the analogous resonsnce structure A' contributes, but not nearly as much; so ester carbonyls are stabilized compared to ketone carbonyls.

In the enols, the ester resonance structure B is not particularly favorable due to the negative charge on carbon and positive charge on oxygen, just the reverse of what we would want based on electronegativities.  At least in the ketone case resonance structure B' has a positive charge on hydrogen, so resonance structure B' contributes more than B (this is why carbon-carbon double bond stability increases with increasing alkylation).  As a result the enol resulting from the ketone is stabilized compared to the enol resulting from the ester.

Just where the carbonyl-enol equilibrium lies will depend on the relative stabilities of the carbonyl compound and the enol.  As we've just discussed, the ester carbonyl is stabilized relative to the ketone carbonyl, this will push the equilibrium further to the carbonyl side for the ester.  Also, the enol from the ketone is somewhat more stabilized than the enol from the ester, this will push the equilibrium to the enol side for the ketone.

So both factors, starting material (the carbonyl compound) stability and product (enol) stability will work to increase the enol content for ketones and reduce the enol content for esters.


Solubility of Iodine in Some Organic Solvents The solubilities of iodine in $\ce{CCl4,n-Hex,Et2O,PhCH3}$ were given respectively as 19 g/kg, 13 g/kg, 337 g/kg, and 182 g/kg. These results puzzle me for the following reasons:

1) Iodine, being a non-polar substance, should dissolve the most in nonpolar solvents. Yet diethyl ether the best solvent for iodine.

2) Toluene and n-Hexane, both being hydrocarbons, should have similar solubilities of iodine. Yet the solubility of Iodine in Toluene almost 20 times that of n-Hexane.

How do we qualitatively predict these results?

Note: This question's source is from HBCSE INCHO 2012, a chemistry test for high school students, without fancy tools such as the Abrahams' equation.

Iodine as a halogen has the strong electron affinity, what is the best met by the free electron pair of diethylether, then by $\pi$ electron density of toluene.

Note that iodine is also well soluble in ethanol, commercially used in medical treatment of skin injuries.


How does standardising a secondary standard solution make it suitable as a standard solution? I know that a standard solution in volumetric analysis must have an accurately known solution. Primary standard solutions (e.g. anhydrous sodium carbonate) can be used straight-away as they follow the list of criteria (high level of purity, unaffected by air exposure, etc.)

I also know that secondary standard solutions (hydrochloric acid) cannot be used directly like primary standard solutions as they don't follow the criteria. But I don't understand how standardising hydrochloric acid with a primary solution (e.g. anhydrous sodium carbonate) will make it able to be used as a standard solution.

When talking about standards, there's always the dilemma: How is a standard standardized? Well, you compare it to another standard you already have, and after you consider all the sources of errors, you have a new standard.

In case of the primary standards, you can always promptly trust the label's purity information, as well as your balance's and other measuring equipment (like glassware) accuracy, and at a much lower degree the molar masses, and so on. Having that concentration "certified", anything you calculate using that as a reference will be "certified" as well. However...

If you're going to follow strict analytical rules, than you won't blindly trust those, which is why you cshould calculate the error (or many of the related statistical measures) by considering the accuracy of, like I said, the balance, the glassware, temperature effect on volume (density), etc.

One common way to do this, called the standardized uncertainty, is take the square root of the sum of the squares of each instrument/measure uncertainty divided by it's quantity. You should define a minimal formula which describes the property you want to account the error for, and go for there. Let's take, for example, the concentration of 100 mL of a 0.01 M solution of potassium hydrogen phthalate (KHP), used as a primary standard to standardize sodium hydroxide.

The concentration of the solution is $[KHP]=\frac{n}{vol}$, but we obviously didn't measure molês, so we write

$$[KHP]=\frac{\frac{m}{MM}}{vol}$$

where $m$ is the mass weighed on the balance, and $MM$ the molar mass. Now that we know what influence in your concentration, we have to use some numbers. Let's consider that:

Weighed mass: 0.20422 g, on a balance with uncertainty of ± 0.001 g (1 mg)
Volume measured: 100 mL, in a volumetric flask with uncertainty of ± 0.1 mL

This means that the uncertainty of the prepared solution is

$$\frac{u_{[KHP]}}{0.01} = \sqrt{\left(\frac{0.001}{0.20422}\right)^2 + \left(\frac{0.1}{100}\right)^2} $$

$$u_{[KHP]} \approx 5\times10^{-5}$$

Now note that I didn't take into account the purity of the salt (to make things simpler), and the error on the molar mass (we could assume it's much lower than the others). And what does all this have to do with your question?

My point is, even standards have a variations in them, and when you standardize something, like a HCl or NaOH solution, you could take that into account. The concentration of NaOH, in this case, will have a relative uncertainty higher than the concentration of KHP, because all of the uncertainty of the KHP preparation must be taken into account, as well as the titration variation. The outcome is that a standard based on  a standard of another standard (...) will eventually have a very high uncertainty, which is why we try to sitck to primary and secondary standards.

PS.: The uncertainties of the equipments/glassware are usually available in the manuals or even engraved on them. This values are usually the calibration precision, and should be transformed into uncertainty using some statistical inference, which could be that the variation follows a retangular distribution, resulting in

$$u = \frac{precision}{\sqrt3}$$

Disclaimer: There's a lot more to this than what I summarized above, starting with being able to indentify what should and shouldn't be taken into account, not to mention replicates and a lot of statistics. I don't recommend blindly following this formulae to calculate uncertainties.


Calling Diamond an element? Is it scientifically correct to call Diamond an element? Carbon has a wide range of allotropes, so would it be correct to actually just call graphite, diamond etc an element. For that matter, could we call any molecule containing only one type of atom, an element?

It might be scientifically correct but it is linguistically misleading

The sentence "diamond is an element" can be seen to be misleading when compared to the sentence "diamond is an allotrope of the element carbon". Or even "diamond consists of the element carbon".

The issue is that clear language should distinguish between the form and the composition of an element with more than one allotrope. Saying "diamond is an element" confuses the fact that diamond is made from carbon with the fact that it is made from an infinite tetrahedral array of carbon atoms. It also subtly excludes the fact that there are other allotropes.

For clarity in language a specific form of an element (an allotrope) should not be referred to as "an element" but should be referred to as one form of that element. As in "diamond is one form of the element carbon, buckminsterfullerene is another form."


Is the equation pH + pOH = 14 valid at different temperatures? $\mathrm{pH} + \mathrm{pOH} = 14$

For water at 60 degrees Celsius,

$K_\mathrm{w} = 1 \times 10^{-13} = [\ce{H+}] \times [\ce{OH-}]$

Hence, $[\ce{H+}] = 1 \times 10^{-6.5} = [\ce{OH-}]$

So $\mathrm{pH} = -\log(1\times 10^{-6.5}) = 6.5 = \mathrm{pOH}$

which gives $\mathrm{pH} + \mathrm{pOH} = 6.5+6.5 = 13$, not $14$

So is my math or chemistry wrong, or are there limitations on $\mathrm{pH} + \mathrm{pOH} = 14$?

$\mathrm{pH + pOH = 14}$

This equation only holds true at around $25~^\circ\mathrm{C}$, where the water autodissociation constant $K_\mathrm{w} = 10^{-14}$.

Mathematically, you can write

$$\begin{align}
K_\mathrm{w} = [\ce{H+}][\ce{OH-}] &= 10^{-14} & &\\
\log([\ce{H+}][\ce{OH-}]) &= \log(10^{-14}) & &\text{(logarithms of both sides)} \\
\log[\ce{H+}] + \log[\ce{OH-}] &= -14 & &(\log(ab) = \log a + \log b) \\
-\log[\ce{H+}] - \log[\ce{OH-}] &= 14 & & \\
\mathrm{pH + pOH} &= 14 & & \text{(definitions of pH and pOH)}
\end{align}$$

It's fairly clear from this that if $K_\mathrm{w}$ is $10^{-x}$ at some other temperature, then you will obtain $\mathrm{pH + pOH} = x$.


Stereoselective enolate formation with different bases: Does the addition of a base NR3 to a ketone affords the cis or the trans enolate? I'm wondering if the addition of a base of type $\ce{NR_3}$ (R: alkyl) to a ketone (e.g. 3-pentanone or 4-heptanone) results in the cis or trans-enolate.

I'd like to ask here about four cases:

$\ce{NMe_3}$ (small)
$\ce{NMeEt_2}$ (medium)
$\ce{NEt_3}$ (big)
$\ce{N(iPr)_2Et}$ (very bulky)

Maybe smaller bases as 1 or 2 usually give the trans-enolate, and bulkier bases such as 3 or even 4 affords the cis-enolate. (That's what I read, but I'm not able to understand why.) Is there a good explanation that rationalizes which enolate is formed, the cis or the trans?

Sterically hindered bases forms the the kinetic enolate. For deprotonation of 3-Methyl-4-heptanone I assume the enolate at the linear unbranched chain site, that's clear for me, but kinetic/thermodynamic enolate has nothing to do with cis/trans enolate or does it?

Related:
Does the addition of base to 3-pentanone give the cis or trans enolate?

The keto-enol tautomerization cannot take place without at least a trace of acid or base, meaning that there is no direct shift of proton from alpha-carbon to oxygen or vice versa.

The two most common enolation mechanisms are:

Base-catalyzed (with organolithium reagent such as LDA (secondary amine))

In the base-catalyzed enol formation, the first step, proton extraction from the alpha-carbon, is slow. (The subsequent enolate formation is fast (due to resonance)). The base is proposed to form a chair-like transition state (TS) with the ketone, extracting the alpha-carbon proton and at the same time donating the Li to the oxygen. The resulting product is the classical lithium enolate.

Based on the structure of TS, you can then predict whether the “side chain” of the base would form a unfavorable syn-pentane interaction.

https://en.wikipedia.org/wiki/File:Ireland_model_enolate1.svg

Acid-catalyzed

In the acid-catalyzed enol formnation, the first step, the protonation of oxygen and subsequent “charge re-localization” to the hydroxyl carbon, is fast. (The subsequent enol formation is slow). Since the protonation can occur from any free proton in the environment, the size of the acid plays little direct role in the stereochemistry of the resulting enolate. (However, the size of the “side chain” in the ketone reactant does).

However, there is a 3rd mechanism when the “base” is a tertiary amine (as you appear to be very interested in) but the reaction is acid-catalyzed. In this mechanism, the amine behaves like a nucleophile and a carbinolamine intermediate is detected.

See Bruice, J. Am. Chem. Soc. 1983, 105, 4982

http://pubs.acs.org/doi/abs/10.1021/ja00353a023

(Also see Bruice, J. Am. Chem. Soc. 1989, 111, 962 and 1990, 112, 7361)

In drawing this carbinolamine intermediate, you can predict a bulky tertiary amine to favor a cis-enolate due to steric hindrance. However, as suggested in the paper, "severely sterically hindered" tertiary amine can be too bulky to follow this 3rd mechanism and just catalyzes enolation via the general base-catalyzed mechanism.


What is the mechanism of the reaction between hexane-2,5-dione and nitrous acid? I don't understand the mechanism of the first reaction; I tried different ways but I can usually find how to make the ring without the double bonds, but it's obviously not the right thing; I think HNO2 reacts as the nitrosonium ion, bonding with one of the oxygens (since it has free electrons) but I can't never make it work.

EDIT:
Probably the acid is formed by nitric acid (like in one of the articles orthocresol linked), that I think is not written because it can be formed by the decomposition of nitrous acid; I still don't understand the mechanism of the formation of the ring.

Image: https://commons.wikimedia.org/wiki/File%3AIsocarboxazid_synthesis.png
by Nuklear (Own work) [CC BY-SA 4.0 (https://creativecommons.org/licenses/by-sa/4.0)], via Wikimedia Commons

As you may know from diazotisation reactions, $\ce{HNO2}$ is a generator of the nitrosonium cation, $\ce{NO+}$. Electrophiles like these can react with the enol tautomer of hexanedione:

Following this, you tautomerise the nitroso compound to the oxime. The mechanism is exactly analogous to keto-enol tautomerism:

Note that I've elected to draw the oxime with a (E)-configuration. Both (E) and (Z) oximes can form, but the (Z)-configured oxime cannot cyclise in the following step to form an isoxazole:

Following this, to get to the compound in your question, you supposedly need to oxidise the acetyl substituent ($\ce{COCH3}$) to a carboxyl substituent ($\ce{COOH}$). Nitrous acid isn't an oxidant, which casts doubts on whether your source is veritable (perhaps it's just a typo).

In the literature, I find that this oxidation is accomplished using nitric acid, $\ce{HNO3}$, which is a perfectly fine oxidant. In one case (J. Heterocycl. Chem. 2009, 46, 909), the diketone is first treated with nitrous acid to get the 3-acetylisoxazole, and is then treated with nitric acid to form the 3-carboxylisoxazole. In a separate case (J. Med. Chem. 2004, 47, 3642), the diketone is simply refluxed with nitric acid to form the 3-carboxylisoxazole. I suspect that this works because nitrous acid is present in the nitric acid, so the mechanistic pathway is exactly the same (a similar effect has been observed with the nitration of phenols using dilute nitric acid).

Exactly how nitric acid oxidises an acetyl group to a carboxyl group, though, I don't know (and I'm not sure if it is known).

Edit: There is, in fact, a mechanism by which nitrous acid may oxidise the acetyl group to a carboxyl; see user55119's answer.


Why is cyclopentadiene more stable than its aromatic counterpart cyclopentadienyl anion? What will be the order of stability?

The answer key says that the stability order is: cyclopentadiene > cyclopentadienyl anion > cyclopentadiene  cation

According the me it should be: cyclopentadienyl anion > cyclopentadiene > cyclopentadienyl cation
because the order of stability is aromatic > non-aromatic > anti-aromatic.

This is a nice question because it forces you to step back and think rather than fall into the obvious trap.

You are entirely correct that aromatic compounds are more stable than non aromatic compounds which are more stable than anti aromatic compounds assuming that all other things are equal. In this case, all other things are not equal because one of the species is neutral while the other two are charged. Having a positive or negative charge on carbon is not very favourable, even if the negative charge does make the species aromatic. The $\text{p}K_\text{a}$ of cyclopentadiene in DMSO is 18[1] which means the equilibrium constant for dissociation of cyclopentadiene into the cyclopentadienyl anion and a hydrogen ion (which will actually be attached to a DMSO molecule) is $10^{-18}$! In other words, cyclopentadiene is much more stable than the cyclopentadienyl anion.


How to calculate atomic number of an atom when wavelength is known? In the emission spectrum of an atom, an emission line in UV region is observed for a transition from third orbit to first orbit with the wavelength of 64.1 Angstrom. What will be the atomic number of the atom?

I tried using the formula to calculate change in energy
$$\Delta E = R_h \left(\frac{1}{n_i}+\frac{1}{n_f}\right)$$
where $\Delta E$ is change in energy, $R_h$ is the Rydberg Constant $\left(2.18 \cdot 10^{-18}\right)$, $n_i$ is the initial orbit (i.e 3rd orbit) and $n_f$ is the final orbit (i.e 1st orbit).

On finding the value of this I equated it with another formula:
$$E = -\frac{R_h Z^2}{n^2}$$
where $E$ is the energy, $R_h$ is the same as above, $Z$ is the atomic number and $n$ is the orbit.

I used this formula to find out the energy at 3rd orbit and then 1st orbit and took $Z^2$ common. Then I subtracted both the obtained values but failed to find out the value of $Z$ ...

Wikipedia (Bohr Model > Rydberg formula) tells us:

  The energy of a photon emitted by a hydrogen atom is given by the difference of two hydrogen energy levels:
  $$E=E_i-E_f=R_\mathrm{E} \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right)$$
  where $n_f$ is the final energy level, and $n_i$ is the initial energy level.

  Since the energy of a photon is
  $$E=\frac{hc}{\lambda}$$
  the wavelength of the photon given off is given by
  $$\frac{1}{\lambda}=R_\infty \left( \frac{1}{n_{f}^2} - \frac{1}{n_{i}^2} \right).$$

  This is known as the Rydberg formula, and the Rydberg constant $R$ is $R_\mathrm{E}/hc$, or $R_\mathrm{E}/2\pi$ in natural units.

Wikipedia also tells that it can be extended with $Z^2$ for hydrogen-like elements:

  The formula above can be extended for use with any hydrogen-like chemical elements with
  $$\frac{1}{\lambda_{\mathrm{vac}}} = R_\infty Z^2 \left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$
  ... where $\lambda_{\mathrm{vac}}$ is the wavelength of the light emitted in vacuum, $R_\infty$ is the Rydberg constant for this element, $Z$ is the atomic number, i.e. the number of protons in the atomic nucleus of this element and, $n_1$ and $n_2$ are integers such that $n_1 < n_2$, corresponding to the principal quantum numbers of the orbitals occupied before and after.

Now you can see, that your first equation can be derived from your second equation. You also can see, that you a) forgot the squares in your first equation and b) made a sign error.

$\lambda = \mathrm{64.1 \cdot 10^{-10}\ m}$
$R_\infty = \mathrm{10973731.568508\ m^{−1}} \left(= \frac{R_E}{h\ c} = \frac{\mathrm{2.18\ 10^{-18}\ J}}{h\ c} \right)$

With all this in mind, you should be able to solve the equation to get $Z$.

   $$Z^2 = \frac{n_f^2\ n_i^2}{\lambda\ R_\infty\ (n_f^2 - n_i^2)} \xrightarrow{Z>0} Z \approx 4$$


Calculating temperature change due to Joule-Thomson effect I have a 10 L compressed argon cylinder with filling pressure of 10 MPa. According to Perry's Chemical Engineer's Handbook, argon's J-T coefficient (unit: K/MPa) is 3.7 at 0.1 MPa, 2.6 at 10 MPa. (Temperature is 300 K.)

When I open up the cylinder valve, 10 MPa argon in the cylinder will expand into 0.1 MPa atmosphere. So the $\Delta P$ is 9.9 MPa.

  How much will the temperature change?
  The initial temperature is 300 K.

$9.9 \times 3.7 = 36.63$

(Which means the temperature of argon outside the cylinder will be 263.37 K)

$9.9 \times 2.6 = 25.74$
Something in between (like when the J-T coefficient in effect changes continuously as the gas expands).

It is going to be something in between.  The definition of the JT coefficient is $$\mu=\left(\frac{\partial T}{\partial P}\right)_H$$If the temperature-dependence can be neglected, then this gives $$\Delta T=-\int_{0.1}^{10}{\mu(P)dP}$$


Why does the quantum number n limit the value of l, and why does l, likewise, limit the value of m? Any quantum number that represents the size of an orbital is a positive integer, $n$.
Any quantum number that represents the shape of an orbital is a non-negative integer, $l : l < n$.
Any quantum number that represents the orientation of an orbital, $m$, is an integer such that $|m| \leq l$.

Why do $n$ and $l$ restrict the ranges of $l$ and $m$, respectively?

To be clearer, I ask why the quantum numbers limit each other, not what they signify, which is already explained in the related question: What do the quantum numbers actually signify?

The shapes of the orbitals, i.e. the solutions to the Schrödinger equation for the hydrogen atom $\hat{H}\Psi_i = E_i\Psi_i $, are primarily due to the requirement that all orbitals are orthogonal: $\langle\Psi_i|\Psi_j\rangle=0$.  Orthogonality is made possible by the fact that the orbitals have "positive" and "negative" regions separated by nodes.

One what to think about $n$ is that $n-1$ is the number of nodes in $\Psi_i$. So there can be no $1p$ orbitals since $p$-shaped orbitals have at least one node. Similarly, there can be no $2d$ orbitals, and so on.

Because $s$ orbitals have a spherical shape, there is only one way to orient them, i.e. $m=0$ if $l=0$. $p$-shaped orbitals lie along one axis and there are three Cartesian axes, so there are three possible - orthogonal - orientations (I'm sacrificing a bit of accuracy here for clearness).

It's a bit harder to visualize why there should be 5 and only 5 $d$-shaped orbitals, but the answer is that there are only 5 ways to construct 3-dimensional shapes with (a minimum of) 2 nodes that are orthogonal to the $s$- and $p$-shaped solution as well as the other $d$-shaped solutions.


Oxidation number and Pourbaix diagrams In a Pourbaix diagram (E-pH diagram) where the regions are given , but the assignment of these to the elements we have is not done yet , how can the oxidation numbers helps me out assigning each element in its appropriate region?

If you do not have a list of the species that belong in the Pourbaix diagram, then it is not possible to accurately predict what species occur where, unless you have extensive experience with the system in question.  If you have a list of species and need to place them in the appropriate regions of a Pourbaix diagram (fill in the blank style), then the higher oxidation state species occur at the top of the diagram and the more basic (e.g. OH containing) species occur towards the right.

If you have Latimer diagrams of all the species in the Pourbaix diagram (this will require the Latimer diagrams for both acidic and basic conditions, typically) then you can more accurately predict the structure of a Pourbaix diagram by (a) writing balanced half reactions and (b) adjusting them for the appropriate pH using the Nernst equation.


Water solubility of different types of Iron III For an experiment I require an aqueous solution of Iron(III). My school has the following compounds, which of them are water soluble? I believe that only Iron(III) oxide is insoluble, but I would like to verify.

Iron(III) oxide
Iron(III) oxalate
Iron(III) chloride-hexahydrate

Many thanks in advance.

Iron(III) oxide is insoluble.
Iron(III) oxalate is, in practical situations, insoluble.
Iron(III) chloride-hexahydrate is soluble.

Aditionally, Iron(III) oxalate is destroyed by visible light, and I intend to complete my experiment during the day, especially considering that involves colour changes.


Solving a quadratic equation for pressure of real gas with unknown volume Question

  Calculate the pressure exerted (in atm) by 1 mole of $\ce{CH4}$ gas at a temperature of $\displaystyle\pu{\frac{24}{0.0821} K}$ if volume occupied by $\ce{CH4}$ molecules is negligible. Given $a = \pu{2 L^2 atm mol^{-2}}$.

Solution:

  $$
\begin{align}
\left(p + \frac{a}{V_\mathrm{m}^2}\right)V_\mathrm{m} &= RT\\
pV_\mathrm{m} + \frac{a}{V_\mathrm{m}} &= RT\\
pV_\mathrm{m}^2 - RTV_\mathrm{m} + a &= 0\\
pV_\mathrm{m}^2 -24V_\mathrm{m} + 2 &= 0\\
\end{align}
$$

  For real gas this equation must have only one root:

  $$D = 0\quad\implies\quad b^2 - 4ac = 0$$
  $$24^2 - 4×p×2 = 0$$
  $$p = \pu{72 atm}$$

My thoughts

I had a doubt whether this solution was really true or not. When I first encountered it, I was convinced that it was true. But, now, I doubt it. I don't have any appropriate reason why and that's what I seek to ask. Is this approach correct? Why?

Assuming the Van der Waals equation with $b=0$, I agree with the solution to the equation:

$$pV_\mathrm{m}^2 - 24V_\mathrm{m} + 2 = 0\label{eqn:1}\tag{1}$$

I'll point out that I'm following the notation of the given solution, but $V_\mathrm{m}$ seems odd to me. I'd think that $V_\mathrm{m}$ would be reserved to mean the molar volume at STP.

But equation \eqref{eqn:1} has two unknowns, $p$ and $V_\mathrm{m}$. Letting $a' = p$, $b' = -24$, and $c' = 2$ then the appropriate quadratic equation is of course

\begin{alignat}
2V_\mathrm{m} &= \dfrac{-b' \pm\sqrt{b'^2 - 4a'c'}}{2a'}\tag{2}\\
V_\mathrm{m} &= \dfrac{-(-24) \pm\sqrt{(-24)^2 - 4×p×2}}{2p}\tag{3}\\
V_\mathrm{m} &= \dfrac{24 \pm\sqrt{(-24)^2 - 8p}}{2p}\tag{4}\\
V_\mathrm{m} &= \dfrac{12 \pm\sqrt{144 - 2p}}{p}\label{eqn:5}\tag{5}
\end{alignat}

So we have two unknowns and one equation. Thus the problem is unsolvable without additional information. Notice that would still be the case if the simple ideal gas equation, $PV=nRT$, had been used. The ideal gas equation could be solved only to $pV_\mathrm{m}=24$.

For equation \eqref{eqn:5} there are three cases for the sqrt term $\sqrt{144 - 2p}$:

The term is negative.

If $144 - 2p$ is negative, then both roots would be imaginary.
The term is equal to zero.

We have $\sqrt{144 - 2p} = 0\implies 144 - 2p = 0,$ so $p = 72$. Therefore, $V_\mathrm{m} = 12/72 = 0.167.$
The term is greater than zero.

Then $72 > p$ also, and mathematically $V_\mathrm{m}$ has two roots.

Now, again there are three possibilities.

(a) $\sqrt{144 - 2p} < 12$

This is impossible since $V_\mathrm{m}$ would have two valid values.

(b) $\sqrt{144 - 2p} = 12$

This is impossible since $V_\mathrm{m} = 0$ and the gas volume can't be $0$.

(c) $\sqrt{144 - 2p} > 12$

This would yield one negative value and one positive value for $V_\mathrm{m}$, which is OK for $V_\mathrm{m}$. However it also means that $p < 0,$ which is nonsensical.

Thus the only reasonable solution is if the square root term is equal to zero.

P.S. The comment by user Poutnik made me look at this again, and I now understand the book's solution…


Can you determine when the orbtials of an atom will hybridize? We know that when it comes to C its orbitals hybridize (they become sp3 , sp2 or sp) and the energy levels of an hybridized orbital of this kind is lower than that of the p orbital that formed it but higher than the s orbital so for example energetically speaking is more convenient for an atom to have its electrons (even if they are non-bonding and they don't do too much) in an hybridized orbital if possible.And I know that O and N also can hybridize so the energy barrier between their orbitals (the s and the p orbitals) is not that high to be unable to do this so why don't they just hybridize first in O2 or N2 and then form the molecular orbitals like is the case with C-C bonds , why in the case of O-O bonds we have no hybridization? I mean ok I get the perspective of being able to form a double bond but O2 can have a double bond even if it was sp2 hybridized and then all the non-bonding electrons would be in lower energy orbitals (the other sp2).

The molecule $O_2$ does not have a double bond O=O. It is a biradical which has a single bond O-O, and two independant electrons.

Anyway in diatomic molecules, it is not necessary to consider hybridization. It does not help. Hybridization is an approach that has been invented by Pauling to explain the geometry of polyatomic molecules. Hybridization is not really necessary. It allows to easily describe the shape of the molecules, but the same result may be obtained without hybridization, although it is much more complicated to get to the same result.


Is a hydrogen molecule smaller than hydrogen atom? Recently I came across a textbook of physical chemistry which stated the following:

  In the formation of a stable hydrogen molecule, two hydrogen atoms share a pair of electrons between them and two $s$-orbitals overlap along the same axis resulting in the formation of sigma $(\sigma)$ bond. As the shared electrons are attracted by both the nuclei, the size of hydrogen molecule is smaller than that of hydrogen atom.

However, the actual radius of hydrogen molecule is $120~\text{pm}$ which is greater than that of hydrogen atom i.e. $53~\text{pm}$. So, what does the author actually mean when he says that the size of hydrogen molecule is smaller than that of hydrogen atom? Moreover, does it imply that the electron density is maximum at the COM of the hydrogen molecule? Any help would be appreciated.

You can define the size of atoms and molecules in various ways. You can, for instance, derive a measure of the size of a hydrogen molecule from the density of solid hydrogen:

  Solid hydrogen has a density of 0.086 g/cm3 making it one of the lowest-density solids.

From the above you can derive an effective radius for $\ce{H2}$ of $\pu{2.10 Å}$, which is eerily similar to $4 \times \pu{0.53 Å}= \pu{2.12 Å}$.

However this is a distraction from the main point of the statement in the book, which is simply that concentration of electron density between the two $\ce{H}$ nuclei results in an "end-to-end" distance (the distance, using some measure to determine the edges of the molecule, along the internuclear axis) that is less$^\dagger$ than $\pu{2.12 Å}$, and that the internuclear distance is less than $\pu{2\times\pu{0.53 Å}= 1.06 Å}$ (which it is: the H-H bond length is $\pu{0.74 Å}$). An internuclear distance shorter than twice the lone H radius is consistent with bonding, a process by which electrons and nuclei are allowed to occupy lower-energy regions of the potential than they would when the atoms are far appart.

But the answer to the question is no, a hydrogen molecule is still bigger than a hydrogen atom.

$\dagger$ But just barely. Define the edge of an atom or molecule as the point where the electron density equals that at a distance of $\pu{1 Bohr}$ ($\pu{0.529 Å}$) from the nucleus in a hydrogen atom. Next find the edges in a simple  MO model of $\ce{H2}$ ([3]). The edges are $\pu{1.056 Å}$ from the center of the molecule, that is, the molecule along the longest axis has a diameter of $\pu{2.11 Å}$, barely less than $4\times$ the Bohr radius.

[3]: Pearson, Ralph G., and Palke, William E.. Proc.Natl.Acad.Sci.USA, Vol. 77, No.4, pp.1725-1727, April 1980.


Can you call a table frozen? Someone has argued that the table sitting in front of us is 'frozen'. I do agree that the table has a freezing point. The table has a melting point, but that is beyond the ignition point and the boiling point is even further beyond that. If an object only exists in one state can you call it frozen? Secondly, I think that there is some disconnect because I was arguing more from a sociological stand point. I think that the definition is more subjective. If I call a wooden bowl frozen, it would hinder the meaning of the sentence from an outsiders stand point. What do you think? Can you say that the table in front of you is frozen?

The technical definition of 'freezing' implies a phase transition from liquid to solid, and therefore one assumes your interpretation requires that a table can at some point be a liquid.

Under normal circumstances, being a solid does not necessarily imply something is frozen, because not all materials can exist as liquid. Heating some materials will simply lead to (i) phase transition through sublimation or (ii) combustion or decomposition (a chemical reaction) leading to a both gaseous and solid products.

Presumably, the table you are pondering is made of wood. Wood is a mixture, composed of mostly cellulose, hemicellulose and lignin. The ratios depend on the wood type, and there are a few other bits in there also. Importantly, wood, even tables, still contain water. Under standard conditions one might expect to find a table (normal pressures), simply heating will not cause your table to melt. Above about 100oC, decomposition products are slowly released from the bulk wood; noncombustible carbon dioxide, water vapour and trace organics. Cellulose decomposes above about 250oC, and hemicellulose somewhat below that. Lignin begins to decompose above about 150oC, but has a large decomposition range.

Now, if you could heat a table under immense pressure, you may be able to influence the decomposition rates to a point where you could possibly get these materials to melt rather than simply decompose. Heating wood under moist high pressure conditions is known to impart an element of plasticity to some woods, so that it can be bent. I'm not confident that you could melt it though.

So I would call your table simply solid, not frozen.


Why does two nails of the same material not produce electricity when placed in saltwater? If I put two the same nails in a saltwater why could it not produce electricity when I put the metal aluminum and copper it produces electricity?

First, you need to understand what is electricity. Electricity is simply the flow of electrons. Now, what I really suspect you are asking is why these electrons flow.

Electrons flow in an electric current due to a potential difference which is the difference in potential energy of the electron at two different places. Like everything else, electrons like to lower their potential energy (imagine water falling in a waterfall) hence in an electric current, electrons move because they are moving from a place of high potential energy to a place of lower potential energy.

Now when you place two of the same nails in a salt solution, there is no potential difference between the two nails as the potential energy of the electron moving from the two nails remain the same (since they are made of the same material).

However when you place two different metals into salt water, such as aluminium and copper, there is now a potential difference between the two metals. Hence this will cause the electrons to move creating electricity (an electric current).

For more detail about the voltage change and what the chemical reactions that are actually occurring and you google electrochemical cells and electrochemical series. Hyperphysics and Chemguide are both very good resources which provide in-depth explanation on what is actually going on.


Metal reagent in dehalogenation of vicinal dihalides In the dehalogenation of vicinal dihalides why can't sodium be taken as a reagent?

I understand that zinc is taken because it is reactive and rate of reaction would be considerably high and it doesn't form complexes. Because of this many other metals can't be used including magnesium as it forms complexes. But I don't understand why sodium can't be taken.

You mean something like the following right
$$
\mathrm{R-X ~+~ Na \longrightarrow R\cdot ~+ ~NaX}
$$
and we add a proton source like an alcohole to transfer an H atom to $\mathrm{R\cdot}$:
$$
\mathrm{R\cdot~ + ~R'-H} \longrightarrow \mathrm{R-H}
$$
Well, its in principle possible. But a) you don't want an explosion right?, so be careful as this will probably be very exothermic b) it will probably be really messy as side products like
$$
2~\mathrm{R\cdot} \longrightarrow \mathrm{R-R}
$$
can be formed. Also the whole approach is not really tolerant to pretty much any functional group and will just tear the substrate apart in many cases. The other alternatives (e.g. Zinc, $\mathrm{SmI_2}$) are just much milder, easier to use and much more versatile. But sure, try it out have fun (less than 0.5 g sodium should still be relatively harmless) ; )


Nuclear Testing and Carbon Dating Will the nuclear testing done during the past decades or if a nuclear war happened in the present day, affect the accuracy of carbon dating in the future? (Hypothetically, we haven't developed a better dating method)

I mean you could just adjust the dating to the amount of Carbon-14 if there will be a sudden change, but will the fossils found in around and near the blast sites be dated differently from the unaffected areas?

I'm sorry not really good in science, I don't know if this should be in Physics STEX and I might be wrong in my understandings. Thanks in advance!

If the amount of extra C14 was unknown this might become a problem. If the change in atmospheric C14 was recorded and the records survive into the future, it would be easy to adjust the dates. At present we have good data on C14 concentration over time for various locations around the world. I don't know the spatial resolution though.


Boiling point trend in group 13 My Theory:
Since atomic mass increases down the group, the van der Waal's forces should also operate to a greater extent, thereby making it difficult to change the phase of the substance. Hence, boiling point should increase down the group from boron to thallium.

But in my book, the answer given is that the forces of attraction decrease down the group because of the increase in size of atoms down the group. Hence, the boiling point decreases down the group.

I'm confused. Even though the size of the atom increases, shouldn't the boiling point follow the same trend of increasing because atomic mass increases?
 Moreover, due to poor shielding of d-electrons the increase in size down the group is very minimal.

Experimental data: (source) shows a regular decrease in boiling point down the group

$$
\begin{array}{|c|c|}\hline
\text{Element}&\text{Boiling point (K)}\\\hline
\ce{B}&\pu{4200K}\\\hline
\ce{Al}&\pu{2743K}\\\hline
\ce{Ga}&\pu{2673K}\\\hline
\ce{In}&\pu{2345K}\\\hline
\ce{Th}&\pu{1746K}\\\hline
\end{array}
$$

EDIT:
Can we explain the above trend with metallic bonding present in these elements? Boron has its highly networked structure so let us leave it alone. As we move down the group, size of the element increases. This increase in size, decreases the strength of metallic bonding as the same number of valence electrons now have to hold a larger cation. Metallic bonding is viable in liquid state too. So could this be the reason?

London forces are the forces that you should always consider in the end. They're like the slow pawns on the chessboard, their effect is negligible when the fast rooks and the industrious queens are around!

In this case, our queens are the different structures of these elements. Let me first tell you the fact that even the melting point of boron ($\pu{2349K}$) is more than the boiling points of thallium and indium!

So, first off, why is the melting point of boron higher than that of all other group thirteen metals? From this Wikipedia page, its α-rhombohedral allotrope has B-12 icosahedral units arranged in a crystalline structure, and connected by strong covalent bonds[a]. As a comparison, the other metals in their solid state have unit cells whose atoms are bonded by only weak metallic bonding. The larger the atoms get, the weaker the metallic bonding gets, and hence, the melting point decreases down the group.[b]

Coming to the trend in boiling points now. Part of the difference in the trend has to arise from the excessively large melting point of boron. You see, thallium boils off even when boron has yet to melt, so that is a contributing factor in the excessively large boiling point of boron over the others.

According to this paper[1], for liquid thallium,

  The coordination number changes from 11.5 atoms at $\pu{350^\circ C}$ to 10.9 atoms at $\pu{700^\circ C}$

This is said to be due to:

  creation of vacant spaces about an atom. Although in liquid thallium the interatomic distances remain unchanged, some of the atoms are moving farther apart within the first coordination shell.

Another paper[2] however shows that, while CN of liquid indium decreases with increasing temperature, their inter-atomic separation also decreases. We can interpret this to be a "reinforcement" that helps increase the boiling point of indium over thallium.

With indium and thallium cleared, notice that the boiling points of indium, gallium, and aluminum are almost identical. Sadly, I've been unable to find any research papers dealing with these three elements in liquid state near their boiling point. For example, this paper[3] deals with liquid aluminium, but only upto $\pu{1273K}$.

We do have some data for boron. You'll be surprised how strong its covalent bonds are. In fact, this paper[4] states that, even at as absurdly high temperatures as $\pu{2600K}$, though the "icosahedra are destroyed", the "atoms still form an open packing with sixfold coordination". Moreover, this paper[5] observed that the density of liquid boron, at temperatures above $\pu{2360K}$, is higher than that of solid boron, which is unlike aluminum and indium (though for gallium the density at melting point is still slightly higher). Another paper[6] has measured the properties of liquid boron, but that sadly only at around $\pu{2400K}$.

I have checked a dozen best-matching research papers, and this was all the relevant data I could pull out. This isn't the complete answer, but again, without any proper data in the first place, there really cannot be a proper qualitative analysis either. So, this answer is really all there is.

References:

Temperature Dependence of the Structure and Transport Properties of Liquid Thallium, N. C. Halder and C. N. J. Wagner, The Journal of Chemical Physics 45, 482 (1966); https://doi.org/10.1063/1.1727593
Temperature Dependence of the Structure of Liquid Indium, H. Ocken and C. N. J. Wagner, Phys. Rev. 149, 122 – Published 9 September 1966
The structures of liquid mercury and liquid aluminum, P. J. Black and J. A. Cundall, Acta Cryst. (1965). 19, 807-814 https://doi.org/10.1107/S0365110X65004383
Structural and electronic properties of liquid boron from a molecular-dynamics simulation; N. Vast, S. Bernard, and G. Zerah; Phys. Rev. B 52, 4123 – Published 1 August 1995
Millot, F., Rifflet, J.C., Sarou-Kanian, V. et al. International Journal of Thermophysics (2002) 23: 1185. https://doi.org/10.1023/A:1019836102776
Structure of Liquid Boron; S. Krishnan, S. Ansell, J. J. Felten, K. J. Volin, and D. L. Price; Phys. Rev. Lett. 81, 586 – Published 20 July 1998; https://doi.org/10.1103/PhysRevLett.81.586

[a]: actually "the icosahedra are connected via a combination of two-center and three-center bonds" as claimed in [6], but suffice to say that they are almost as strong as the covalent bonds, if not more
[b]: except a few exceptions


Is there a way of finding solubility/solvent from basic information? Let's say I have a compound and I want to find a solvent for it.
Looking on some where like Wikipedia, it is quite easy to find solubility, in various solvents for common compounds. However it can be quite lacking for less well know compounds (For example Glycin).
Although one can quite often find a list of some basic metrics (Polarizability,  log D, etc) for the compound. Say on:
Chemspider
Given those basic metrics is it possible to infer what some likely solvents would be? Or What would be the name of those techniques?
*Assuming normal temperature and pressure etc

This is a broad question as solubility equally depends e.g., on the temperature of the solvent(s), presence of other (already dissolved) compounds. A database like Reaxys used by research institutes may indicate the publications with these (measured) values, or contain these values itself. Based on known data, programs may be trained to anticipate solubility of unknowns based on a structure drawn. E.g., for X in water background of a relevant program, Marvin, in function of the pH value, or an implementation of ESOL.


Chemical equilibrium for simultaneous dissociation reactions While calculating the adiabatic flame temperature for the following combustion reaction:

$$\ce{\Phi\, C2H6 + $3.5$\,(O2 + $3.76$\,N2) -> $a$\,CO2 + $b$\,CO + $d$\,H2O + $e$\,H2 + $f$\,O2 + $g$\,N2}$$

Where $\Phi$ is the equivalence ratio.

To determine the product composition ratios $a$, $b$, $d$, $e$, $f$ I need to make mass balance on $\ce{C}$, $\ce{O}$ and $\ce{H}$ and then use two $K_{p}$ (Chemical equilibrium constant) relations. Then solve this system of equation simultaneously.

My question is about the elementary reactions of dissociation I should use to apply $K_{p}$ equations, As I could have the following scenarios happening (regarding dissociation):
\begin{align}
\ce{CO + H2O &<=> CO2 + H2}\\
\ce{CO2      &<=> CO + $0.5$\,O2}\\
\ce{H2O      &<=> H2 + $0.5$\,O2}\\
\end{align}

I only need two equations for $K_{p}$ which two should I use and why?

Short answer (After some trials): It does not matter which dissociation reaction you use, The system of equation will be solved anyway since the reactions are not independent.

However, using the first reaction as a dissociation mechanism would generate a Kp equation that's a lot easier to solve than the others.


Is light a reactant in photochemical reactions? According to IUPAC a photochemical reaction is a reaction caused by absorption of light. I still can't understand how we should consider light as a part of a reaction. Is a catalyst or a reactant?
In some reactions there is the $hν$ notation above the reaction arrow. For example we may have the following reaction:
$$\ce{A ->[$hν$] B}$$
I want to understand how light affect reactions and chemical equilibriums. I read somewhere about photoisomerization and that favors the less thermodynamic stable product and that make me think of how is it possible for the value of equilbrium constant to change (considering interoconversion as a "reaction").
Light in photoabsorption
When a molecule absorbs a photon it moves to an excited state. This process is associated with a probability factor, lets call it $P$. We can write such a transition as:
$$\ce{A + $hν$ -> A^*}$$
Suppose we have a box with molecules $\ce{A}$ that initially is a closed system. That is photons can enter the box. Now we use a laser and pump photons of specific wavelength into the box and after that we isolate the system, that is no energy can go in or out of the system. Does the above arrow $\ce{->}$ implies a reaction? I was thinking that if $P = 0.2$ then after the light enters the box we will have $[\ce{A}^*] = 0.2$ and $[\ce{A}] = 0.8,$ so the equilibrium constant will be $\displaystyle K = \frac{0.2}{0.8} = 0.25$ (I have omitted concentration units for simplicity).
But this doesn't make sense because first a new thermal equilibrium will be reached and the relative concentrations (populations) will be in accordance with Boltzmann distribution. Secondly if it is indeed a reaction then we must include light also. Also, the concentration $[\ce{A^*}]$ should depend on the incident light intensity.
So is photon absorption just a physical process and therefore the chemical equilibrium concept doesn't apply? I thought that if photoabsorption reaches an equilibrium such that:
$$\ce{A <=> A^*}$$
then for interconversion from cis to trans of a compound $\ce{A}:$
$$\ce{A_\textit{cis} <=> A_\textit{trans}}$$
by populating the excited state of the one isomer the equilibrium position should change. But still this doesn't make sense because Gibbs free energy change of the reaction take into account both ground and excited states of both products and reactants.
Light in chemical reactions
First the notation $hν$ can't be thought to be a catalyst as it doesn't make sense because isn't regenerated. For example, the chlorination of methane to yield chloromethane
$$\ce{CH4 + Cl2 ->[$hν$] CH3Cl +  HCl}$$
uses light to initiate the reaction. But it isn't regenerated in any other step so it shouldn't be a catalyst. Then how one should think about light in a chemical reaction? It is a reactant? In other words could we write the chlorination of methane in the following way?
$$\ce{CH4 + Cl2 + $hν$ -> CH3Cl +  HCl}$$
If that is the case can we find an equilibrium constant that includes the concentration of photons? Because for every reaction there must be a corresponding equilibrium constant according to thermodynamics.
I am asking the above because as I said I read about photoisomerization and I couldn't understand how is it possible to favor a thermodynamically less stable product via radiation.

It is neither reactant nor catalyst, and equilibrium concepts do not apply
There are processes other than photochemistry that to go in a direction away from equilibrium. They require mechanical or electrical work, and there is no established way to incorporate these into a chemical equation.
If we write a conceptual equation for charging a battery, it could look like this:
$$\text{empty battery} \ce{->} \text{charged battery}$$
The external power source is neither a reactant nor a catalyst; it is work done on the system, which serves to move the system away from chemical equilibrium.
If we write a conceptual equation for a refrigerator (or more generally, a heat pump), it could look like this:
$$\text{warm body + warm body} \ce{->} \text{cold body + hot body}$$
Again, the mechanical work done by the compressor is neither a reactant nor a catalyst, and it serves to move the system away from thermal equilibrium.
For both processes, you could not say that in the presence of work, an equilibrium is reached. It is just the opposite - we are moving away from the equilibrium.

[OP] So is photon absorption just a physical process and therefore the chemical equilibrium concept doesn't apply?

Photochemical reactions can move a reaction away from equilibrium (as in the example of cis/trans isomerization the OP mentions). This makes it a non-equilibrium process, so equilibrium concepts need to be expanded. If you want to describe the situation under certain irradiation conditions, you might use the term photostationary cis:trans ratio (see section on stilbene in https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/photchem.htm).

[OP] I read somewhere about photoisomerization and that favors the less thermodynamic stable product and that make me think of how is it possible for the value of equilbrium constant to change (considering interoconversion as a "reaction").

The equilibrium constant does not change. The reaction just does not go to equilibrium (or in fact moves away from equilibrium).

[OP] First the notation hν can't be thought to be a catalyst as it doesn't make sense because isn't regenerated.

Things written on top of the reaction arrow are not necessarily catalysts. In general, that spot is used for reaction conditions, such as solvent, high temperature, or "reflux".

[OP] Then how one should think about light in a chemical reaction? It is a reactant?

Some textbooks write "heat" as reactant or product, which already does not make sense (but is used in combination with Le Chatelier principle to memorize how equilibrium constant change with temperature). For photochemical reactions, however, there is a stoichiometric aspect to the role of photons; for every reaction, one photon has to be captured.
A comprehensive published treatment
The abstract of this paper (Perspective – life and death of a photon: an intuitive non-equilibrium thermodynamic distinction between photochemistry and thermochemistry) addresses some of the misconceptions for photochemical reactions. It is a bit long, so I am breaking it up into smaller parts.
First, it states that photons are not chemical reactants in photochemical reactions:

Neither the thermodynamically determined probability isotherm nor its kinetically manifest rate isotherm can be applied to photo-absorptive reactions such that the participants, including photons, may be treated as if they were chemical reactants. Photons and chemical reactants differ from each other fundamentally: firstly, a photon's energy is absolute and, in all instances of practical relevance to the present paper, independent of its surrounding electrochemical field, while the energy of a chemical reactant is relative and defined by its surrounding field; secondly, while both photons and chemical reactants can and do engage in entropy creation, only chemical reactants can engage in entropy exchange.

It then goes on to address the mistake of treating photochemical processes using equilibrium concepts:

Clarification of these differences requires identification and abandonment of fundamental historical errors in photochemical thought deriving from inappropriate overreach of analogies drawn between light and ideal gases, and including: treatment of photo-absorption as a reversible chemical reaction; attribution to light of thermal potential, or temperature (as distinct from the idealised abstraction of a ‘temperature signature’); attribution to light of exchangeable entropy content.

Then, it addresses how entropy plays a role in these processes:

We begin by addressing widespread misapprehensions concerning the perennially misunderstood concept of entropy and the frequently overlooked distinction between entropy creation and entropy exchange. Armed with these clarifications, we arrive at a useful perspective for understanding energy absorption and transfer in photosynthetic processes which, through the chemical ‘kidnapping’ of metastable excited states within structured metabolic pathways, achieves outcomes which the Second Law denies to thermal chemical reactions


Chemical in (s) mixed with chemical in (aq) vs. chemical in (aq) mixed with chemical in (aq) Hopefully the title is comprehensive enough.

But I am wondering, does it make a difference if you take a powder and mix it with an aqueous solution or take a solution with the very same chemical formula as the powder and mix it with the same aqueous solution?

Short answer is yes. A lot of times you are trying to control the exothermicity of a reaction by doing a slow addition. This is kind of hard with a powder, and if you dissolve it in a lot of solvent, you can add it slower than you would be able to with a powder.

Without the slow addition, the temperature may increase and decrease your yield or selectivity.


Could an ester react with magnesium oxide? Esters undergo saponification in the presence of hydroxides. Assuming anhydrous conditions, would they react with magnesium oxide?

Magnesium oxide can be an effective catalyst in transesterification reaction of vegetable oil samples mostly containing high carbon esters:

$\ce{MgO}$ is used as catalyst to increase the rate of transesterification reaction of sunflower oil in supercritical methanol.
The reaction in case of soybean oil has also been reported. This method was developed using metal-chitosan complex leading to higher yield of fatty acids and ethyl ester (biodiesel).
Sometimes alkaline supported catalyst like $\ce{MgO-K2CO3}$ ($\ce{K2CO3}$ fused $\ce{MgO}$) has also been used in the transesterification reaction of canola oil.

Another reaction using magnesium oxide as catalyst is the carbon-acylation of malonates with either acyl chlorides or chloroformates. Various malonic esters (methyl, ethyl, isopropyl and isobutyl) were easily alkoxycarbonylated to give the corresponding methanetricarboxylic esters.

References (in the order of mentioned points)

A. Demirbas, "Biodiesel from Vegetable Oils with MgO Catalytic Transesterification in Supercritical Methanol" Pages 1645-1651, 22 Jul 2008 DOI: https://doi.org/10.1080/15567030701268401
Gizelle I. Almerindoa, Luiz F., D.Probsta Carlos E. ,M. Campos Rusiene , M.de Almeidac , Simoni M. P.Meneghettic, Mario R.Meneghettic , Jean-Marc Clacens, Humberto V.Fajardoe "Magnesium oxide prepared via metal–chitosan complexation method: Application as catalyst for transesterification of soybean oil and catalyst deactivation studies" Journal of Power Sources Volume 196, Issue 19, 1 October 2011, Pages 8057-8063 DOI: https://doi.org/10.1016/j.jpowsour.2011.05.030
http://www.asianjournalofchemistry.co.in/User/ViewFreeArticle.aspx?ArticleID=28_2_6
Jacek Skarżewski "Carbon-acylations in the presence of magnesium oxide. A simple synthesis of methanetricarboxylic esters"Tetrahedron Volume 45, Issue 14, 1989, Pages 4593-4598 DOI: https://doi.org/10.1016/S0040-4020(01)89094-3


Does pure HCl (without water) is an acid? Hydrochloric acid is always discussed as dissolved in water, and acidity is measured in terms of pH by the concentration of the hydronium ion.
Is pure HCl still an acid, and if so, how can we measure its acidity?

If we look at the classic case of ammonia and hydrogen chloride in the gas phase, the answer is yes -- but it's more complicated than one might think.  The salt does form, but it requires either a proton-transfer medium such as water vapor or the formation of particles out of the gas phase to actually get the ions.
The acid-base reaction is formally,
$$\ce{HCl + NH3->(NH4)^+Cl^-}$$
but according to an an ab initio calculation by Cazar et al. [1] a single pair of molecules do not actually form the indicated ions.  They form instead a hydrogen-bonded species in which the would-be transferred proton remains primarily bonded to the chlorine atom.  Only with the introduction of water molecules do we find the formation of ammonium and chloride ions in the gas phase.
An empirical study by Countess et al. [2] does show the formation of ammonium chloride salt as condensed particles in anhydrous nitrogen.  They give a rate constant in terms of molecules per unit volume for formation of the monomeric gas-phase species, which is polar and attracts other monomers or reactant molecules to nucleate the particles.  It is during this polymerization phase where the ions are definitively formed; Chaban et al. [3] compute that even two ammonium chloride entities are sufficient to form an ionic structure.
References

Robert A. Cazar, Alan J. Jamka, and Fu-Ming Tao,
"Ab Initio Investigation of Proton Transfer in Ammonia−Hydrogen Chloride and the Effect of Water Molecules in the Gas Phase", J. Phys. Chem. A., 1998, 102, 26, 5117–5123.

Richard J. Countess and Julian Heicklen, "Kinetics of particle growth. II. Kinetics of the reaction of ammonia with hydrogen chloride and the growth of pariculate ammonium chloride", J. Phys. Chem. ,1973, 77, 4, 444–447.

Galina M. Chaban, R. Benny Gerber, and Kenneth C. Janda, "Transition from Hydrogen Bonding to Ionization in (HCl)n(NH3)n and (HCl)n(H2O)n Clusters:  Consequences for Anharmonic Vibrational Spectroscopy", J. Phys. Chem. A 2001, 105, 36, 8323–8332.


Hydrolysis of salts containing amphiprotic anion I am trying to understand the derivation of pH of salt containing amphiprotic anions, but I am facing some difficulty:

Ionisation:

$\ce{HCO3^- + H2O <=> CO3^2- + H3O+};\quad K_\mathrm{a_2} $

Hydrolysis:

$\displaystyle\ce{HCO3- + H2O <=> H2CO3 +OH-};\quad \frac{K_\mathrm{w}}{K_\mathrm{a_1}}$

Taking the assumption: $\text{Degree of ionisation} = \text{degree of hydrolysis}$, or, $[\ce{CO3^2-}]=[\ce{H2CO3}]$, an approximate calculation of pH can be done by the relation:
$$\mathrm{pH}(\ce{HCO3-})=\frac{\mathrm{p}K_\mathrm{a_1}+\mathrm{p}K_\mathrm{a_2}}{2}$$

I don't understand the assumption $[\ce{CO3^2-}]=[\ce{H2CO3}]$. We know that $[\ce{CO3^2-}]=[\ce{H3O+}]$ and $[\ce{OH-}]=[\ce{H2CO3}]$. But if $[\ce{CO3^2-}]=[\ce{H2CO3}]$ then $[\ce{H3O+}]=[\ce{OH-}]$. Hence pH should be 7. I don't know where my logic is going wrong and it would be great if someone could help me clarify the derivation.

If the degree of ionisation is equal to the degree of hydrolysis, then (assuming you take the same initial molar concentration of [HCO3-]):

Let's say the initial concentration was 'c' (with no [H2CO3] and no [CO32-]), the degree of dissociation to be α and the degree of hydrolysis to be η

At equilibrium:

[CO32-] = c (1 - α)

[H2CO3] = c (1 - η)

Since we know α = η, we can say that [CO32-] = [H2CO3]

Only because you made the assumption that degree of dissociation = degree of ionisation, do you get pH = 7.

But, since we assume water to always be in excess, even in a real-world situation, difference in pH (from 7) would not be significant.


Spontaneous inequilibrium in a gas in an isolated system? Can an isolated system(of several phases) in equilibrium by itself move out of equilibrium?
if a system contains for example an inflammable gas dispersed with another combustible gas such that the molecular speed are insufficient to cause the reaction. but as the molecular speeds keep fluctuating, at some point, it may cross the energy barrier and may start the reaction which will eventually spread evenly but till then the temperature there would be high and thus the system in inequilibrium.
isn't this also an example that although the free energy is minimum and entropy maximum, some useful work would've been done if the temperature gradient had been utilised?

The short answer to the question "Can an isolated system(of several phases) in equilibrium by itself move out of equilibrium?" is NO.  Once the system is in equilibrium it will remain there.

If it is possible, as is asked in the second part of the question, to have some reaction be triggered at a higher temperature reached by fluctuation, then as has been commented, the answer is that the system was never in equilibrium.

We should never forget that there is no "absolute" starting temperature for a reaction.  If the reaction is physically possible, it will take place, but the rate of that reaction can be horribly slow.


How do I cool my water bottle immediately for an emergency use? What are the chemicals I can use to cool my water-bottle at once if I'm stranded in a desert and I have only chemicals to cool water?

(I'm keeping those chemicals outside my water bottle and not inside it.)

Ammonium nitrate, calcium ammonium nitrate, and urea are each used in instant-cold packs. The dissolution of any of these compounds in water is an endothermic process and will thus cool your container of water.

In a practical sense, it doesn't seem likely that you would sacrifice some of your water so the rest of it could be cold in an emergency situation. Of course, the water you use for cooling does not have to be clean and potable.


Will group 2 chlorides dissolve in water in water or react to produce distinct products? How does this differ by the type of salt? I thought it'd dissolve... but am i wrong? I added calcium chloride to water, and it was smoking, i was so confused. Can someone explain happened? I am doing an experiment where i dissolve salts in  water, but im very confused now.
PS - my other salts are barium chloride, strontium chloride and magnesium chloride. Are these safe to use?

Have you used anhydrous $\ce{CaCl2}$, dihydrate $\ce{CaCl2 . 2 H2O}$  or hexahydrate $\ce{CaCl2 . 6 H2O}$ ( 3 most common forms) ? It is a big difference in the resulting thermal effect.
Hexahydrate causes cooling down of the solution while being dissolved. If ice is used instead of water,  as the mixture hexahydrate : ice  2 : 1, it forms the famous freezing mixture, cooling itself down to $\pu{ -50 ^{\circ}C}$.
Anhydrous calcium chloride causes heating up while being dissolved. It is because of releasing energy of ion hydration. Rather then smoking, it was condensed water vapor.
Dihydrate is somewhere between, with roughly neutral thermal effect.
Interaction of the group 2 element chlorides with water is generally safe. There is very slight hydrolysis, growing progressively upwards along the group. The beryllium chloride hydrolysis is significant, releasing hydrogen chloride. Note that barium and especially beryllium are poisonous.


How to compare stability of following two alkenes How to find which alkene is more stable ?. As same number of hydrogens are added across the double bond so I am not able to use the fact that more substituted alkene is more stable.

I was initially thinking in terms of steric effect (Repulsion Between the two methyl groups) which led me to believe the second alkene should be less stable but in answer given the second alkene is shown more stable .

Can somebody explain on what basis this result is deduced. Any help is appreciated.

The second alkene is more stable since it is symmetric. Achiral compounds are always more stable than chiral compounds due to the Gauche Effect. The explanation for why the Gauche Effect takes place is here.


When to permute two times to find optical configuration I'm learning about S and R configuration, and I know that in case where the 4th substituant is not behind we should permute it with the one which is behind. then we find the new configuration and use the opposit, if we find it R then the original configuration is S.

My question concerns the case when should I make 2 permutations ? Is that even possible ? because this will give back the original configuration but I dont know what are the cases which I use only one permutation and what are the cases where I apply more than one permutation, this seems very confusing.

If you switch any two substituents, that will change the stereochemistry. If you do a second switch (any two), you are back to the original stereochemistry. You can try it out with substituent 4 in the back, and seeing what switching around 1, 2, and 3 do to the clockwise or counterclockwise orientation (i.e. whether it is R or S).

Two switches (for example switching 2 and 4 and then switching 3 and 2, see animated GIF) amounts to a rotation around the bond of the "unswitched substituent (here substituent 1). This does not change the stereochemistry.

If you are spatially gifted, you can just do the rotation in your head to get substituent 4 in the back. If not, you can do one switch (realizing you just interchanged R and S) or two switches (R stays R, and S stays S). It is personal preference (oops, I think I just said the question should be closed because the answer is opinion-based).

If you want to try it with 3D models yourself, check out http://proteopedia.org/wiki/index.php/R/S_nomenclature.


Compound formed from concentrated sulfuric acid and sulfur In the reaction below, I know what A and B are, but I'm not sure what compound C is.

All I know about it is the following:

Deep yellow solid
Dication contains only sulfur
Raman and infra-red spectra of the dication show no common features

I think that the dication might be $\ce{[S4]^{2+}}$ since I think that this would be square planar and therefore have $D_{\mathrm{4h}}$ symmetry, which has no irreps that are both Raman and IR active, but I don't know what the anion would be.

The other likely options for the sulfur dication would be $\ce{[S8]^{2+}}$ or $\ce{[S19]^{2+}}$ but I think the fact that it's yellow would make it $\ce{[S4]^{2+}}$.

The reaction is given here.

$$\ce{S + 2H2SO4 → 3SO2 + 2H2O}$$

Sulfur react with sulfuric acid to produce sulfur dioxide and water. Sulfuric acid should be concentrated solution. The reaction takes place in a boiling solution.

$\ce{[S4]^2+}$, $\ce{[S8]^2+}$ or $\ce{[S19]^2+}$ is formed when sulfur is dissolved in oleum. The following is a paraphrase of an inorganic textbook.


Placement of Electrodes for Electrochemistry? I'm kind of curious as to the relationship between resistance and the yield of electrochemical methods. If I have my electrodes rather close together in a conductive solution the resistance will be smaller than if I had them further apart and more current can flow. Would there be any penalty for having the electrodes close together like this? Will I miss some ions in solution, or are they distributed equally in solution and so will always move towards the electrodes as some will be taken out of the solution as gases (such as electrolysis of (H2O -> 2H2 + O2))?

It really depends on the process and how
homogeneous electric filed is needed. The more are electrodes close together, the inhomogeneity will be greater.

The second problem will be with the homogenity of the solution. You can assume, that electrolyte is kinda homogeneous in a bulk. But more close you get to the electrode, the more is the electrolyte separated.

Ions, as a charged particles, are transferred to near electrode surface.


Why must polymers have a repeating unit? In organic chemistry, we learned that small molecules can form a polymer via a process called polymerization. For example, $\ce{CH2=CH-Br}$ molecules can form the polymer

\begin{align}
\ce{nCH2=CH-Br-> -[-CH2 -}&\ce{CH -]_n -}.\\
&\;\ce{|}\\
&\ce{Br}
\end{align}

So the polymer is a periodic chain $\ce{-CH2-CHBr-CH2-CHBr-\cdots}$. But since every monomer can have two orientations ($\ce{-CH2-CHBr -}$ or $\ce{-CHBr-CH2 -}\!$), there is no requirement that all monomers must be in the same orientation, does the polymer have to be a peroidic chain? Can it be a random chain with a structure that looks like

$$\ce{-CH2-CHBr-CHBr-CH2-CHBr-CH2-CH2-CHBr-CH2-\cdots}\,?$$

Most textbooks emphasize that $n$ is random, but still assume that the unit repeats.

Entropy would favour a random orientation of each monomer unit when building up the chain, however whenever you have two different substituents at the ethylene unit, then electrostatics and sterical hindrance will make one side be preferred to make the connection to the growing chain end.

Polymers that do grow via chain growth therefore have a regular structure, but there can be defects. The defect density depends on how strong the preference for one side is during the polymerisation reaction. You can have as little as one in ten thousand. That's actually a bit of a problem, because the regularity often allows polymers to crystallise, which is important for the technical application. You want the number of defects always to be the same, because too high crystalinity makes your material brittle, too low makes it soft.

Polymers that make step growth (any two molecules randomly connect with their end groups, until you have a long chain) could be different, but also there often a regular structure occurs, because of A-B + A-B , either the A+B like to react, or A+A and B+B.


Which will be the most acidic hydrogen in the following organic compounds? Which will be the most acidic hydrogen in both cases? Please explain.

According to me, in the first compound 2 should be most acidic as in both 1 and 2, resonance occurs but 2’s carbon is closer to the oxygen, which can stabilize the negative charge on carbon.

In the second compound, 1 should be the most acidic as there is $+I$ at 2 by carbon 3, which destabilizes the negative charge on it after removing hydrogen from carbon 2.

Am I correct?

Resonance effect is a more dominating effect than inductive effect. Here I think in both the structures hydrogen attached to carbon 1 will be more acidic. The reason is the negative charge formed on the carbon after removal of hydrogen is resonance stabilised to a greater extent.

In the first structure though after removal of hydrogen from the second carbon, the negative charge formed will be resonance stabilised, but its delocalisation is less compared to the delocalisation of the negative charged which would have been formed from the removal of hydrogen from the first carbon. So, if the negative charge forms from the first carbon ( from the removal of hydrogen), then it is more stabilised. Hence the hydrogen attached to the first carbon is more acidic.

In the second structure also the hydrogen attached to the first carbon is more acidic. This is because if negative charge forms on the second carbon, there is no stabilisation effect. But if the negative charge forms on the first carbon, then again by resonance it will be stabilised. So in the second structure also the hydrogen attached to the first carbon is more acidic.


Plant Coloration Stabilizer Question

What chemical is used to stabilize berry juices used as wood stains?

Aim & tries

I am trying to make a stain for a snowboard design and coloration of the wood. I used pure pokeberry juice, but almost all of the color vanished after sending through the heat compressor. Now I need to find a way to stabilize the berry substance and I am unsure of where to start.

You're looking for a mordant or fixative, such as potassium alum, $\ce{KAl(SO4)2·12H2O}$.

Though I do not know how well this will work on wood, mordants are often used on textiles. Check for cotton and linen mordants in particular, since they are cellulosic material, like wood (though wood also has lignin, which may inhibit uptake of the stain).

You might ask your teacher if she thought you are too young to dye.


Why the dilution of solvent doesn't affect the solid phase, in comparison to the liquid phase? Dilution of the solvent adds new energy states to the liquid, but does
  not affect the solid phase.

The above line appeared on the following page under the section "5: Colligative properties and entropy" and the last paragraph

Actually, I can understand a little bit. According to me, while freezing the solvent freezes separately from the solute( ie If we freeze salt and water solution, the salt will be separated out from ice). Therefore, whatever the presence of the solute maybe, the solid form of solution consists of solid 'ice', which is identical in every case and hence have identical entropy.

Am I going right? Is there an alternative explanation?

I am not clear with the concepts of what actually happens when we freeze a solution. Any Links regarding it or any other related stuff will be welcomed.

Instead of "making the jump" from liquid to gas, the water molecules are trying to stack together into an ice crystal. The salt molecules again get in the way.
Think of making a solid as like stacking shoeboxes, but the shoeboxes are vibrating slightly. The higher the temperature, the more they vibrate. At some temperature (the melting point), the vibration gets so powerful that the whole shoebox stack collapses. Now imagine that instead of a stack of just shoeboxes (pure ice), you start putting golf balls in between some of the shoeboxes. Those are salt molecules. They get in the way of the ordered stack, they mess everything up, they make the stack collapse really easily. There has to be a lot less vibration (lower temperature) to stack the boxes when there are golf balls getting up in everyone's business. So the freezing point is reduced.


Deriving The Triple Point Temperature and Pressure as the following expression Using the usual notations of $\Delta H_\text{fus}$, $\Delta H_\text{vap}$, $\Delta H_\text{sub}$, as well as ($T_\mathrm l,p_\mathrm l$) and ($T_\mathrm s,p_\mathrm s$), which corresponds to the temperature and vapor pressure of liquid and solid phase, I have to derive the triple point temperature and pressure.
My strategy is to use the Clausius-Clapeyron Relation:
$$\frac{\mathrm dp}{\mathrm dT}= \frac{\Delta S}{\Delta V}$$
but the rest is history...
Could someone help me with
$\Delta H_\text{fus}$ and ($T_\mathrm s,p_\mathrm s$)?
So that I know that I'm moving in the right direction?

On a phase diagram there are three 1-dimensional curves, the melting curve, the liquid vaporization curve, and the sublimation curve.

It is sufficient to consider just two of the three curve, and determine the point at which they intersect.

Given the limited information availible to you, (no information about how volume changes upon going from the solid to liquid phase), it is wise to choose the liquid vaporization curve and sublimation curve, rather than the melting curve, as $\Delta V$ can be approximated as the volume of gas, the volume of solid or liquid being approximately zero compare to the much larger volume of gas.

From the Clapeyron equation in the question, you need to change variables from  $\Delta S$ which you don't know to $\Delta H$ which you do know.

When two phases are in equilibrium, such as along any of the three curves discussed above, $\Delta G$ = 0.

$G = H-TS$

$0 = dG = dH-TdS -SdT$

Considering that $T$ is constant during a phase change:

$\Delta H = T \Delta S$

So substituting:

$\frac{dP}{dT}= \frac{ΔH}{TΔv}$

$\frac{dP}{dT}= \frac{ΔH_{vap}}{TΔv}$ and $\frac{dP}{dT}= \frac{ΔH_{sub}}{TΔv}$

Then, as mentioned above the approximation $\Delta V = V_{gas}$ can be made and the ideal gas law (or other equation of state) can be used to express $V$ as a function of $T$ and $P$.  Then (after intergrating) you have two equations with two unknowns, $T$ an $P$, the temperature and pressure at the triple point.