Why are there no cheap diamond equivalents? We recently learned in school that diamonds sparkle as the are very optically dense, meaning that it takes longer for light to pass through them, thus meaning that the light totally internally reflects when you look at a diamond, whereas in glass it would pass through at that angle.

Why have we not been able to create a solid material that is so optically dense? Is it because such materials are rare, expensive or hard to produce, or because no-one has looked into it?

There are four properties that make a diamond look like a diamond. I will compare them with the two best diamond simulants at the moment:

cubic zirconia $\ce{ZrO2}$ - a special ("cubic") form of zirconium oxide. Don't confuse it with zircon, a different mineral with the forumula $\ce{ZrSiO4}$.
moissanite $\ce{SiC}$ - the name for the naturally occurring form of silicon carbide, also used for the synthetic gemstones varieties of it.

So what are the differences and why none of them is an all-round replacement for a diamond?

Refractive index: determines the brilliance. This is what you were talking about when you said "optical density". This basically means the amount of light the diamond reflects to the observer. If you change the refractive index, you also have to slightly change the cut of the diamond to achieve optimal brilliance, and then your "diamond" will not look exactly like a diamond. The refractive index of diamond is 2.42. Moissanite is higher: 2.65 and cubic zirconia is lower: 2.17.
Dispersion: determines the fire. This is what you wanted to say when you mentioned that diamonds are sparkling. Fire is the property of diamonds to break the light into different wavelengths thus creating the characteristic play of colors that you get in diamonds. Here's the interesting thing: The dispersion of diamond is 0.044 and interestingly the dispersion of both cubic zirconia and moissanite is higher: 0.060 and 0.104, respectively. So theoretically, if you consider only fire (the play of colors), then they both look better than diamonds.
Birefringence: (the lack of) determines the clarity. When you look through a diamond at all angles, you can clearly see what's behind. This is also true for cubic zirconia. Both minerals are isotropic. On the other hand, moissanite is birefrigent - it will double the image you see through the mineral, causing it to look blurred and lose the visual clarity.
Hardness: while not directly related to optical properties, a mineral with higher hardness will retain it's sharp cut. If the cut is damaged, the gemstone will not look as good and lose its value. Diamond has a hardness of 10, moissanite 9.25 and cubic zirconia 8.25. Note that this scale is not linear: 10 is much harder than 9 compared to 9 and 8, for example.

As you can see, none of the alternatives fully satisfies the entire set of properties that make a diamond. Moissanite may be slightly better, but it's also more expensive. And remember - one of the most appealing things about diamonds is their rarity and natural occurrence. No synthetic simulant can replace that.


What polymer can be dissolved in ethanol and is insoluble in water? I need a polymer with parameters noted above. How to determine the solubility theoretically? Hansen/Hildebrand solubility parameters? Wypych is not the ultimate solution...

As a preface, I'm not sure what the exact solubility levels are, so this answer is qualitative.  Hopefully someone else can improve upon it or suggest something better, but for now:

Poly(1,2butylene glycol) (PBG) of moderate molar mass (Mn ~1000) is soluble in ethanol but sparingly soluble in water. Most solids will wind up being soluble in water if they're soluble in ethanol, owing to the two solvents' similar intermollecular forces but water's smaller size and greater overall solvent strength. This said, materials with the ethyleneoxy- repeating unit like Polyethyleneglycol (PEG), Poly(1,2butylene glycol) (PBG), or crown ethers have enhanced solubility in ethanol due to ethanol's ability to line up along the backbone of the polymer. Meanwhile, the ethyl pendant group hanging off the backbone of Poly(1,2butylene glycol) strongly decreases water solubility. The end of the ethyl group is far enough away from the oxygen that it's completely non-polar and the shape winds up causing something called clathrate formation in the water.

This polymer is mostly used as a Solgel or a coating. It could also be used for chelating some metal salts in solution. However, it would make a rather terrible structural polymer.


How to calculate the number of waters of hydration? I have three questions about waters of hydration, the first two of which I'm really stuck on:

A sample of cobalt(II) nitrate hydrate was heated to remove all the water of hydration. The hydrate was found to be $65.96\%$ oxygen. Calculate the number of water molecules associated with each formula unit of cobalt(II) nitrate.
Epsom salts is $\ce{MgSO4.xH2O}$. The hydrate was found to contain $71.4\%$ oxygen. Calculate the  value of $x$.
Zinc nitrate $\ce{Zn(NO3)2.xH2O}$ contains $21.98\%$ zinc by mass. What is the value of $x$?

I have absolutely no idea for the first two questions.

For the third:

Assume $\pu{100 g}$ of material
$\pu{21.98 g}$ of Zn = $\pu{0.3284 mol}$ of $\ce{Zn}$

I couldn't get any further than this.

All of the questions involve the same reasoning, so let's look at question #2.

What is the % oxygen in $\ce{MgSO4}$ when there is no water of hydration (x=0)?  It would be given by $$\mathrm{\%~ oxygen~ =~ \frac{(4 \cdot atomic~ wgt~ O)}{(atomic~wgt~ Mg + atomic~ wgt~S + (4 \cdot atomic~ wgt~ O))}}\\\mathrm{=~\frac{(4 \cdot 16)}{(24_{.}31+32+(4 \cdot 16))}}\\=53.2\%$$  How would the equation change if we added one $\ce{H2O}$ of hydration to the molecular formula? How about if we added 2 or 3?


Understanding Gibbs free energy I want to verify a thought I had about Gibbs free energy. In some ways, I'm actually wondering why we are using enthalpy instead of the internal energy $U$. Right now, I'm thinking that enthalpy is mostly useful because reactions are done at constant pressure (in an open beaker). If we were to do our reactions at constant volume, we would actually use Helmholtz's free energy instead of Gibbs. My logic:

The definition of internal energy:
$\Delta U = Q_\text{heat} - p\Delta V$

If we inject $Q_\text{heat}$ without holding the volume constant, we might change $U$ or $V$, which is a problem because we'll want to be able to describe our system as fully as possible. If we define the enthalpy:

$\Delta H = Q + V\Delta p$

Then, by keeping the pressure constant, we know that the heat added goes straight into this "enthalpy".

Afterwards, to know if a reaction is spontaneous, we need to see if the enthalpy-heat from a reaction overthrows the energy associated with the change in entropy.

Example: a molecule of two atoms wants to break apart. The bond has potential energy $Q_\text{heat} = \Delta H_0$ but it will increase the entropy by $Q_\text{entropy} = T \Delta S_0$. So, we check if:

$ Q_\text{heat} - Q_\text{entropy} > 0 $

$ \Delta H - T \Delta S > 0$

And thus Gibbs free energy was invented as:

$\Delta G = \Delta H - T \Delta S$

BUT, if we wanted to use a closed beaker, we would keep the internal energy U, and then to see if we had a spontaneous reaction, we would use Helmholtz's free energy:

$ \Delta F = \Delta U - T \Delta S $

I'm not sure about any of this and I'm mostly looking for confirmation/information of my logic. Please tell me if my intuition is wrong somewhere.

Yes your logic is basically correct.

Gibbs free energy is really derived is from the Clausius in equality which is: $$dS\geq \frac{dq}{T}$$ Basically is states that in a reaction, the change in entropy must always be equal (in the case of an ideal gas) or greater than the transfer of heat energy divided by temperature. Therefore, for a reaction to be spontaneous, the following relation must be satisfied: $$dS-\frac{dq}{T}\geq0$$ Now, lets consider two cases so we can simplify this relation.

Constant Pressure

Enthalpy is given by: $$\Delta H_p = \Delta U + p\Delta V$$
and internal energy by: $$\Delta U = q + w$$
Now by substituting $\Delta U$ in the expression for enthalpy we get: $$\Delta H _p= q + w + p\Delta V$$ Now, since w = $-p\Delta V$, the above equation simplified so that: $$\Delta H_p = q$$ So, now in the Clausius inequality, we can substitute dq with dH and multiply by $T$ to get: $$TdS \geq dH$$ Now by integrating both sides and putting all the variables on one side, we get the question for gibbs free energy: $$\Delta G = \Delta H - T\Delta S$$ and hence this is why $\Delta G$ has to be negative for a reaction to be spontaneous

Constant Volume

At a constant volume, there is no expansion work done, therefore $dU = dq$
Therefore by using the same processes as before, we are able to get the expression for Helmholtz energy, A, which is defined as: $$\Delta A = \Delta U - T\Delta S$$ Here, for a reaction to be spontaneous, $\Delta A$ must be less than 0


Friedel-Crafts synthesis - How is HCl produced? In the mechanism for the acetylation of bromobenzene, how is $\ce{HCl}$ gas produced?

The reactants are acetic anhydride, bromobenzene, $\ce{AlCl3}$ and dichloromethane.

Initially, 4-bromoacetophenone is made, accompanied with (as far as I know) acetic acid and the regeneration of $\ce{AlCl3}$.

How then is $\ce{HCl}$ produced?

To remove the $\ce{AlCl3}$ at the end, 3 $\ce{H2O}$ are added and $\ce{HCl}$ gas is produced then.

I'm trying to think of ways that $\ce{HCl}$ gas is produced prior to that, however. Could a chlorine of the aluminate ion deprotonate the arenium ion?

Could dichloromethane (solvent) donate a $\ce{Cl-}$ to $\ce{AlCl3}$?

You could perform the electrophilic aromatic substitution with acetyl chloride as well instead of acetic anhydride - in the first case it is evident where $\ce{HCl}$ comes from. However, the mechanism is similar if you use acetic anhydride. The first step consists in the electrophile activation by $\ce{AlCl3}$, which generates in situ the "acylium" ion ($\ce{H3C-C#O+}$) - the actual acylating agent. The second step is the electrophilic aromatic substitution, in which the electrophile is preferentially "attacked" by the aromatic ring at the ortho- or para-position (ortho- and para-orienting effect of halides). The driving force for the reaction is the last step, which leads to restore the aromaticity and it is assisted by $\ce{(AcO)(Al)Cl2}$ with liberation of $\ce{HCl}$.

Here I have drawn the mechanism:


How to compare the energies of sigma and pi bonds in molecular orbitals of diatomic heteronuclear species? There is a rule for diatomic homo-nuclear species that says that if $Z<8$ then $$E(\sigma_{2p})>[E(\pi_{2p_y})=E(\pi_{2p_z})]$$ and the inverse if $Z\geq8$ and in both case we have energy of $$E(\sigma_{2s}^{*})>[E(\pi_{2p_y}^{*})=E(\pi_{2p_z}^{*})].$$

This is understandable but what if we have diatomic heteronuclear species like $\ce{NO}$ or $\ce{CO}$?

How can we determine which of the molecular orbitals $\sigma$ or $\pi$ is higher in energy? I.e. how should we know which to place below which in the molecular orbital diagram (sigma or pi orbitals)?

Actually, the gist idea is the difference of energy between the  orbitals $E_{2s(\mathrm{A})}-E_{2p(\mathrm{B})}$ and $E_{2s(\mathrm{B})}-E_{2p(\mathrm{A})}$.
If the difference is less than 10 eV in absolute value, there are interactions between the orbitals $2s$ and $2p$ and the diagram for diatomic homo-nuclear species with $Z<8$ is valid. (with a slight difference is that the levels of energy of the atomic orbitals of the  more electronegative atom are more negative than the those of the other one). This is the case for the molecules you propose, i.e. $\ce{CO}$ and $\ce{NO}$.


Can recent developments improve the total synthesis of B-12? The  total synthesis  of vitamin B-12 by Robert Burns Woodward and Albert Eschenmoser, is over 30 years old. At its time, it was considered a landmark in the field.

  With current developments (e.g. olefin metathesis, palladium catalysis, etc.) is it possible to optimize this synthesis?

By "optimize" I mean reduce the number of steps in the synthesis.

Almost certainly, yes. Applications of modern organic synthesis methods have drastically shortened the synthetic routes for such previously "daunting" targets such as strychnine see Vanderwal, tetracyclines Myers  and pyrroloindole alkaloids Movassaghi, to name but a few - and recent efforts by Baran and White on C-H activation of alkanes have already yielded more efficient syntheses of many complex polycyclic terpenes. There is no reason to think that significant improvement could not occur for chlorophyll or B12. In fact, the absence of chlorophyll or Vitamin B12 from modern synthetic efforts represents a bit of a blind spot; it could well be that new chemistry could be discovered in the process. Although that 1,16 hydride shift by Eschenmoser will be hard to beat :-)


Confused Reading of Pressure on Vacuum Gauge I have a gauge used to measure vacuum but I'm a bit confused as the measurements state "Hg" (I would have supposed inches of mercury except it goes to -30) and "Bar" going to minus 1.
I need the pressure in Torr, how can I properly convert my readings?
I made a google conversion but it gives me a negative result in Torr which cannot be right.

Pressure gauges typically measure pressure relative to ambient, so "0" on a gauge means equal to ambient (~ 1 bar absolute pressure depending on where you are). In some cases, the units have a "g" at the end, for example "psig" to indicate that the measure is "gauge pressure" rather than absolute, but that isn't always the case.


What is the smallest molar volume? I wondered how small a volume a mole of material could occupy, so I started with carbon, which would need 12 grams. That's 60 carats, and there happens to be a famous 60 carat diamond.

If my density/atomic mass calculations are correct, the substances with the smallest molar volume would be nickel, carbon, beryllium, and boron, with boron being the smallest. But I had trouble finding images of large-enough beryllium or boron crystals.

Are there any compounds of sufficient density and low enough atomic mass to beat boron? Can a maximal density boron crystal be made that large?

What substance has the smallest molar volume?

EDIT -- The Noor-ul-Ain diamond is closer to being 12 grams.

Boron is a covalent solid with high melting point, like diamond (though not quite), and hence its crystals are hard to make. Unlike diamond crystals, they are not nice and probably wouldn't make a great display.

The table on http://periodictable.com/Properties/A/MolarVolume.v.log.html seems to corroborate your findings about boron molar volume being the smallest among all elements. Pity it's wrong, and so are you. (Or rather, it is technically right, but in a way that conveys a wrong impression.) Some elements just tend to have multiple polymorphs (called allotropes in this case), and carbon is one of them. All data in the standard tables are for the standard polymorph, which is graphite. But diamond at $3.5\;\ce{g/cm^3}$ is much denser, and decisively beats boron in the contest for the smallest molar volume.

Sometimes it takes walking around the world to realize that the aim of your quest has been in your pocket all along. The picture of the "smallest mole" is the one you brought here.

So it goes.


Why was Nobel Prize given to Baeyer for his "faulty" Baeyer's Strain Theory? Baeyer Strain Theory said that using the formula

$\frac{(109^ \circ 28' - \alpha)}{2}$, where $\alpha$ is the angle between 2 sides of the cycloalkane, the most stable cycloalkane could be found. Thus he proved cyclopentane to be the most stable.

But later using the value of heat evolved during combustion, cyclohexane was found to be the most stable.

Then why was the Nobel Prize given to Baeyer for a faulty theory, still resides beside his name?

For a theory which no longer could explain the maximum stability of cyclohexane?

Scientists are human beings after all. Nobody gets a Nobel Prize for inventing a "formula". It is the idea that makes an impact and new ideas are respected. Scientists like normal human beings make honest mistakes, but try their best to be correct. The guy who got the Nobel Prize in polarography, Hervosky, in the 1950s made errors in other electrochemistry papers, does this mean the prize be snatched back from him as well? Max Planck when he derived his blackbody radiation law, he did not believe much in atomic/molecular theory (later on he changed his views). He also received a Nobel prize.

If you read the citation/biographer by the Nobel Prize committee, the award was not due to some cycloalkane strain angle but the prize motivation was "in recognition of his services in the advancement of organic chemistry and the chemical industry, through his work on organic dyes and hydroaromatic compounds."

The Nobel Prize clearly says "Von Baeyer’s work was at once pioneering and many-sided. With admirable penetration and extraordinary experimental skill he combined dogged perseverance and, even at 70 years old, a youthful buoyancy in his work. He was careful never to overestimate the value of a theory. While Kekulé sometimes approached Nature with preconceived opinions, von Baeyer would say: “I have never set up an experiment to see whether I was right, but to see how the materials behave”. Even in old age his views did not become fixed, and his mind remained open to new developments in chemical science." (https://www.nobelprize.org/prizes/chemistry/1905/baeyer/biographical/)

You can see here, (good) scientists are open to change and are willing to listen to others. I found very good scientists to be very humble (with a small fraction of outliers), but only the mediocre ones are at times rigid and arrogant.


How does increase in volume change the speed of reaction in production of NO2? $$\ce{2 NO + O2 <=> 2 NO2}$$

I understand that increasing the volume $×2$ means lower pressure so production of $\ce{NO2}$ is slowed down, but when asked how to express this mathematicaly I can't get to the correct value, which is either $4$ or $8.$ I don't see how inserting $1/2$ in $v = Δc/(2×Δt)$ can give anything close to the offered solutions.

I'm not an expert but this is how I would do it.

Determine the reaction rate formula from the reaction equation

$$r = K[\ce{NO}]^2[\ce{O2}]$$
Changing the volume by a factor $2$ means the concentration of $\ce{NO}$ and $\ce{O2}$ will become half

$$r' = K\cdot\frac{[\ce{NO}]}{2}^2\cdot\frac{[\ce{O2}]}{2}$$
Compare the reaction rates

$$r' = K[\ce{NO}]^2\cdot\frac{1}{4}\cdot [\ce{O2}]\cdot\frac{1}{2}$$

$$\frac{r}{r'} = \frac{K[\ce{NO}]^2[\ce{O2}]}{K[\ce{NO}]^2[\ce{O2}]\cdot\frac{1}{8}} = 8$$

Conclusion: increasing the volume by $2$ will decrease the concentration (or pressure) of $\ce{NO}$ and $\ce{O2}.$ This causes the reaction rate to decrease by a factor of $8$.


What are the magnetic quantum numbers for specific d orbitals? I mean I know they’re $\ce{d_{-2}}$, $\ce{d_{-1}}$, $\ce{d_{0}}$, $\ce{d_{1}}$, $\ce{d_{2}}$ but how do these numbers relate to each one of $\ce{d}$ orbitals $\ce{d_{z_2}}$, $\ce{d_{xz}}$, $\ce{d_{yz}}$, $\ce{d_{xy}}$, $\ce{d_{x_2 - y_2}}$)?
I couldn't find a definitive answer anywhere.

On p. 238 of my copy of McQuarrie (1983; his Table 6-6), he lists the following for the hydrogenlike atomic wave functions (expressed as real functions), describing these as "commonly used":
\begin{align*}
n=3, l = 2, m=0:     & \ \psi_{3d_{z^2}} \ ... \\
n=3, l = 2, m=\pm 1: & \ \psi_{3d_{xz }} \ ... \\
       : & \ \psi_{3d_{yz }} \ ... \\
n=3, l = 2, m=\pm 2: & \ \psi_{3d_{x^2-y^2 }} \ ... \\
       : & \ \psi_{3d_{xy }} \ ... \\
\end{align*}


Why does IR spectroscopy specifically require a change in dipole moment, and why doesn't this apply to other types of spectroscopy? I would really appreciate any amount of detail/depth. I do not have a background in quantum, which is probably why I don't know the answer to this. However, any answer referencing quantum concepts is welcome and appreciated; I will follow up on any resources provided.

My question is basically as follows: Why does IR absorption require a change in dipole moment, and why is it that other types of spectroscopy (UV Vis, microwave) don't have this requirement? What causes this requirement specifically for IR?

Thank you!

All the types of spectroscopy you mention have a similar condition for a spectrum to be observed.  This is that the molecule has to be able to interact with the radiation in the first place and then the radiation has to be of the correct energy to change the energy of the molecule from one level to another.

A plain (classical) explanation is as follows. (We implicitly assume that the radiation has the correct energy). As radiation is electric dipole in nature (it consists of oscillating electric and magnetic fields) if the molecule also generates an oscillating electric field it can interact with the radiation; energy is taken from the radiation field and enters the molecule. Thus a molecule may gain a quantum of vibrational or rotational energy or an excited state may be produced.

In the case of a vibrating diatomic molecule only if it is hetero-nuclear e.g. HCl, will it have a dipole and thus an oscillating dipole and oscillating electric field as it vibrates. Thus hetero-nuclear diatomics show an IR spectrum, homo-nuclear diatomics do not. In a polyatomic molecule vibrations can distort the molecule, e.g. benzene is distorted so that it is no longer 6 fold symmetric, so that a  dipole is produced and an IR spectrum may result. What form this takes depends on the molecule's symmetry and group theory point groups are used to determine this.

In a rotating diatomic molecule we usually use a rigid-rotor model. If the molecule has a dipole then this clearly changes as the molecule rotates, produced an oscillating electric field, and so a microwave (rotational) spectrum is produced. A homo-nuclear diatomic e.g. $\ce{O2}$ has no microwave spectrum.

In a UV/visible transition an excited state may be produced but only if the electron distribution in the excited state is different from that of the ground state, i.e. a dipole is produced that can interact with the radiation. A molecule such a benzene has a forbidden first electronically excited state due to its symmetry (which is however observed weakly due to other effects) and so absorbs very little compared to aniline (amino-benzene) or many dye molecules where symmetry does not limit any electronic dipole formation.

Thus although the details vary, the underlying ideas are the same. You should understand now that the statements about spectra in your question are not all correct. The rules used to determine what transitions are allowed or not allowed (called forbidden) are called 'selection rules'.

You can read about this in more detail, as a quantum version, in most undergraduate phys. chem. textbooks.

You may realise that the vibrational frequencies of homo-nuclear molecules are known even though they have no ir spectrum and this is because they can be measured by Raman spectroscopy. This depends on the change in shape/size of the 'electron cloud' (polarisability) a molecule possesses. Electro-magnetic radiation also possesses an oscillating magnetic field but this interacts far more weakly with molecules than its electric counterpart and can often but not always be ignored, the exception being NMR spectroscopy.


In Reimer-Tiemann reaction why does phenol attack the carbene from ortho position? From the Wikipedia article on Reimer-Tiemann reation:

In step 5, why doesn't the oxygen attack the carbene. The way I see it: upon attacking the carbene from ortho position, the mechanism proceeds via a non aromatic intermediate. Oxygen bearing a negative charge is a good nucleophile too.

So I proceeded with oxygen acting as an nucleophile and came up with this mechanism:

The final product is not a major product (I wasn't even able to find it listed as a product after a cursory google search), so why does phenoxide attack the carbene from ortho position.

EDIT

The product I'm asking about has not even been mentioned in the linked question.

First of all, in the phenoxide ion, the lone pair on the oxygen atom doesn't remain fully available, it remains in continuous conjugation with the double bond of the benzene ring. In the resonance hybrid, there is also electron density on the Carbon atom in the ortho and para position,and also definitely on the oxygen atom itself. But, Carbon being less electronegative acts as a better nucleophile in this case rather than Oxygen.  So, attack from the Carbon atom rather than Oxygen atom is more probable.  More over, attack from the ortho position, though creates a chance of Steric crowding, but the intermidiate generated from the attack by ortho-position creates an ordered transition state.The metal ion(say, $\ce{Na+}$) can easily interact with the electronegative $\ce{O}$ and $\ce{Cl}$ atoms, and can increase the stability of the transition state. Here is a picture of the intermidiate and a 3D view of it,

This ordered transition state will not appear in case of attack by the Oxygen atom. This is also another reason why that transition state is more favoured, and the reaction rate through that step is higher than any other possible steps.


How do I know which is the "correct" way to write the reaction? We need to calculate the heat produced by a set of $\mathrm{N_R}$ reactions in our reactor model.

\begin{align}
\mathrm{Q_r = \sum_{j=1}^{Nr} Rr_j*\Delta H_{R,j}}
\end{align}

$\mathrm{Rr_j}$ is the reaction rate of reaction $\mathrm{j}$, in $\mathrm{\frac{kmol}{kg_{cat} s}}$

$\mathrm{\Delta H_{R,j}}$, in  $\mathrm{\frac{kJ}{kmol}}$, is constant and calculated as:

\begin{align}
\mathrm{\Delta H_{R,j} = \sum_{i=1}^{Nc}\nu_{i,j}*\Delta h_{i,j}^f}
\end{align}

Now consider these:
\begin{align}
\mathrm{R1:}\ \ce{H2 + 1/2O2 \rightarrow H_2O}
\end{align}
\begin{align}
\mathrm{R2:}\ \ce{2H2 + O2 \rightarrow 2H_2O}
\end{align}

They are undoubtedly the same reaction, but the R2 gives a  $\mathrm{\Delta H_R}$ that is twice the $\mathrm{\Delta H_R}$ in R1. It is naturally so, but greatly affects my energy balance.

How do I know which is the "correct" way to write the reaction?

Does it mean that the reaction rate should be specific to R1 or R2?

How do I know which is the "correct" way to write the reaction?

There is no single correct way.  Picking a different way to write the reaction affects both $\Delta H_{R_j}$ and $Rr_j$ in opposite ways that cancel out.

  Does it mean that the reaction rate should be specific to R1 or R2?

Yes.  The unit of your reaction rate $Rr_j$ is $\frac{\textrm{kmol}}{\textrm{kg}_{cat}}$.  But the key question is kmol of what?  A good way to think about the answer is "kmol of things that have a stoichiometric coefficient of 1 in the reaction".

So if the reaction is R2, then the rate of the reaction will be the rate of disappearance of $\ce{O2}$.  The rate of formation of water will be twice the reaction rate $Rr_2$.  But if the reaction is R1, then the reaction rate is equal to the rate of formation of water (or the rate of depletion of hydrogen).

But you don't directly measure "reaction" rates, you can only measure the formation or disappearance of products or reactants.  So say you measured the formation rate of water.  If you decide the reaction is R2, then $Rr_2 = \frac{1}{2}R_{\ce{H2O}}$.  If you decide the reaction is R1, then $Rr_1 = R_{\ce{H2O}}$.  The factor of $\frac{1}{2}$ will cancel out the factor of two in $\Delta H_R$.

So in the end it shouldn't matter what choice you make.

For the exact same idea in a totally different chemical context, see this se.chem question on electrochemistry.


pH of 0.05 M NaH2PO3 What is the pH of a $\pu{0.050 M}$ solution of $\ce{NaH2PO3}$? I am given that $\mathrm{p}K_\mathrm{a1} = 2.0$ and $\mathrm{p}K_\mathrm{a2} = 6.58$.

I am getting two different answers, $3.94$ and $4.30$, using two different methods that I have learned:

Method 1

$\ce{H2PO3-}$ can either be protonated or deprotonated. The protonation reaction has

$$K = \frac{10^{-14}}{10^{-2}} = 10^{-12},$$

while the deprotonation reaction has

$$K = 10^{-6.58} = 2.63\times 10^{-7}.$$

Since the protonation reaction is significantly less favorable than the deprotonation ($K$ ratio is less than $1000$), we can ignore the protonation and only focus on the deprotonation. This yields:

$$10^{-6.58} = \frac{x^2}{0.05-x} \to x = [\ce{H+}] = \pu{1.15e-4 M},$$

$$\mathrm{pH} = -\log(x) = 3.94.$$

Method 2

With a diprotic acid that can either be protonated or deprotonated, the $\mathrm{pH}$ is given by the following approximation:

$$pH = 1/2(\mathrm{p}K_\mathrm{a1} + \mathrm{p}K_\mathrm{a1}),$$

$$\mathrm{pH} = 1/2 \times (2.0 + 6.58) = 4.29.$$

This is the correct answer as presented in the text. Why is the first method invalid?

The problem with method 1 is that you discard the protonation process, which effectively would increase pH as it produces $\ce{OH-}$ ions.

The criteria of $K_1 > 1000K_2$ is applicable when you have a polyprotic acid in its completely (de)protonated state. The ionisation of H2A produces only a small amount of HA- and K2 is small as well, so ionisation of HA- would be negligible and we can ignore it.

However with an ampholytic species, while K2 is smaller than K1, the initial reactant is the same (HA-) and it is in high concentration, so both processes occur. The formula in method 2 is the result of this consideration.


What solvent would have the fewest ions? We know that water auto-ionizes (to a rather small extent).

My question is which solvent would have the fewest ions (or at least fewer ions than water).

I'm thinking that perhaps some kind of hydrocarbon, as these are both weaker acids and weaker bases than water and thus even less reactive. I've seen computers submerged in oils (as well as distilled water) due to their low conductivities.

2-Methylpropane has a $\mathrm{p}K_\mathrm{a}$ of 53 ($\mathrm{p}K_\mathrm{a}$ table).  That sounds pretty non-ionic.  Can't think of any common all-carbon (no substituents of any kind) liquid compounds.


Which invariant reactions are congruent? I am studying the concept of incongruent melting and invariant reactions. I was wondering, would all the invariant reactions that I show below be incongruent? As I understand it, we can only have congruent melting when the liquid to which our solid melts to has the exact same composition, and by the definition of these invariant reactions, this seems not to be the case in any of them.

You are correct, none of the reactions are congruent reactions.

Per IUPAC GoldBook:

  Congruent Transition: A transition in which the two-phase equilibrium of melting, vaporization or allotropism of a compound involves phases of the same composition.

The key term here is "phases of the same composition." Since all of the reactions have two phases transforming into one phase (or vise-versa) the reaction is not congruent.

See the phase diagram for silicon and titanium. Congruent melting occurs for $\ce{Ti5Si3}$ and $\ce{TiSi2}$ where you observe a $\ce{\color{red}{comp A (s)} -> \color{red}{comp A (l)}}$ transition, whereas at the eutectics you have $\ce{\color{blue}{comp B (s)} + \color{green}{comp C (s)} -> \color{red}{comp A (l)}} $ transition where $\mathrm{B \ne C \ne A \ne B}$. Thus the eutectics are not congruent transitions.


How does Ammonia and water mixture works in an absorption refrigeration? As per one of the questions in this site $$\ce{NH3 + H2O -> NH4 + OH}$$

However, per another perspective in absorption refrigeration

  The mixture is placed in the same container. Then when it is heated the mixture is separated and $\ce{NH3}$ changes phase, becomes a gas which moves further to cooling, while water is trapped down in the reservoir.

Which one is correct?

The difference between the absorption refrigerator and the more common refrigerator is that there is no compressor in the absorption refrigerator.  The Absorption refrigerator still needs a way to convert the evaporated refrigerant back to a liquid state.  Instead of compressing it in one step, the evaporated refrigerant is absorbed by a liquid.  The liquid is heated and the refrigerant evaporates from the liquid.  The refrigerant is cooled in a heat exchanger and condensed.  It is now a liquid and ready to go back through the cycle.

The equation you have written is a reaction between ammonia and water.  This reaction will occur in an absorption refrigerator that uses water as the absorber and ammonia as the refrigerant.  The equilibrium constant of the reaction is low so it is not accounted for in a discussion of how an absorption refrigerator works.


Is alcohol Lewis base or Brønsted base? I know that Lewis' concept of acid base deals in terms of electrons and Brønsted in terms of $\ce{H+}$ and base which can provide electrons or $\ce{OH-}$. If I am wrong (partially or fully) please explain Lewis and Brønsted-Lowry concept along.

According to me it is a Lewis base because of the lone pair on oxygen of the alcohol and a Brønsted acid.

Alcohols can act as both.  Many reactions of alcohols are believed to start with their protonations or deprotonation. However, isolating salts where they act as bases is much harder, while alcoholates are quite easy both to make and isolate. Protonated alcohols are very active alkylating agents and if fitting $\beta$-hydrogen is present, are prone to beta-elimination.


What software can calculate aqueous solution equilibria? What software is available out there to calculate the equilibrium in a set of reactions in aqueous solution? In particular, I'm interested in software general enough to simulate things like titration curves involving acido-basicity, complex formation, precipitation…

A web search turns up plenty of academic papers giving algorithms to find the equilibrium of such systems, but I'd like to find software that actually performs it. In particular, it would be really helpful if there existed such software that included a decent database of reaction constants for some well-known reactions (pKa for common acide-base couples, complexation and precipitation constants, etc.).

Edit: Apparently I was unclear: I'm not looking for molecular simulation codes, but for software that can calculate chemical equilibrium in solution from initial concentrations and a database of equilibrium constants for each possible reaction. The kind of thing teachers can use to predict titration curve and demonstrate the effect of various initial concentrations to students.

ChemEQL is pretty nice in some instances and free, but it looks like it has been mostly abandoned.

I'm aware of Titrator, but it only runs on Windows and I've never had  a chance to run it. It's also free.

I've heard good things about Visual MINTEQ, but it is Windows only and I've never had a chance to run it. Supposedly well maintained.

There is also PHREEQC, which sort of works on stuff other than Windows. I hear it is harder to use, but probably worth it.


What is reaction between isopropyl alcohol and acetone? What would be the probable reaction between acetone, $\ce{(CH3)2CO}$, and isopropyl alcohol, $\ce{CH3CHOHCH3}$, at room temperature without any catalyst in different concentrations?

I've used acetone/ipa mix to flush chromatography columns. There is no reaction between them without another reagent present


How difficult is it to produce diatomic hydrogen? Just to preface this, I have a very basic understanding of chemistry, so forgive me if what I'm asking is super basic.

How difficult is it to produce diatomic hydrogen? I've been doing some reading on the internet about diatomic hydrogen and a fair amount of the stuff I've come across are health articles relating to the benefits of drinking hydrogen rich water, so I assume that it's not terribly difficult to come across.

Diatomic hydrogen is hydrogen’s stable elemental modification under standard temperature and pressure — so if you generate hydrogen you will be generating diatomic hydrogen.

The simplest way to do this with household means is probably to use acetic acid concentrate (the stronger the better) and zinc powder. (Solid zinc works but much slower.) The reaction may be sluggish; if that is the case, very carefully and slowly heat the acetic acid. Capture the hydrogen in some glass, cup or test tube held upside down (hydrogen is the least dense gas with each molecule only weighing $2~\mathrm{u}$ (atomic mass units) so it will always rise).

If you have collected some amounts of hydrogen gas you can proove its presence. This was a common experiment in my high school chemistry courses so I’m going to tell you how to do it. However, exercise caution! If you didn’t do this at school yourself or didn’t watch someone do it before, I would advise you to not attempt it. Only do this with very small amounts of hydrogen, no more than a test tube of $2~\mathrm{cm}$ diameter. Make sure the test tube is full (by collecting enough hydrogen). Carefully bring the opening towards a candle flame. Do not use a pocket lighter where your thumb is right next to the flame for safety reasons. The hydrogen should burn with a faint ‘pop’ creating water. Considering how slurrish the reaction of zinc and acetic acid probably is, it may be too slow to actually collect enough hydrogen. Don’t attempt the burning if that is the case.

The reason for being cautious is the following: $2\,:\,1$ mixtures of hydrogen and oxygen are explosive. And with explosive I mean that they give a very loud bang. Luckily, air is mainly nitrogen so the probability of accidentally creating an explosive mixture is rather low. However, if you did, you will regret it.

Finally, on the topic of ‘hydrogen-enriched water’: Hydrogen, like all other gases, is not well soluble in water. However, it is hardly present in standard water since it is almost non-abundant in the atmosphere so it cannot dissolve. (Carbon dioxide, nitrogen and oxygen are far more abundant and therefore spontaneously dissolve in water to a certain extent — especially carbon dioxide which, in fact, reacts.) It is possible to create a hydrogen-saturated aquaeous solution or something less hydrogen-containing and sell it as ‘hydrogen-enriched water’. It would have no positive or negative effects on you or your surrounding. (Unless maybe your voice got slightly lighter in the same way as when you breathe helium — however, the amount of hydrogen is very likely too low for this to happen.) So it sounds like my dream of life: Making money with other people’s stupidity. It’s a shame that I’m not ruthless enough to do such a thing …


Can we goldify metals? I've read in some book that one of the old chemists was able to transform cheap metals into gold.

I know that gold is an element (Au) so we can't transform a metal into Au. But as I read further it says that the chemist doesn't actually transform metals into gold but into something very similar to gold (similar physical and chemical characteristics )
such that you can't differentiate between it and between gold.

Is this chemically possible?

In the days before chemistry, alchemists searched for the Philosopher's stone which would enable them to transform base metals into gold. They spent a lot of effort searching for it but we know now that their search was futile as no chemical process can cause elemental transformation. Only nuclear processes can do that but they cost far, far more than mining the element ever will so this doesn't really count as a solution.

But plenty of people faked a result or were conned by their lack of understanding. Fool's gold, for example looks a lot like gold, but doesn't share many other properties but, presumably, the name arises because some people are gullible. There are also many metal alloys that look like gold but are not pure (including many alloys that contain some, but not much, gold). It is also possible to use a variety of chemical or electrochemical processes to coat the surface of many metals with a thin film of gold.

But careful analysis can reveal most of these methods of fakery (including the alloying of gold with other metals which was very tempting when coins were made of gold). One of the most famous stories of Greek science is based on one of the most reliable methods. Archimedes was faced with the task of distinguishing debased coinage or other gold objects and, in the legend, realised the answer while having a bath, leapt out the bath and ran down the street naked yelling "eureka". What he realised was that an object immersed in water displaces a volume of water equal to its volume and this can be measured by weighing the displaced fluid. Combined with the weight of the object this can be used to estimate the density of the object. Since the only other metal with a density close to gold is tungsten (gold is 19.28 g/cm3 and tungsten is 19.25 g/cm3) knowing the density is a good way to test the purity of the metal. Common metals used in alloys, like silver (density 10.5 g/cm3) will substantially lower the density.

So, in short, you can make a lot of things that look like gold, but it is very hard to make something that shares enough properties to convince a skeptical observer who will test more than the appearance.


Why does aluminum foil "electrocute" dental fillings? This is probably one of the weirdest questions on here. When I was a kid, I put aluminum foil in my mouth (maybe it was still attached to some food or something, I don't remember.)

Anyway, it made my mouth feel like I'd licked a 9-volt battery. Being the natural born scientist, I repeated the procedure with similar results.

Why does this happen? I really think it has something to do with my dental fillings since the sensation occurred near them.

Ouch...  actually a known phenomenon.  Between the foil of aluminum/Al, and the metal filling of amalgam in your tooth, you created a little electrochemical cell.  Your saliva served as an electrolyte and your nerves around sensibly detected the current. This already "works" with little splinters of Al foil wrapping chocolate (obviously, it was not a small Hershey bar), yet is less powerful than the said 9 V block.

(one reference)


Chemical reactions as exponential functions Are all graphs of molar concentration vs. time, for the reactants, exponential decay functions? (Until equilibrium is reached, if it goes to equilibrium at all.)

If so, intuitively, why is that the case?

EDIT: @ShankRam has brought it to my attention that zero-order reactions do not entail exponential decay functions. However, I would still like to know if first-order and second-order reactions do, and why they do.

Intuitively, we can understand why only a first order reaction (at least for noncomplex reaction orders) displays exponential decay because a first order reaction implies that the rate of decay ($d[A]\over dt$) is proportional to the concentration ($[A]$). This relationship precisely describes an exponential function.


Solubility determination of Ca(OH)₂ in water In one problem it's asking me to find the solubility of $\ce{Ca(OH)2}$ in water. In that problem it assumes the starting concentration $\ce{OH}$ to be zero. But for other problems like $\ce{BaSO4}$ in $0.1~M~~ \ce{Na2SO4}$, we account for the $\ce{SO4}$ in the $\ce{NaSO4}$. So in the first problem, why don't we worry about the $\ce{OH}$ from the self ionization of water? Is it just that extremely low of an amount?

The self ionization of $\ce{OH}$ is $10^{-7}$. If the concentration of $\ce{Ca(OH)2}$ is greater by a few magnitudes, then the self ionization should not matter much. In fact, $\ce{Ca(OH)2}$ is considered soluble, so if not in miniscule amounts, the self ionization should not matter much


Can geometrical isomers be enantiomers? I have just been pondering upon the question above. I have found this example and I am sure they are enantiomers, but do they also count as geometrical isomers?

The enantiomers 1a and 1b can have their double bonds defined by CIP rules 1a and 5. Rule 1a dictates that CH3>H while Rule 5 has R>S. Thus, 1a has an E-double bond  and 1b is of the Z-configuration as exemplified by the red bonds. This issue has been addressed previously on this site.

Addendum: While the enantiomers 1a and 1b (vide supra) have their double bonds labeled in upper case E/Z (ChemDraw), JSmol takes into account the enantiomorphic nature of the double bonds by using lower case e/z (vide infra). See the answer of @orthocresol.


What is Markovnikov's addition? The electrophilic addition of $\ce{HBr}$ across an unsymmetrical alkene follows Markovnikov's addition, but sometimes if the addition occurs via rearrangement then should we call it a Markovnikov addition?

Earlier, in an entrance exam, a similar question was asked and they have given the correct answer which states Markovnikov's addition for a reaction that occurs without a rearrangement.

In my opinion, if addition occurs without rearrangement then it should be Markovnikov's addition.

The idea of Markivnikov or anti-Markovnikov addition fundamentally isn't related to the mechanism by which a reaction took place, rather, the term is a way of describing the observed products obtained from certain kinds of addition.

The original statement of  Markovnikov addition is as follows, and applies specifically to the addition of hydrogen halides ($\ce{HX}$) across an alkene:

  In the addition of hydrogen halides to unsymmetrically constituted [unsaturated] hydrocarbons, the halogen atom becomes attached to the carbon bearing the lesser number of hydrogen atoms.

The statement was a purely empirical one (i.e. it was based on the results of experiment, rather than any consideration of the mechanism/pathway that the reaction had taken). It can also be expanded to consider any addition of a polar reagent ($\ce{RX}$) across alkenes and alkenes. The following definition is taken from the IUPAC Gold Book (a standard reference for definition of chemical terminology):

  In the heterolytic addition of a polar molecule to an alkene or alkyne, the more electronegative (nucleophilic) atom (or part) of the polar molecule becomes attached to the carbon atom bearing the smaller number of hydrogen atoms.

Thinking about the situation mechanistically, the Markovnikov rule basically just says that the less electronegative atom in the polar reagent $\ce{RX}$ ends up bound to the end of the alkene (or alkyne) most able to stabilise a developing positive charge (i.e. the most hindered end of the alkene/alkyne).

In terms of labelling the products of reactions, as discussed in your question, its perfectly valid to describe the products as those of Markovnikov addition (as stated previously, the term itself is saying nothing directly about mechanism, even if implied).


What is a simple ratio in which atoms combine? According to Modern Atomic Theory:

  The ratio in which the different atoms combine may be fixed and integral but may not always be simple.

The example given is that in sugar molecule the ratio of $\ce{C,H}$ and $\ce{O}$ atoms is $12:22:11$ which is not simple.

Well, what I see here is that 12,22 and 11 are whole numbers, then why is this not a simple ratio?

The ratio of C, H and O atoms of 12:22:11 is likely to be sucrose. Glucose has a simpler ratio of 1:2:1 (even though it has 6 carbons). The usual context for presenting the theory is with inorganic compounds with no more than a 2:3 or 3:5 ratio. With this limitation, they use the words simple ratio since the numbers are small and can easily be found. The ratio of 12:22:11 could be confused with another ratio if your calculations are off by 20%.


On the relative ease of E2 dehydration Concerning alcohol dehydration, I get that it's E1 in secondary and tertiary alcohols, that it's E2 in primary alcohols, and also why E1 reaction is easier for tertiary alcohols as compared to secondary ones.
Why is the ease of dehydration for primary alcohols less than both of these? It would have been obvious it was reacting through E1, but it's doing E2, which is favourable for primary alcohols.
Why then, is it harder to dehydrate primary alcohols than either of the other two?

The rate of reaction depends on how stable the transition state is. In both E1 and E2 good leaving group is present along with presence of a weak and strong base ,respectively. This leads to an alkene type of transition state in E2 mechanism and a cationic type transition state in E1 mechanism . Both these transition states gets stabilized by the electron donating ability of the attached alkyl groups (i.e By Hyperconjugation and Inductive Effect).This leads to the increase in rate of reaction and thus the secondary and tertiary alcohols are easier to dehydrate than the primary alcohols.
Hope this helps!!


Volume of sample for quantitative NMR I’m interested in quantitating lithium via NMR using a Magritek Spinsolve 60MHz bench top NMR spectrometer. Is there any evidence that suggests the volume of the liquid sample used in the NMR tube has any effect on the area of the spectra peaks? I’m interested in hearing opinions on the matter before running tests verifying if such correlations in-fact do exist.
I’ve seen in relevant literature that 0.6-1mL is generally a sufficient amount of solution for the 5mm tubes. However it’s not clear to me yet if this volume should be consistent between the calibrants and experimental solutions for analytical NMR. Thanks!

No, sample volume will not effect quantitation if the following conditions are met: 1) you are comparing samples in tubes with the same inside diameter, 2) you have enough sample volume so that the sample liquid extends at least a few diameters past the ends of the RF coil, 3) your shim is good enough so that the peaks you want to measure do not significantly overlap other peaks. Field distortion due to susceptibility discontinuity at the ends of the sample liquid can degrade shim, and people often use samples that extend far beyond the RF coil for that reason.  However, provided the above conditions are met, shim does not change peak integrals. This is because the integral of the line shape is exactly equal to the initial amplitude that the line contributes to the FID, and is independent of how it decays in time.


Disproportionation Reactions of the Halogens: Writing balanced net ionic equations for four reactions. See details I first conducted the following reactions, each in separate test tubes:

$\ce{NaBr}$ + $\ce{H2SO4}$ + $\ce{NaBrO3}$
$\ce{KI}$ + $\ce{H2SO4}$ + $\ce{KIO3}$

...with water and dichloroethane added to each as well.
(In the experiment, I put a small quantity of sodium bromide in one test tube and a similar quantity of potassium iodide in the other test tube. I then added 1 mL of water, 3 drops of 9M sulfuric acid, and 2 mL of dichloroethane to each. I then added a small quantity of sodium bromate to the first test tube and a similar quantity of potassium iodate to the other.)
Then, I removed the aqueous layer from each test tube and added $\ce{NaOH}$ to the dichloroethane layer in each.

(#1) $\ce{NaOH}$ added to dichloroethane
(#2) $\ce{NaOH}$ added to dichloroethane

So far, I have the following equations written:

$\ce{NaBrO3 + 5NaBr + 3H2SO4 -> 3Na2SO4 + 3H2O + 3Br2}$
$\ce{KIO3 + 5KI + 3H2SO4 -> 3K2SO4 + 3H2O + 3I2}$
$\ce{3Br2 + 6OH^{-} -> 5Br^{-} + BrO3^{-} + 3H2O}$
$\ce{3I2 + 6OH -> 5I^{-} + IO3^{-} + 3H2O}$

Please check my work. If without error, I just need to convert these to equations to net ionic.

(answer to original question)

No, you are not on the right track.

If bromate is a starting material, it should not appear as a product in reaction 1. Instead, consider how bromide and bromate can react with each other.

If iodate is a starting material, it should not appear as a product in reaction 2. Instead, consider how iodide and iodate can react with each other.

For reactions 3 and 4, a hint would be "disproportionation".


Reactivity of different isomers of N2F2 $\ce{ N2F2}$ exists in two forms cis and trans. Today my chemistry teacher tells me that among the two only cis form takes part in reactions and that it is more reactive. But I have read elsewhere that trans is less stable. As far as I understand less stable compounds are usually more reactive. So why this discrepancy?

According to Dinitrogen Difluoride Chemistry Inorg. Chem., 2010, 49 (15), pp 6823–6833:

  This increased reactivity of the cis-isomer can be readily
  explained. In both $\ce{N2F2}$ isomers, each nitrogen atom
  possesses a sterically active free valence electron pair.
  The transformation of $\ce{N2F2}$ to $\ce{N2F+}$ requires the conversion
  of the $\ce{N=N}$ double bond into a $\ce{N#N}$ triple bond
  by an overlap of electrons from these two free valence electron pairs. Obviously, this overlap is greatly facilitated
  in cis-$\ce{N2F2}$, where these free pairs are on the same
  side of the molecule.


Why does sulfate have this structure? Look at the structure for sulfate:

Why does sulfur form so many covalent bonds. Its valency is only $-2$, it only needs two electrons, yet here it's getting $6$.

The same thing happens with phosphate.

Phosphorus has a valency of $-3$, yet, it forms $5$ covalent bonds.

Ultimately those structures are wrong.  Most of the community of chemistry educators teaches that formal charge reduction creates hypervalency.  All modern research and models shows this not to be the case.  You can't hybridize a d orbital and even get the proper tetrahedral geometry here.  Reducing formal charges to 0 makes no real world sense, when sulfur has a measured empirical charge of around 1.7 in sulfate.

Sulfate is 4 single bonds.  The bond length difference is due to the fact that the sulfur carries extra charge, as does the oxygens.  This ionic like bond is of course going to be shorter than more neutral forms of S-O.  Further, why would two double and two single hybrid bonds average to be the same length as a double bond?

Essentially any general chemistry solution that invokes d-orbitals as it's solution, outside of transition metal chemistry, is wrong.  Sulfur hexafluoride is not in fact 6 single bonds.

Unfortunately, biochemists are not going to update their picture of phosphate anytime in the next fifty years because no one cares.

The fact that basically every science major in the world learns this wrong is abhorrent.


How does the smell of a compound come about, and is it possible to define a smell? Colour - and eyesight in general - arises because objects reflect/transmit certain wavelengths of colour, which is detected by our eyes.

On the other hand, what gives rise to smell? Is there a branch of chemistry associated with this? Is it possible to define the smell of any substance, using some parameters analogous to wavelength in the case of sight?

We know that $\ce{H2S}$ has a foul odor, but distilled $\ce{H2O}$ does not have any strong smell. Does $\ce{H2Se}$ have a similar foul smell?

Smell is what you feel when molecules of some compound (and not any kind of radiation, mind you) touch the olfactory nerves deep inside your nose. It is just another property of a compound, like molecular weight, or melting point, or color, only different.

Unlike molecular weight, it can't be calculated from molecular formula by simple rules.
Unlike molecular weight or melting point, it can't even be adequately described with a single number.
Unlike all of the above, it can't be reliably predicted for a new compound, though much can be done by the way of analogy.

$\ce{H2Se}$ smells worse than $\ce{H2S}$, and $\ce{H2Te}$ is worse still. A disturbing characteristic of the latter, given by an anonymous expert in the field, I'd better keep hidden so as not to offend the public taste:

   Smell it, and you'll vomit farther than you can see.


How to run a B2PLYP-D3 calculation in Gaussian 09? Can anyone help me to run a geometry optimization and frequency calculation using B2PLYP-D3 functional including the DFT-D3 dispersion correction? I went through Gaussian manual, however, it was not very clear to me.

%NProcshared=4
%Mem=4gb
%chk=fohsdft3
# B3PLYP/aug-cc-pVDZ opt freq scf=tight

Input file taken from fohsdft

0 1
 O                  -3.46278   2.54045   0.
 H                  -2.50278   2.54045   0.
 H                  -3.78324   3.44539   0.

We have the Gaussian 09 Revision C.01 version.

Search for B2PLYP on the Moller-Plesset page:

  B2PLYPD3 requests the same but with Grimme's D3BJ dispersion [Grimme11, Goerigk11].

The correct D3 parameters will be set for you automatically.

%nprocshared=4
%mem=4gb
%chk=fohsdft3
# opt freq b2plypd3/aug-cc-pvdz scf=tight

Input file taken from fohsdft

0 1
O                  -3.46278   2.54045   0.
H                  -2.50278   2.54045   0.
H                  -3.78324   3.44539   0.

Since B2PLYP-D3 is a method, it goes where the density functional is normally specified.


How can I interpolate between two densities at different temperatures? I'm trying to find the temperature of methanol ($\ce{CH3OH}$) when its density is equal to $\pu{780 kg m-3}$.

I know that when $T = \pu{30^\circ C}$ the density is equal to $\pu{783 kg m-3}$, and when $T = \pu{40^\circ C}$ the density is equal to $\pu{774  kg m-3}$.

Is there any formula or a trick to find the temperature corresponding to an intermediate density? Likewise is it possible to find the density at a temperature intermediate between the two?

You could use tabulated literature data and then interpolate between the data points. For example, the following plot was created using data from NIST Reference Fluid Thermodynamic and Transport Properties Database (REFPROP) – NIST Standard Reference Database 23, Version 9. It shows the liquid saturation line as well as the vapour saturation line.

In this case, interpolation for a liquid density of $780.00\ \mathrm{kg/m^3}$ yields a temperature of $31.643\ \mathrm{^\circ C}$.

By way of comparison, the calculated liquid density is $781.55\ \mathrm{kg/m^3}$ for a temperature of $30.000\ \mathrm{^\circ C}$, and $772.10\ \mathrm{kg/m^3}$ for a temperature of $40.000\ \mathrm{^\circ C}$.


Why do we write electrode potentials next to chemical reactions? We might have a half cell consisting of the redox couple $\ce{Ag+}/\ce{Ag}$, which for example's sake might be fixed to be at the cathode (i.e. undergoing reduction).

The reaction at the cathode is then

$$\ce{Ag+ + e- -> Ag}$$

The standard electrode potential of the $\ce{Ag+}/\ce{Ag}$ couple is $+0.80\ \mathrm V$ with respect to SHE. I've seen the above reaction written as such:

$$\ce{Ag+ + e- <=> Ag} \quad E = +0.80\ \mathrm V$$

I'm confused as to why we write this electrode potential next to the reaction, considering that the electrode potential of a redox couple is reasonably disjoint to the concept of the chemical reaction. Although, at the end of the day, the chemical reaction results in the interfacial potential difference which causes the electrode potential, $E$ is a property of the electrode, not the chemical reaction.

Is it perhaps just because we want to write the equation of the reaction occurring at a certain electrode and the electrode potential of the electrode on the same line?

I'm aware that for associating potentials with written reactions, there exist two different conventions: European, where all reactions are labelled with the reduction potentials of the couple, and American, where oxidation reactions are labelled with the oxidation potentials of the redox couple and likewise reductions with reduction potentials. However, my question is why we need to attach the electrode potential to the chemical reaction in the first place!

That is, ultimately, aren't redox potentials associated with redox couples/electrodes, not reactions?

Your question is a valid question, and ignore downvotes. They don't mean anything.

Your understanding is very good and that you realized that the electrode potential is a property of the electrode and it really does not care how the reaction is written. However, a equation is $needed$ to keep track of the electrons lost or gained in the Nernst equation. So as a tradition and a matter of convenience, electrode potentials are quoted along with a balanced half-cell.

Another very important thing to remember (even some university teachers and PhDs lack this understanding), that the sign of the electrode potential in invariant. It does not care how the reaction is written. It is the electrostatic sign of the electrode. This part takes care of the European and American conventions. American convention is due to Gibbs and Latimer and no longer used. Most international electrochemists have agreed to stick to the European convention i.e., write all half reactions as reduction.


Grocery store crystalline citric acid Which type of citric acid crystalline powder do they sell in grocery stores?

Is the concentration of citric acid in the crystals 100%?Is it anhydrous or a monohydrate?

The box in my kitchen is labelled "reine Zitronensäure" (= pure citric acid) and "lebensmittelsauber".

According to the manufacturer, the latter means that citric acid in this quality can be used to descale (decalcify) kitchen items, but it is not intended to be used as a food supplement and should not be digested!

This may have real technical reasons, i.e. the product may in fact contain impurities resulting from production, processing and packaging. The manufacturer however does not specify the kind or amount of impurities.

On the other hand, this kind of labelling may just reflect different legal treatment (certification, monitoring, etc.) of products intended as food supplement and those only used to clean kitchen equipment.

Concerning waterfree vs monohydrate, neither the package in my kitchen, nor the website of the manufacturer provide any information.

If you know a chemist, this question is easily solved: Let him measure the melting point of a sample. Waterfree citric acid melts at 153 °C, the monohydrate at 100 °C.


Calculating bond energy given the enthalpy value of burning reaction The burning enthalpy (exothermic) of $\ce{NO +O2 ->NO2}$ is $\pu{57 kJ mol^-1}$ and $\ce{O=O}$ bond energy is $\ce{494 kJ mol^{-1}}$, then what is the bond energy of $\ce{N-O}$ ?

So I balanced the equation

$$\ce{2NO + O2 ->2NO2}$$

wrote down Lewis structures and got

$$(2\Delta H(\ce{N=O}) + \Delta H(\ce{O=O}) - 2(\Delta H(\ce{N-O}) + \Delta H(\ce{N=O})) =  \pu{-57 kJ mol^-1}$$

($\pu{-57 kJ mol^-1}$ since it is an exothermic reaction), so we are left only with

$$\Delta H(\ce{O=O}) - 2\Delta H(\ce{N-O}) = \pu{-57 kJ mol^-1}$$

thus

$$\Delta H(\ce{N-O}) = \frac{494+57}{2}\pu{kJ mol^-1} = \pu{275.5 kJ mol^-1}$$

However the right answer is

$$\frac{494-2\cdot 57}{2} \pu{kJ mol^-1}= \pu{190 kJ mol^-1}$$

and I really don't get what am I doing wrong?

How can I calculate the enthalpy for those balanced reaction given the data of $\ce{NO +O2 -> NO2}$ enthalpy?

The way the enthalpy data is presented is sloppy (is this the original text of the exercise?). For some types of enthalpy, it is unnecessary to specify the chemical reaction equation because there is a convention. For example, for the enthalpy of formation, it is understood that the substance in question is the only product, the stoichiometry coefficient is one, and that the reactants are elements with coefficients so that the equation is balanced. I am not aware of any such convention for the enthalpy of combustion.

What is the enthalpy of $\ce{2NO(g) + O2(g) -> 2NO2(g)}$?

We can look up the enthalpies of formation for the nitrogen-containining compounds:

$$\Delta H_f(\ce{NO}) = \pu {90.25 kJ/mol}$$
$$\Delta H_f(\ce{NO2}) = \pu {31.18 kJ/mol}$$

From this, we can calculate the reaction enthalpy:

$$\Delta H_r = 2 \cdot \pu {31.18 kJ/mol} - 2 \cdot \pu {90.25 kJ/mol} =  \pu {-118.14 kJ/mol}$$

Which combustion reaction has an enthalpy of -57 kJ/mol?

If we divide the balanced chemical equation by 2, we get:

$$\ce{NO + 1/2 O2 -> NO2}$$

The reaction enthalpy is one half of that of the original reaction, -59.07 kJ/mol. So for the reaction describing the combustion of one mole of NO, the reaction enthalpy is about -57 kJ/mol.

What about the bond dissociation energy?

Everything else in the OP's calculation is correct. If you plug in the correct combustion enthalpy, the calculated bond energy will be correct.


Reaction of conjugated alkene with KNH2 I could not think of this reaction fitting into any of the reactions I know except the aryne mechanism. And hence I came up with the following mechanism after seeing the answer (which I've written in the picture next to question). Please tell if my proposed mechanism is correct.

Also, I am highly unsure of the fact that the last species formed will be able to extract a hydrogen atom from somewhere in the solution.

Question:

This is my proposed mechanism:

Source: Joint Entrance Exam (JEE) 1997 India

Here is a flowchart of what the commenters have stated. An α-elimination occurs stepwise 1 --> 2 --> 3. There is a phenyl migration (bridged?) to afford tolane 4. Look here for this chemistry.


How to quantify the adsorption affinity of gases? Is there a term/quantity which shows how 'sticky/adsorptive' a molecule is? I am interested in gas adsorption on steel surfaces in our mass spectrometer and would like to estimate which gases have a higher propensity to adsorb.

The term you are looking for is "adsorption enthalpy". However calculating that value will depend on a variety of factors mainly the absorbate and substrate, but it is not that simple. The surface structure of substrates matters and a $[111]$ oriented crystallographic face will have a different adsorption enthalpy with a given absorbate that say a $[001]$ crystallographic face.

I suggest you study this more, namely the Langmuir adsorption model and Brunauer–Emmett–Teller Theory (BET). Especially since BET is a popular technique to measure adsorption.


The percentage of ingredients in a glass of tea When a pot, pitcher, glass, etc of tea is brewed, how do I find the exact concentrations of tea and water in the end result. I realize it depends on the amount tea and water used when brewing. Is it still called tea, albeit weak tea, when the water is more prevalent in the water to tea ratio?

You can't really talk about the 'concentration' of tea, because 'tea' is not a well defined compound. It contains a lot of different chemicals, some of which are dissolved into water when brewing.

You could talk about the concentration of caffeine, epigallocatechin gallate, or manganese in a particular cup of tea, but they depend on the type of tea, the temperature of the water, the infusion time, etc.

The minimum strength of a cup of tea that still allows it to be called tea is a fuzzy question without an answer. Take a pile of sand. Remove grains of sand one by one. When does it stop being a pile? A reasonable limit for the tea would be when a normal human being can't taste the tea anymore.


Reduction of only double bonds In the conversion shown below I need to find B . Initially I thought of protecting the carbonyl group using ethylene glycol and then carry out reduction of double bonds using hydrogenation but it is required to be a single reagent and cannot be multi step. What are the other ways in which it can be done?

You can do a normal hydrogenation reaction(H2,Pd) under low pressures as suggested by @Waylander. Alternatively you can use the Wilkinson's catalyst which also is a way to do catalytic hydrogenation but is capable of only reducing the double bond.


Alllylic bromination of 1-butene My book asks me to consider reacting 1-butene with NBS and light.

I know that NBS and light leads to radical formation, and this is an allylic/benzylic bromination problem.

However, my book only provides one product.

Aren't both products expected? Actually, I would expect one product in excess (the book actually listed my minor product as the sole product). There are at least two resonance contributors, and the more major one has the more substituted radical.

Yes you are correct, at least two products are formed (ref. 1, ref. 2).  Unfortunately, neither of these references provide the actual product distribution between the isomers.

I say that "at least two products are formed" because depending on your reaction conditions 3 products might be expected.  NBS brominations can be carried out over a range of temperatures, from cold to somewhat above room temperature.  The activation energy required to rotate about an allylic bond is only ~15-16 kcal/m (see here).  Therefore, your 1-bromo isomer could exist as a cis-trans mixture, unless the reaction is performed at very cold conditions.

In the analogous case of NBS bromination of 1-octene, a mixture of 3-bromo-1-octene (~18%) and 1-bromo-2-octene (~82%) - both cis and trans are produced.

$\ce{RCH2CH=CH2  +  NBS \rightarrow RCHBrCH=CH2  +  RCH=CHCH2Br}$ (cis & trans)


How does strong acids conduct electricity in water? How do strong acids conduct electricity in water?
I have just studied ionization of water and I'm confused about some points.
First: I know that water ionizes  in a very small amount ($10^{-14}$) so it doesn't conduct electricity and if we add strong acid (or base) to it , it becomes a good conductor of electricity. My question here is how it conduct electricity although if a strong acid is in water its $\ce{H+}$ ions concentration  in water would range from $10^{-1}$ to $10^{-6}$ (which is avery small amount) to keep $K_w$ constant.
I'm confused so if you understood me please explain and tell me if I have something wrong

In water at 25 ºC $[H+]=[OH-]=10^{-7}M$, so a concentration of $[H+]=10^{-6}M$ is very low. With a strong acid like HCl you have to dissolve only around 35 micrograms in a liter of water to get this concentration.
Sea water on the other hand has a high conductivity and, in average, 35 grams per liter of salts dissolved in it. If you dissolve 35 grams of HCl en 1 liter of water you have $[H+]=1M$ and surely you can expect a high conductivity.
These are gross numbers, but I think they help to get the idea.
Conductivity of a solution depends on the free ions present on the mixture. With a strong acid you can get a high concentration of ions using a little quantity of acid. With a weak acid you need much more.


In the periodic table, why doesn't the 2nd row have exactly 2 elements? Row 1: 2 elements

Row 2: 8 elements

Row 3: 8 elements

Row 4: 18 elements

Row 5: 18 elements

Row 6: 32 elements

Row 7: 32 elements

In other words: 2, 8, 8, 18, 18, 32, 32

Why does the first row have only 2 elements, but all the next rows have a pattern? There's a pattern of two rows having the same number of elements that only starts after the first row.

EDIT: I just noticed this alternate periodic table! It's first two rows have two elements each! Why does it work there? (I can't interpret the s, p, d, f part on the left). Which one is the right periodic table?

The pattern is better expressed this way:

Row 1: 2 elements
Row 2: 2+6 elements
Row 3: 2+6 elements
Row 4: 2+6+10 elements
Row 5: 2+6+10 elements
Row 6: 2+6+10+14 elements
Row 7: 2+6+10+14 elements

The reason comes down to how the electrons fill the available energy levels.  The thing that differentiates one element from another is the proton number, and each time a proton is added and a new element defined it requires one more electron to neutralise the charge.  That electron naturally occupies the lowest available energy level in the atom.

The energy levels available are defined by the quantum state, including the quantum numbers $n, l,$ and $m_l$.  $n$ is the principle quantum number and relates to the period the element is in, or the shell.  $l$ is the angular momentum quantum number which defines the sub shell s, p, d, f, of which there are $n$ subshells whose values are $l=0,\dots {n-1}$.  The magnetic quantum number $m_l$ further subdivides the subshell into orbitals, of which there are $2l+1$ orbitals whose values are $m_l=-l,\dots {+l}$.

subshell                 number of orbitals    subshell
label     l value       (number of ml values)  electron capacity
 s         0               1                   2
 p         1               3                   6
 d         2               5                   10
 f         3               7                   14

Each orbital (i.e. each value of $m_l$) may contain 2 electrons  The available quantum numbers are:

n    l values   (subshells)   ml values        total shell electron capacity
1    0           (s)           0                 2
2    0,1         (s,p)        -1,0,1             2+6
3    0,1,2       (s,p,d)      -2,-1,0,1,2        2+6+10
4    0,1,2,3     (s,p,d,f)    -3,-2,-1,0,1,2,3   2+6+10+14

The pattern is there, but it doesn't seem to match up because by row 7 you would have 98 electrons in the shell and might expect the row to contain 98 elements.  This is not the case!  The energy levels for the $l$ values with large deviations from $0$ (i.e. orbitals with high angular momentum) become increasingly far apart, so even though 3d orbitals exist in the third row, they are so much higher in energy than the 3p orbitals that they are higher even than the 4s orbitals.  This happens again in row 4 where 4f orbitals are so much higher in energy than 4d orbitals, they are even higher than 5s.

The different in energy is not always so large in specific cases, so there are examples where bonding or ligands or symmetry cause the energy levels to switch around, but otherwise the trend is true. If they can produce (or discover) some elements in the next row, we might expect to see the 5g shell finally start to get filled and have a new block in the table, but realistically it doesn't look like they will be stable atoms.

I think I'm missing the reason why we arrange the elements into this table in the first place.  The periodic table is arranged so that all elements in the same column have the same number of outer electrons.  This is useful because elements with the same outer shell configuration of electrons react in similar ways, so they are grouped together in columns. This has the effect of creating 'periods' that begin with a reactive metal and end with an inert gas. Row 1 has two elements, and it's special in a way because it's the first in the series.  Hydrogen has 1 electron, and a maximum capacity of 2 electrons.  This means it behaves in a similar way as other elements that only have 1 electron in their outer shell (group 1), and also in a similar way to elements that only need 1 more electron to complete their outer shell (group 7).  Where do we position it then?  Often it is placed somewhere in the middle.  Helium then has a full outer shell, which means it has many similarities to other elements with a full outer shell and goes in group 8 (or group 0) on the right.

Edit:
The alternative period table that you mention, the left-hand version, is laid out so as you read it from left to right you fill up the orbitals.  This takes in to account that situation I describe where 3d orbitals are higher than 4s, and using this left-hand table you clearly see that. However this alternative table does not allow you to easily read the valance shell configuration like the Mendelev one does.  Other types of table highlight different kinds of properties or relationships, which can be more useful in certain situations.  The labels on the left (1s; 2s; 2p 3s; etc) relate to the outer subshell in that row, so Carbon is in the "2p" section of that chart while Magnesium is in the "3s" section on the same row.


What factors could lower the voltage of a galvanic cell Recently I conducted a lab in which the voltage of a galvanic cell between two metals was recorded. Both metals were placed in an solution of their own ions with nitrate and a salt bridge was made by dipping some paper in salt water. The only issue is the voltage recorded in the cell was lower than that calculated using standard reduction and oxidation potentials. Does anyone have any ideas as to why this might be? I known the metals were not super clean although I don't think that would have an affect on the voltage.

Although it seems simple to make a galvanic cell ( i.e. a battery) it is really fairly complex.

The theoretical voltage is given by the Nernst equation.
The actual measured voltage also depends on the current. The equation
$V=iR$ still applies. So various restance sources in the cell cause voltage drops.

    The current must be very low so a high impedance voltmeter must be    used. (Not really a problem these days with modern semiconductor
    voltmeters as opposed to galvanometers.)
    The salt bridge offers resistance, so it has a voltage drop.
    Impure metals for the anode or cathode can cause a voltage change.
    Surface coating can cause a voltage drop.
    The acid or base concentration can cause a voltage change.
    The ionic strength of the solution can change the voltage.


What happens to magnesium in water with acid? I have a pure hunk of magnesium metal submerged in water. After several hours/days, the metal becomes coated with something.

Question 1:
What is the coating?

I've been told it's magnesium oxide. This would make sense to me if it was exposed to air, but it's submerged in water. Is it still MgO? I'll assume there is enough free oxygen in the water to react slowly.

If I add an acid to the mix, the coating gradually disappears. I use some malic acid crystals. Where does the MgO go?

Question 2:
What is in the water when the coating disappears?

As the coating disappears to expose the metal, the magnesium reacts more and more vigorously offgassing hydrogen, which I understand would leave magnesium malate in the water. But what about the MgO? Does the acid react with MgO? And either way, where does it go? Is the MgO just dissolved invisibly in the water or is it reacting to form something else?

In water Mg oxidizes and the half-reactions are:
$$\ce{Mg -> Mg^{2+} + 2e-}$$
$$\ce{2H2O + 2e- → H2 + 2OH-}$$

As the E($\ce{Mg}$/$\ce{Mg^{2+}}$)>E($\ce{H2O}$/$\ce{H2}$), Mg oxidizes and water reduces.

Metal ions can change their speciation depending on the pH. In acidic conditions $\ce{Mg^{2+}}$ dissolves and associates with malate (which in solution is deprotonated) and end up precipitating because the malate is in high concentration and the solution is saturated. And, in neutral o basic conditions they precipitate as metal hydroxides which are quite insoluble.

What you have to take into account is that metal ions can precipitate in basic conditions and dissolve in acidic ones.
Also, water has a reduction potential and has to be taken into account in rusting processes with metals. You can look for more information with Pourbaix diagrams where you can see the changes in speciation of metals in front of pH and the reduction potential of several metals.

Here's the Pourbaix diagram of Mg:

I hope this helps!


Darzens reaction on allylic position Question

Attempt

I thought the products would be from simple darzens reaction. So   I replaced the OH by Cl. But the answer of mine seems way wrong. Do give some tips on what might be happening in the reacton.
Please help

This reaction is a special case of Darzen's halogenation. I think you are aware of mechanism of Darzen's reaction or else you can check it here. It occurs through a 6 membererd cyclic intermediate. I'll show you the first reaction you would be able to answer second one in succession.


How does lead toxicity from jewelry happen vs. occupational exposure? It is known that lead levels in jewelry is a problem, especially for children (https://dtsc.ca.gov/toxics-in-products/lead-in-jewelry/). As a parent with jewelry-loving children I am aware that ingestion or inhalation of dust or any part of the jewelry will be toxic, while simply touching it would not be. However, several sites claim it is dangerous if one touches the jewelry and then touches their eyes, nose, mouth, food, etc and propose hand washing as a solution.
First, I am wondering what form the lead is in when it's found in jewelry metal alloys (not painted) and how exactly it gets to be toxic from there or what happens when you touch it (ions, oxidation?). I can't find any other information online about the form of lead in jewelry. I know it's inorganic lead and was told by friends it's elemental lead*.
Second, I am wondering what is the best way to remove the lead from jewelry exposure from skin? I've read that hand washing with soap and water is insufficient to remove lead from skin according to CDC, NIOSH, and such studies. But these were all focused on PbO from occupational exposure, and not on jewelry exposure (which I presume is not PbO). Is there some difference here that makes regular soap and water effective in this case?
*After reading similar posts on here and here, elemental and metallic lead (what's the difference?) even if ingested does not seem to be as toxic as ionic Pb(II) and covalent compounds of Pb, but there have been and continue to be cases of lead poisonings involving jewelry items, so either it is not elemental lead or elemental lead is more toxic than indicated here.

Metallic lead is comparatively inert, but after exposure to air it quickly develops a film of lead compounds, including litharge, $\ce{PbO}$, basic lead sulfate, $\ce{Pb2SO4(OH)2}$, etc. Though these are not very soluble, sweat and bacterial action can convert them to much more soluble compounds.
Lead is also very soft, so touching an article made of lead leaves a residue on the fingers, similar to the way using an aluminum-handled tool leaves the hands covered in gray aluminum and its compounds.
The danger is usually not immediate, but rather the accumulation of toxic lead in the body. However, lead worn next to the skin is steadily abraded. Also, children (and pets) put things in their mouths or even swallow them. Stomach acids do dissolve lead, and there is high mortality in waterfowl from swallowing lead pellets.
This is why lead has no place in play things (or perhaps in most products, since it will almost invariably leach out to poison the environment).
That said, this is an abstract comment, and not to be construed as a way to "safely" use lead in toys.


Forcing a steel pipe to rust completely I understand rust.  Fe + H20 → OH- → Fe(OH)2 or 3 and/or FenOm
I understand rust prevention and steels that are resistant to rusting.  Trying to google this question gives thousands of answers on prevention.
But what I'm interested in is how to force steel to rust.  And not in an artsy way, so my steel stuff looks weathered and cool (these are the other thousands of search results for forcing iron/steel to rust).  I want to dispose of steel by turning it into iron oxide.  I don't actually really want to do this since it's a lot easier to just landfill or recycle, but I thought of it as an alternative and realized it's really not that easy.
The trouble seems to me that the passivation layer stops rust from completely going completely through the metal.  Can this be forced by running active corrosion circuit in reverse, or will the equation still slow to a crawl once a layer of rust builds up on the surface?

As we in the trade say "rust never sleeps". Rust develops on the surface , rust is somewhat protective in some "weathering" steels such as Corten. Small additions of Cu ,Cr, and P , mostly provide limited slowing of rust . Otherwise there are many tables , based on conditions and location of exposure . The tables will give rust rates like 5 mpy ( mils per year ; mil is a thousandth of an inch , pardon my US units). The loss rates will be much higher on seacoasts or where rain often occurs. So with no other information ,you can guesstimate your steel will disappear in (THICKNESS divided by 0.005 ) years. Half that time for two sides exposed.


Why doesn't separating the layers change the equilibrium point? I'm following this tutorial for synthesizing esters. I don't understand why adding water to the separation funnel doesn't push the equilibrium reaction point back to the left, or why draining it doesn't push the reaction to the right, given the tutorial's explanation of Le Chatelier's Principle:

When the reaction reaches equilibrium there is still a large amount of reactants left in the mixture resulting in a poor yield of the ester.
The yield of ester can be improved by increasing the concentration of one of the reactants (either the alcohol or the carboxylic acid).
By Le Chatelier's Principle an excess of one reactant will drive the reaction to the right, increasing the production of ester, and therefore increasing the yield of ester.
In our experiment we are using an excess of carboxylic aicd (acetic acid which is also known as ethanoic acid).
From the balanced chemical equation, you will also appreciate that the presence of water in the reaction mixture will drive the equilibrium to the left, favouring the formation of reactants rather than ester.

Draining the aqueous phase cannot change anything. If the phases are in equillibrium before (both internally and with each other), they stay that way if they are separated.
Now if you add water to the aqueous phase, the reaction in the unpolar phase (which is saturated with water, as the phases are in equillibrium) is in the first moment not affected.
However the acid, alcohol and ester have different solubilities in the aqueous phase. That means the concentrations in the unpolar phase will change a bit, and so does the reaction equillibrium.
But: This is all irrelevant, because the reaction happens under reflux. While you are standing there with the cooled mixture in the separation funnel, the reaction rate is much lower, you will never notice a change.


Three stacked arrows, what does it signify? I was checking the mechanism for the Lossen Rearrangement, when I came across the chemical reaction denoted as follows:

I couldn't find any resource where this notation was talked about.
What do the three arrows stacked as follows mean? A possible example where this is applied in literature would be appreciated.

There is an explicit answer in section 3. Other Horizontal Arrows of prof. Santiago Alvarez' published work [1, p. 594]:

[…]
Finally, multiple arrows (7) are used as an ellipsis for several intermediate steps in a chemical reaction.

Reference

Alvarez, S. Chemistry: A Panoply of Arrows. Angew. Chem. Int. Ed. 2012, 51 (3), 590–600. DOI: 10.1002/anie.201101767.


A simple question about the direction of a chemical reaction I do not know much about Chemistry and whatever I knew has become rusty. I have a very simple query. I take a beaker containing pure $\ce{H2SO4}$ solution and immerse a rod of pure $\rm Zn$ into it. Now, the $\rm Zn$ will react with $\ce{H2SO4}$ to form $\rm ZnSO_4$ and hydrogen gas.
$$\ce{H2SO4(aq) + Zn(s) -> ZnSO4(aq) + H2(g)}$$
Why does not the reaction spontaneously reverse once again forming pure $\rm Zn$ and $\ce{H2SO4}$ molecules?

$$\ce{H2SO4(aq) + Zn(s) <- ZnSO4(aq) + H2(g) ?}$$

If your answer contains any technical jargon of Chemistry, please explain that in simple terms.

Before I talk about the chemistry, I will use an analogy from physics. If I drop a piece of chalk, it falls down. It does not at some point later "fall up" again. However, the atoms inside the piece of chalk do go up and down a bit because of thermal vibration. The chalk does not start jumping up because these vibrations are random rather than synchronized.

  Why does not the reaction spontaneously reverse once again forming pure Zn and H2SO4 molecules?

If you are asking why at no time the mass of elemental Zn (metal) increases again, it is because there is a strong driving force toward turning the reactants into products for this specific reaction. If you are asking about individual particles, it turns out every now and then, the reverse reaction does occur (this is true for all reactions and is called micro-reversibility).

For some reactions, this back and forth results in the reaction not going to completion. Instead, these reactions may reach an equilibrium where reactant and products are present, and the forward and reverse reactions go on at the same speed so nothing seems to change (this state is called a dynamic equilibrium).

In the reaction you are asking about the driving force forward is too large, and one of the products continuously "escapes" from the reaction mix, so the reaction goes on until one of the reactants is used up (goes to completion rather than reach equilibrium).

In any reaction, reversible or not, once the reaction reaches equilibrium or goes to completion, it does not spontaneously go in the opposite direction unless something changes (add materials, change temperature, do work).


Why does the photoelectric effect depend on the frequency of the light? In experiments to detect the photoelectric effect, a clean metal was irradiated by monochromatic light and electrons were emitted.

Why was monochromatic light used in the experiment, and why does the frequency of the light have to be above a threshold frequency?

The photoelectric effect is described by the following equation

$$E_\mathrm{max} = h\nu - \mathrm{WF_M}$$

where $E_\mathrm{max}$ is the maximum kinetic energy of the electron escaping from the metal surface, $\nu$ is the frequency of the incoming photon and $\mathrm{WF_M}$ is the workfunction for the particular metal. The kinetic energies of all electrons emitted are distributed from $0$ to $E_\mathrm{max}$.

  Why was monochromatic light used in the experiment? Why is a proper
  frequency used and not any other frequency?

The experiment is typically performed by scanning through a continuous range of monochromatic wavelengths from lower to higher energy.  At some specific wavelength, the observer will notice that electrons start to be emitted (the threshold).  As the scan continues to wavelengths with even more energy, the emitted electrons will increase in kinetic energy.  From the threshold energy and the above equation, the experimenter can determine the workfunction of the metal.  If the experiment had been run with light containing many different wavelengths (non-monochromatic light), electrons would still be ejected, but you wouldn't know what the threshold wavelength was and you wouldn't be able to determine the workfunction of the metal being studied.

If you'd like more information on the photoelectric effect, here's a good, concise reference.


Why CO2 has 4 vibrational degrees of freedom? If I have a diatomic molecule, so a linear molecule, $3N-5=6-5=1$ holds, since this vibrational degree of freedom corresponds to the stretching coordinate beetwen the two nuclei.

If I consider water for example, I have a nonlinear triatomic molecule, so the formula says $3N-6=9-6=3$ vibrational degrees of freedom. Also in this case it makes sense since they correspond to the bending angle and to the $2$ stretching lenghts.

While, If I consider $\ce{CO2}$, in this case I have a triatomic linear molecule, so $3N-5=9-5=4$ vibrational degrees of freedom. Of course $2$ of these correspond to the two stretching lenghts, but I really don't understand what the other two correspond to. Since the molecule is linear, there shouldn't be angle to consider, otherwise it would not be linear anymore.

Molecular vibrations can certainly occur by changing the O-C-O bond angle away from 180º. When we call the molecule linear we are saying that the geometric minimum is a linear configuration; however, vibrational modes can (and do) break that symmetry. What's important is that the average nuclear position is a linear molecule (or if you're doing a geometry calculation you would say that it's the 0 K positions).

So the two vibrational modes you're missing are the two O-C-O bends at 90º from each other.

Also as a note: you seem to be suggesting that the two stretches correspond with the two bonds stretching. That's not quite true. One of them corresponds to both C-O bonds stretching outward at the same time (in phase) and the other will correspond one compressing while the other stretches (out of phase).

For a full derivation of the normal modes of CO$_2$ there are many sites, but this one looked good: http://www.colby.edu/chemistry/PChem/notes/NormalModesText.pdf


What happens when you mix Pine Sol with Clorox I have read all these rumors online that mixing these two solutions will produce the chloride gas. I accidentally mixed these two while doing berthing cleaners in the communal restroom and felt very light headed afterwards.

Pine-sol contains glycolic acid (1.75%) and the other ingredients are C10-12 alcohol ethoxylates, dimethicone/silica/PEG distearates, sodium C14-17 secondary alkyl sulfonates.

My Clorox contains 8.75% sodium hypchlorite.

Are the culprits in producing chloride gas glycolic acid and sodium hypochlorite? I'm not too familiar with chemistry and I need someone who took college organic chemistry to explain this to me. I'm in the military so I can't take physical classes. I would also like to know the mechanism.

There is no reason to believe that there would be a chemical reaction between any of the components of the Pine Sol with sodium hypochlorite that would produce noxious compounds.

The most likely culprit in this case is using Pine Sol without adequate ventilation. As you've indicated, it is loaded with volatile and semi-volatile organics, which could cause the type of central nervous system (CNS) symptoms you described.  This is the reason for the warning on the bottle to only use in well ventilated areas.

Chlorine gas on the other hand attacks the eyes, throat and lungs.  You would not feel CNS effects from chlorine without having noticed these things first.  Also, though mixing anything with bleach is a bad idea, Pine Sol contains only a pretty small amount of a weak acid, which results in the release of a bit more chlorine gas from the sodium hypochlorite.

Please don't heasitate to ask for clarifications in the comments below. I hope I understood the question correctly.