Bulging Li-ion battery – is it an immediate danger? What’s the likelihood of explosion? I have a macbook battery that shows as completely dead (doesn’t hold the laptop on even for a second). It now started bulging, and in last week it inflated to about 50 % extra thickness, like this – I can see the inflated battery packs filled with gas.

How big of a danger is to keep a battery like this at home? Will the reaction stop, or will it continue and will the pressure rip through the battery packs?

If that happens, will the battery combust? Internet is full of videos of exploding batteries, but what’s the real likelihood? One in a million? Or one out of two if it swells like this?

I found an advice to keep the battery in a bucket of salty water, does that make sense (I remember that water + lithium = intensive fire)?

How are Li-ion batteries disposed of when I take them into store?

I ask here, because I couldn’t find a qualified answers on the internet, only scare articles about explosions and vague conflicting advices. Thanks!

The battery may have already ruptured due to thermal runaway, meaning that despite it being bulged, pressure from boiling electrolyte has already escaped.  In this case it is also likely that the lithium graphite electrode has already degraded through reaction with air.  If this is true, you are lucky that the previous thermal runaway did not result in a fire.

However, it is impossible to say that the the battery is completely dead without more information, information that would require making measurements that are difficult or impossible for most home chemists.  Thus there may well still be appreciable amounts of flammable electrolyte and air-reactive lithium graphite remaining in the battery. If heated or allowed to react with air, any such material could re-pressurize the system, resulting in a fire or small explosion.  In awareness of this possibility, you should still treat the battery as dangerous.


How does the Mentos-Coke explosion work? We all might have seen this in some videos or in real life. But how exactly does this work? What is the chemical reaction involved?
And why is the reaction of Mentos with Diet Coke more vigorous?

From left to right: action of five Mentos candies (per bottle) with Perrier, classic Coke, Sprite and Diet Coke
Image source

It's actually not a chemical reaction, but a physical reaction.  The surface of mentos acts as a nucleation site for the dissolved carbon dioxide in soda, allowing the gas to come out of solution quite rapidly.  This rapid release of gas is what causes the coke geyser.  Despite mentos appearing to have a smooth surface to the naked eye, there are actually a multitude of microscopic crevices, providing a large amount of nucleation sites.  As far as diet coke working better than regular coke, I am unsure (perhaps diet coke has more carbonation?  Or perhaps there are chemicals in diet coke which make it form foam much better than the other sodas).  I always thought that it was the mint vs fruity mentos that made the largest difference (mint seems to have a lot more nucleation sites which explains the larger geyser in comparison to fruity mentos).


How do we predict the order of melting points of group 1 and group 2 halides? I get really confused by this:

The order of melting points of group 2 chlorides is: $$\ce{BaCl2 > SrCl2 > CaCl2 > MgCl2 > BeCl2}$$ This is completely understandable by Fajan's Rules.

But, the order for group 1 halides is: $$\ce{NaCl > KCl > RbCl > CsCl > FrCl > LiCl}$$

This is what I don't get! Considering Fajan's Rules, $\ce{NaCl}$ is the most covalent, least cationic size, so shouldn"t it have the lowest melting point?

For group 1 halides polarisation is not a dominating factor(except for lithium) since the size of cations is relatively larger than any other cation therefore we must consider lattice energy for determining melting point. As interionic distance increases lattice energy decreases hence melting point decreases down the group.
Lithium is an exception because of its very small size that brings high degree of polarisation. you may want to refer to lother meyer's curve on atomic volumes of elements to get relative idea about the size of the elements.


How to determine the greatest d orbital splitting? This question comes specifically from an IB Chemistry HL Paper 1 in May 2018 TZ1, namely question 8.

Which complex has the greatest d orbital splitting?

It gives 4 Complexes $\ce{[Fe(H_2O)_6]^{2+}}$, $\ce{[Fe(H_2O)_6]^{3+}}$, $\ce{[Co(H_2O)_6]^{3+}}$, $\ce{[Cr(NH_3)_6]^{3+}}$ and it says that they give the colours green, orange, blue and violet respectively.

Initially I thought that the answer would be $\ce{[Cr(NH_3)_6]^{3+}}$ because it gives the highest energy light, being violet. However, the answer is given as $\ce{[Fe(H_2O)_6]^{3+}}$, why is this?

The colour at which the complex absorbs reflects the wavelength of the d–d* electronic transitions. However, this colour is not the same as the transmitted colour (which you see), but is instead complementary to the transmitted colour. Therefore, a complex that appears purple is actually absorbing lower-energy light than a complex that appears red.


Polysytrene factories and hypothetical health concerns Background/Research so far

I'm actually not much of a chemist at all; but I'm doing research for a book I'm writing.

The events of the story begin from a factory that produces/supplies jewel cases for cheap CDs. This factory then pollutes the city's water by reckless dumping. Basically I'm a psychopath and I want to give my city's citizens harmful affects that are revealed after 15 years. Maybe their livers melt or something.

I learned that these jewel cases are made from a plastic called polystyrene.

From this website, I found this:

  Polystyrene (PS, #6): styrene, ethylbenzene, benzene, ethylene, carbon
  tetrachloride, polyvinyl alcohol, antimony oxide, and tert-butyl
  hydroperoxide, bensoquinone

I thought maybe the pollution could be caused by leftover ingredients for the plastic used in the process.

I then examined carbon tetracholoride ($\ce{CCl4}$). This seems really harmful and may cause long term damage if accidentally ingested, especially to the liver. So this sounded exactly what I am looking for. However, I'm not sure how realistic/true this is because further research showed that $\ce{CCl4}$ is, in fact, so harmful that it's rarely used anymore. So it seems like it may be unrealistic to use in the story, because maybe the above article is inaccurate and $\ce{CCl4}$ is not that prevalent.

Benzene also is an obvious one, but it seems to mostly cause blood cancer (leukemia) through ingestion. For reasons that occur later in the story, isn't really isn't going to work. I was kind of looking for a specific organ failure/problems/cancer and not those that effect blood and/or skin.

Ethylene also looks promising because it's so close to ethylene gylcol, and ethylene glycol would really mess people up (it is my understanding is that it's basically antifreeze). But further research indicates that ethylene isn't very harmful at all.

I also thought of cleaning agents for polystyrene instead of actual ingredients. And it sounds like $\ce{CCl4}$ comes into play again because it's used as a "dry solvent," according to Wikipedia in "Historical Uses".

Question

I'm starting to fall short in research and I want my story to be as facutal as possible. So I'm wondering which chemicals would be the most realistic in terms of use in the practical manufacture of polystyrene that could be dumped a really cause problems health-wise for a city.

It is good to see the some authors want to build realism into their fiction: this clearly isn't common. And you have done extensive research already.

The problem is that industrial chemistry is very heavily regulated. So, while it is common for dangerous substances to be used, it is rare to find places where they can plausibly be misused. And the standards for harmfulness are very low. Even some things at the harmful end of the official scale are not that harmful.

The biggest problem you will have when it comes to realism isn't really to do with polystyrene manufacture at all. A factory that makes products with polystyrene won't usually make its own polystyrene but will buy in the plastic raw material as polystyrene beads which can be mixed with other ingredients (like colours) and melted in moulding machines which produce the product. So none of the chemicals used to manufacture the plastic will be used anywhere while making products from the plastic.

If you were to assume that the raw polystyrene was made locally (which isn't that realistic) then you might find plenty of chemicals that could be misused as you have already realised. CCl4 is one of the best candidates but i rarely used any more. Benzene is used in manufacture so is also plausible; styrene (the monomer that is used to make the plastic) is also nasty and possibly the best candidate you haven't considered.

However, if you want to stick with broad realism and a story that uses plausible things associated with a jewel case factory, then you need to consider a different approach. But it might be easier because you can make some things up that even experienced experts can't question. You could, for example, imagine that your factory adds some special ingredient to its products to create unusual colours or effects (maybe the cases glow in the dark, are unusually strong or contain inks that allow patterns to be printed deep into the plastic by lasers). The ingredients to create this effect could be anything you want (and you probably don't have to be too explicit about exactly what it is: give it an obscure acronym). This ingredient could plausibly have any bad properties you want without being unrealistic. Possibly the product will have been invented recently to give the products a special edge and nobody else has yet observed the side effects.


How can I find the stoichiometry of a reaction from the volumes of the reactants and products given? I have volume $V$ of gas $\ce{A}$ and an equal volume $V$ of gas $\ce{B}$ at room temperature and pressure. I mix, ignite, and find I have a gas (mixture of $\ce{A}$ and product $\ce{C}$) at room pressure, $325~ \mathrm{^\circ C}$, and total volume $3V$. What was the stoichiometry of the reaction?
  \begin{align}
 \text{a)}&&\ce{2A + B &-> C}\\
 \text{b)}&&\ce{A + 2B &-> 2C}\\
 \text{c)}&&\ce{2A + B &-> 2C}\\
 \text{d)}&&\ce{A – 2B &-> C}\\
\end{align}

The answer is $\text{b)}$ but I can't seem to get the logic.
$V$ vol of $\ce{A}$ plus $V$ vol of $\ce{B}$ reacts to give $2V$ vol of $\ce{C}$. How did we find stoichiometry from this? What about some unreacted $\ce{A}$ ?

In a problem like this I think that the best way to answer the question is to set up the equations and try all the answers. However, in this case we can infer from the text that $B$ need to be the limiting reacting, as all of it reacts but some of $A$ is left unreacted. The only answers providing this are answer b and answer d. However, the latter contains a minus sign which doesn't makes sense chemically, so let's focus on the former.

We can compute the moles involved in the reaction using the ideal gas equation $PV = nRT$. Let's assume the volume is $1 \mathrm{L}$, we can calculate the amount of moles for both $A$ and $B$ (as they have the same volume and are present in the same conditions):

$$n = \dfrac{PV}{RT} = 40.88 \mathrm{mmol}$$

If the stoichiometry is the one proposed in answer b ($\ce{A + 2B -> 2C}$)
then the whole quantitity of $B$ will react with half the quantity of $A$ to give the same amount of moles of compound $C$, which will be $n = 40.88 \mathrm{mmol}$. Half of $A$ will be left unreacted, which is $n = 20.44 \mathrm{mmol}$. Using the aforementioned equation we can now compute their volumes:

$$V = \dfrac{nRT}{P}$$

Which is $V_a = 1 \mathrm{L}$ and $V_c = 2 \mathrm{L}$, which in total is $V_{tot} = 3 \mathrm{L}$, exactly three times more than the original volume.


What is the relation between molarity and molality? Is the following equation correct:

$$m = \frac{1000 \times M}{M \times M_b - 1000 \times d},$$

where $M$ - molarity, $m$ - molality, $d$ -density of solution, $M_b$ - molar mass of solute?

$$m = \frac{M}{\rho - (M \times wt. of solute)}$$

$m=$molality

$M=$Molarity

$\rho=$density

all quantities in S.I. units


Chemical Spill in Charleston, West Virginia I live in Charleston, WV.  On 1/9/2014 we were told not to use our tap water for anything other than flushing the toilet after a storage facility leaked over 7,400 gallons of 4-methylcyclohexane methanol into the Elk river, less than two miles up-river from the water company's main in-take.

The CDC said that concentrations under 1ppm were safe.  After a week of bathing in my sink and drinking bottled water, my area was determined to be safe, however my water still has a strong odor.

There is not much information about this chemical's health effects.  While the chemical itself is low-toxicity, I wonder if it might react with items in the water supply to create other compounds which may be more toxic.

Specifically, would this chemical react with chlorine in the water, or cast iron, PVC, and copper in the pipes?  Does anyone have any experience with this compound?

The name MCHM or "4-methycycloheaxe methanol" (as newspapers reported it) is not entirely a systematic name and is a little ambiguous. The actual compound is better called  4-methylcyclohexylmethanol and the structure is shown below:

Specific details about its toxicity are not easy to find, though it has been patented as an air freshener ingredient. It has also been around for a long time and is used in the coal industry as a frothing agent in significant quantities so if it had some particularly nasty property we would probably know by now. Some safety information is available (pdf) on the related 1,4-cyclohexanedimethanol and that isn't particularly toxic. So there doesn't seem to be any strong reason to worry about it, though it will taint the flavour of water. There are no obvious paths for it to known nasties in low concentrations in water, so the biggest worry is probably the taint.

Having said all this, the company should not have let it get into the water supply.


Gas solubility in water and alcohol I know that a certain gas is soluble in water but not in alcohol. How can I find the solubility of the gas in mixture of water and alcohol? Do I use a simple linear relation? The gas in question is not a noble gas, but does not react with either liquid.

So long as we assume that this is an ideal situation in which there is really insignificant solubility of the gas in ethanol but significant solubility in water, then yes, you can assume that the concentration in the mixture is linear with the water fraction of the liquid. In other words:

$${H_m = H_w * x_{w}}$$

Where:
${H_m}$ = Henry's law solubility of the gas in the water-ethanol mixture
${H_w}$ = Henry's law solubility of the gas in water
${x_{w}}$ = mol fraction of water in the liquid phase


Why is 1-ethylidene-4-methylcyclohexane chiral? As far as I can tell, there doesn't seem to be a chiral carbon in this compound. C-4 of the cyclohexane ring has two groups with exactly the same connectivity, and the exocyclic double bond can't give rise to optical isomerism. What am I missing?

The strict criterion for a compound to display chirality is that it must not be superimposable upon its mirror image. Let's ignore the chair conformation of the ring for a while, and assume it adopts a planar conformation. You could draw a side-on view of the ring like this:

We can see that if we reflect the molecule (in the plane of the ring + double bond), its mirror image is not superimposable on itself, which makes it chiral:

In practice, the cyclohexane ring does adopt a chair conformation. That does not fundamentally affect the fact that the compound is chiral, as the presence of a chair conformation cannot "remove" this existing source of chirality (for example, there is no way for the hydrogen and methyl groups to swap places by virtue of a cyclohexane ring flip or a similar process).
Actually, this is very similar to the case of an allene, which you may or may not be familiar with. Notice how the conformation drawn above looks almost like this allene:

with the planar ring taking the place of the second double bond. These are examples of axial chirality (Wikipedia; IUPAC), where the chirality stems from the disposition of groups about an axis (in both cases, the axis in question is the C=C bond axis). Further explanation of axial chirality in allenes can be found in this question.

A note on nomenclature
The tetrahedral chiral centre can be named with the usual (R) and (S) stereodescriptors, as described in user55119's answer.
When it comes to the double bond, ChemDraw (and likely some other software) suggest that they can be named with the familiar stereodescriptors (E) and (Z) (the process is also described in user55119's answer). However, this is not fully appropriate.

The exact rules are complicated, but it boils down to the fact that the double bond in question is enantiomorphic: that means that when it is reflected, the configuration of the double bond is reversed. We can see that from the names generated above: the two mirror images have supposedly different configurations at the C=C double bond.
The descriptors (E) and (Z) are supposed to be used only for diastereomorphic double bonds, where the configuration is invariant (i.e. does not change) upon reflection. The large majority of stereogenic double bonds fall into this category, but this one does not.
In place of (E) and (Z), the preferred descriptors are respectively seqTrans and seqCis (see mykhal's answer, and P-92.1.1 (f) of the 2013 IUPAC Blue Book) .


What is difference between intramolecular redox and disproportionation redox I can not understand difference between the intramolecular redox reaction and the Disproportionation redox reaction. Both involve the same molecule as substrate, but what is difference?

Disproportionation is a redox reaction where one element works both ways, with part of it being oxidized and another part reduced. It does not have to be intramolecular in the strict sense, since the oxidizing and reducing agents do not necessarily come from the same molecule (though they come from similar molecules, of course). Think of the well-known reaction:
$$\ce{3KClO -> KClO3 + 2KCl}$$
It is certainly a disproportionation, but one may hardly call it intramolecular; there is not much going on within one molecule (or rather within one ion, for all of these are ionic compounds). Chlorine oxidizes chlorine; now, one ion contains one atom of chlorine, so to oxidize anything it needs to find another ion in the first place.

Intramolecular redox reaction occurs when we have the oxidizing and reducing agents within one molecule. They may or may not be of the same element. In the former case it is going to be a disproportionation, but not in the latter. Think of TNT:
$$\ce{2 C6H2(NO2)3CH3 → 3 N2 + 5 H2O + 7 CO + 7 C}$$
See what's going on? Basically, N oxidizes C. They are indeed from the same molecule, so this may be intramolecular, but they are not the same element, hence this is not a disproportionation.


Hydrogen Fuel Cell and Battery For hydrogen fuel cell to produce electricity, hydrogen  will need to be supplied to the anode. To split oxygen and hydrogen from water and obtain hydrogen, we need to use electrolysis. To use electrolysis, we need to use a battery.

My question is, why do we bother splitting up hydrogen and oxygen and then feed it into the fuel cell instead of just using battery to generate electricity? Or is it not practical to obtain hydrogen and oxygen from electrolysis and use it to generate electricity from fuel cell? Thank you

If you have a fuel cell, only hydrogen needs to be carried with the car; the air provides all the oxygen you need. Hydrogen is an extremely light fuel, but there are some issues with current means to contain it (e.g. heavy high pressure container, or massive hydride tanks).

According to Iceland is Bullish on Hydrogen, they've started to use their plentiful geothermal and hydroelectric power to electrolyze water, using the hydrogen to power cars and the oxygen byproduct for other uses.


Solubility of Ag3AsO4 in 0.02 M K3AsO4 Calculate the solubility of $\ce{Ag3AsO4}$ in $\pu{0.02M}~\ce{K3AsO4}$ neglecting the activity coefficients. Find the relative error. $K_\mathrm{sp}(\ce{Ag3AsO4}) = \pu{6.0e-23}$

I know how to calculate the relative error but I get a very complicated equation finding the concentration solubility product constant ($K'_\mathrm{sp}$). There should be a quicker way to solve this since it is a midterm question.
I tried this:
$$\ce{Ag3AsO4 -> 3Ag+ + AsO4^3-}$$
$$K_\mathrm{sp} = 27x^4$$
$$x = \pu{2.78e-6}$$
$$[\ce{Ag+}] = 3x = \pu{8.34e-6M}$$
$$[\ce{AsO4^3-}] = x = \pu{2.78e-6M}$$
$0.02~\mathrm{M}\ \ce{AsO4^3-}$ comes from $\ce{K2AsO4}$. So there should be an equation like:
$$K_\mathrm{sp} = (\pu{8.34e-6} - 3x)^3 \times (\pu{2.78e-6} + 0.02 - x) = \pu{6.0e-6}$$
And things get complicated. After finding $x$, I will also have found the final concentrations of silver and $\ce{AsO4^3-}$ ions. Then I will read the activity coefficients of them from the appendix table.
I have two questions:

Is my method true or false?

What is an easier way of solving this problem?

You start of by calculating solubility with one approximation: that all $\ce{AsO4^3-}$ ions are obtained by dissolving the potassium salt:
Let $S$ be the molar solubility:
$$K_\mathrm{sp} = 0.02 \times 27 S^3$$
$$S = \pu{0.0223 M}$$
Then, you calculate the solubility by taking in account the concentration of arsenate ion obtained by dissolution of silver arsenate:
$$K_\mathrm{sp} = 27 S^3 \times (S+0.02) = 27 S^4 + 0.54S^3$$
Solving this equation by applying the iteration method gives the value of $S = \pu{0.018M}$.
So the relative error is $23.9~\%$.


Hydrogen fuel cell - why do the H+ ions move through the electrolyte Consider the following hydrogen fuel cell

http://butane.chem.uiuc.edu/pshapley/Enlist/Labs/FuelCellLab/FuelCell.html

At the anode, hydrogen is oxidised (losing electrons). My first question is this: why is the anode negative if it is attracting electrons?

My second question refers to the fact that once the hydrogen atoms are iodised, the H+ ions move through the electrolyte towards the oxygen ions at the cathode, reacting to form water.
If the anode is negative, why would positive H+ ions move away from it?

In a galvanic cell, the anode is positively charged.  The cathode is negatively charged.  A good mnemonic I learned a while ago is that anions go to the anode, and cations go to the cathode.  This is because opposite electric charges attract.  So negatively charged anions are attracted to the positively charged anode.  And positive cations are attracted to the negative cathode.

The electron itself was originally discovered as the component particles of cathode rays from cathode ray tubes.  That electrons come out of the cathode is a hint about cathode charge: the buildup of electrons on the cathode eventually results in the emission of an electron beam.  If electrons are building up, the cathode must be negative.


Why are higher oxides of non-metals more acidic? I have studied that if the electronegativity difference between O- and element M is more then it is a basic oxide and if the difference is less then it is an acidic oxide. However, it is a fact that higher oxides of Non metals are more acidic than lower oxides. If the theory concerns polarisation, is it not contradictory?

It is expected oxides with more polar $\ce{M-O}$ bonds are stronger Lewis bases than those with less polar bonds.

Furthermore, for the same element, the polarization gets weaker/Lewis alkality weaker when more $\ce{O}$ atoms are attached.

When hydrated to acids, another phenomena plays a role in acidity: a charge delocalization.

It is well observable that acids like $\ce{HNO3}$, $\ce{HClO3}$, $\ce{HClO4}$, $\ce{H2SO4}$, $\ce{HBF4}$ are strong, as the delocalization  has a good sandbox to play in.

It also plays role in solubility of salts in organic polar solvents, where there is a large  organic cation with one of the above small anions.


What is the physical explanation driving current flow in a concentration cell? In a galvanic cell, current is driven because the species at the cathode is reduced by grabbing some electrons from the cathode. The cathode now lacks electrons so it pulls them from the anode. Now the anode needs electrons, so it pulls them from the species that will be oxidized in the other half cell. So, now we have a current!
Now here is a concentration cell:

As you can see. the anode and cathode are made of the same metal. Only the concentrations of the solutions are different. Using my logic from above, since the metals are the same, they should be pulling on the electrons with equal magnitude. So how come current still flows? In other words, why are the nickel ions at the cathode more "hungry" for electrons than the nickel ions at the anode?
My attempt: It has to do with the rates of the metal dissolving. If I were to cut the wire connecting the two half-cells, I would notice that the beaker with the lower concentration solution would be increasing in concentration (from solvating the metal) faster than the beaker with the higher concentration. This is because the lower concentration one is "further" from the solubility equilibrium.
When I connect them back together the metal at the anode dissolves faster, thus demanding electron recompense from the cathode metal.
Attempt 2: The solution with higher Ni ion concentration has overall more ions dissolved in it. Maybe all these ions create an electric field greater than the one on the other side, thus prompting electron movement.
Attempt 3: It has to do with kinetics. Lets assume that everytime a nickel cation collides with an electrode, it steals some electrons. The half-cell with the higher concentration will have a higher amount of collisions per second. Thus, more electrons are being "consumed" at the cathode than anode per second, creating a charge differential, which in turn generates a current.

Start with a simple thought experiment: pour 100 mL of 1 M nickel (II) sulfate solution into a beaker and very carefully layer 100 mL of 0.01 M nickel (II) sulfate solution on top of the more concentrated layer. Then, even without convection or deliberate mixing, diffusion will, sooner or later, result in the solution having concentration of 0.55 M. In what follows, it is assumed that evaporation is negligible, even on long time scales.
Now consider Fig. 1 below:

In this concentration cell, it is assumed, for simplicity, that the solution volumes are equal in the two electrode solution reservoirs. The salt bridge is 0.55 M nickel (II) sulfate solution, so its concentration is exactly half-way between that of the anode solution and cathode solution.
With the concentrations shown in the figure, the cell potential is +0.02958 V, and the anode is the nickel electrode in the more dilute solution at the left. On the cathode side, the log term is zero, in the Nernst ewuation at upper right in Fig. 1, so the potential is -0.236 V. But on the anode side, the potential is lower by 29.58 mV, due to the log term. So the anode is more negative, the cathode is more positive and electron flow is always from more negative to more positive. Therefore, Ni is spontaneously oxidized at the anode, yielding the necessary electrons, and nickel ions are spontaneously reduced at the cathode.
As a direct consequence of having an external connection, i.e., load or voltmeter, between the electrodes, net oxidation will spontaneously take place at the anode, resulting in an increase in the nickel ion concentration in that solution reservoir. Likewise, nickel ions in cathode solution reservoir will be spontaneously reduced at the nickel cathode and the nickel ion concentration will decrease. The external connection affords the opportunity for something useful to happen, i.e., using the cell as a power source. Otherwise, only diffusion happens (see below).
The cell will be "dead", i.e., have no more free energy to tap, when all three solutions are 0.55 M. There will be no concentration gradient left.
Now consider Fig. 2:

This is the same as Fig. 1 except that the cell is open circuited, i.e., there is no way for electrons to transfer from one electrode to the other. So all that is going to happen is spontaneous diffusion through the salt bridge. Eventually, there will be just one final concentration at 0.55 M.
During this process, no net oxidation occurs at the left electrode and no net reduction occurs at the right electrode. If a voltmeter is subsequently attached, the reading will be less than +0.02958 V because the concentrations are no longer the original ones: both concentrations are closer, thanks to diffusion, to the 0.55 M center value.
The measured voltage, even using a high impedance voltmeter, is not actually the open circuit voltage. It may be very close, but there is a conceptual issue. In the open circuit situation, no electrons can flow between the electrodes and no net redox processes occur at the electrodes. So the open circuit potential is an ideal potential and the measurement goal is to estimate accurately the open circuit potential while allowing negligible current to flow.
With a high input impedance voltmeter attached to the electrodes, a very small current, i.e., nA or pA, can flow. This only slightly loads down ("perturbs") the cell potential, so it provides an accurate estimate of the true open circuit potential. And it means that oxidation occurs at the anode, to a very slight extent, and reduction at the cathode, likewise to a very slight extent. Solution concentrations are negligibly perturbed.
So there are two parallel processes taking place. First, diffusion spontaneously acts to equalize the solution concentrations. Second, the concentration cell provides a way of hurrying the equalization process along while extracting useful electrical energy. So the former wastes the available free energy while the latter extracts much of it.
Finally, consider Fig. 3 below:

Now the salt bridge is gone and nothing happens: there is no DC cell potential and the solution concentrations are constant. In figures 2 and 3, there is no net redox happening: for every Ni atom that might get oxidized, there is a nickel ion that gets reduced. Everything is balanced at both electrodes. (So if the electrodes start with highly polished "mirror" finishes, they may get surface roughened by this zero net redox process.)
But once a load or voltmeter is attached to the two electrodes in Fig. 2, this opportunity causes the equilibria to become unbalanced: the solution concentrations must change by increasing (in the anode reservoir) and decreasing (in the cathode reservoir), and that immediately implies that electrons flow from the anode (at the left), through the load or voltmeter, to the cathode (at the right).
Perhaps the following will also help clarify matters. Consider Fig. 4 below:

This shows two beakers, each made of solid nickel, and containing the indicated nickel (II) sulfate solutions. Obviously, this does nothing: it is just two fancy beakers of solutions. Next consider Fig. 5:

Now the two nickel beakers are touching, i.e., in electrical contact. Does this change anything? No. Nothing happens. Next, consider Fig. 6 below:

Now there is simply a wide glass beaker with a solid nickel partition that entirely prevents the two solutions from making contact. Does this change anything? No. Nothing happens. Finally, consider Fig. 7 below:

This differs from the previous figure in having a salt bridge. It is exactly equivalent to using a nickel wire to short the two nickel electrodes in Fig. 2. So the cell is shorted out, there is no cell potential, and the cell will try to supply its short circuit (i.e., maximum) current, limited by the kinetics at the electrodes. So the nickel ion concentration will increase in the anode reservoir and decrease in the cathode reservoir, as expected.


How can one tell from the MO diagram of the cyanide ion that the HOMO is carbon-centred? The cyanide ion, when acting as a nucleophile, typically attacks via carbon as its HOMO has a greater contribution from carbon. But how can we deduce this from the MO diagram?

To me it appears that the nitrogen $\mathrm{2p}$ orbitals are energetically closest to the $3\sigma$ HOMO, so its coefficient should be larger in the HOMO. What is wrong with this line of reasoning?

.

The 2s and 2pz atomic orbitals are of the same symmetry in the molecule. That means that the 2 sigma* and 3 sigma are actually hybrids. The former is a hybrid were most of the electron density points to the bond (making it partially bonding), while in the latter most of the electron density points outside, i.e. a lone pair in the carbon.


What defines a proton as "labile" in NMR? What makes some protons labile and others not? For example the OH and NH2 protons get exchanged but CH ones don't.
When we use the D2O test, we say the labile protons get exchanged for D so their peaks disappear. But can't the H be re-exchanged? And why doesn't the H now in the solvent give its own peak?

For your first question you may want to look up "proton exchange NMR", as partly addressed in for instance this post:

Proton attached to N or O in HNMR spectroscopy

$\ce{^1H/^2H}$ exchange is a solvent catalyzed process that results in rupture of the original O-H or N-H bond with formation of a new bond to previously H-bonded solvent $\ce{^2H/^1H}$. What keeps the exchange process from occuring is an activation energy barrier.  Available nonbonding electrons on O or N atoms in the solute or solvent which can interact with $\ce{^1H}$ or $\ce{^2H}$  affect the process, as does the pH (exchange can proceed through acid or based-catalyzed mechanisms). Catalysis of C-H bond rupture with formation of a new C-H bond has a larger activation energy barrier or may lead to chemically nonidentical products and reactants.

As regards the second question, if the (protonated) solute is dilute and the solvent is deuterated, there isn't a significant concentration of solvent $\ce{^1H}$ to drive reversal to the protonated state. For instance, if your molecule is present in water at 1 mM concentration and contains one exchangeable proton, and if all of the protons exchange and the solvent water was originally 100% deuterated, then the final concentration of protons in the solvent is $\pu{\rho = ( 0.001/110) mole/mole = 9 \mu mole/mole (or 9 ppm)}$ and the solute will exhibit a similar final isotopic concentration ratio (ignoring isotope effects).


Intuition on Gibbs Free Energy In many places, I read that Gibbs Free Energy is called "free energy" because some of the enthalpy that comes from the chemical reaction becomes "waste heat" due to a change in the entropy of the system, and only the rest is usable as work. However, I cannot imagine how this is the case. Take an exothermic reaction with a positive change in entropy for example. Because $\Delta G = \Delta H - T \Delta S$, then we find that $\Delta G$ is more negative than $\Delta H.$ That is, more energy is released by the system than the enthalpy of reaction. How does this fit in with the "waste heat" explanation, that not all of the energy released is usable? There is more energy released now.

From Second Law of Thermodynamics viz. $$\Delta S_\textrm{sys}+ \Delta S_\textrm{res}\geq 0\,,$$ it can be concluded that $$W_\textrm{sys}\leq (U_1- U_2)- T(S_1 -S_2) \tag{1}\;.$$

Now, $$W_\textrm{sys}= \underbrace{W_T}_\textrm{pressure-volume/compression work}+ \underbrace{A_T}_\textrm{non-compression work}\;.$$

Putting the the first term on the RHS of $(1)\,,$ it yields,

\begin{align}A_T &\leq (U_1- U_2)- T(S_1 -S_2) - W_T\\ &\leq (U_1- U_2)- T(S_1 -S_2)+ P(V_1 -V_2)\;.\end{align}

Now, this is equal to the negative change of Gibbs energy $G$ provided the thermodynamic process is done at constant pressure and temperature.

So, that means,  $$A_T \leq G_1 -G_2\;.$$ Note the $\leq$ sign; this implies $(G_1-G_2)$ is the maximum value $A_T$ can attain.

Since, $\Delta G \lt 0\;,$ this implies $A_T$ is positive. Now, if the process is reversible, then the non-compression work $A_T$ is maximum viz: $$A_T= G_1 -G_2\;.$$

This definitely supports the term free energy as the decrease in $G$ gives the amount of energy "that can be 'freed' and made available for work".

  That is, more energy is released by the system than the enthalpy of reaction.

Of course, the extra energy came from heat energy absorbed from the environment as the entropy of the products was greater than that of the reactants by $\Delta S\;.$


is it possible for covalent compounds to produce colour during flame test? I have heard it is the metal ions that are the reason for flame colour during a flame test. so in covalent bond, there are no metal ions does that mean covalent compounds don't produce flame colour. if not what is the reason, if yes why doesn't glucose produce colour. Also is it possible to identify whether a substance is ionic, or covalent with just the flame test. I have searched the web but there are no clear explanations.

The colors produced in flame tests are the result of electrons becoming excited and returning to their ground states, releasing photons of characteristic colors. Different colors from flame tests result from different energy gaps between the electron shells in different atoms. Since flame tests determine properties of individual atoms, they cannot be used to determine if a compound is ionic or covalent. In fact, bonds are normally broken when a compound is placed into a flame, so it would be nearly impossible to use a flame test to find anything about the bonding nature of a compound. Glucose does not produce a color when it is burned because it just undergoes combustion - more or less the same reaction that is creating the flame in the first place.

On the other hand, there are other ways to determine if a compound is ionic or covalent. Ionic compounds normally have ridiculously high melting points, and if you put an ionic compound into distilled water, it will conduct electricity. There are lots of tools to determine if a compound is covalent or ionic; however, a flame test is not one of them.


Stability by using charge/radius rule I found a question

Which is more stable among the following?
(a) $\ce{CaCO3}$
(b) $\ce{MgCO3}$

Answer : (a)
Explanation was given as:

$\ce{Mg^2+}$ has smaller radius than $\ce{Ca^2+}$ and thus charge/radius of $\ce{Mg^2+}$ is greater , hence it is less stable.

While when reading about the coordination compound, I found the sentence

Higher the charge of central metal atom ion i.e., greater ionic potential
charge/radius then greater is the stability.

In first case higher the ratio less is the stability.
In second case higher the ratio more is the stability.
Is the difference in the answers in both cases due to a difference in the salts (i.e. simple and  complex salt) or is any one of them wrong?

Please allow me to answer the question in two parts.

Stability of carbonates

We observe this order when we compare the thermal stability of group I-A carbonates: $$\ce{Li2CO3 \lt Na2CO3 \lt K2CO3 \lt Rb2CO3 \lt Cs2CO3}$$

On similar grounds, the thermal stability of group II-A carbonates can be predicted.

This order is in accordance with Fajans' Rules. Here's the table we need to glance at from the Wikipedia page.

  $$\begin{array}{|c|c|} \hline
\mathrm{Ionic} & \text{Covalent}\  \\ \hline
\text{Low positive charge} & \text{High positive charge} \\
\text{Large cation} & \text{Small cation} \\
\text{Small anion} & \text{Large anion} \\ \hline \end{array}$$
  Polarization will be increased by high charge and small size of the cation.

Thus, larger the ratio of charge and radius, higher is the polarization power of the cation, more is the covalent character in the compound and lesser is the ionic character; Ionic compounds are more stable when exposed to heat. A higher ratio will therefore mean lesser stability.

Stability of complex compounds

Thermodynamic stability of complexes, on the other hand, is determined by the Irving–Williams series, according to which general stability sequence of high spin octahedral metal complexes for the replacement of water by other ligands is:
$$\ce{Mn^2+ < Fe^2+ < Co^2+ < Ni^2+ < Cu^2+ > Zn^2+}$$
Owing to this observation, a lot of theories were formulated to explain this phenomenon. One such theory was that since Crystal Field Theory (CFT) was based on the idea that a purely electrostatic interaction existed between the central metal ion and the ligands, the stability of complexes should be somehow related to the size by charge ratio (otherwise known as the ionic potential) of the central metal atom. [Source]

An interesting thing to note here is that Wikipedia has stated in its article that:

  None of the explanations can satisfactorily explain the success of the Irving–Williams series in predicting the relative stabilities of transition metal complexes.

Thus from manganese till copper, it can be said that higher the ionic potential, more is the stability.


How to get the freeze point of mixed liquids? I recently bought in Italy a bottle of "Alcool puro" which is in fact 95% Alcohol and 5 % Water. So alcohol has a freezing point of -114°C and water a freezing point of 0°C. How can I calculate then the freezing point of the "Alcool puro"?

In general it is extremely hard to get the freezing point of a mixture purely by calculation. The reason is that the specific molecular interactions of the two substances you have mixed cause pretty much every mixture to need a unique formula for the calculation.

That said, there are some ways to get a rough estimate but these estimates mainly work for (linear) hydrocarbons because those are (fairly) simple to generalize. (google for "freezing point hydrocarbon mixture")

In your specific case, because it is ethanol-water your best way to find out the freezing point is to look up the experimental results of someone else. The engineering toolbox has a dataset on ethanol-water which you could use to get a reasonable estimate for your case (they don't list 95v%, but they do have the whole range from 0-100v% in steps of 10v%).


Multiple "/" in catalysts What does it mean if a catalyst has multiple "/" symbols, such as $\ce{Mn/Fe/Al2O3}$? Does it mean either $\ce{Mn}$ or $\ce{Fe}$, or does it mean both are supported on $\ce{Al2O3}$?

There should be only one "/" symbol. When more are used, it should be unambiguous from the context what is the carrier, e.g. in your case it would be more appropriate to use a dash:

$$\ce{Mn-Fe/Al2O3}$$

From section 1.3 Composition, structure and texture of catalysts of 1975 IUPAC recommendations for heterogeneous catalysis [1, p. 79] (emphasis mine):

  Support. In multiphase catalysts, the active catalytic material is often present as the minor component dispersed upon a support sometimes called a carrier. The support
  may be catalytically inert but it may contribute to the overall catalytic activity. Certain bifunctional catalysts (§1.2.8) constitute an extreme example of this. In naming
  such a catalyst, the active component should be listed first, the support second and the two words or phrases should be separated by a solidus, for example, platinum/silica or
  platinum/silica-alumina. The solidus is sometimes replaced by the word "on", for example, platinum on alumina.

  Promoter. In some cases, a relatively small quantity of one or more substances, the promoter or promoters, when added to a catalyst improves the activity, the selectivity, or the useful lifetime of the catalyst. In general, a promoter may either augment a desired reaction or suppress an undesired one. There is no formal system of nomenclature
  for designating promoted catalysts. One may, however, for example, employ the phrase "iron promoted with alumina and potassium oxide".

References

Burwell, R. L. Manual of Symbols and Terminology for Physicochemical Quantities and Units. Appendix II. Heterogeneous Catalysis; Pure Appl. Chem. 46, 71, 1976. https://doi.org/10.1351/pac197646010071 (Open Access)


Why is tert-butoxide often used in elimination reactions when it is not necessary? I understand that tert-butoxide is used to favor the Hofmann product in elimination reactions due to its steric bulk (and thus inability to abstract a proton in a more substituted position); however, I have often seen proposed organic synthesis involving the use of tert-butoxide when it is seemingly irrelevant. For example, 1-bromobutane will undergo an elimination reaction with or without a sterically hindered base (such as tert-butoxide); however, anytime I see reactions involving elimination reaction with primary substrates, such as the bromo group in 1-bromobutane, tert-butoxide is used rather than ethoxide. Is tert-butoxide simply used to disfavor the formation of substitution products. Is its steric bulk enough to not allow it to act as a good nucleophile in substitution reactions?

Several reasons:

t-BuOK is especially known as a strong base, and a poor nucleophile. Its large, bulky structure causes it to perform exceptionally poorly in substitution, literally eliminating any side reactions when the desired product is the elimination product.
It is easily available, like Raphaël insists.
It's basic strength depends on the medium. It is very strong in DMSO, where the solvent complexates well with $\ce{K+}$ ions. However it is relatively weaker in benzene. This flexibility gives it inevitable popularity.
Many condensation reactions (Stobe's, Darzen's) are seen to have greater yield when t-BuOK is used.


What general uses do beta-lactam functional groups have? I am currently working on a mechanistic study that deals with stereoselective $\beta$-lactam formation. I am well aware that there is a class of antibiotics, that involve this particular functional group. However, the wikipedia article of this compound class does not offer very much to continue, apart from the antibiotics.

Are there any other uses of the $\beta$-lactam moiety in general, apart from drug design?

I would like to be able to view my research in a broader scope. Since my part is more focussed on the reaction mechanism of the synthesis, I have not been too much concerned about the main purpose of this particular functional group, yet. This is of course apart from the obvious drug design process. For me this is just a question of interest.

As I am resurrecting this question, I am in particular looking for some interesting publications about the above mentioned question. However, I am willing to recognise, that the most common use is in drug design. If someone from this area of expertise could provide a source as a starting point for more research in this area.

β-lactams, also known as 2-azetidinones, have experienced a resurgence of interest in the last decade or so, with a number of new applications being exploited. Of course, antibiotic activity is well-known, but I hope you don't limit your acceptance of drug design just to this class of chemical.

β-Lactams have found use in a number of fields of medical application, inlcuding gene activation, enzyme inhibition and cholesterol absorption inhibition, and recently there has been growing interest in the use of these molecules as synthetic templates for accessing other target moelcules.

There are lots of sources of references if you go digging, but some key reviews that may interest you include:

TETRAHEDRON Volume 56, Issue 31, Pages 5553-5742 (July 2000). The entire issue is concerned with developments in β-lactam chemistry;
EUROPEAN JOURNAL OF MEDICINAL CHEMISTRY  Volume 45, Issue 12, Pages 5541-5560   (DEC 2010) "2-Azetidinone - A new profile of various pharmacological activities"  gives some very good insights into new medical applications of this class of compound, including anticancer, antiparkinson, antidiabetic, antiinflammatory activities. I think they can also do the washing up and buy your groceries.
TETRAHEDRON  Volume 59, Issue 39, Pages 7631-7649 (Sep 2003) "Recent progress in the synthesis and chemistry of azetidinones" gives a pretty good summary of synthetic approaches to these molecules.
Alcaide and Almendros do a lot of work using 2-azetidones as reactive synthons in catalytic cyclisations. A recent review by them is

ACCOUNTS OF CHEMICAL RESEARCH  Volume 47, Issue 3, Pages 939-952 (Mar 2014) "Gold-Catalyzed Cyclization Reactions of Allenol and Alkynol Derivatives", but they have a heap of other papers also.

Another example, probably somewhat similar, of their use as a reactive synthon is
BULLETIN OF THE KOREAN CHEMICAL SOCIETY  Volume 28, Issue 1, Pages 17-28 (2007) "Indium and gallium-mediated addition reactions" - I have to admit I only read the abstract of this paper, but they appear to use azetidones as synthons in metal-mediated syntheses of a range of compounds.

Finally, the August edition of Chemical Reviews Vol 114, Issue 16 looks at small heterocycles in synthesis and has some interesing papers, inparticular:

CHEMICAL REVIEWS Volume 114, Issue 16, Pages 7930-7953 "Chemical Synthesis of beta-Lactams: Asymmetric Catalysis and Other Recent Advances". This is a very good paper for you, I think.
CHEMICAL REVIEWS Volume 114, Issue 16, Pages 8257-8322 "Four-Membered Ring-Containing Spirocycles: Synthetic Strategies and Opportunities". An interesting application of lactams in spiro molecule formation.
CHEMICAL REVIEWS Volume 114, Issue 16, Pages 7954-8015 "Synthesis and Reactivity of 2-(Carboxymethyl)aziridine Derivatives". This looks at lactams being used as synthons for a range of molecules, here focussing on aziridines.

I hope some of these point you in the direction of what you are looking for.


Mechanism of acid-catalyzed addition of water across an alkene My book tells me that when 1-propene reacts with water and sulfuric acid, I get 2-propanol. The proposed mechanism is that one of the electron pairs from the double bond of 1-propene travels to sulfuric acid, de-protonating it. At the next step, water comes in as a nucleophile and attacks the carbocation intermediate - the terminal carbon gets hydrogen added on it because its the carbon with the most hydrogens. We get CH3CH(H2O)CH3. Now water comes in as a nucleophile again and de-protonate a hydrogen to form 2-propanol.

Now when I asked someone on my ship who used to study organic chemistry at a university, she said that she remembered that it was one of the electron pairs from the double bond of 1-propene that travels to H3O+ instead.

Is either proposed mechanism acceptable?

No, the person on your ship is incorrect.

An electron pair can only be transferred to electrophiles, which are species having a vacant orbital present on them. $\ce{H3O+}$ doesn't have a vacant orbital, so that can't happen. Here is a description as to why:

However,  $\ce{H+}$ ions, which have vacant orbitals can accept the pair and have it shift, while the other carbon sharing the double bond loses its electron, making it electrophilic, and can be approached by a nucleophile, in this case, $\ce{H2O}$:


Why are aprotic solvents unable to solvate negative ions well? Why are aprotic solvents unable to solvate negative ions well? I know that the reason is due to the absence of very positive hydrogen atoms, but then why are they able to solvate positive ions well? Shouldn't its ability to solvate both types of ions be the same? I thought there their ability to solvate ions is related to whether they are partially positively or negatively charged, but shouldn't the extent to which they are partially positively or negatively charged be the same as it depends on the relative positions of the electrons?

This can be attributed to the lewis acid/base properties of the solute/solvent. Anions (negative ions) and aprotic solvents both function as lewis bases. They 'repel' each other. However, a cation (positive ion) functions as a lewis acid and is 'attracted', resulting in greater dissolution.
How well a compound dissolves in a given solvent depends on a multitude of factors. This includes the two very important ones discusses here, polarity and lewis acid/base character.


Determine # of molecules in an unknown sample How might one determine the # of molecules in an unknown sample? If the sample is a gas,  we might use $PV=nRT$ to estimate # of molecules. If the sample is a solid or liquid and the chemical formula is known, we simply need to measure the mass. But can we determine the # of molecules without first determining the chemical formula?

There are attempts to redefine the mol through measuring spheres of monocrystalline silicon.  If you have the macroscopic dimension of your sample and the dimensions of your unit cell (from x-ray diffraction) you know how many atoms if silicon are there.


Why is nitric acid such a strong oxidizing agent? In my teaching-lab experiments I've seen that nitric acid solutions are able to facilitate the dissolution of metals such as silver, even though they are more active than hydrogen. Does anyone know why nitric acid is so special? Why is it such a strong oxidizing agent?

Because, unlike in other metal dissolution reactions, the $\ce{H+}$ of $\ce{HNO3}$ isn't reduced—the $\ce{NO3-}$ is. The following data and balanced reactions are taken from Wikipedia:

\begin{align}
  \ce{NO3- + 4 H+ + 3 e- &-> NO  + 2 H2O} &
    E^\circ_\mathrm{red} &= \pu{0.96 V} \\
  \ce{NO3- + 2 H+ +   e- &-> NO2 +   H2O} &
    E^\circ_\mathrm{red} &= \pu{0.79 V} \\
  \ce{Ag+(aq) + e-       &-> Ag(s)} &
    E^\circ_\mathrm{red} &= \pu{0.799 V}
\end{align}

Since the standard reduction potential (SRP) of the $\ce{NO2}$ reaction looks smaller than that of $\ce{Ag+}$ (I may be wrong, but by significant digits it can't be greater than $0.799$), one can conclude that it's the $\ce{NO}$ reaction that's occurring here. And the $\ce{NO}$ reaction has a large enough SRP to oxidise $\ce{Ag+}$.

Usually nitrogen compounds are pretty versatile when it comes to redox reactions, since nitrogen shows many oxidation states. So the simple reason for why $\ce{HNO3}$ is so strong an oxidising agent (with respect to other acids) is that it has a different, better path available to it to get reduced.

Note that the exact reduction path (i.e. final reduction products/oxidation state) depends upon the concentration of nitric acid—so much that copper can be oxidised in three different ways.


Do polymerized cooking oils release free radicals I am seasoning a carbon steel wok and from what I understand it basically involves heating cooking oil on it until it gets to the smoke point and then forms a polymerized layer that makes it like a non stick pan.  I have also heard that as oil gets heated close to and past the smoke point it breaks down releasing free radicals and other unhealthy compounds.  My question is that when the heated oil polymerizes do these compounds get permanently locked up in the polymer or will they eventually get into the food cooked in the pan?

If you use fresh cooking oil containing unsaturated fats (like olive or canola oil) to season your pan, it will thermally polymerize to form a stable layer that gives the cooking surface its nonstick properties. Note that a saturated fat like coconut oil, lard, or ghee will not work very well or at all for this, since it does not have any double bonds to undergo polymerization. The radical formation you are concerned about results over time from air oxidation of unsaturated fats to form peroxides--these in turn break down to give aldehydes and acids that cause the bad flavor of rancid fat, and have adverse health effects. (Not a problem for coconut oil.) Heating fresh oil in a pan to create a nonstick coating may result in some air oxidation in addition to polymer formation, but I think it is not a big cause for concern. High polymers are generally nontoxic because the body does not absorb them.


Do alkali metals react with ammonia to form hydrogen gas? I was recently studying about s-block alkali metals and I found the following reaction:
$$\ce{\underset{alkali\\ metal}{M\phantom{_y}} + \underset{solution}{$(x +y)$ NH3} -> \underset{ammoniated\\ metal cation}{[M(NH3)_x]+} + \underset{ammoniated\\ electron (blue)}{[e(NH3)_y]-}}.$$
It is also known that nitrogen has electronegativity of 3.0 on Pauling's scale. So, the hydrogen attached to it is an active hydrogen. Also, it shows H-bonding, so the hydrogen is an active-H.
Alkali metals react with active-H compounds to release hydrogen gas. But the reaction that takes place here is in contradiction to the aforementioned fact.
Do we have an explanation for this? Or is it a  mistake on my part?

What you have written is right, but it is just the first step.
The release and solvation of an electron is fast. Reaction of an electron with ammonia to release hydrogen $$\ce{2 e-(solv) + 2 NH3 -> H2 + 2 NH2-(solv)}$$ is slow. Deep blue (diluted) or bronze (concentrated) "sodium electride" solution in liquid ammonia slowly converts itself to the colorless ammonia solution of $\ce{NaNH2}$.

For the water case, all is extremely fast, as water is a much stronger acid than liquid ammonia. Blue-like color can be catched by 10000 fps camera, what is commented in SE  Chem Q - Reaction of potassium with water
BBC article, referring the Nature article by Pavel Jungwirth and colleagues (ref.1),

ABSTRACT: Alkali metals can react explosively with water and it is textbook knowledge that this vigorous behaviour results from heat release, steam formation and ignition of the hydrogen gas that is produced. Here we suggest that the initial process enabling the alkali metal explosion in water is, however, of a completely different nature. High-speed camera imaging of liquid drops of a sodium/potassium alloy in water reveals submillisecond formation of metal spikes that protrude from the surface of the drop. Molecular dynamics simulations demonstrate that on immersion in water there is an almost immediate release of electrons from the metal surface. The system thus quickly reaches the Rayleigh instability limit, which leads to a ‘coulomb explosion’ of the alkali metal drop. Consequently, a new metal surface in contact with water is formed, which explains why the reaction does not become self-quenched by its products, but can rather lead to explosive behaviour.

with high speed (cca 10000 fps) camera photos shows the blue colour near 0.3 ms after the metal/water contact (ref. 2). It may be rather more toward violet, it seems to me. The abstract of ref. 2 explains the blue color:

Alkali metals in water are always at the brink of explosion. Herein, we show that this vigorous reaction can be kept in a non‐exploding regime, revealing a fascinating richness of hitherto unexplored chemical processes. A combination of high‐speed camera imaging and visible/near‐infrared/infrared spectroscopy allowed us to catch and characterize the system at each stage of the reaction. After gently placing a drop of a sodium/potassium alloy on water under an inert atmosphere, the production of solvated electrons became so strong that their characteristic blue color could be observed with the naked eye. The exoergic reaction leading to the formation of hydrogen and hydroxide eventually heated the alkali metal drop such that it became glowing red, and part of the metal evaporated. As a result of the reaction, a perfectly transparent drop consisting of molten hydroxide was temporarily stabilized on water through the Leidenfrost effect, bursting spectacularly after it had cooled sufficiently.

References:

Philip E. Mason, Frank Uhlig, Václav Vaněk, Tillmann Buttersack, Sigurd Bauerecker, Pavel Jungwirth, “Coulomb explosion during the early stages of the reaction of alkali metals with water,” Nature Chemistry 2015, 7, 250–254 (https://doi.org/10.1038/nchem.2161).
Philip E. Mason, Tillmann Buttersack, Sigurd Bauerecker, Pavel Jungwirth, “A Non‐Exploding Alkali Metal Drop on Water: From Blue Solvated Electrons to Bursting Molten Hydroxide,” Angew. Chem., Internat. Edn. Engl. 2016, 55(42), 13019-13022 (https://doi.org/10.1002/anie.201605986).


Why does rearrangement not take place in Wurtz reaction? The Wurtz reaction is $$\ce{R-X + Na ->[ether] R-R}$$ where $\ce{R}$ is alkyl. But if we take butylchloride with $\ce{Na}$, the product is n-octane and not 3,4-dimethylhexane. I am unable to understand the fact that rearrangement does not happen even when it's a free radical mechanism.

The Wurtz Reaction mechanism involves radical anions. Neither radicals nor carbanions rearrange the same way that carbocations do.
The first step of the mechanism is a single electron transfer from $\ce{Na}$ to $\ce{R-X}$:
$$\ce{Na + R-X -> Na+ +[R-X].^{-}}$$
The alkyl halide radical anion fragments to a carbon radical and a halide anion:
$$\ce{[R-X].^{-}->R. + X-}$$
The primary alkyl radical is the least stable radical, but is not as self-destructive as the primary carbocation. Most importantly, primary radicals decompose through disproportionation ...
$$\ce{R-CH2-CH2. +.CH2-CH2-R->R-CH=CH2 + CH3-CH2-R}$$
... or coupling:
$$\ce{R-CH2-CH2. + .CH2-CH2-R -> R-CH2-CH2-CH2-CH2-R}$$
The coupling pathway is a perfectly good explanation of the Wurtz reaction, but another pathway is possible. $\ce{Na}$ is a good reducing agent, and that primary radical is readily reduced to an anion:
$$\ce{R. + Na -> R:^{-} + Na+}$$
The carbanion then reacts with a molecule of alkyl halide in a SN2 reaction:
$$\ce{R:^{-} + X-R -> R-R + X-}$$
The formation of the anion is actually problematic, especially if $\ce{R-X}$ is secondary or tertiary, since E2 elimination will compete.
$$\ce{(CH3)3C:^{-} + (CH3)3C-X -> (CH3)3C-H + (CH3)C=CH2 + X-}$$


Why is nitric acid often used to digest any solid sample? Why is $\ce{HNO3}$ often used to digest any solid sample?

I heard many solid samples often digested with $\ce{HNO3}$ solution because nitrate compound always dissolve in water.

Is the concept correct or not?

This is pretty much correct. Nitric acid is a convenient oxidizing acid which will dissolve most of the metals (except for $\ce{Au}$, $\ce{Ta}$ and elements of platinum group), forming either oxides or nitrates; sulfur is oxidized to sulfate; phosphorus to phosphate; organics is oxidized and nitrated. This assures consistent predictable products in probe preparation.

Most oxides can be separated by filtering, and the majority of inorganic nitrates is soluble in aqueous solutions, which paves the way for the further classic analytical methods (titration, UV-Vis spectroscopy, atomic emission spectroscopy, etc.). Nitrate-ion is not prone to hydrolysis, and formed nitrates are no longer demonstrating oxidizing properties (unless they are solidified and heated), which is overall quite convenient.

Also, nitric acid is cheap and doesn't react with glass or PTFE, which allows to built various lab apparatus suitable for particular analytical method.


Does melanostatin peptide whiten the skin in a similar manner as to how melanotan darkens it? I was reading about peptides and came across an interesting peptide that bodybuilders use to darken their skin.

How can peptides change the genetic makeup of someone's skin and darken it?

I also found another peptide called melanostatin that inhibits alpha-MSH. So my question is, can it whiten the skin?

Because bodybuilders aren't very much interested in whitening their skin, there were no trials and experiences with that peptide.

So does melanostatin work similarly to how Melanotan darkens the skin?

Disclaimer: I do not intend on using this peptide on animal experiment, nor humans for that matter. My question is purely theoretical and for educational purposes.

Melanotan

The "Melanotan" you're referring to is probably Melanotan I, also known as afamelanotide, which is an analogue of α-melanocyte stimulating hormone (α-MSH).  In other words, it activates (some of) the same receptors as α-MSH.

α-MSH has several functions, one of which is to stimulate the cells that darken the skin (melanocytes), so they produce more melatonin.  It activates these cells via the MC1 receptor.  Afamelanotide is more specific to that receptor, so it darkens the skin without some of the other effects.

It acts at the hormone level, so it won't be changing anyone's genetic makeup - stop using it and you'll go back to your normal colour.

Melanostatin

Confusingly, the name Melanostatin has been applied to more than one compound.  I've found two short peptides that might fit the bill.

The most likely candidate is described in this leaflet, which says it is used for skin lightening.  It's a receptor antagonist - it prevents α-MSH from binding to the MC1 receptor, so reducing melanin production.  So yes, it's working at the same site as Melanotan I but having the opposite effect.

The other possibility is also known as Melanocyte-inhibiting factor, which inhibits production of (natural) α-MSH, so it would be expected to lighten the skin.  However, since α-MSH has several other functions, reducing its production would also affect other systems - there would be side-effects, and skin lightening doesn't seem to be the main thing it's known for.  More here: Wikepedia


Polarising Power and Covalent Character Higher the positive oxidation state of the central atom, more will be its polarising power which, in turn, increases the covalent character of bond formed between the central atom and the other atom.

This seems counter-intuitive to me. When the polarising power of the central atom increases, shouldn't the bond be more ionic instead of covalent? If not, can you tell me where I am wrong?

I believe it is about perspective. Note that "covalent molecule with ionic character" and "ionic compound with covalent character" are different.

If we are looking at a covalent molecule and assessing its ionic character, we cannot talk about polarising power, because in a covalent molecule, there would be no cation and anion to polarise in the first place. Thus, we would go by the rule: the bigger the electronegativity difference, the higher the ionic character and therefore, the more polarised the bond.

If we are looking at a ionic compound and assessing its covalent character, then we will go by the rule: the more polarising the central atom, the higher the covalent character.

Now, from the way you phrased your question, I believe that you are conflating the concepts "big electronegativity difference" and "polarising power". They are distinct concepts and should be thought of separately.

Although bond polarity is associated with electronegativity difference, "polarising power" is about the polarisation of the electron cloud of the anion by the cation and not the polarisation of the bond. This is also why "polarising power" is a concept associated with ionic compounds with covalent character.

Hope this clarifies your doubt! If you are still unclear and require some examples for clearer understanding, please leave a comment on my post.


Superscript and subscript together after the same atom In the chemical formula of some ion compounds, there is the necessity to write the charge in the superscript and the numbers of atoms in the subscript, but together at the right of the same symbol like in the PO43− ion:

Does IUPAC provide universal guidelines saying that I have to typewrite the two numbers in a stacked or staggered way?

From IUPAC Green Book [1, p. 51]:

In writing the formula for a complex ion, spacing for charge number can be added (staggered arrangement), as well as parentheses: $\ce{SO4^2-}$, $\ce{(SO4)^2-}$. The staggered arrangement is now recommended.

Also, as you are apparently a $\mathrm{\LaTeX}$-user, the same convention is used by default in chemformula package. From the manual:

If a compound does not start with a sub- or superscript and there is both a sub- and a superscript, the superscript is shifted additionally by a length determined from the option charge-hshift = {❬dim❭}, also see page 13f.
The second point follows IUPAC’s recommendations.

References:

IUPAC “Green Book” Quantities, Units, and Symbols in Physical Chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.


Full equation when using vinegar, hydrogen peroxide, and salt to etch copper What is the full chemical equation when you use the following reactants to etch copper:

vinegar / acetic acid (5%) $\ce{CH3COOH (aq)}$
hydrogen peroxide (3%) $\ce{H2O2 (aq)}$
salt $\ce{NaCl (s)}$
copper $\ce{Cu (s)}$

Yes there are previous questions asking the same thing (one and two). However, the answers to both give only parts of the reaction, not the full equation.

Looking elsewhere on the web, I've come across this Instructables article that says copper(II) chloride $\ce{CuCl2}$ is formed when the copper ions bond with chloride ions from the salt. And therefore the solution is greenish in color instead of the blue it would be if copper(II) acetate $\ce{Cu(CH3COO)2}$ was formed as answered in question two. (When doing this at home, I also got a distinctly greenish color when salt was used, and a distinctly blue color when it was omitted).

  Table salt, or $\ce{NaCl}$, brings chloride ions $\ce{Cl-}$ to which $\ce{Cu^2+}$ ions will bond to form cupric chloride or $\ce{CuCl2}$ instead of being left in solution (those ions would endlessly come back and forth from $\ce{Cu}$ to $\ce{Cu^2+)}$. You can see this during the reaction (if you leave the reaction with no salt it will turn blue which is the color of copper(II) acetate and stop, whereas if you put salt it will turn green, the color of cupric chloride, and carry on).

Another article on the web says both copper(II) acetate and copper(II) chloride are produced, but no equation is provided:

  Equal parts vinegar and peroxide worked fine, dump in plenty of salt (to the point of having undissolved salt in the bottom of jar) since it is consumed in the reaction. You should see a vigorous fizzing at the copper surface. Add more salt as the reaction slows.

  The solution will turn green as the copper is converted to copper acetate and/or copper (II) chloride

Can someone please give the complete chemical equation of the reaction that occurs when acetic acid, hydrogen peroxide, and salt are used to etch copper? I can't seem to find it anywhere on the web.

Context:

I want to dispose the products of the etching process safely. If copper(II) acetate is the product as suggested in answer two, then following the recommended approach for disposal is safe:

  You can bypass the problem by collecting the etch solution, and submerge aluminum foil.

  $$\ce{3Cu(CH3COO)2 (aq) + 2Al (s) -> 2Al(CH3COO)3 (aq) + 3Cu (s)}$$

  React with the aluminum until the bluish color disappears, then collect the solid copper + remaining aluminum to throw in the trash. You can now flush the solution down the drain.

However, if the product is instead copper(II) chloride, the approach is not safe as the $\ce{AlCl3}$ formed is considered a hazardous waste and shouldn't be flushed down the drain.

$$\ce{2Al (s) + 3CuCl2 (aq) -> 3Cu (s) + 2AlCl3 (aq)}$$

The acetic Acid is used to dissolve the copper, however it cannot do so directly as the reaction of copper with protons is not spontaneous.

$$\require{cancel} \ce{Cu_{(s)} + 2H+_{(aq)} \cancel{->} Cu^2+_{(aq)} + H2_{(g)}}$$
The hydrogen peroxide is used to oxidize the copper since copper does not spontaneously react with acids directly.
$$\ce{\frac12 H2\color{red}{O2} +e- -> OH-}\\ \ce{Cu_{(s)} -> Cu^2+_{(aq)} + 2e-}$$
However, hydrogen peroxide is unstable in acid so the acetic acid is used because it reacts with hydrogen peroxide to form a peroxyacid which is more stable.

$$\ce{CH3C(O)OH + H2\color{red}{O2} -> CH3C(O)\color{red}{OO}H + H2O}$$

This peroxyacid can react directly with copper to dissolve it.

$$\ce{CH3C(O)\color{red}{OO}H_{(aq)} + Cu_{(s)} -> Cu[CH3C(O)O]OH_{(aq)}} \\ \ce{Cu[CH3C(O)O](OH)_{(aq)} + CH3C(O)OH_{(aq)} -> Cu[CH3C(O)O]2_{(aq)} +H2O}$$

As far as the sodium chloride goes, I'm not sure what that is for. My best guess is that it is catalytic in dissolving the copper similar to how salt corrodes steel

$$\ce{CH3C(O)\color{red}{OO}H_{(aq)} + Cu_{(s)} ->[\text{cat.} Cl-_{(aq)}] Cu[CH3C(O)O]OH_{(aq)}}$$

As far as the final species goes, you probably have a complex from the excess chloride

$$\ce{Cu[AcO]2 (aq) + 2NaCl (aq) -> [Cu(AcO)2Cl2]^2- (aq) + 2Na+ (aq)}$$
which is in dynamic equilibrium
$$\ce{[Cu(AcO)2Cl2]^{2-} (aq) <=>[Cl-][AcO-] [Cu(AcO)Cl3]^{2-} (aq) <=>[Cl-][AcO-] [CuCl4]^{2-} (aq)}$$

The color comes from the excess chloride that acts as a ligand to the copper From Wikipedia:

  Aqueous solutions of copper(II) chloride. Greenish when high in [$\ce{Cl−}$], more blue when lower in [$\ce{Cl−}$].


What causes oxygen cylinders to explode when in contact with oil (oiled cloth)? Perhaps I did not put it correctly, but from the welders I know that if you leave the oiled rag on the cylinder valve, it can explode. At least, I saw how greasy oily rags that just lay in the trash can are smoldering. It is interesting to me to understand the principle of this phenomenon: whether it depends on the amount of oil, oxygen, pressure in the cylinder. If possible, I would like to see the chemical reactions occurring.

Summary
⤑ Oils are mostly made up of Fatty Acids ("FAs")
⤑ FAs oxidize and release a lot of energy
⤑ The reaction is proportional to the O2 content of the gas in contact with FAs
⤑ The reaction is also proportional to the contact surface area between FAs & surrounding gas
⤑ Rags offer a high contact surface area between the liquid soaked into them and the surrounding atmosphere
⤑ Maybe that increasing the surface area between air and oil (via rags) also increases volatile FAs which yields an extreme increase in contact area between FAs & O2

Tentative Explanation
I think that oils are made up of mon-/di-/tri-glycerides (at least natural oils): One glycerine bound to 1, 2 or (most often) 3 fatty acid molecules. Fatty acid chains can be oxidised, using up O2 and releasing water + a lot of energy (that's part of our metabolism, and yields much more energy per gram than the metabolism of carbohydrates). I think that the less saturated a fatty acid chain is, the more readily it can be oxidised, which is why linseed oil, which has a very high Alpha-linolenic acid content (a polyunsaturated fatty acid), can undergo spontaneous combustion. A rag exposes oil to a high contact surface with the surrounding atmosphere. Now if in addition to all of that the atmosphere is high in O2, you have perfect conditions for a high-energy-yielding reaction to massively occur.

The local temperature, oil-to-air surface area, air pressure, and air O2 content are the factors which most affect the likeliness of spontaneous combustion. The amount of oil will mostly affect how much of the reaction can occur / how long it can occur (it's the fuel in the reaction). It has a minor impact on the reaction speed only proportional to the fact that there has to be some correlation between oil volume and contact surface area.


Adjusting PBS with HCl: pH drift after dilution of phosphate buffer It is some kind of everyday lab science and with a feel of shame I cannot answer this observation.

I noticed a pH drift (~0.5 units) of my 10X phosphate buffer (PBS), when I adjust the pH with $\ce{HCl}$ and dilute the 10X buffer afterwards.
When I prepare another buffer by mixing $\ce{Na2HPO4}$ and $\ce{NaH2PO4}$ to achieve the desired pH value, there is virtually no drift after dilution.
If using carbonate buffer, $\ce{CO2}$ is created if I set the pH with $\ce{HCl}$, but none with the also acidic $\ce{NaH2CO3}$ is used.

I just want to know how the phosphate buffer system is working and why there is a pH drift after dilution if I use $\ce{HCl}$ to set the pH. This might answer the general question, why adjusting the pH with $\ce{HCl}$ in PBS is not suggested.

I feel there could be something wrong with the counter ion maybe?!

The differences between PBS and the buffer obtained by mixing $\ce{Na2HPO4}$ and
 $\ce{NaH2PO4}$ are in ionic strength ( because of the presence of $\ce{KCl}$ d $\ce{NaCl}$ in PBS) and in the ratio of [acid]/[conjugated base]. We have to signal that the buffer capacity (buffer effectiveness) is maximum when $\ce{pH=pKa}$.
The best way to dilute 10x PBS into 1PBS (let's assume we want to prepare 1L of 1xPBS from 10xPBs) is to begin by diluting 100 mL of 10xPBS solution with 800mL water, then adjusting the pH and finally completig the volume to 1000mL. In doing so, there will be no drift in pH.


Synthesize coumarin from 4-allyl-2-methoxyphenol (eugenol) I'm trying to figure out how to get the following reaction: (coumarin from eugenol)

My immediate thoughts are

$\ce{OH}$ and $\ce{OCH3}$are bad leaving groups.
Could I just add $\ce{HI}$ to the $\ce{C10H12O2}$ to convert the methoxy ($\ce{OCH3}$) group to an $\ce{OH}$group? Then, I could add $\ce{Zn}$ to remove the $\ce{OH}$groups?
Or, is there a way to make the methoxy ($\ce{OCH3}$) group into methanol, which would be a better leaving group?
Once the methoxy and hydroxy groups are removed, I'm left with the benzene ring and the attached butene. I would need to add the $\ce{CO2}$ to form the second benzene ring, but I'm not sure how this would be done.
I might be thinking about this all the wrong way. I was thinking that I need to remove the methoxy and hydroxy groups since they're on a benzene ring, and my product's benzene ring doesn't have these groups. Maybe I don't need to get rid of the methoxy and hydroxy groups at all. Maybe I just need to shift the bonds around?

Since no-one has offered an answer I'm going to try but as I have no database access some of this is a bit flaky.

Heat (maybe microwave) to give multiple Claisen rearrangement products then separate to isolate 2-allyl, 6-methoxyphenol. This is the flakiest step.
Hydroborate with oxidative work-up to take the allyl to propionic acid.
Ring close (PTSA, Toluene)
Reflux over activated MnO2 to introduce double bond.
Treat with BBr3 to remove Me from MeO group
Zn reduction of OH to H


Physical Interpretation of Gibbs Free Energy I know that $\Delta G = \Delta H - T\Delta S$ and that if $\Delta G$ less than $0$ a reaction will be spontaneous. What I am wondering is what is the actually physical interpretation of Gibbs free energy. I mean, if the Gibbs free energy of a particular reaction $20\ \mathrm{J\ mol^{-1}}$, what does that actually mean?

Gibbs free energy is a  measure of the total non-expansionary work that a system can do. So in the case of your example, that system has the potential to do 20 joules of non-expansionary work per a mole of substance. Non-expansionary work refers to work other than pressure-volume work, such as electrical work. Hence not only is Gibbs free energy is important to tell whether a reaction will be spontaneous or not, it is also important to engineers as they tell how much non-expansionary work a system, such as motor engine, can do.

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

Constant Pressure

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


How did chemists in the 18th century learn that magnesium gain mass when it's burned? Crucial to disprove the phlogiston hypothesis was the observation that magnesium gains mass when burns.

But how was this observation made in the 18th century? Is there any way this property of magnesium would be obvious to a layman today?

The phlogiston hypothesis is about a substance, phlogiston, which was supposed to exist within matter that be consumed or released when it is combusted.  If this were true, then when magnesium is combusted the mass would decrease.

To someone who believed this to be true, it would come as a great surprise to find that if they combusted 100g of magnesium the mass would increase by about 65%.  This would be easy to measure with a set of balances, even with much smaller masses of a few grams.

A layman might not come across many day-to-day examples of a combustion reaction with a mass increase, since most things that burn release a lot of mass as gases, and that effect is what phlogiston was intended to explain.  One possibility, aside from burning metal, is the oxidation of metal in air.  Someone could weigh an iron nail or other object, allow it to rust, and weigh again to show it has gained mass.  This could easily be demonstrated by any merchant with a set of scales, but it's not obvious the rust is an oxide related to the combustion products.


Why no thermal expansion for camping tents? I know that solid materials expand when heated, which is called thermal expansion.

But what is happening to tent square or tent canvas of (plastic?) camping tents? I have noticed, that during the day, when it’s hot, there is a lot of tension in the skin of a tent. But when it gets colder (when raining or during the night), the tension is less and the fabric gets sloppy.

What is the reason?

The surface of a tent consists mostly of water-resistant polymers such as latex or plastic. These polymers are disordered molecules such as spaghetti, so they have unlike crystals a high entropy.

Gough (1805) and Joule (1859) studied the thermal behavior of rubber. They made the following observations, which are known as Gough-Joule effects:
• a taut with constant weight rubber shrinks when it is heated
• rubber emits heat when it is stretched

The cause of the rubber elasticity is fundamentally different than the elasticity of crystalline materials. In contrast to a crystal, the atoms are not regularly arranged in a rubber, but they form long chains (polymers) which are interconnected and form a network. Clearly, one can think of it as a plate of cooked spaghetti. This difference in microstructure are reflected in the mechanical properties of rubber. When the rubber network is loose, very many different configurations (configurations) of the chain between two connection points are possible - more disorder (larger entropy S).


How do you determine which entity gains or loses an H+ in an equation? $$\ce{HCO3- (aq) + H2O(l) <=> H2CO3(aq) +OH- (aq)}$$

I do not understand why this makes an Arrhenius base and why it doesn't form:
$$\ce{HCO3- (aq) + H2O(l) <=> CO3^2- (aq) +H3O+ (aq)}$$

Could someone explain this to me?

Both of these equations are valid, as $\ce{HCO3-}$ is amphoteric (i.e. both an acid and a base).

To answer your question, one has to consider that the $\mathrm{p}K_\text{a}$ value for $\ce{H2CO3}$ is 6.37 and that for $\ce{HCO3-}$ is 10.32. So if we started from "pure" $\ce{HCO3-}$ (let's assume sodium bicarbonate) and add it to water, what would happen?

The $\mathrm{p}K_\text{b}$ for $\ce{HCO3-}$ is of $14 - 6.37 = 7.63$, it is larger than the $\mathrm{p}K_\text{a}$ for the acidic function (10.32). Therefore although both a weak base and a weak acid, it is a stronger base than acid, increasing the pH when added in water.


Why do heavier elements have smaller specific heat values. In the periodic table the trend is that as the mass of each element increase's the specific heat tends to go down. This seems to be counter-intuitive because if I am not mistaken as mass increase's the amount of heat that the substance can hold goes up. So why does the specific heat of these elements go down?

The more particles in a system, the more modes of vibration (and rotation), so one gram of $\ce{H2}$ has about $235 \times$ more particles than one gram of U. The heat is "stored" in the motions of the atoms and electrons, until reaching millions of kelvin, at which point dissociation of the nucleus would be significant in specific heat. See these links and  for more information.


During electrolysis, why are the products attracted to the cathode? Take the electrolysis of Lead(II) bromide:

We can write it as two half-reactions:

$$\ce{Pb^{2+}(l) + 2e^{-} \rightarrow Pb(l)}$$

$$\ce{Br^{-} \rightarrow Br + e^{-}}$$

In the electrolysis reaction, lead is formed at the cathode and bromine is liberated at the anode. But why is it so? I mean, before gaining or losing electrons to become neutral, lead is positive and bromide is negative. So, naturally they should be attracted to the cathode and the anode respectively.

But why do they become neutral after they reach the electrodes. Why do they gain electrons at only the electrodes to become neutral?

You have to think about what exactly a cathode and anode are. You are falling victim to this:

A widespread misconception is that anode polarity is always positive (+). This is often incorrectly inferred from the correct fact that in all electrochemical devices, negatively charged anions move towards the anode and positively charged cations move away from it. In fact anode polarity depends on the device type, and sometimes even in which mode it operates, as per the above electric current direction-based universal definition. Consequently, as can be seen from the following examples, the anode is positive in a device that consumes power, and the anode is negative in a device that provides power. - Wikipedia

It's your conditioning from other chemical nomenclature--"cation" and "anion"--which are absolute about their charges.

So beyond misleading nomenclature, what does cause this to happen?

Take this picture for example: http://img.sparknotes.com/figures/0/02480ae8fc1a41b131a3fdb5a698e9a3/compare.gif

If E°cell > 0, then the process is spontaneous (galvanic cell)
If E°cell < 0, then the process is nonspontaneous (electrolytic cell)
where E°cell = E°(V) reduction + E°(V) oxidation
In your case,

Pb → Pb2+ + 2e- = 0.13

2Br- → Br2 + 2e- = -1.06

E°cell = .13 - 1.06

E°cell = -.93

In your case, the process is nonspontaneous (makes sense, think about trying to reduce ionized bromine), so you're looking at an electrolytic cell, which is battery driven (it needs work put in to operate.)

So, again, to directly answer your question:

  In the electrolysis reaction, lead is formed at the cathode and bromine is liberated at the anode. But why is it so? I mean, before gaining or losing electrons to become neutral, lead is positive and bromide is negative. So, naturally they should be attracted to the cathode and the anode respectively.

Lead is formed at the anode because work from a battery was put in that pushed the reaction in that direction. Before work was put in, the reaction would have proceeded in the opposite direction (think exothermic vs endothermic, but in this case spontaneous cell potential vs nonspontaneous cell potential.)

  But why do they become neutral after they reach the electrodes.

They become neutral at the electrodes because that is where the reaction happens, and that is where the electrons are transferred.

  Why do they gain electrons at only the electrodes to become neutral?

Again, where else could they gain electrons? They are sitting in a solution contain themselves and their respective anions, which they have already reacted with, and will no longer be transferring electrons with.


The effect of pressure on equilibrium constant What will happen to the $K_\text{eq}$ in the following reaction if volume is increased?

$$\ce{2NO(g) + O2(g) <=> 2NO2(g) + energy}$$

From le Chatelier's principle I know that it must be constant unless temperature is changed, however it doesn't make sense because if

$$K_\text{eq}= \frac{[\ce{NO2}]^2}{[\ce{O2}][\ce{NO}]^2},$$

then by increasing the volume we know that equilibrium will shift left and favor reactants. Hence concentration of reactants would increase while concentration of products would decrease. If we plug that back into the equilibrium constant equation we see that if numerator decrease and denominator increase that must yield a smaller number than the initial number. Hence the equilibrium is not constant. Why this occurs?

First step: Don't use Kc... use Kp!

You are focusing on concentration, but as we're working with gases it becomes easier to work with partial pressures.

$K_\text{p}= \frac{\ce{p(NO2)}^2}{\ce{p(O2)}\ce{p(NO)}^2}$

Now, if we remember that partial pressure is pressure $\times$ mole fraction ($p_{a}=P \times \chi_{a}$), and substitute these in we get (and cancelling out the pressures):

$K_\text{p}= \frac{\chi_{(NO_{2})}^2}{\chi_{(O_{2})}\chi_{(NO)}^{2}P}$

Now, if we consider increasing the volume. By doing so (and assuming that temperature is kept constant), then we have decreased the pressure (ideal gas law: PV=nRT).

Now we can see that although the number of moles changes (and hence the mole fractions), our equilibrium constant will remain unchanged thanks to the effect of pressure.

A good guide is here.


Oxidation of alcohols by Ceric Ammonium Nitrate I have realized that reaction with ceric ammonium nitrate essentially results in coordination of alcohol with $\ce{Ce(IV)}$ and my teacher referred that we should be able to proceed it to form ketone by dissociating $\ce{C-H}$ of the alcohol.I have the following doubts:
A) Precise mechanism as I am literally lost as to why the $\ce{C-H}$ being as strong bond is dissociated. what is the driving force here?
B) What is the final oxidation state of cerium as it seems that $\ce{Ce(III)}$ is produced but we initially had $\ce{Ce(IV)}$. Is there an electron transfer mechanism here?

In this reaction, the ion $\ce{Ce^{4+}}$ is a powerful oxidant. It needs one electron to be transformed into the more stable ion $\ce{Ce^{3+}}$. This electron may come from the central function $\ce{-CHOH-}$ of any secondary alcohol $\ce{R-CHOH-R'}$. The bond between $\ce{O}$ and $\ce{H}$ is able to loose $1$ electron and $1$ $\ce{H+}$ ion. Same thing happens with the bond $\ce{C-H}$ belonging to the same group $\ce{-CHOH-}$. As a consequence, two electrons are emitted with two $\ce{H+}$ ions, and the remaining organic molecule becomes a ketone $\ce{R-CO-R'}$. The two electrons react with two ions $\ce{Ce^{4+}}$. The two half-equations are : $$\ce{Ce^{4+} + e^- -> Ce^{3+}}$$  $$\ce{R-CHOH-R'->  R-CO-R' + 2 H+ + 2 e-}$$ so that the final equation is : $$\ce{2 Ce^{4+} + R-CHOH-R' -> 2 Ce^{3+} + R-CO-R' + 2 H^+}$$


Acid catalysed dehydration of alcohol with a bicyclo group Pathway 1: Rate determining step is the formation of carbocation. In this pathway, a conjugated tertiary carbocation is generated as an intermediate.

Pathway 2: One of the resonance form of intermediate carbocation proceeds to produce a more stable  product (due to benzene ring) through alkyl migration. Though, the resonating form of carbocation is conjugated secondary carbocation which is less stable than the carbocation formed through pathway 1.

My questions are:
1- Is tert-butyl (don't know the correct term) migration possible in pathway- 2?
2- What are the major and minor products? Which factor will decide the major product: stability of final compound or the most stable intermediate?

Pathway 2 is unlikely. The tert-butyl group stays put, i.e., it remains attached to the same carbon throughout the reaction. It is the rest of the molecule that rearranges. The mechanism of this reaction was demonstrated 70 years ago. Compare this reaction with Possible nonclassical ion from a bicyclic system on ChemSE. There is information at the link on labeling to elaborate the mechanism.


Formation of strontium peroxide Wikipedia states:

  Strontium peroxide is produced by passing oxygen over heated strontium oxide. Upon heating in the absence of $\ce{O2}$, it degrades to $\ce{SrO}$ and $\ce{O2}$.

I found some literature about what temperature is needed but I did not find any literature about the condition of oxygen pressure. If I do annealing of $\ce{SrO}$ at $\pu{800 ^\circ C}$ in $\pu{0.5 atm}$ oxygen, will I form $\ce{SrO2}$?

There are a few method to synthesize $\ce{SrO2}$ from $\ce{SrO}$ and $\ce{O2}$. One of them uses $\ce{KClO3}$ as the source of $\ce{O2}$ [Ref. 1]. The paper is in German but has English abstract which states:

  Abstract. Single crystals of $\ce{SrO2}$ have been oblained after high
  pressure/high temperature reaction of a $\ce{SrO}$/$\ce{KClO3}$ mtxture at $\pu{20 kbar}$, $\pu{1400 ^\circ C}$. ...  From a refinement of the site occupation factor for oxygen a composition $\ce{SrO_{1.95(2)}}$ has been found for the crystal investigated.

Note that $\pu{20 kbar}$ pressure was inserted by the $\ce{O2}$ released by the following reaction at high temperature (Ref. 2): $\ce{2KClO3(s) -> 2KCl(s) + 3O2(g)}$.

Experimental information on the $\ce{Sr-O}$ has been reviewed in Ref. 2 and stability of $\ce{SrO2}$ and phase-diagram of $\ce{Sr-O}$ is graphically presented (see Figure 1 & 2, respectively). Accordingly, $\ce{SrO}$ under $\pu{0.5 atm}$ of $\ce{O2}$ pressure at $\pu{800 ^\circ C}$ would not produce $\ce{SrO2}$ (see Figure 1: Only $\ce{SrO}$ exists under the conditions, $\log_{10} p = \log_{10} 0.5 = -0.30$ and $\frac{1000}{\mathrm{T}} = \frac{1000}{1073.15} = 0.93$).

References:

K.-J. Kange, F. Rau, U. SchieBl, U. Klement, Verfeinerung der Kristallstruktur von $\ce{SrO2}$ (Refinement of the Crystal Structure of $\ce{SrO2}$), Z. anorg. allg. Chem., 1994, 620, 879-881 (https://onlinelibrary.wiley.com/doi/pdf/10.1002/zaac.19946200521).
D. Risold, B. Hallstedt, L. J. Gauckler, The strontium-oxygen system, Calphad, 1996, 20(3), 353-361 (https://doi.org/10.1016/S0364-5916(96)00037-5).


How can some enzymes work faster than the diffusion rates of the reactants allow? Quoting Wikipedia: "Some enzymes operate with kinetics which are faster than diffusion rates, which would seem to be impossible." Which are those enzymes and how can they be so fast?

One example is catalase in which peroxide is catalyzed faster than the diffusion rate of peroxide allows. Lionel Milgrom discusses this case in Water Journal no. 7.

The conceptually easiest case is that of a positively charged active site with a negatively charged substrate. The substrate (i.e. reactant) enters the active site with kinetics that are faster than diffusion because there is a long-range electrostatic interaction.
Catalase works near the diffusion limit. The document the OP cites is a hypothesis, without experimental data. It suggests that catalase acts at a distance to catalyze reactions outside of its active site. I would not worry about this hypothesis until there is some experimental evidence supporting it.
Wikipedia has a list of 9 examples of diffusion limited enzymes:

Acetylcholinesterase
β-lactamase
Catalase
Carbonic anhydrase
Carbon monoxide dehydrogenase
Cytochrome c peroxidase
Fumarase
Superoxide dismutase
Triosephosphate isomerase

Some of them have neutral substrates, so the electrostatic attraction would not apply to them. There are other hypothetical mechanisms, though, like weak binding sites on the surface of the enzyme that increase the local concentration of substrate, for example.


Why only one peak is observed in NMR spectrum of H2? Assuming a downward magnetic field is applied and the nuclear spin of two equivalent protons in $\ce{H2}$ are all in ground state $\downarrow\downarrow$ initially. Would the required energy of one of the protons excited to spin up state $\uparrow\downarrow$ be different from the energy of proton excited form state $\uparrow\downarrow$ to state $\uparrow\uparrow$ ?

And if it's true why only one peak is observed in NMR spectrum?

For starters, it is incorrect to think of the ground state being substantially more populated in NMR spectroscopy. Because the energy difference between aligned and misaligned nuclear spins in the external magnetic field is very low, we can say that both states are almost equally populated due to thermal excitation alone. For a $500~\mathrm{MHz}$ experiment, the energy difference can be calculated to be $0.21~\mathrm{J \cdot mol^{-1}}$ which corresponds to a population ratio as follows:

$$\frac{n_\mathrm{ex}}{n_\mathrm{g}} = \exp \left(\frac{\Delta E}{RT}\right) = 0.999916$$

Next, consider the extremely high symmetry of the $\ce{H2}$ molecule: It has the point group $D_{\infty\mathrm h}$, one of the most symmetric ones around. Due to this high symmetry, we cannot distinguish between either proton at all — both can be transformed into each other by a rotation as defined by the point group. Thus, the two nuclei are magnetically equivalent — not only must their chemical shift be identical but also the coupling to the other one.

Think of it this way: Assume, that the spin transition of one proton happened at a slightly different frequency than the other’s. That would mean that one proton had to be different from the other in some way (a different transition energy is equivalent to a different chemical shift is equivalent to a different environment). But how would you explain a different environment on one side of the $\ce{H2}$ molecule compared to the other? Correct: You cannot.

Now we could still assume that the energy required to excite a nuclear spin whose neighbour is parallel be different from one whose neighbour is antiparallel. However, no matter how we excite, the transition is always $\ce{parallel <=> antiparallel}$. And since we cannot distinguish between the hydrogens, we also cannot distinguish which one pointed in which direction, so those two energies must be identical. Therefore, we only see one peak and no $^1J_\ce{HH}$ coupling.