Why does carbon monoxide have a greater affinity for hemoglobin than oxygen? Hemoglobin is an iron-containing oxygen transport metalloprotein in the red blood cells of most mammals. Simply put, it's a carrier protein. Interestingly it doesn't carry carbon dioxide in the same way it does for oxygen $\ce{O2}$. Oxygen binds to the iron atoms in the protein whereas carbon dioxide $\ce{CO2}$ is bound to the protein chains of the structure. Carbon dioxide doesn't compete with oxygen in this binding process.

However, carbon monoxide $\ce{CO}$ is a very aggressive molecule. It's a colourless, odourless, and tasteless gas that is lighter than air and can be fatal to life. It has a greater affinity for hemoglobin than oxygen does. It displaces oxygen and quickly binds, so very little oxygen is transported through the body cells.

There are two equilibrium reactions of binding oxygen and becoming oxygenated hemoglobin:

$$\ce{Hb (aq) + 4O2 (g) -> Hb(O2)4 (aq)}$$

$$\ce{Hb (aq) + 4O2 (g) <- Hb(O2)4 (aq)}$$

And carbon monoxide binding equation at equilibrium:

$$\ce{Hb (aq) + 4CO (g) ⇌ Hb(CO)4 (aq)}$$

It is said
 the equation is shifted towards right, generating $\ce{Hb(CO)4(aq)}$, since its bond is much stronger.

I have two main related questions:

Why is the carboxyhemoglobin bond stronger relative to that of oxygenated hemoglobin?
Why is carbon monoxide highly attracted to hemoglobin?

Does it have anything to do with the oxidation state of oxygen in each molecule? Please show some reaction equations associating $\ce{Fe}$.

Excursion into simple coordination chemistry: Bonding, backbonding and simple orbital schemes

Please refer to Breaking Bioinformatic’s answer for the MO scheme of carbon monoxide, it is very helpful. You might also look at the orbital pictures in this answer by Martin. Carbon monoxide can bind to metal centres via a σ coordinative bond where the HOMO of $\ce{CO}$ interacts with metal orbitals and also by the π backbonding, Breaking Bioinformatics mentioned. I’ll start by touching the σ bond so we can later better understand the π bond. In figure 1 you can see the molecular orbital scheme of a complex composed of a central metal ion and six ligands that donate in a σ manner exclusively.

Figure 1: Molecular orbital scheme of an octahedral complex with six σ donors around a central metal. Copied from this site and first used in this answer of mine. Metal orbitals are 3d, 4s, 4p from bottom to top; ligand orbitals are of s-type.

You will notice that figure 1 contains the irreducible representations of the orbitals beneath them. Orbitals can only interact if they have identical irreducible representations; otherwise, their interactions will sum up to zero. We can see that the metal d orbitals are split up into a $\mathrm{t_{2g}}$ set and and $\mathrm{e_g}^*$ set. This is the effect of σ bonding and why it stabilises the entire entity.

π bonding can only occur if the ligands have available orbitals of π symmetry. The 2π orbitals of $\ce{CO}$ are a nice example, but one could also simply assume a halide with its p-orbitals for the same effect. Further down in the internet scriptum I originally copied the picture from, you can see a set of two pictures that introduce p-orbitals. Twelve ligand p-type orbitals will transform as $\mathrm{t_{1g} + t_{1u} + t_{2g} + t_{2u}}$, thus a further stabilising/destabilising interaction is introduced for the $\mathrm{t_{2g}}$ orbitals. Due to the $2\unicode[Times]{x3c0}^*(\ce{CO})$ orbitals being similar in energy to the metal’s $\mathrm{t_{2g}}$ and empty, they can stabilise the former well, creating a possibly strong stabilisation for the overall system. Since this is an interaction of two orbitals creating both bonding and antibonding orbitals, and since the resulting molecular orbital is filled with two electrons that stem from the metal centre this is termed π-backbonding.

The binding mode of carbon monoxide to haemoglobin

After this extensive background discussion, it may be clear that $\ce{CO}$ is generally a good ligand. For reasons I didn’t go into, a carbonyl complex of metal ions isn’t that stable, but a single carbonyl ligand will almost always be beneficial. Such is the case for the porphyrin-iron(II) system that forms the heart of haemoglobin: The central iron(II) ion is coordinated well from five directions (four from the porphyrin ring and one histidine of haemoglobin) and has a weakly bound water atom in the ground state in its sixth coordination slot, sometimes displaced by a distal histidine. Carbon monoxide can diffuse in and bind very well to this system, displacing the weakly bound water and histidine. In fact, isolated haem can bind carbon monoxide $10^5$ times better that it can oxygen.

This is something that nature could not really inhibit, because it is based on fundamental properties, but also something nature didn’t really care for, because the natural concentration of carbon monoxide is very low, and nature rarely had to deal with it competitively inhibiting oxygen binding to haemoglobin.

The binding mode of oxygen to haemoglobin

Oxygen, the $\ce{O2}$ molecule, is an absolutely lowsy σ donor — especially when compared to $\ce{CO}$. Its molecular orbital scheme is generally that of carbon monoxide except that it is entirely symmetric and two more electrons are included: They populate the 2π orbitals to give a triplet ground state. These orbitals are now the HOMOs and they hardly extend into space in any meaningful way — plus the still existant lone pair on each oxygen atoms is now at a much lower energy and also does not extend into space and thus cannot bind to a metal centre in a σ manner.

What happens here is rather complex, and the last lecture I heard on the topic basically said that final conclusive evidence has not yet been provided. Orthocresol discusses the different viewpoints in detail in this question. The diamagnetic properties of the resulting complex are, however, unquestioned and thus one must assume a singlet ground state or one where antiferromagnetic coupling cancels any spins at molecular levels. Since the ground state of haemoglobin has a high-spin iron(II) centre and the ground state of oxygen is a paramagnetic triplet, it makes sense to assume those two to be the initial competitors.

Professor Klüfers now states the steps to be the following:

  Iron(II) is in a high-spin state with four unpaired electrons;
  Paramagnetic triplet oxygen approaches and one of its $\unicode[Times]{x3c0}^*$ orbitals coordinates to the iron(II) centre in a σ fashion.
  This induces a high-spin low-spin transition on iron and slightly reorganises the ligand sphere (pulling oxygen closer to the iron centre). We now have a low-spin singlet iron(II) centre and a coordinative σ half-bond from oxygen. We can attribute that electron to the iron centre.
  Through linear combination, we can adjust the spin-carrying σ orbital and the populated iron orbitals so that an orbital is generated which can interact with the other $\unicode[Times]{x3c0}^*$ of oxygen in a π fashion.
  We thus expect an antiferromagnetic coupling and an overall state that can be described best as $\ce{Fe^{III}-^2O2^{.-}}$ — a formal one-electron oxidation of iron to iron(III), reducing oxygen to superoxide ($\ce{O2-}$).
  If the spin-carrying orbitals are all treated as being metal-centred, we gain a $\ce{Fe^{II}-^1O2}$ state.

(Content in the link I quoted from are in German; translation mine and shortened from the original.)

Only due to the complexly tuned ligand sphere and also due to the stabilising high-spin low-spin transition (plus reorganising) is oxygen able to bind to iron at all. The distal histidine further stabilises the complex by a hydrogen bond, alleviating the charge slightly. It is to be assumed that nature did a great deal of tuning throughout the evolution since the entire process is rather complex and well-adjusted for collecting oxygen where it is plentiful (in the lungs) and liberating it in tissue where it is scarce.

Comparison of the oxygen and carbon monoxide binding modes

The simpler picture I drew above for carbon monoxide is not correct. Inside haemoglobin, carbon monoxide also binds in an angular fashion as if it were oxygen — see bonCodigo’s answer for space-filling atom models. This is because the entire binding pocket is made to allow oxygen to bind (as I stated) thus attempting everything to make oxygen a comfortable home. Carbon monoxide is rather strained in there, its binding affinity is reduced by a factor of $1000$. However, since we started with a binding affinity difference of $10^5$, carbon monoxide can still bind $100$ times better than oxygen can.

Summary

Carbon monoxide is generally a good ligand that can bind to metal centres well.
Oxygen is generally a poor ligand.
Nature did everything to make oxygen a comfortable home in haemoglobin.
In doing so, the binding pocket became substantially less comfortable for $\ce{CO}$.
But since carbon monoxide was so good and oxygen so poor to start with, the former still binds better than the latter.
Nature didn’t really care, because individuals seldomly come in contact with carbon monoxide so the collateral damage was taken into account.


Stability comparison on the basis of violation of Bredts rule between 2 bicyclo carbocations In the following question ,

Though in part A ,
both the compounds I and II seem to be unstable because of violation of bredt's rule ; that is ; they are sp2 hybridized .

Then how to compare the stability between the two bicyclo compounds?

The carbocation's energy would be effectively minimised when it adopts an $\ce {sp^2}$ hybridisation scheme, that is, when it adopts a trigonal planar molecular shape. This would ensure that all the precious s orbital character of the carbon atom would be used for bonding with the other three surrounding atoms, leaving a vacant p orbital. However, this bicyclic structure prevents the carbocation from adopting this scheme, destabilising the carbocation formed. This is the basis of Bredt's rule.

On p. 435, Carey & Sundberg (2007) wrote the following:

  ... bridgehead carbocations become accessible in more flexible structures. The relative solvolysis rates of the bridgehead bromides 1-bromoadamantane, 1-bromobicyclo[2.2.2]octane, and 1-bromobicyclo[2.2.1]heptane illustrate this trend. The relative rates for solvolysis in 80% ethanol at 25 degrees Celsius are shown.

When there is greater flexibility in the structure, the carbocation is better able to adopt a structure that is closer to the preferred planar structure, thus it is better able to be minimise its energy. This results in a lower activation energy and an increased rate of reaction, explaining the data above.

Reference

Carey, F. A., & Sundberg, R. J. (2007). Advanced Organic Chemistry Part A. Structure and Mechanisms (5th ed.). Springer.


Why glucose and other similar carbohydrates are oxidised fully by HIO4? $\ce{HIO4_{(aq)}}$ requires the vicinal diols to have syn configuration, as it forms a cyclic periodate ester.

But glucose does not have syn configuration at C3, and if you consider cyclic forms of glucose it shouldn't oxidise fully at all because alternate hydroxy groups with anti configuration will be unreactive. But it still gets oxidised by $\ce{HIO4}$ to give 5 molecules of methanoic acid, $\ce{HCOOH}$, and one molecule of formaldehyde, $\ce{HCHO}$.

Glucose can react with $\ce{HIO4}$, which can be understood easily by looking at chair conformation of glucose.

Here is example of a diol, for more information looks at this answer https://chemistry.stackexchange.com/a/114836/67254


How do orbitals coexist with a nucleus? Many diagrams of orbitals I've seen involve electrons moving through a center point—where the nucleus is. How can this be? Clearly they don't actually pass through the nucleus, so what does happen?

In the definition of an s shell, you will find that its $\ell$ number is zero. In classical terms, that corresponds to an orbit with zero orbital or angular momentum — which for a large object is a clear impossibility. For an electron, it gives the peculiar result that any electron in any s shell is, classically speaking, moving back and forth through the nucleus, rather than around it. So, in one curious sense, what you just asked is exactly what happens: the classical analogy is that the electrons do go through the nucleus, which is why they have such a lovely spherical symmetry.

The second part of the answer, however, is that electrons can't go through the nucleus unless they are hugely more energetic than the ones found in a typical small-nucleus atom. Clearly there's a bit of a paradox going on there!

The resolution of the paradox is that very-low-mass charged particles must be treated by quantum rules. So, for example, rather than the electron behaving like a well-defined particle, it behaves like a standing wave. That standing wave can in turn be thought of as two simultaneous versions of the electron, one going (for example) clockwise and the other counterclockwise around. (The real situation has an infinite number of such components; I'm just picking out a single pair of them that demonstrate the principle.)

Each of these components can furthermore be thought of as being refracted by the powerful spherical charge field of the nucleus, curving around it without every striking it. This refraction is not the same as an attraction. In fact, it is this refraction effect that prevents the density of the electron cloud from reaching infinity at the nucleus — that is, from striking the nucleus. If you think of how a tank of water can cause a beam of light to bounce off the surface instead of entering the tank — and that's a terrible analogy, I know, I know — you can at least get some idea of how an increasing "optical density" towards a central point might keep light away rather than bringing it closer.

So for an electron that behaves "like" two waves going both clockwise and counterclockwise, the combined waves curve around the nucleus rather than striking it. This is a very quantum sort of event, since for a classical object such "splitting" of the object is simply not possible and the object simply dives straight into the source of attraction. But if objects are light enough, that kind of particle-like behavior simply ceases to be available to the object. Instead, you get waves that neatly and with perfect spherical symmetry curve around the nucleus, never obtaining enough energy (which makes it more particle-like) to connect directly with that nucleus.

Finally, notice that electrons in s (and other) shells necessarily combine multiple paths all at the same time. For each "image" of the electron that is traveling clockwise, there must also be an exactly balancing "image" of the same electron traveling counterclockwise, so that the two images always balance out to zero orbital momentum. What an amazing thing that is! And an important thing, too, since it's what makes chemistry possible.

So, good question, even if it really is more of a physics question per se than a chemistry question. But it's such an important chemistry question! It's like asking how the engine that powers the car works. You can accept it as a given that all cars and vehicles have engines, and that they all work in a certain way. However, sometimes it's nice to dive in a little deeper and try to understand why these peculiar things do the things that make chemistry possible — that is, how the engine really works.


Does the diiodide(1−) anion exist? With respect to chemical bonding does the $\ce{I2-}$ ion exist?

I found a paper$\ce{^{[1]}}$ regarding the formation of diiodide anion species($\ce{I2^-}$). It is assumed to be an unstable species which forms during the formation of triiodide anion (excess of iodide in solution which makes it brown). This is a part of the paper which describes the formation of the species. For more information, read the full paper. Also, you can check the abstract which @andselisk linked.

The $\ce{I2^-}$ species is formed by the 248-nm laser photolysis of
iodide through the following reactions:
$$\ce{I- + hν -> I + e_s-}$$
$$\ce{I + I- ->[k2]I2^-}$$
$$\ce{I2- + I2- ->[k3]I3- + I-}$$
As discussed later, we found that photodetachment of electrons from
$\ce{I-}$ ions produces solvated electrons ($\ce{e_s-}$) in ionic
liquids as well as in molecular solvents. After photodetachment,
iodine atoms react with $\ce{I-}$ to form diiodide anion radicals. In
aqueous solution, the transient absorption maxima of $\ce{I2-}$ are
located around 400 and 720 nm. The extinction coefficients of
$\ce{I2-}$ in aqueous solution at 385 and 725 nm are 10000 and 2560
M-1cm-1, respectively.[...]

$\ce{^{[1]}}$: J. Phys. Chem. B, 2007, 111 (18), pp 4807–4811


Sugden's concept of singlet linkage I was reading about the octet rule and saw exceptions in it. Different theories I read further and one of those was Sugden's concept of singlet linkage which state that the octet rule is never violated.

My question is that why Sugden's concept is not taught us or is not as popular as covalent bond or dative bond? Is there any problem with this concept? Does it have any limitations?

What he said
He did not say that the octet rule is never violated.
On the contrary, he suggested that "the maximum number of electrons in the valency orbit can not [sic] exceed eight". (source)
This means, while he denies that the octet rule is violated by molecules like chlorine pentafluoride, he did not challenge whether molecules traditionally labeled as octet-deficient such as beryllium difluoride indeed violates the octet rule.
He argued for the existence of a two-center one-electron bond.
Why his theory was poorly accepted
Preconceptions?
The success of his theory
To quote Wikipedia:

In the 1940s and 1950s, Rundle and Pimentel popularized the idea of the three-center four-electron bond, which is essentially the same concept which Sugden attempted to advance decades earlier; the three-center four-electron bond can be alternatively viewed as consisting of two collinear two-center one-electron bonds, with the remaining two nonbonding electrons localized to the ligands.

Indeed, the three-center-four-electron explains the bonding in the triiodide anion well.
Also, another success is that this theory, unlike the expanded-octet theory, does not rule out the existence of hypervalent second-period element, such as carbon ("Frozen" Transition States: Pentavalent Carbon et al., J. C. Martin, Science, 05 Aug 1983, Vol. 221, Issue 4610, pp. 509-514, DOI: 10.1126/science.221.4610.509).
The problems with this theory
He was reluctant to accept that ionic characters do indeed play a larger in some molecules traditionally labelled as hypervalent, such as the sulfate anion.
The failures of this theory
In the sulfate anion, there is no two-center one-electron bond (Wikipedia):

For sulfuric acid, computational analysis (with natural bond orbitals) confirms a clear positive charge on sulfur (theoretically +2.45) and a low 3d occupancy.

Also, some of the traditionally-labeled hypervalent molecules involve very electronegative elements such as fluorine for sulfur hexafluoride, oxygen for sulfate, and chlorine for phosphorus pentachloride. A better explanation for these molecules would need to invoke ionic characters instead of the covalent singlet linkage.
For sulfur hexafluoride (source):

The sulfur d orbitals are found to contribute very strongly to the binding energy ($250$ kcal/mol) and to have a  total occupancy of around $0.25\mathrm{e}$, the valence electron distribution on sulfur being about $32\%$ in $3\mathrm{s}$, $59\%$ in $3\mathrm{p}$, $8\%$ in $3\mathrm{d}$, and $1\%$ in $4\mathrm{p}$. The occupancy of the two sulfur $3\mathrm{d}_{\sigma}$ orbitals ($0.16\mathrm{e}$), however, is only one-sixth of what would be required for $\mathrm{s}\mathrm{p}^3\mathrm{d}^2$ hybridization, and the energetic contribution of these orbitals is only two to three times larger than that of the sulfur $3\mathrm{d}_{\pi}$ orbitals. The sulfur $\mathrm{d}$ orbitals are important because they allow strong back transfer from the negatively charged fluorine ligands to the strongly positively charged ($+2.9\mathrm{e}$) sulfur, in turn allowing significant contraction of the $\ce{S-F}$ bonds and greatly increased molecular stability.


How to find the alcohol content of a probiotic liquid? Probiotics are microorganisms that are believed to provide health benefits when consumed. Water Kefir is one such probiotic drink. This has blend of bacteria, yeast (BY), sugar water and the result of action of BY on sugar water. There is alcohol. One of my problem is to find out the alcohol content in the solution. Less expensive, simpler methods/protocols are much appreciated.

Complex instrumental analytics (HPLC, photometry using a $\ce{NAD+/\,NADH}$ testing kit, etc.) is out of question.

In principle, the most simple approach would be the use of a hydrometer to measure the density before and after fermentation.


Relative lowering of vapour pressure Why is lowering of vapour pressure (Po - Ps) temperature dependent? And why is relative lowering of vapour pressure (1 - Ps/Po) independant of temperature?

In the simplest terms and most convenient definitions, the lowering of vapor pressure is not a colligative property while the relative lowering of vapor pressure is indeed a colligative property.

A bit more explanation: A colligative property is dependent on the moles of solute only. The ratio in the definition you provided eliminates the temperature dependence of formulation (assuming ideal solutions) by making the measurement relative to another quantity, which in this case is the vapor pressure of the pure solution. Otherwise, the influences on vapor pressure are multifactorial. I.e. external pressure may affect it, as well as temperature.

Here is a somewhat hand-wavy out-of-the-pocket calculation to demonstrate. Assume the vapor pressure can be expressed as an ideal gas. Therefore pressure may be expressed using the ideal gas law, $P = \frac{nRT}{V}$. Also know Raoult's law for vapor pressure, $P_s = \chi_s P_o$ where $\chi_s$ is the mole fraction. Using your definitions, we have for absolute pressure difference
$$P_o - P_s = P_o - \chi_s P_o = \left(1-\chi_s\right)P_o = \left(1-\chi_s\right)\frac{n_oRT}{V}$$
This is therefore dependent upon temperature. Now do the same for the relative difference in pressures
$$1-\frac{P_s}{P_o} = 1 - \frac{\chi_sP_o}{P_o} = 1-\chi_s$$
The expression is now dependent upon the moles only, i.e. a colligative property. Of course, in reality this depends on the assumptions we made under ideal Raoult conditions.


Why can we disregard stoichiometric coefficients when computing reduction potentials? A $\ce{Cr(s) | Cr^3+ || Cd^2+ | Cd(s)}$ galvanic cell has redox half reactions:
$$
\begin{align}
\ce{Cd^2+ + 2e- &-> Cd(s)} &\quad E^\circ &= \pu{-0.4 V}\tag{R1}\\
\ce{Cr^3+ + 3e- &-> Cr(s)} &\quad E^\circ &= \pu{-0.7 V}\tag{R2}
\end{align}
$$
Balanced reaction equation then is
$$\ce{3 Cd^2+ + 2 Cr(s) -> 2 Cr^3+ + 3 Cd(s)}\tag{R3}$$
$$E_\mathrm{cell} = E_\mathrm{red} + E_\mathrm{ox} = \pu{-0.4 V} + \pu{0.7 V} = \pu{0.3 V}$$
Why is $E_\mathrm{cell}^\circ$ independent of molar coefficients $(3,2)?$

You have to be careful of the difference between $E^\circ$ and $E$. $E^\circ$ is the standard electrode potential at defined conditions, some of which are mentioned in GODPARTICLE's answer. Standard thermodynamic conditions are usually

A temperature of 298 K
A pressure of gaseous components of 1 atm (or 1 bar)
A concentration of 1 M

The reference reduction potentials you quote are all standard potentials. These are the contributions to the cell potential when the half-cells meet the criteria above.

$E$ is the cell potential at all other conditions. The correction of $E^\circ$ to $E$ involves temperature, pressure, concentration, and stoichiometry. Behold the Nernst equation:

$$E=E^\circ - \dfrac{RT}{nF}\ln{Q}$$

$R$ is the ideal gas constant (this is thermo after all)
$n$ is the number of electrons transferred - here is the stoichiometry!
$F$ is the Faraday constant
$Q$ is the reaction quotient - which accounts for changes in pressure and concentration


Is it possible to store moist air at high pressure? Is there any way to store moist air inside a high pressure tank?
Let's suppose that I have humid air (30%) at room temperature and I want to store it at 300Bar in a tank. when I extract the fluid from the tank (isothermal expansion) to bring it back to ambient pressure will I keep the same percentage of humidity?
In most cases where it is necessary to store high pressure air to be used in a system that requires humid air, a cylinder is used where ambient air is stored which then pass into a humidifier. I'd like to know if it is possible, conceptually, to remove the humidifier and directly store moist air in some way

There is one importing principle to consider.
Air does not store vapor. Space does. Less space means less vapor capacity, no matter how much of air is there. (*)
If air is being compressed, the partial vapor pressure increases. It it reaches the saturated vapor pressure at given temperature, water vapor starts to condensate. When air is then expanded, its  absolute and relative humidity decreases, according to the expansion ratio.
E.g. Lets have 1 cubic metre of compressed air with 100% relative humidity. If it expands to 2 cubic metres and temperature raises back to the initial temperature, its relative humidity will be 50 %.
So compression provides liquid water and expansion needs liquid water to restore humidity.
If there is compressed air with cold water and if pressure drop e.g. 20 to 1 atm) then its relative humidity suddenly drops from 100% to 5% (when air warms up back after adiabatic cooling down.). Water exposed to low humidity air will start slow evaporation.
It will not help much in humidity restoration, unless some active evaporation technique is involved a/o the air release is slow, passing evaporation unit.

(*) Exactly speaking, saturated vapor pressure and therefore vapor capacity of space very slighly increases with total pressure. But it is negligible in context of your question.

Note that absolute humidity is in g/m3 - it is intensive, not extensive variable. Rather, the vapour content  = absolute humidity * volume.


Is delta S the same in reversible and irreversible processes? I've read that change in entropy of a system is the same for a system in reversible and irreversible processes since it is a state function and does not depend on the pathway. Does this only apply to the system since I understand change in entropy of the universe is always positive for irreversible processes?

For an irreverislbe process with specified initial and final states of both the system and of the surroundings, calculating the change in entropy  between the same two end states requires you to separate the system from the surroundings, and to devise a separate alternative reversible path for each to determine the integral of dq/T for each.  This will give you the separate changes in entropy of the system, the surroundings, and the sum (i.e., the universe).  So the reversible paths between the initial and final states will not typically bear much resemblance to the actual irreversible path that the combination of system and surroundings had suffered in the irreversible process.


Which has higher rate of hydrogenation - methyl propene or trans-2-butene? Which has higher rate of hydrogenation- methyl propene or trans-2-butene?

I know that cis-2-butene has a higher rate of hydrogenation than both of these due to steric reasons, but here, both of these compounds have the same steric hindrance.

So, when they come in contact with the hydrogen adsorbed catalyst, they both should behave in a similar manner. Is this correct?

We should be able to determine which molecule will hydrogenate faster by comparing the activation energies for hydrogenation of these two molecules.  A catalyst does not alter the reaction mechanism, it merely reduces the activation energy for the process.

The heat of formation of 2-methyl-propene is ca. 1.3 kcal/mol higher than that of trans-2-butene, presumably because of the adverse steric interaction between the geminal methyl groups in 2-methyl-propene.  The transition state for this reaction will start to resemble the corresponding alkane in its eclipsed conformation (due to the cis-hydrogen delivery).  If we only consider methyl-methyl steric interactions, then for 2-methyl-propene as we approach the transition state the two geminal methyl groups, initially separated by a 120 degree ($\ce{sp^2}$) angle, will displace towards a 109 degree ($\ce{sp^3}$) angular separation.  Therefore, the methyl-methyl steric interaction should increase in 2-methyl-propene as we approach the transition state.  2-methyl-propene's  transition state will be more destabilized than the ground state, consequently increasing the activation energy required to reach it.  With trans-2-butene, there are no methyl-methyl steric interactions in the ground state or transition state; its activation energy, the energy separating the ground state and transition state, should not be increased or decreased.

Based on this analysis, the ground state - transition state energy separation would be expected to increase (slightly) for 2-methyl-propene as compared to trans-2-butene.  Consequently, faster hydrogenation of trans-2-butene would be predicted.  However the effect is expected to be very small.


What could cause a peak to split into 2 resonances as temperature increases in NMR spectroscopy I understand that some peaks may coalesce upon an increase in temperature due to an increase in free rotation, this does occur with some peaks in my data. Why might you see a peak split into two other peaks? I have thought it may be to do with an equilibrium shift in a rapid conversion between species but really I would expect that to show as one peak shrinking and another one growing. The observed effect is one peak clearly splitting into two. Any ideas why this could happen?

Edit:
The sample is a diamagnetic complex with aromatic ligands with t-Bu groups in the meta and para positions. It is hard to assign peaks as I believe there may be an equilibrium between different conformations in solution. The peak that splits is in the alkyl region of the NMR (around 1.5 ppm) and probably corresponds to some part of the t-Bu groups. The 2 peaks produced do appear to get sharper with temperature and the two peaks are the same size, however the difference in chemical shift (between the 2 peaks) at higher temperature is much greater than the linewidth at the lower temperature (where it is one peak), so I'm not sure if this could be J-coupling being resolved.

If it was the case of 2 conformers being present, would you not expect the single peak to reduce in intensity as another separate peak appears and increases in temperature? Rather than a splitting into 2 new peaks?

Since I do not know which compound it is, my guess is that you have two different  $\ce{H^1}$ that have almost the same chemical shift at room temperature. The chemical shifts change with temperature as for instance in sucrose:

Here  are represented mainly the  $\ce{H^1}$  of the $\ce{-OH}$ but the $\ce{C-H}$ (the peak of 1) shifts also.

In your case,  one is almost fixed, while the other one moves to higher $\delta$  and, therefore, they split into two.

More info on the sucrose spectrum here.


Heisenberg's uncertainity principle(HUC) I know the mathematical form of HUC and its statement i.e.It   is   impossible     to   determine   simultaneously. , the exact  position  and  momentum  of  an  electron. i don't know why this principle exists. Moreover, i read that this all is not due to technique inadequancies, but due to the nature of electron. What all is this? I tried to understand derivation for the mathematical form of HUC (just to get the idea why such thing exist), but it used advanced mathematics (I'm in class 11). Please give me the idea about why this principle exists. Why it is so that we are unable to measure velocity and position at the same time to high accuracy. (No usage of earlier established principles and maths please, in the answer.)

I can provide some physical intuition, but it's still going to require math and previous work to make it happen (aside on this at the end).

Let's consider a particle in just one dimension. There are two quantities of interest: (1) the position of the particle along this coordinate, (2) the scalar momentum (related to the velocity) along this coordinate.

I'm going to represent this particle as a plot on a graph. The horizontal axis will measure position. The vertical axis represents some notion of how likely it is to find the particle at this position (formally, it would be squared and normalized). Finally, we also stipulate that momentum of the particle is going to be the frequency of what I draw. This may seem strange when we're talking about an arbitrary plot, but it's actually completely reasonable from the perspective of representing functions from as Fourier series. You're just going to have to assume that this is true because a formal analysis does require math (see other answers).

Now, let's make this more concrete with some plots.

Here's a standard sine function. It has a very well defined frequency, so we know the momentum exactly. Where is the particle? Well, the sine wave stretches out to infinity in both directions, so we have no idea where it "is."

Here's a different graph. Where is the particle? It's very close to $x=0$. What is it's momentum? It's kind of hard to pick out single frequency here, so you cannot easily establish what the momentum is.

And that's basically it. Identifying frequency (momentum) and position for a plot like this are traded-off against one another. You'll probably end up in the intermediate case where you know some information about both. This is wave packet.

This particle is roughly centered around $x = 0$, and we can roughly say what its momentum is.

The uncertainty principle simply formalizes this idea except using the formal mathematical structure of quantum mechanics.

Now, for the aside I promised. Your question was entirely reasonable until you got to the last line, and I quote: "No usage of earlier established principles and maths please."

This is impossible. Science does exist in a vacuum. It is based on the past observations and structure built by previous generations of scientists. If we were to start from scratch, would we arrive at the same science that we have today? Possibly, possibly not. But the understanding we are working with is built upon foundations that we are stuck with, for better or for worse.

You also can't really do this without math. Part of science is to describe relationships, and mathematics allows us to do so unambiguously. Words suck at conveying ideas because words mean different things.

Lastly, it's fine to try to build intuition on a concept by relating it to the macroscopic world. However, you should not assume that this is possible. It may be philosophically satisfying that objects behave in similar ways regardless of size, but that's not the way the world is.


Why when I add jam to my tea it stops producing thin "membrane" on top? Thin "membrane" is created if I leave cup of tea for hour or more. And it usually sticks to the inner sides of the cup. But I noticed that when I drink tea with jam or syrup then the cup always clean inside as if the syrup contains some kind of acid that remove that tea-thingy.

The parallel process to deposition of limescale ( that is inhibited by acidic jam) is deposition of products of oxidation of epigallocatechine gallate(EGCG) and similar compounds, that are getting adsorbed on limescale particles and cup walls.

Epigallocatechin gallate (EGCG), also known as epigallocatechin-3-gallate, is the ester of epigallocatechin and gallic acid, and is a type of catechin.
It is found in high content in the dried leaves of green tea (7380 mg per 100 g), white tea(4245 mg per 100 g), and in smaller quantities, black tea (936 mg per 100 g).During black tea production, the catechins are mostly converted to theaflavins and thearubigins via polyphenol oxidases.

EGCG is a polyphenol, vulnerable to oxidation on neutral/alkalic environment, forming insoluble product of intense brown color.
Acidic environment, caused by addition of jam or lemon juice, stabilizes the compound.
Minor natural content of ascorbic acid in fruit juice is protected by abundance of organic acids by similar way.


If a gas always occupies the volume of its container, will its volume always be 22.4L at STP? According to the Avogadro's law, an equal volume of gases have a equal number  of moles (with constant pressure and temperature), therefore STP molar volume is $\pu{22.414 dm3 mol-1}$.

Think this: There is a container of volume $\pu{40 dm3}$ without any gases. I fill it with 1 mole of $\ce{O2}$ gas ($\pu{32 g}$), therefore according to the STP molar volume, the volume should be $\pu{22.414 dm3}$.

But we know that a gas always take the volume of its container. In above problem, the container volume is $\pu{40 dm3}$. So the volume of the gas should be $\pu{40 dm3}$. But why $\pu{22.414 dm3}$? I am confused with it.

According to Avogadro's law, equal volumes of gases at the same temperature and pressure have an equal number of particles. If you had one mole of particles at STP it would occupy $\pu{22.4dm3}$

However according to Boyle's law, pressure is inversely proportional to volume. Hence, by increasing the volume like you did, the vessel's pressure will decrease, thereby not making the measurement to be at STP. Hence, it will not be $\pu{22.4dm3}$.


Common etchant for Stainless Steel 304L and Titanium Grade 2 I have Stainless Steel welded to Titanium. I need a common etchant to etch both S.S. and $\ce{Ti}$ so that I can do further analysis (like SEM etc). Right now I'm etching Ti Side separately and S.S. side separately and doing analysis twice, once for each metal. The etchants I'm using are mentioned below:

$\ce{HNO3 + 3HCl}$ - for S.S. 304L
$5\%~\ce{HCl} + 6\%~\ce{HNO3} + 16\%~\ce{HF} + 74\%~\ce{H2O}$ - for $\ce{Ti}$

Is it possible to combine these two etchants to form a common etchant, or are there any standard etchants available to etch both S.S. and $\ce{Ti}$?

Common etchant for stainless steel is $\ce{FeCl3}$ with $\ce{HCl}$ (up to 3%) heated at 45°C.

The etching solution for $\ce{Ti}$ looks fine.


Nomenclature of benzene derivatives with two functional groups In several places, I have seen that while naming benzene derivatives, no priority for substituents is followed. For example, I've seen both 2-hydroxyaniline and 2-aminophenol being used.

Is there an order? Or is the order different from aliphatic compounds?

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the retained names “aniline” and “phenol” are preferred IUPAC names (PINs). Parent structures having retained names are called functional parent compounds.

For both retained names, substitution is permitted at any position. However, substitution of a functional parent compound is limited to substituent groups cited as prefixes in accordance with the seniority of the functional group that is explicitly expressed or implied in the functional parent compound name. (In systematic names, the presence of this characteristic group would normally be expressed as a suffix.)

  P-15.1.8.2.1 Substitution rules for Type 2a retained names

  Type 2a retained names include functional parent compounds whose name expresses or implies a characteristic group expressed as suffix in systematic names.

  P-15.1.8.2.1.1 Substitution by substituent groups, expressed as prefixes, having a lower seniority than that denoted by the suffix is allowed.

  (…)

  P-15.1.8.2.1.2 The senior characteristic group or functional class name must be
  expressed in the name.

Therefore, the correct name for the compound given in the question is “2-aminophenol” (not “2-hydroxyaniline”) since hydroxy compounds are senior to amines.


Will the bond order of CO be greater in its existence as carbon monoxide or in metal carbonyl? $$\text{Bond order} = \frac{N(b) - N(a)}{2}$$
Where $N(b)$ is the number of electrons in bonding orbitals and $N(a)$ is the number of electrons in antibonding orbitals.
As in metal carbonyls, there is synergic effect during formation of the π back bond from metal to $\ce{CO}$, so $\ce{CO}$ accepts electrons in its π* anti bonding orbital. As $N(a)$ increases, Bond order decreases. So, $\ce{CO}$ in its free state has greater bond order in comparison and hence is more stable.
Would the above concept be correct?

In figure 1 you can see the molecular orbital scheme of a $\ce{[ML6]}$ complex including σ and π interactions. The class of ligands is not important; however, care should be taken when correctly assigning orbitals.

Figure 1: Octahedral $\ce{[ML6]}$ complex including σ and π ligand-metal interactions.

The first thing we should be doing is assigning the original, pre-bonding orbitals. The metal orbitals on the left are obvious. The ligand group orbitals of lower energy, $\mathrm{a_{1g} + e_g + t_{1u}}$ are the six ligand orbitals that interact with the metal in a σ fashion; in carbon monoxide, these represent the lone pair on carbon (the HOMO). The twelve p type orbitals, of which nine are omitted and the remaining transform as $\mathrm{t_{2g}}$ are π symmetric with respect to the $\ce{L\bond{->}M}$ bond and represent the $\pi^*_{\ce{C=O}}$ bonds of carbon monoxide; the LUMOs.

In a metal carbonyl complex, the left-hand side is usually higher in energy due to the metal’s lower oxidation state. Thus, the lower an orbital in the centre of the scheme is, the more ligand-centred it is.

While filling in the final complex’ electrons, we start from the bottom; first the twelve ligand electrons that represent the carbon monoxide lone pairs. Once we are done with this, the next orbitals happen to be $\mathrm{t_{2g}}$ — the bonding ones which have more ligand contribution. These can not only be seen as bonding with respect to the $\ce{M\bond{<-}L}$ bond but also antibonding with respect to the $\ce{C=O}$ bond. The maximum we can fill in here is 6 electrons. These six electrons reduce the $\ce{C#O}$ bond order as can be seen in the formula:

$$\text{B.O.} = \frac{(12 + 2\times12) - 6}{2\times6} = \frac{30}{12} = 2.5$$

The additional factor 6 in the denominator is because I am analysing six $\ce{C#O}$ bonds at the same time.

The result ($2.5$) clearly shows that the bond order is reduced with respect to noncoordinating $\ce{CO}$.

The discussion I did with an octahedral complex can be done with any coordination geometry; octahedra are just a very typical case.


Relationship between the ideal gas constant and entropy So it dawned on me the other day that the ideal gas constant $$R = \pu{8.31 J mol-1 K-1}$$ has the same units as molar entropy. Is there some deeper meaning behind this or is it just a coincidence that this occurs?

The gas constant is equal to Avogadro's constant times Boltzmann's constant, the latter serving as a proportionality constant between the average thermal (kinetic) energy of the particles in an ideal gas and the temperature:

$$\left(\frac{\partial \bar U}{\partial T}\right)_p=\frac{3}{2}k_\mathrm{B}$$

The entropy can be regarded as a proportionality constant between the change in free energy $G$ with change in temperature of a system at constant pressure, since

$$\left(\frac{\partial G}{\partial T}\right)_p=-S$$

Note also that the average entropy of a particle can be written as

$$S=k_\mathrm{B} \log\Omega$$

So both $R$ and $S$ can be regarded as proportionality constants between energy and temperature.

The Wikipedia page on the equipartition theorem may provide some enlightenment on the origin of these proportionalities.


Why is a tin can that appears like galvanised steel rusting? I bought a canned pineapple and finished the content. The part of the can that is in contact with the liquid appears like galvanized steel (crystalline surface or spangles sort of appearance), while the part that are not looks pretty shiny like the outside of the can.

I then washed the can and added water to it, and heated the bottom of the can with a butane torch. After the water was heated up, I removed the torch and left it out for a while.

When I came back later, I emptied the water but noticed that the bottom rim of can had quite a bit of rusting. After inspecting it closely, I found that there are spots of (concentrated) rust on the cylindrical wall of the can as well. There are also light patches of rust on the inside bottom of the can. The seam on the cylindrical wall that is in contact with water is also badly rusted.

I thought galvanised steel only rusts when the layer of galvanisation wears out.
So, is the tin made with galvanised steel?
If yes, why are there concentrated spots of rust on the wall and other rusting?
If not, why does the inside of the can look like galvanised steel?
And BTW, why does the emptied and washed can smell "metallish"?

First of all, any 'tin' can you are likely to find is actually made of steel. Tin is too expensive. Also, tin does not 'rust', although it oxidizes. Your rust is iron oxide. Galvanized steel is steel with a thin zinc coating, likely hot-dip galvanization. What you describe (crystalline surface or spangles sort of appearance) does sound like galvanization, although I am not aware of any steel cans that are galvanized.

I thought galvanised steel only rust when the layer of galvanisation wears out. This is true. If your can is indeed galvanized, then it has likely chipped off, possibly on the rims of the can where the lid it crimped or pressed together, or by cutting the can open. Strong acids will dissolve zinc  Also interesting to note: sodium chloride (salt) will dissolve zinc over time by converting the galvanization to zinc chloride, but that's not applicable here.

So, is the tin made with galvanised steel?
  Well, that does sound like what you were describing, but galvanized steel is not fit for foodstuffs. Zinc is mildly toxic and ingesting it can make you ill. I like feetwet's suggestion; that universally the inside of metal food cans are lined, by actually an epoxy that is sprayed on. This is almost certainly to be the case. Perhaps this makes sense: if your can was indeed galvanized, then the rational by the manufacturer was that the lining will keep the zinc separated and the food safe. Sounds legit. It might be good to be aware that the type of epoxy being used here is the one that uses as its plasticizer Bisphenol A and Bisphenol S (BPA/BPS). Ill leave it up to you to decide if that should bother you or not.

If yes, why are there concentrated spots of rust on the wall and other rusting?
Well, if there is rust, then we can deduce that water must have be coming in contact with the iron in the steel, so whatever was protecting the steel has some failure.

And BTW, why does the emptied and washed can smell "metallish"?
While zinc does have a metallic smell to it, iron oxide's is much stronger. Since you visually identified rust, this is what you are smelling.


Pinacole Rearrangement through epoxide This reaction supposedly passes through an epoxide intermediate. Can someone please explain why this is necessary?

Source

As I mentioned above, during an isotope labelled mechanistic study, the formation of epoxide intermediate is suggested to explain the scramble isotopic carbon (between carbonyl carbon and neighboring carbon). The detailed arrow pushing mechanism is given bellow:

These steps are subjected to change according to many factors including the reaction conditions and nature of the diol used as correctly pointed out by Zhe and aforementioned MSU website in your comments, which states:

Many factors must be considered when analyzing the course of a pinacol rearrangement. These include:
• Which hydroxyl group is lost as water? or Which intermediate carbocation is more stable?
• What is the inherent shifting tendency (migratory aptitude) of different substituent groups?
• What is the influence of steric hindrance and other strain factors on the rearrangement?
• Are epoxides formed as intermediates in the pinacol rearrangement?
• Does product stability govern the outcome of competing rearrangements?
• Do the reaction conditions (i.e. type of acid, concentration, solvent and temperature) influence the course of rearrangement?

A reference for the rearrangement through an epoxy intermediate:
C. A. Bunton,  T. Hadwick,  D. R. Llewellyn,  Y. Pocker, Tracer studies in alcohols. Part III. Intermediates in the pinacol–pinacone rearrangement, J. Chem. Soc., 1958, 403-408 [DOI: 10.1039/JR9580000403] (http://pubs.rsc.org/en/Content/ArticleLanding/1958/JR/JR9580000403#!divAbstract).


How dried up detergent water led to precipitation of snowflake like shapes? I forgot to drain the water off the bucket (after cleaning stain from a piece of clothing with detergent) and left it for few days. After few days when I needed it back, I found these glittery very small snowflake like shapes at the bottom. Pardon if it's very basic thing, I haven't studied chemistry after school. Thanks for the help.

There are many water soluble molecules in soap. I believe that as the water evaporated, the concentration of these molecules became great enough for them to start crystallizing. This process is used in chemistry as a means of purification and is known as recrystallization.

Are you considering the X type shape of the crystals to be snowflake like? I am not quite sure why they all seem to have similar geometries, I want to guess that they are crystals of citric acid.


Sodium Sulfate or Sodium Peroxysulfate? In an attempt to clean drain cleaner, I reacted it with Sodium percarbonate to clean up some hardware store variety sulphuric acid.

At first the reaction appeared successful, yielding a Nice Clear Syrupy sulfuric acid with a heavy crystallization at the bottom of the flask, but I don't think the salt that I collected is sodium sulfate.

The crystals are very long resembling fiberglass and based on images I've seen online, sodium sulfate's structure seems to be a lot like sodium chloride.

I'm not sure of the conjugate base formed during the reaction ($\ce{SO4-, SO5- or HSO4-}$) can anyone explain?

EDIT

I've since learned that when basic sodium is reacted with $\ce{H2SO4}$ in a ratio favoring $\ce{H2SO4}$, the conjugate base produced is $\ce{HSO4-}$

Ok, let’s take it one step at a time:
Percarbonate is actually not a single molecule, but a double-salt of sodium carbonate and hydrogen peroxide.
The sodium carbonate reacts with sulfuric acid, resulting in carbon dioxide, sodium sulfate and water. The reaction scheme is: Na2CO3 + H2SO4 -> Na2SO4 + H2O + CO2

For safety reasons I will strongly discourage you from adding carbonate to concentrated sulfuric acid. It will most likely explode in your face!

But in theory, after having added sufficiently carbonate, Sodium sulfate will precipitate.
But getting a drying effect will be uphill, since water is produced alongside the sulfate, as you can see in the reaction scheme, on top of this is any water that comes from hydrogen peroxide, not to mention the fact that concentrated sulfuric acid is strongly hygroscopic itself.
As for the cleaning of impurities, I fail to see what neither carbonate nor hydrogen peroxide can do. HP can oxidize some organic materials, but most of these will probably already have been destroyed by the sulfuric acid. It can also oxidize some inorganic ions like sulfite (SO3), but do you suspect these to be a problem?
I don’t have detailed knowledge of the conditions needed to form peroxo(mono)sulfuric acid. Undoubtedly some will be formed when HP is added to sulfuric acid, but I am also convinced that only a fraction of the sulfuric acid/HP will react.


What are appropriate techniques to ensure homogeneous solidification of an agar-based aqueous solution? I am making a T1 weighted phantom for an MRI project. I'm trying to see if I can calculate how much a persons head heats up during an MRI scan. It consists of distilled boiling water, $\pu{7 g/L}$ agar, $\pu{10 g/L}$ $\ce{NaCl}$ and $\pu{1 g/L}$ $\ce{CuSO4}$. As the phantom cools it will get a jelly like texture. What I would like is for the phantom to be as homogeneous as reasonably possible when it solidifies. What do you think is the best way to do so?

I was thinking of having the mixture on a vibrating plate, or maybe a centrifuge or even just inverting the mixture at regular intervals. Any ideas would be greatly appreciated.

Let me suggest two avenues :

Vertical set up à la potter's wheel: At school we once did something related to your aim: using an old $\pu{75 rpm}$ record player, obviously no longer needed, with a beaker containing previously molten bees wax in the centre.  Under constant rotation (rather than the revolution by a centrifuge), the surface of the wax formed a nice meniscus which frooze by the slow cooling. While I am not aware if the alread-gel particles would keep nicely floating in your suspension, until all is solidified, it may be worth to giv it a try.  Centering of the whole ensemble is important (unbalance and spill), especially if the sample to be generated should really become as large as a human head.  Scaling up appears possible (larger beaker / bucket, and potter's wheel).

Horizontal set up: If you were interested in a smalller sample, a roller mixer would be an alternative:

(screen photo)

A wide mouth bottle (like the still cylindrical $\pu{1 L}$ polymer centrifuge bottles used in biochem) would be filled by immersion with liquid and -- still immersed in solution -- closed with the lid to ensure absence of air bubbles in the bottle.  After external cleaning of the bottle, they would be layed along the rolls.  Subsequently, you would turn on and crank up the rolling and hopefully the gel would form sooner than the liquid dropping out from the bottles.  (One may orient the bottles with the lids facing you, and place a bucket underneath them to collect potential spill.)  You may find such a roller mixer in biochem groups cultivating their own cell cultures.


How are reactants ad-/ -desorbed during Haber process? Collecting information for a presentation [academic high-school] about catalysts, I stumbled upon the terms desorption and adsorption as important parts of the Haber process.

However, I could neither find a good explanation for these processes (why and how do they occur?) nor the difference between ad- / desorption and the process
of the reactants "diffusing through the phase boundary to the active site", as it's stated on German Wikipedia.

All chemical catalysis can be grouped into 1 of 2 categories: homogeneous or heterogeneous catalysis. Homogeneous catalysts are ones that are in the same phase (i.e. solid, liquid, gas BUT USUALLY IN SOLUTION) as the reactants they are helping to react. Heterogeneous catalysts are in a different phase than the reactants they are working on. Industrial processes (like the Haber-Bosch process for making ammonia) frequently use heterogeneous catalysts due to their comparatively inexpensive nature, and there ease of handling on a large scale. Doing it this way can cause some added problems, namely the reactants need to bind to the catalyst's surface (adsorption) for the catalyst to work. In the H-B process this is gaseous N2 and H2 binding to a mainly iron catalyst. To help this process along the iron is made with a high surface area with holes and such (think like the texture of a sponge except on a much smaller scale). After the reaction happens the products then need to release from the catalyst (desorption) so they can be separated from the catalyst and be collected.  The active site simply refers to the place on the catalyst where the reaction happens. Adsorption is helped by the high temperature and massive pressures in the H-B process, as the gasses are flowed through a chamber at high pressure containing the solid heterogeneous catalyst. Not all of the molecules stick and not all that do stick react, so the underacted gas is recycled and passed over the catalyst again. Long story long...the very question you asked won the Gerhard Ertl the 2007 Nobel Prize in Chemistry.


How much sodium hydroxide would it take to dissolve one tonne of bauxite? Generally speaking what proportion of NaOH would be typically required to dissolve one tonne of crushed bauxite of average purity?

Also are varying quantities required for gibbsite, böhmite and diaspore respectively and how do they differ?

Bauxite contains usually 45 - 60 % Al2O3. To get an order of magnitude, one ton bauxite contains about 500 kg Al2O3, or about 5000 mole Al2O3. If 2 NaOH are needed to dissolve 1 Al2O3, it means that a minimum of 10'000 mol NaOH is needed to dissolve Al2O3, which weighs 400 kg. But more NaOH is necessary, because the bauxite contains also some silica, which is also dissolved in NaOH. So the amount of NaOH must be higher than 400 kg.


Why super glue does not stick inside their own bottle? We usually use super glue to stick any two objects but I was wondering like why don't they stick inside their own bottle ?

From explainthatstuff.com:

  Why doesn't glue stick to the tube?

  Adhesives are designed to work when they leave the tube—and not
  before. Different adhesives achieve this in different ways. Some are
  dissolved in chemicals called solvents that keep them stable and
  non-sticky in the tube.

  When you squeeze them out, the solvents
  quickly evaporate in the air or get absorbed by the surfaces you're
  sticking to, freeing the adhesives themselves to do their job.

  Plastic modeling glue works like this. It contains molecules of
  polystyrene in
  an acetone solvent. When you squeeze the tube, the glue spurts out and
  you can usually smell the very strong acetone as it evaporates.

  Once it's gone, the polystyrene molecules lock together to make strong
  chemical bonds.
  Glue doesn't smell when it's dry because all the
  solvent has vanished into the air.

  Some glues (such as synthetic,
  epoxy resins) have to be mixed together before they work. They come in
  two different tubes, one containing the synthetic resin and the other
  containing a chemical that makes the resin harden. The two chemicals
  are useless by themselves but, mixed together, form a tough, permanent
  adhesive.

From sciencefocus.com:

  How super glue works

  The main ingredient in super glue is a chemical called cyanoacrylate, an acrylic resin that forms a bond (cures) almost
  instantly;
  The only trigger it requires are the hydroxide ions found in water;
  Traces of water are found on almost everything and in the surrounding air, which causes the glue to become sticky;
  It works best in moist conditions, although it is not advised to start gluing on a soaking wet surface.

Therefore the lack of water in the tube, or the chance of solvents evaporation, doesn't allow the glue to stick to the tube.

In other words: that strong smell that comes out when we open the tube's cap is due to the solvents evaporation, so once they're gone the glue sticks. If a superglue tube is left open for a long time (exposed to air, allowing the water to come in contact with the glue), it's impossible (or almost) to squeeze it out since it got stuck to the tube.


Sacrificial Anodes and Pit Corrosion I have a slab of aluminum and one part of the slab will be exposed to a very corrosive environment. If I put a sacrificial anode elsewhere on the slab will the section that is exposed still be protected from pit corrosion? I guess I'm unsure because the pit corrosion is a result of a localized reaction and if an anode is added will it prevent the local reaction from taking place.

Generally speaking, there is no protection unless the sacrificial anode is immersed in the electrolyte. It might be better to anodize and seal the aluminum, which should protect it against most acids. There are many aluminum anti-corrosion treatments, but in any case the area under attack would have to be covered. See http://en.wikipedia.org/wiki/Anodizing.


Explanation for why nickel turns green in hydrochloric acid My daughter has a chemistry test that I am helping her prepare for. One question is

  Nickel sulfate, $\ce{NiSO4(aq)}$ is a green solution. Nickel chloride, $\ce{NiCl2(aq)},$ is a yellow solution. If you add nickel, $\ce{Ni(s)},$ to hydrochloric acid, $\ce{HCl(aq)},$ what color solution do you expect?

Wikipedia states that it will turn green but doesn't give an explanation for why. Can anyone fill in the blanks for us? (This is for high school intro to chem so try to keep it at that level.)

The question is really badly worded. For starters, let’s look at solutions of nickel(II):

Figure 1: Nickel(II) solutions. From left to right: $\ce{[Ni(NH3)6]^2+}$, $\ce{[Ni(en)3]^2+}$, $\ce{[NiCl4]^2-}$, $\ce{[Ni(H2O)6]^2+}$. Image taken from Wikipedia, where a full list of authors is available.

You can ignore the left two but the rightmost is a standard nickel(II) solution (could be $\ce{NiSO4}$) while the second from the right is a chlorido-complex—basically $\ce{NiCl2}$ but dissolved in hydrochloric acid. I wouldn’t call the chloride solution ‘yellow’, more like ‘yellowish-green’ but your mileage may vary. Next, let’s look at solids.

Figure 2: salts of nickel(II) chloride. Left or top: hexahydrate. Right or bottom: anhydrous. Images taken from Wikipedia (hexahydrate, anhydrous), where a full list of authors is available.

The picture on the right is dry, i.e. completely water-free nickel(II) chloride (anhydrous). On the left, the salt contains tightly bound water molecules which can be removed by further drying although the salt itself would be dry in everyday terms; this is the hexahydrate of nickel(II) chloride.

It might seem easy that the salt on the left gives a greenish solution while the salt on the right gives a yellowish one. Here’s the catch: they should both give a greenish solution, because there is no excess of chloride as was required in the solution in figure 1.

But what about the actual question? Well, dissolving nickel in hydrochloric acid means that there should in fact be excess chloride, meaning that yellow (or yellowish green) would be the correct answer.

tl;dr: That question is a perfect opportunity to complain if it was marked as ‘incorrect’.


Stability of alkenes by hyper conjugation I have learnt that for carbocations, electrons are delocalized from the $\ce{C-H}$ $\sigma$-bond to the empty p-orbital of carbon.

Moreover, alkenes are stabilized by the delocalization of electron from $\ce{C-H}$ $\sigma$-bond to the empty anti bonding $\pi$ molecular orbital.

But, don't electrons in anti bonding molecular orbitals lead to instability? So, how does hyper conjugation lead to stability of alkenes?

I believe this question has already been asked before, but I couldn't find any satisfactory and/or relevant answers.

You are confusing the mixing of orbitals with the promotion of electrons from one orbital to another. You're thinking of this:

$\hspace{50 mm}$

I don't blame you, because the phrasing "delocalisation into an antibonding orbital" implies that, but that is not actually what happens.

In what actually happens, the two original orbitals are said to mix. They cease to exist. They overlap to give two new MOs, one which is even lower in energy and one which is higher in energy, like in this diagram:

$\hspace{45 mm}$

The "new MO 1" has characteristics of both the original $\ce{C-H}$ $\sigma$ as well as the original $\ce{C=C}$ $\pi^*$, which is why we loosely say "donation into the $\pi^*$ orbital". However, don't take it too literally. The phrase means that two orbitals are mixed. It does not mean that electrons are promoted from one orbital to another.


Can ionic and atomic radii be compared? Can ionic and atomic radii be compared?  For example, this is a common problem in some intro chem courses:

Rank the following from smallest to largest:  sodium ion, magnesium ion, neon, fluoride, oxide.

Any student who has been trained in trends should be able to get the "answer", but if one looks at the actual figures, there are data that show that sodium ion is larger than neon atom.

I am of the opinion that it is comparing apples to oranges and that the problem can be rescued by removing neon from the list.

A periodic table coupled with simple electrostatic arguments should suffice to provide at least a rough ordering of ions and atoms. For the ions:

We expect POSITIVE ions to be SMALLER than the corresponding atom because there are fewer electrons than protons => an excess of POSITIVE charge => the electron cloud is more attracted to the nucleus on average and becomes more 'concentrated' => $\ce{Na+}$ is 'small' and $\ce{Mg^{2+}}$ is 'very small'.
We expect NEGATIVE ions to be LARGER than the corresponding atom because there are more electrons than protons => an excess of NEGATIVE charge => the electron cloud becomes more diffuse to compensate, lowering the Coulombic interactions => $\ce{F-}$ is 'large', $\ce{O^{2-}}$ is 'very large'.

Even though $\ce{Na+}$ has a lower principal quantum number for its outer shell than $\ce{Mg^{2+}}$, Mg is twice as charged and I believe this can be more important. $\ce{F-}$ and $\ce{O^{2-}}$ have the same principal quantum number for their outer shells, but $\ce{O^{2-}}$ is more charged and accordingly, we expect that $\ce{O^{2-}}$ has a larger radius.

$\ce{Ne}$ carries no unbalanced charge and has comparable valence principal quantum number to the ions, so we expect its radius to fall somewhere between the positive ions (which are smaller) and the negative ions (which are larger). Go over these arguments in your head and try to put the pieces together yourself from here.

(Note: I wouldn't be willing to stake my life on this, so some community validation or corrections would be much appreciated)


How to correctly write "7 apples" according to the international system of units (SI) According to international system of units (SI), we can write "7 kg of apples" to refer to the mass of these apples.

However, if we want refer to the amount of apples, that is, the number of entities, the unit should be the mole. So, what's the correct way to denote "7 apples" in accordance with the SI convention?

Additionally, is it correct (according to SI) to say "7 atoms of hydrogen"? Or must we use mole?

In accordance with the International System of Units (SI) [Brochure in English, 8th edition, 2006; updated in 2014] and the corresponding International System of Quantities (ISQ) [ISO/IEC 80000 Quantities and units (14 parts)], you can define a suitable new quantity, for example with the quantity name “number of apples” and the quantity symbol “$N_\text{apples}$”.

The number of apples $N_\text{apples}$ is a quantity of dimension one (for historical reasons, a quantity of dimension one is often called dimensionless):

$$\dim N_\text{apples} = 1$$

A quantity of dimension one acquires the unit one, (symbol: $1$); i.e. the coherent SI unit for the number of apples is the unit one.

Generally, the unit one is an SI derived unit; for example, the derived SI unit for friction factor is newton per newton equal to one, (symbol: $N/N = 1$). However, the unit one for counting numbers, e.g. number of protons in an atom or number of apples, is considered as a base quantity because it cannot be expressed in terms of any other base quantities. Hence, in this case, the unit one is usually considered as a base unit, although the CGPM has not yet adopted it as an SI base unit.

The name and symbol of the measurement unit one are generally not indicated. Therefore, you may write: “The number of apples is $N_\text{apples}=7$.”

The unit one or its symbol $1$ may not be combined with SI prefixes. For example, if you have 2000 apples, you must not write “$N_\text{apples}=2\ \mathrm k$” for $N_\text{apples}=2000$. (And by the way, when you see something like “10K reputation” mentioned on any stackexchange site, you are looking at at least three nonconformities at the same time.)

Any attachment to a unit symbol as a means of giving information about the special nature of the quantity or context of measurement under consideration is not permitted. Expressions for units shall contain nothing else than unit symbols and mathematical symbols. Therefore, write

“the maximum electric potential difference is $U_\text{max}=1000\ \mathrm V$”, not “$U=1000\ \mathrm V_\text{max}$”
“the gauge pressure is $p_\mathrm e=0.5\ \text{bar}$”, not “$p=0.5\ \text{bar(g)}$”
“the electric power is $P_\text{el}=1300\ \mathrm{MW}$”, not “$P=1300\ \mathrm{MW_{el}}$”
“the water content is $170\ \mathrm{g/l}$”, not “$170\ \mathrm{g\ \ce{H2O}/l}$”

and also

“the number of apples is $N_\text{apples}=7$”, not “$N=7\ \text{apples}$”


How do you determine the NIE equation of a buffer? For example, I have formic acid and sodium formate. How do I know the NIE(net ionic equation) equation is $\ce{HCHO_2 + H_2O <=> H_3O+ + CHO_2-}$ and not $\ce{HCHO_2 + OH- <=> H_2O + CHO_2-}$ ? If I added $\ce{HCl}$ or $\ce{NaOH}$, would that help decide which equation? Is a buffer always added to water? Why or why not?

$\ce{HCHO_2 + H_2O <=> H_3O+ + CHO_2-}$ and $\ce{HCHO_2 + OH- <=> H_2O + CHO_2-}$ are both reactions that can happen.
If you added $\ce{HCl}$ (of high concentration), you added a strong acid, and there will be extra hydrogen ions. Therefore, it will drive your first reaction backwards due to the extra hydronium: $\ce{H_3O+ + CHO_2- <=> HCHO_2 + H_2O}$.
If you added $\ce{NaOH}$ (of high concentration), you added a strong base, and there will be extra hydroxide ions. Therefore, it will drive your second reaction forwards: $\ce{HCHO_2 + OH- <=> H_2O + CHO_2-}$.
Distilled water has no buffer added to it, and it is not a buffer itself because of extremely low concentrations of $\ce{H_3O+}$ and $\ce{OH-}$. Tap water might have buffers in it. Lakewater have buffers to prevent sudden changes in pH. For example, it might have limestone at the bottom of the lake, which serves as a buffer against strong acids. As an example, when acid rain falls, it brings with it sulfuric acid, and the buffers prevent the pH from dramatically decreasing, by the equation $\ce{H2SO4 + CaCO3 => CaSO4 + H2O + CO2}$. With the buffers, the fish living in the lake will not be affected by the sudden changes in pH, otherwise, they might stress or even die (but that's a question for Biology SE).


Whats the difference between entropy and the (dis)order of a system? Entropy is often verbally described as the order/disorder of the thermodynamic system. However, I've been told that this description is a vague "hand-waving" attempt at describing what entropy is. For example, a messy bedroom doesn't have greater entropy than a tidy room

My question is why is this the case? Also, what would better describe entropy verbally?

Briefly, spontaneous processes tend to proceed from states of low probability to states of higher probability. The higher-probability states tend to be those that can be realized in many different ways.

Entropy is a measure of the number of different ways a state with a particular energy can be realized. Specifically, $$S=k\ln W$$ where $k$ is Boltzmann's constant and $W$ is the number of equivalent ways to distribute energy in the system. If there are many ways to realize a state with a given energy, we say it has high entropy. Often the many ways to realize a high entropy state might be described as "disorder", but the lack of order is beside the point; the state has high entropy because it can be realized in many different ways, not because it's "messy".

Here's an analogy: if energy were money, entropy would be related to the number of different ways of counting it out. For example, there, there are only two ways of counting out two dollars with American paper money (2 1-dollar bills, or 1 two-dollar bill). But there are five ways of counting out two dollars using 50-cent or 25-cent coins (4 50-cent pieces, 3 50-cent pieces and 2 quarters, and so on). You could say that the "entropy" of a system that dealt in coins was higher than that of a system that dealt only in paper money.

Let's look the change in entropy for a reaction $\rm A\rightarrow B$, where A molecules can take on energies that are multiples of 10 energy units, and B molecules can take on energies that are multiples of 5 units.

Suppose that the total energy of the reacting mixture is 20 units. If we have 3 molecules of A, there are 2 ways to distribute our 20 units among energy levels with 0, 10, and 20 units:

If we have 3 molecules of B, there are 4 ways to distribute 20 units among energy levels with 0, 5, 10, 15, and 20 units:

The entropy of B is higher than the entropy of A because there are more ways to distribute the same amount of energy in B than in A. Therefore, $\Delta S$ for the reaction $\rm A\rightarrow B$ will be positive.


Alcohol spectrophotometry A couple of weeks ago we did a lab where we had a bunch of standard solutions of alcohol, dichromate, and sulfuric acid. We also had one unknown concentration.

We heated them so they would all react, and the initial orange dichromate turned slightly green (the ion $\ce{Cr^{3+}}$ formed). Then we placed each reacted soln in a calorimeter which can emit light of certain frequencies. I believe we chose 635, which should have emitted orange light through the solution. the absorbance/transmittence were calculated for each standard soln until we had a line of best fit. Then the absorbance of the unknown concentration was given, so we used to line to match it to it's concentration (of alcohol).

I'm confused as to why we had to use the color orange in our emission and why a reaction was even needed in the first place.

Since green is the absorption of orange light, you need to supply the orange light to measure how much is absorbed. As the alcohols are oxidized to the acids, the dichromate is reduced, so you can measure how much acid has been formed. BTW, what alcohol was used? Butyl alcohol oxidation has a distinct aroma that can be detected without spectroscopy.


Can cationic species form micelles? Most of the books dive into formation of micelles by soaps. They go on talking about alkyl sulphate and carboxylates, but never do they mention anything about cations. My textbook says that at critical micelle concentration, molecules of colloid aggregate to form micelles.

Can things like $\ce{RNH4+}$ or $\ce{R4N+}$ form micelles, why/why not?

Cationic micelles similar to your examples can and do exist.  For example cetyl trimethylammonium bromide (Growth of Cationic Micelles in the Presence of Organic Additives, P. A. Hassan and J. V. Yakhmi, Langmuir, 2000, 16 (18), pp 7187–7191, DOI: 10.1021/la000517o).

Additionally, there are cationic micelle detergents (Ionization of cationic micelles: Effect of the detergent structure, Journal of Colloid and Interface Science, Volume 78, Issue 2, December 1980, Pages 330-337).

The bottom line is that there are a variety of use cases for cationic micelles, in detergents and otherwise.


Covalent Bonding in odd electron sharing molecules Why in odd electron sharing molecules (like NO ) the bond containing 3 electrons is considired to be half covalent bond?

I will answer to your question, but first let me show you that NO has a $2.5$ bonds, and I will use what I made for NO to answer your question. Here we go ! :)

Proof that NO has a $2.5$ bonds :

First the electron configurations of these two elements are :

$\ce{{}_7 N :1s^2 2s^2 2p^3}$ which have $5$ valence electrons in $4$ orbitals (one $\ce{s}$ and three $\ce{p}$)

$\ce{{}_8 O :1s^2 2s^2 2p^4}$ which have $6$ valence electrons in $4$ orbitals (one $\ce{s}$ and three $\ce{p}$)

Then we can build the orbitals-diagram of $\ce{NO}$ and I will suppose that its not a correlated diagram, and because of electronegativity of these elements verifies $\ce{\chi(N)<\chi(O)}$ then we have:

Then the electon configuration of $\ce{NO}$ is $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)^2 (1 \pi_{x})^2 (1 \pi_{y})^2 (1 \pi_{x}^{\star})}$

The bond order $i$ is : $$i=\frac{\text{number of electrons in bonding MOs}-\text{number of electrons in anti-bonding MOs}}{2}$$

Here we have $$i=\frac{2\times 4-3}{2}=\frac{5}{2}=2.5$$

Quod Erat Demonstrandum.

So now I did it for $\ce{NO}$ I have this configuration : $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)^2 (1 \pi_{x})^2 (1 \pi_{y})^2 (1 \pi_{x}^{\star})}$.

Then you said $3$ electrons bond considered has $0.5$ bond. As I understand your question, its the case if the configuration is $\ce{(1\sigma)^2 (1 \sigma^{\star})^2 (2\sigma)}$ (for example but it could be an other) then you have $3$ electrons in bonding bonds and $2$ in anti-bonding bond.

Then the bond order of your molecule is : $$i=\frac{3-2}{2}=0.5$$

Which conclude my answer.

NB:I'm seeking common examples of this kind.


Bromonium ion or Mesomeric effect ( intermediate during addition to alkenes) It's mentioned in Peter Sykes , A Guidebook To Mechanism In Organic Chemistry , 6th edition, 13th impression by Pearson, Pg. No. 185 , that on addition of $\ce{HBr}$ to bromoethene the only product obtained is 1,1-dibromoethane. It's due to the positve mesomeric effect of bromine to stabilize the carbocation in this case, however the rate is remarkably slower as compared to addition to ethene.

1

Another reaction corresponding to addition of $\ce{Br2}$ to ethene proceeds with the formation of a bromonium ion intermediate.

2

My question is, why can't the first reaction just proceed with a bromonium ion intermediate, like (28) above can form a bromonium ion and give 1,2-dibromoethane as a product (there may be a possibility that the carbocation may be formed away from bromine to counteract the strong negative inductive effect and in that process it may interact with the lone pairs of bromine to form bromonium ion)? And if bromonium ion is unstable (that is if it can't be formed in 1st reaction) why does it form at all in reaction 2? Why can't the Bromonium ion in the 2nd reaction just undergo a 1,2- hydride shift and convert into the "more stable" ion like 29b. It should also lead to formation  of geminal dibromide .It's a circular argument.

It's clearly mentioned that bromonium ion and carbocation are canonical forms of each other , which means you expect them to convert into bromonium ion whenever there's a positively charged carbon near a bromine atom.

3

Here's a similar example of phenonium ion.

[4]: https://i.stack.imgur.com/0j5Ww.jpg .

The carbocation doesn't stay the way it is in structure 4. It rather forms 3. Notice the bond between phenyl ring and the carbon atom on the right. It's not a completely solid bond.

You are not going to see the formation of a primary carbocation—the activation energy is too high. Even though it is true the following are resonance structures of one another:

the geometry in which the primary carbocation is formed does not allow for the simultaneous formation of the bromonium ion:

The bromonium would only be able to form after a rotation of the $\ce{C-C}$ bond, so we need to consider the primary carbocation sans-bromonium as the real product of the protonation of the alkene. This is a very high energy product, so it will be slow to form at low temperatures and is out-competed by the other reaction. Since positive charge on carbon-1 can immediately be delocalized to the bromine, this product of protonation is more stable (meaning it forms faster).


Why the DMM (multimeter) register negative resistance when measure a drop of hydrochloric acid I noticed that when I put the multimeter electrodes (in the ohmmeter mode) in a drop of hydrochloric acid with mild concentration (15%) the multimeter displays negative values. Can someone explain this behavior?

The multimeter probe tips are dissolving and setting up a galvanic cell that augments the current flow thru the ohmmeter.
The probe tips may be of the same metal, but the ohmmeter voltage will favor the dissolution of one over the other, and the cell thus formed will assist the applied voltage.
Quickly switching to voltmeter (or milliammeter) mode will show a cell voltage (current).
Using the ohmmeter to measure resistance of a non-corrosive solution (e.g., NaHCO3) will show some (positive) resistance.
Inert electrodes (Pt, Au) will show a positive resistance.


Why energy of some intermediates of glycolysis increases? It is possible that I do not understand the concept of energy, but according to the diagram in my book, the energy between steps 4 and 5 and between 6 and 8 increases. How is this possible? Also, as ATP is invested in steps 1 and 3, shouldn't the energy increase there?

Step 1:

Phosphorylation activates glucose for the following reactions in the pathway. However, the formation
of such a phosphoester is thermodynamically unfavourable and requires
energy input to operate in the forward direction. The energy comes
from ATP, a requirement that at first seems counterproductive.

  $$\ce{-D-Glucose + ATP^4- → -D-glucose-6-phosphate^2- + ADP3  H+}$$
  $\Delta G$ = -16.7 kJ/mol

and energy is released.

Step 3:

Once again, the substrate that provides the phosphoryl group is ATP. Like the hexokinase/
glucokinase reaction, the phosphorylation of fructose-6-phosphate is a priming
reaction and is endergonic(energy absorbed):

  Fructose-6-P + Pi +fructose-1,6-bisphosphate   $\Delta G$ = 16.3
  kJ/mol

When coupled (by phosphofructokinase) with the hydrolysis of ATP, the overall reaction
becomes exergonic:

  Fructose-6-P + ATP → fructose-1,6-bisphosphate + ADP

  $\Delta G$ = -14.2 kJ/mol

Step 4:

Fructose bisphosphate aldolase cleaves fructose-1,6-bisphosphate between the
C-3 and C-4 carbons to yield two triose phosphates. The products are dihydroxyacetone
phosphate (DHAP) and glyceraldehyde-3-phosphate. The reaction has an
equilibrium constant of approximately 10^-4 M, and a corresponding $\Delta G$  of +23.9 kJ/mol.

(the equilibrium is thus greatly influenced by concentration.)

Step 6:

In the first glycolytic reaction to involve oxidation–reduction, glyceraldehyde-3-
phosphate is oxidized to 1,3-bisphosphoglycerate by glyceraldehyde-3-phosphate dehydrogenase.

Although the oxidation of an aldehyde to a carboxylic acid is a highly
exergonic reaction, the overall reaction involves both formation of a carboxylic–
phosphoric anhydride and the reduction of NAD to NADH and is therefore slightly
endergonic at standard state, with a $\Delta G$ of + 6.30 kJ/mol.

(The free energy that might otherwise be released as heat in this reaction is directed into the formation of
a high-energy phosphate compound, 1,3-bisphosphoglycerate, and the reduction of
NAD.)

Step 7:

Here the glycolytic pathway breaks even in terms of ATPs consumed and produced
with this reaction.The enzyme phosphoglycerate kinase transfers a phosphoryl
group from 1,3-bisphosphoglycerate to ADP to form an ATP.

The phosphoglycerate kinase reaction is sufficiently exergonic at
standard state to pull the G-3-P dehydrogenase reaction along. (In fact, the
aldolase and triose phosphate isomerase are also pulled forward by phosphoglycerate
kinase.) The net result of these coupled reactions is

  Glyceraldehyde-3-phosphate + ADP + Pi + NAD → 3-phosphoglycerate + ATP
  + NADH + H+ $\Delta G$ = -12.6 kJ/mol

(Important: Another reflection of the coupling between these reactions lies in their values
of $\Delta G$ under cellular conditions, despite its strongly negative  the phosphoglycerate kinase reaction operates at equilibrium in the erythrocyte ($\Delta G$ =0.1 kJ/mol).

So in this case you'ill notice factors that influence energy expenditure and gain. Step 1 and 3 (exegonic reactions have net -G hence ) will see energy being released (and accroding to your diagram
there is a decrease. Conversely in Step in steps 4-5 (reaction not favoured but due to concentration proceeds and is aided by energy and 6-7( energy is absorbed to form a high energy compound
1,3-bisphosphoglycerate) and 7-8 is mainly determined by reaction environment and relative concentrations of substrates and is slightly endergonic in above case.

Hope this helps

Reference

Biochemistry 4th Ed (Grisham)


Valence Shell for Transition Metals In transition metals, is the shell with the highest energy considered the valence shell? For example, in copper the electronic configuration is ${[Ar]\text{ } 3d^{10}\text{ } 4s^{1}}$. However, in accordance with the afbau principle the $3d$ subshell actually has higher energy than the $4s$ one. Does this mean the $3d$ subshell is considered the valence?

Second, why is gold not considered to be a reactive substance. It's electronic configuration is not stable: $[Xe]\text{ } {4f}^{14} {5d}^{10} 6s^1$, as it has an unfilled $6s$ orbital. So why is it not highly reactive like some other transition metals?

Nope, as wiki tells us, only outermost electron shell is considered
valence shell, though d-electrons for transition metals are
considered as valence electrons. No logic here, only tradition.
as you can see, there are three electron subshells above obviously inert Xe core. 5d-electons, that most of the time are far from the nuclei because of the form of their orbitals, feel dramatic increase in effective core charge through the third d-row. For example, for Ti effective charge, keeping its d-electrons is 4. For gold it is 11, so the electrons are well-held.

For s-electrons the form of the orbital puts very significant part of electronic density near the nuclei. This means, that for heavy elements (staring from as far as Ga) a so-known 'inert electron pair effect' is observed: in comparison with p- and d-electrons, due the form of the s-orbital, s-electrons are much less shielded from the charge of the nuclei by underlying, leading to much more strong bonding of said electrons. As effect, Bi is highly unstable in valence state +5, perbromates are much stronger oxidizing agent than perchlorates, but same is not true for bromates and chlorates, and Ag, Au and Hg have very tightly bound s-electons, leading to relative inertness of said metals.

For Au and Ag, however, strongly bound s-electrons participate in extremely powerful (due to relative small size of s-orbitals) s-s interactions, while for Hg only relatively weak s-p and d-p interactions are possible because of filled d- and s- subshells. Thus, metallic Ag and especially Au in metallic state have s-electrons involved into effective covalent interactions, leading to relatively inert nature. Beware, however, that they are far from unreactive: Ag have high affinity to sulfur, and both Ag and Au dissolve in cyanide solutions when oxygen is present, not saying about such agressive reagents as chlorine and bromine.

Still, Au, Ag and Cu share one trait with Ia (I main subgroup) elements. Due to only spherical s-orbitals involvement in atom bonding in metal state, all three metals have high plasticity. Other d-elements have significant d-d bonding in metal states, and d-orbitals are not spherically symmetrical, leading to much less plastic behavior.


Hydrogen as a fuel really worth it? Is hydrogen good to be used in cars? How much of traveling can a car do with medium sized balloon filled with hydrogen? Is it worth it?

The energy density of hydrogen at atmospheric pressure (which it roughly is inside a balloon) per unit volume is a mere 0.01005 MJ/L, which is roughly a factor 3600 less then that of gasoline (at 36 MJ/L).

A typical car does 15 km/L so where a gasoline filled balloon (say 3 L) does 45 km for you, a hydrogen filled balloon will get you only 12.5 meters, i.e. just out of the parking lot.

Therefore, any realistic application of hydrogen is to store it either at high pressure (e.g. 70 bar, where the energy density is about 6 MJ/L) or in so called metal hydrides which allow much higher energy densities. Another option which to me seems to most viable, is to use the Fischer-Tropsch process to make methane or even higher alkanes out of hydrogen and $\ce{CO}$. The big advantage of this route is that our current infrastructure of gas stations can still be used


Photoisomerization of Azobenzene Rotation around the double bond of azobenzene is restricted because it would distort the P orbital overlap between the nitrogen atoms. However, in the $n \rightarrow \pi^*$ excited state ($S_1$), the rotation is no longer hindered and it can isomerize.

I don't understand why this is the case. You are taking an electron from the non-bonding orbital, and moving it into the pi-antibonding. However, the $\pi$-bonding is still present, so I see no reason why the rotation would be unrestrained. Is it something about the interaction between the electrons in the $\pi^*$ with the $\pi$ that weakens the double bond? And if so, how?

Azobenzenes are one of the most widely used moietes in photochemically  switchable systems. Nevertheless, the pathways for the E-Z isomerisation of azobenzene in the $S_1$ and the $S_2$ state have been in debate for decades and I'm not quite sure whether this is finally solved.

  However, in the $n \rightarrow \pi^*$ excited state ($S_1$), the rotation is no longer hindered and it can isomerize.

Yes and no.

It is true that the $S_1$ state arises from a $n \rightarrow \pi^*$ transition in a weak, broad band with $\lambda_{\mathrm{max}} = 432\,\mathrm{nm}$ and $\epsilon \approx 400\,\mathrm{L\,mol^{-1}\,cm^{-1}}$ in hexane.

By the way, E-azobenzene exhibits another, much stronger absorption at $\lambda_{\mathrm{max}} = 319\,\mathrm{nm}$ with $\epsilon \approx 22800\,\mathrm{L\,mol^{-1}\,cm^{-1}}$ in the same solvent. This $\pi\rightarrow\pi^*$ transition to the $S_2$ state would indeed allow for an isomerisation through an one-bond-flip rotational mechanism.

(For the UV data, see 1 and 2).

But - you saw that coming after this lengthy introduction - the situation is much more complicated!

You have neglected the fact that E-Z isomerisations can proceed by other mechanisms than just the simple one-bond-flip that is usually postulated for alkenes!

Forgive me for not getting into all details on this, but for polyenes, to give an example, the so-called Bicycle Pedal (BP) and Hula twist (HT) pathways have been proposed and confirmed as space-conserving alternatives to the one-bon-flip mechanism. The latter can be visualized as "let's just flip the other side of the former double bond around like a rotor".

Back to the azobenzenes. Even from more recent works (2, 3, 4, 5), it seems that the E-Z isomerisation is still not fully explained, although a plain rotation around the former double bond can be excluded.

The two main fractions in the debate support mechanisms in which a contribution of inversion through a change of the $\ce{C-N-N}$ angle in the excited state is assumed. Whether this is the only effect and the reaction proceeds through a transition state with a linear $\ce{C-N-N}$ arrangement or whether an inversion-assisted torsional pathway provided the beter explanation is still unclear.

References

1 George Zimmerman, Lue-Yung Chow, and Un-Jin Paik, J. Am. Chem. Soc., 1958, 80, 3528-3531.

2  Alessandro Cembran, Fernando Bernardi, Marco Garavelli, Laura Gagliardi, and Giorgio Orlandi, J. Am. Chem. Soc., 2004, 126, 3234-3243.

3 Eric Wei-Guang Diau, J. Phys. Chem. A, 2004, 108, 950-956.

4 Eric M. M. Tan, Saeed Amirjalayer, Szymon Smolarek, Alexander Vdovin,    Francesco Zerbetto, and Wybren Jan Buma, Nature Communications, 2015, 6, Article number: 5860.

5 Junfeng Shao, Yibo Lei, Zhenyi Wen, Yusheng Dou, and Zhisong Wang,
J. Chem. Phys., 2008, 129, 164111


Predicting 260/280 of DNA or RNA from sequence? I asked this in biology, but I think they don't want to deal with the chemistry behind UV-Vis.

My question has to do with the Poly A tails on mRNA. I've been doing in vitro transcription and tailing for a while now and have noticed that the untailed RNA has a 260/280 of about 2.0, like you'd expect, but successful tailing results in ratios from 2.15 to 2.45, with higher ratios correlating to better band shifts on agarose gels, implying a longer tail. I suspect the reason for this is because the adenine has a much higher 260/280 than the other bases, so if you add a lot of adenines to the 3' end of the RNA you add much more 260 absorbance than 280 absorbance.

After seeing this I tried to come up with a mathematical method to estimate tail length based on 260/280, since gels are annoying, while I can get the ratio directly from the nanodrop.

I started by using the 260/280 for each nucleotide, A: 4.50, G: 1.15, C: 1.51, U: 4.00, T: 1.47 source here: 1. I then took the sequence for the Luciferase gene I'm using for mRNA and counted the nucleotides, 449 A, 566 C, 470 G, 374 U. I tried to get the weighted average of the 260/280 for my RNA.

A: 449 x 4.50 = 2020.50
C: 566 x 1.51 =  854.60
G: 470 x 1.15 =  540.50
U: 374 x 4.00 = 1496.00
Sum:            4911.60

Nucleotides: 449 + 566 + 470 + 374 = 1859
Sum / Nucleotides : 4911.60 / 1859 = 2.64

2.64 is clearly not 2.0. Even accounting for the potential effect of pH and ionic strength on RNA 260/280 ( acidic solutions have lower ratios ) I can't explain why the calculated ratio is so high. Even if we assumed an RNA with 25% of each base, the weighted average is still (1.15 + 4.50 + 1.51 + 4.00)/4 = 2.79, even higher than for my real sequence.

Doing the same calculation for DNA, (1.15 + 4.50 + 1.51 + 1.47)/4 = 2.16.

So why don't my calculated ratios for RNA and DNA 260/280 match the "ideal" ratios of 2.0 and 1.8? It probably has something to do with base stacking interactions, but I don't enough to compensate for it.

The problem with doing this, especially with RNA, is that secondary structure can have a large impact on the absorption properties of the molecule. Interactions between bases will change the molar absorptivities in ways that may not be easily predictable. The "ideal" ratios are only guidelines and I wouldn't worry too much that a long string of a single base produces a deviation from them. If very accurate measurements are required, hydrolysis into nucleotides before measurement is usually done to avoid these interactions. For your specific case, it may be possible to construct a calibration curve of sorts to get an estimate if you can measure the tail length by some other means. So long as the tail is sort relative to the rest of the molecule, I would expect to see some predictable change in ratios.


Why does Nitrogen form N₂ (and not N₄) while but Phosphorus forms P₄ (and not P₂)? $\ce{N2}$ is a gas
$\ce{P4}$ is a solid
$\ce{N}$ and $\ce{P}$ are in the same group in the periodic table

The bond enthalpies are:
$\ce{N#N}$ triple bond is $949~ \mathrm{kJ/mol}$

$\ce{P#P}$ triple bond is $490~ \mathrm{kJ/mol}$

$\ce{P-P}$ single bond is $200 ~ \mathrm{kJ/mol}$

$\ce{N-N}$ single bond is $159-296 ~ \mathrm{kJ/mol}$

So why does $\ce{P}$ go to $\ce{P4}$ not $\ce{P2}$ and why does $\ce{N}$ go to $\ce{N2}$ and not $\ce{N4}$?

Due to larger atomic size $\ce{P}$ is unable to form pi bonds and so it is tetra-atomic in which each $\ce{P}$ atom is linked with 3 other $\ce{P}$ atoms by 3 sigma bond. But, due to smaller atomic size $\ce{N}$ forms 1 sigma and 2 pi bonds i.e. triple bonds with other $\ce{N}$ atom and exists as diatomic molecule.

Reference: http://philmintz.tripod.com/Chemistry/page4.html


Can covalently bonded hydrocarbons be influenced by a magnetic field? I am doing a University project on Magnetically Assisted Combustion and am curious to know more about how magnetic fields may or may not affect gases such as butane and propane.

I think I am right in saying that due to the gas molecules being covalently bonded, there are no free electrons to be influenced in anyway by an induced magnetic field.

Someone briefly mentioned something about that depending on whether they are saturated or unsaturated but the conversation didn't go any further as he then promptly left.. so I am still unsure.

If anyone could shed any light on this or has any thoughts at all I'd be very grateful for your input.

The method of some experiments I have done was attach a neodynium magnetic arrangement around the fuel line of a gas burner and measuring heat transfer to water.  So far the results have been fairly inconclusive as I am using a camping stove which has problems sustaining a constant gas flow as well as a number of inconsistonsies with heat loss and human error etc .. but that's another issue.

Just because there are no unpaired electrons doesn't mean magnetic fields can't influence something. Unpaired electrons tend to lead to strong interactions with magnetic fields (the dioxygen in air has two and is paramagnetic, an effect best seen in the liquid phase where a magnet will attract the cold liquid).

But compounds with paired electrons also interact with magnetic fields, though the effects are usually weak. But the repulsive effect is sometimes strong enough to levitate a diamagnetic substance (see this youtube video showing how strong magnets can levitate plates of graphite which is strongly paramagnetic).

It is unlikely that such effects would be easy to see in gases. But you are talking about combustion and the most likely effect will not be on the neutral gas but on the flame where the gas is not neutral but a plasma of charged particles. Charged particles or molecules will have a much stronger interaction with a magnetic field though it might still require sensitive apparatus to detect how the field makes a difference.


Reaction of chlorine with hot/cold solution of hydroxide When chlorine reacts with cold solution of say potasium hydroxide, the disproportionation goes to lower oxidation states:

$\ce{Cl2 + 2 KOH -> KCl + KClO + H2O} $

Whilst with hot solution the oxidation state of chlorine goes up to $+V$:

$\ce{3 Cl2 + 6 KOH -> KClO3 + 5 KCl + 3 H2O} $

Why is that? Does the heat make the oxidation to higher ox. states easier?

The chlorate ($\ce{ClO3-}$) is formed by disproportionation of hypochlorite and hypochlorous acid:

$$\ce{ClO- + 2 HClO -> ClO3- + 2 HCl}$$

Higher temperature promotes the formation of hypochlorous acid through hydrolysis of hypochlorite and therefore speeds up the reaction.