Why 1-octanol is immiscible with water? Octan-1-ol is polar due to the presence of the $\ce{-OH}$ group, but it is immiscible with water. How is it possible?

I just thought I'd add a visualisation of this concept. Electrostatic maps show the electrostatic potential at a given point of the molecule. We know water is very polar, and this is evident from its electrostatic map:

If we look at octanol though, we see that the vast majority of the molecule is non-polar (evenly distributed electrostatic charge):

The hydroxy-group of octanol is indeed hydrophilic (polar), but the majority of the molecule wants to stay in the non-polar phase.

Molecules with strongly hydrophilic "heads" and non-polar tails have unique properties in aqueous solutions, including some very important for life. Take phospholipids. In water, they self assemble into the basic structure of the cell membrane, the so-called phospholipid bilayer:

You can see how the phospholipids arrange themselves to have their polar portion in contact with water (hence "hydrophilic" - "loving water"), while their non-polar portion is in contact only with the non-polar portions of the surrounding phospholipids (hence "hydrophobic" - "fearing water").


Why is heroin a more potent drug than morphine, despite having a similar structure? The structures of heroin and morphine are quite similar, with heroin being formed by acetylation of morphine:

Why is heroin so much more potent than morphine, when their structures are so similar? What biochemical effects lead to heroin's prevalence as a recreational drug, and consequently, the much higher number of fatalities from overdosing on heroin as compared to morphine.
This Source has an attached reference1, which claims the following:

It has been reported that Diamorphine is twice as potent as morphine .
...
We found no significant differences between the groups over a 24-hour period in the degree of euphoria, nausea and vomiting or sedation, as assessed by subjective visual analogue scores, or in pain relief, nausea and euphoria as assessed by verbal rating. There was, in addition, no significant difference in the requirements for antiemetics between the two groups. Both groups demonstrated relatively low nausea scores. The mean amount of morphine demanded was approximately twice that of diamorphine.
...
The scientific basis for the alleged superiority of diamorphine over morphine is dubious and since diamorphine is metabolised to monacetyl morphine, and morphine itself, it would be surprising if there were a marked difference. Very few studies authenticate this claim of superiority and the present study is no exception.

This does not explain the prevalence of heroin compared to morphine. Over 15,000 people died in the US from heroin overdose2, whereas the number for morphine is quite less. Are the reasons behind this purely sociological (easier access to heroin compared to morphine), or is there some chemistry that I am missing out on here?
References:

Robinson, S. L., et al. “Morphine Compared with Diamorphine.: A Comparison of Dose Requirements and Side-Effects after Hip Surgery.” Anaesthesia, vol. 46, no. 7, July 1991, pp. 538–40. doi:10.1111/j.1365-2044.1991.tb09650.x.
CDC Heroin Overdose data. Link.

The answer to this question lies in the pharmacokinetics of these two drugs: the acetyl groups cause Heroin to be 200 times more lipid soluble than Morphine1. Another example of a common drug would be Fentanyl, which is 580 times more lipid soluble than morphine, with a structure very similar to that of Morphine.
Increased lipid solubility allows these drugs to pass more rapidly across the blood-brain barrier into the Central Nervous System, giving them more analgesic potency and a more rapid onset of action than morphine (the peak blood level of heroin when used intravenously is around 5 minutes2, whereas that of fentanyl is 1.7 minutes3).
As for the sociological aspects, the high potency of these drugs acts as a positive feedback loop. The heroin/fentanyl dose is smaller than morphine, which is preferable for drug smugglers, as they have to smuggle less of the drug, reducing their risk. This explains the high potency and usage of heroin as a recreational drug as compared to morphine.
References:

Mahesh Trivedi, Shafee Shaikh, Carl Gwinnut: "Pharmacology of Opioids", Updates in Anaesthesia, 2007, pp. 118-124. Link.
Oelhaf RC, Azadfard M. Heroin Toxicity. In: StatPearls [Internet]. Treasure Island (FL): StatPearls Publishing; Jan 2020. Available from: https://www.ncbi.nlm.nih.gov/books/NBK430736/
Fentanyl Citrate Injection, USP. FDA Label.


What happens if you cool water in a container too small for it to freeze? Freezing a full bottle of water tends to shatter the glass bottle. What if you used something tougher than glass, like diamond? What would happen if you kept dropping the temperature, but restrained the liquid volume so it couldn't freeze and what sort of force would the liquid exert on the container's walls?

Good question.  Let's assume the container is infinitely strong, non-deformable, and constant in volume.  Let's also assume that cooling the water is an equilibrium process -- that way, we won't have any supercooling.

At equilibrium, the first tiny bit of ice that freezes will take up more volume than the water it froze from.  This will raise the pressure on the rest of the water.  Eventually the pressure may get so high that additional freezing of more water is not thermodynamically favored.

Of course, as the pressure is raised, even the solid ice compresses a bit, freeing up a bit more volume for the liquid water.  According to this paper from 2004, ice is less compressible than water, so as a starting assumption, it may be approximately true to neglect the ice compression effect.

Figure 4 from that same paper gives the freezing point depression of water as a function of pressure:

To fully answer your question, in addition to that data, an equation that gives pressure as a function of ice volume would also be needed.  If we make the assumption I was talking about above -- i.e. that ice is incompressible, then from the data point that water has a constant compressibility of 46.4 ppm per atm we can come up with a very simple version of that equation.

$\frac{\Delta V_{water}}{V_{water}}=46.4 \times 10^{-6} \times P$, where P is the pressure in atmospheres.

Before freezing of a fraction $X$ of the water:

$$V_{ice} = X V_{tot}$$
$$V_{water} = (1-X) V_{tot}$$

After freezing:

$$ V_{ice} = X V_{tot} 1.11 $$
$$ V_{water} = (1-X) V_{tot} - \Delta V_{ice} $$

Combining those equations, you can get

$$0.11 \frac{X}{k(1-X)}= P$$, where $k$ is the compressibility of water.  If even 1% of the water in the container freezes (and all our assumptions are true), then the pressure will be 24 atmospheres!  Freezing 10% of the water would mean a pressure of 260 atmospheres.  Looking at the chart above, reaching this point would require a temperature of only 271 or 272 K, i.e. only -1 °C or -2°C.  Freezing 45% of the water would reach a pressure of 2000 atm, already off the chart above -- but the temperature required to reach that point would only be 253K or -20 °C, the setting of the average home residential freezer!  ((Of course, at these extreme pressures, (i) ice is actually compressible, and (ii) the compressibility of liquid water is not constant but also a function of pressure, so the calculations would get quite a bit more complicated.))

The lesson is that for even moderate degrees of cooling, you'd need a very, very strong container.


Why active mass of solids and pure liquids taken as unity There is a similar question already asked here and answers provided say that activity of solid is taken as unity because their density dosn't changes... but I still have some doubts in it.
Suppose we have a solid for example $\ce{NH4Cl}$ which decomposes into gases $\ce{NH3}$ and $\ce{HCl}$ then $K_{\mathrm{eq}} = [\ce{NH3}][\ce{HCl}]$.
Now if we take in a container which initially has no $\ce{HCl}$ and $\ce{NH3}$ and put some amount of $\ce{NH4Cl}_\mathrm{(s)}$ in it then after some time it will achieve equilibrium. Let the amount of solid left be $x$ mol. In equilibrium the  rate of production of $\ce{NH4Cl}$ is equal to the rate of its decomposition.
Question:
Now if we add another $x$ mol of $\ce{NH4Cl}_\mathrm{(s)}$ in the container then the rate of production of $\ce{NH3}$ and $\ce{HCl}$ should be double and hence would exceed the rate of production of $\ce{NH4Cl}_\mathrm{(s)}$ which will alter the equilibrium but the equilibrium constant equation predicts this would not be the case. Where is the mistake?

Maybe the following reasoning may help Param 1729. The thermal decomposition of $\ce{NH_4Cl}$ may be pedagogically seen as occurring in two steps. First, solid $\ce{NH_4Cl}$ is simply vaporized producing the hypothetical gaseous species $\ce{NH_4Cl(g)}$
$$\ce{NH_4Cl(s) <=> NH_4Cl(g)}\label{rxn:nh4cl}\tag{1}$$
And this transition is described by an equilibrium constant $\ce{K_1}$ which is similar to Henry's law
$$\ce{K_1 = \frac{p(NH_4Cl)(g)}{[NH_4Cl(s)]}}$$
Immediately after this vaporisation phenomena, a gas phase decomposition happens with the following equation :
$$\ce{NH_4Cl(g) <=> NH_3(g) + HCl(g)}\label{run:nh3+hcl}\tag{2}$$
with an equilibrium constant $\ce{K_2}$ defined by :
$$\ce{K_2 = \frac{p(NH_3)·p(HCl)}{p{(NH_4Cl)(g)}}}$$
So what is really observed is the sum $(1) + (2)= (3)$ $$\ce{NH_4Cl(s) <=>  NH_3(g) + HCl(g)} \label{run:dnh4cl}\tag{3}$$ with a constant $\ce{K_3}$ equal to
$$\ce{K_3 =  \frac{p(NH_3)·p(HCl)}{[NH_4Cl](s)} = K_1·K_2}$$
This last expression shows that the pressure of the gases $\ce{NH_3}$ and $\ce{HCl}$ does not depend on the amount of $\ce{NH_4Cl}$ in the solid state. And, as $\ce{[NH_4Cl](s)}$ is a constant, it may be included in $\ce{K_3}$ giving a new constant $\ce{K_3'}$
$$\ce{K_3' = p(NH_3)·p(HCl) = K_1·K_2·[NH_4Cl](s)}$$ This is "equivalent" to stating that the concentration of $\ce{NH_4Cl(s)}$ is equal to $1$.


Difference between exothermic and exergonic In High School I learned that an exothermic reactions releases energy, while an endothermic reaction needs energy to occur. Now I learned that there is a separate, somewhat similar classification scheme of exergonic and endergonic reactions.

What is the difference between these two classification schemes? Are exothermic reactions always exergonic, and if not, can you give me an example?

The classifications endothermic and exothermic refer to transfer of heat $q$ or changes in enthalpy $\Delta_\mathrm{R} H$. The classifications endergonic and exergonic refer to changes in free energy (usually the Gibbs Free Energy) $\Delta_\mathrm{R} G$.

If reactions are characterized and balanced by solely by heat transfer (or change in enthalpy), then you're going to use reaction enthalpy $\Delta{}_{\mathrm{R}}H$.

Then there are three cases to distinguish:

$\Delta{}_{\mathrm{R}}H < 0$, an exothermic reaction that releases heat to the surroundings (temperature increases)
$\Delta{}_{\mathrm{R}}H = 0$, no net exchange of heat
$\Delta{}_{\mathrm{R}}H > 0$, an endothermic reaction that absorbs heat from the surroundings (temperature decreases)

In 1876, Thomson and Berthelot described this driving force in a principle regarding affinities of reactions. According to them, only exothermic reactions were possible.

Yet how would you explain, for example, wet cloths being suspended on a cloth-line -- dry, even during cold winter? Thanks to works by von Helmholtz,  van't Hoff, Boltzmann (and others) we may do. Entropy $S$, depending on the number of accessible realisations of the reactants ("describing the degree of order") necessarily is to be taken into account, too.

These two contribute to the maximum work a reaction may produce, described by the Gibbs free energy $G$. This is of particular importance considering reactions with gases, because the number of accessible realisations of the reactants ("degree or order") may change ($\Delta_\mathrm{R} S$ may be large).  For a given reaction, the change in reaction Gibbs free energy is $\Delta{}_{\mathrm{R}}G = \Delta{}_{\mathrm{R}}H - T\Delta{}_{\mathrm{}R}S$.

Then there are three cases to distinguish:

$\Delta{}_{\mathrm{R}}G < 0$, an exergonic reaction, "running voluntarily" from the left to the right side of the reaction equation (react is spontaneous as written)
$\Delta{}_{\mathrm{R}}G = 0$, the state of thermodynamic equilibrium, i.e. on a macroscopic level, there is no net reaction or
$\Delta{}_{\mathrm{R}}G > 0$, an endergonic reaction, which either needs energy input from outside to run from the left to the right side of the reaction equation or otherwise runs backwards, from the right to the left side (reaction is spontaneous in the reverse direction)

Reactions may be classified according to reaction enthalpy, reaction entropy, free reaction enthalpy -- even simultaneously -- always favouring an exergonic reaction:

Example, combustion of propane with oxygen, $\ce{5 O2 + C3H8 -> 4H2O + 3CO2}$. Since both heat dissipation ($\Delta_{\mathrm{R}}H < 0$, exothermic) and increase of the number of particles ($\Delta_{\mathrm{R}}S > 0$) favour the reaction, it is an exergonic reaction ($\Delta_{\mathrm{R}}G < 0$).
Example, reaction of dioxygen to ozone, $\ce{3 O2 -> 2 O3}$. This is an endergonic reaction ($\Delta_{\mathrm{R}}G > 0$), because the number of molecules decreases ($\Delta_{\mathrm{R}}S < 0$) and simultaneously it is endothermic ($\Delta_{\mathrm{R}}H > 0$), too.
Water gas reaction, where water vapour is guided over solid carbon $\ce{H2O + C <=> CO + H2}$. Only at temperatures $T$ yielding an entropic contribution $T \cdot \Delta_{\mathrm{R}}S > \Delta_{\mathrm{R}}H$, an endothermic reaction may become exergonic.
Reaction of hydrogen and oxygen to yield water vapour, $\ce{2 H2 + O2 -> 2 H2O}$.  This is an exothermic reaction ($\Delta_{\mathrm{R}}H < 0$) with decreasing number of particles ($\Delta_{\mathrm{R}}S < 0$). Only at temperatures at or below $T$ with $|T \cdot \Delta_{\mathrm{R}}S| < |\Delta_{\mathrm{R}}H|$ there is a macroscopic reaction.  In other words, while the reaction works fine at room temperature, at high temperatures (e.g. 6000 K), this reaction does not run.

After all, please keep in mind this is about thermodynamics, and not kinetics. There are also indications of spontaneity of a reaction.


Isn't the reaction of chloral with sodium hydroxide a disproportionation reaction? Chloral undergoes a haloform reaction with sodium hydroxide. Now the carbon atoms in chloral are in the $\mathrm{+III}$ and $\mathrm{+I}$ oxidation state. Chloroform’s carbon is in the $\mathrm{+II}$ oxidation state, as is sodium formiate’s. Now, carbon has both been reduced and oxidised. Then why is it not a disproportionation reaction ?

You should not call the haloform reaction of trichloroacetaldehyde with sodium hydroxide a comproportionation. (And even less a disproportionation.)

Disproportionations are defined such that you start at two species of the same type and oxidation state (e.g. two formal $\ce{Cr^4+}$ species) and end up with two different species (e.g. a formal $\ce{Cr^3+}$ and a formal $\ce{Cr^6+}$ species). This is not the case here, the oxidation states start out differently.

Comproportionations are defined oppositely: You start at two different species and create two of the same. But in no way can chloroform and sodium formiate be called ‘the same’, even though their oxidation states are. Carbon monoxide (also $\mathrm{+II}$) is not the same as either, either.


Can any compound be a Lewis acid but not be a Brønsted acid? Is it possible for a compound to act as a Lewis acid but NOT act as a Brønsted acid? When would this happen? Could you please give a few examples where/why this happens ?

Certainly.  There are many compounds that can accept a pair of electrons from a Lewis base, yet don't have any Brønsted-Lowry acid (proton) moiety. The two are really rather different things, although there are of course many compounds that are both Lewis and Brønsted-Lowry acids ($\ce{Al(OH)3}$ for example).

Some examples of Lewis acids that are not also Brønsted-Lowry include:

$\ce{BF3}$
$\ce{(CH3)3B}$
$\ce{AlCl3}$
Metal cations like $\ce{Li+}$, $\ce{Na+}$, $\ce{Mg^2+}$, etc.


Reactivity of Benzaldehyde between aldehydes Are Benzaldehydes less reactive than aldehydes due to the -R Effect of the -CHO group present on the benzene ring?

For a carbonyl group to show reactivity toward nucleophilic reagents there should be positive charge on the carbon of the carbonyl group. The existence of positive charge on the carbon allows the reagent to donate electrons much easily.

Alternatively, A compound with more stabilized positive charge will show less reactivity towards nucleophilic reagents. The positive charge could either be stabilized by resonance or electron donating groups. In case of benzaldehyde, the positive charge is highly stabilized (due to resonance) than in other simpler aliphatic aldehydes.


Why does sodium hydroxide react with aluminum oxide? I have seen all the chemical formulas for the reaction between sodium hydroxide, water, and aluminium. I understand that the hydrogen gas comes from the aluminum reacting with the water and the sodium hydroxide just breaks down the aluminum oxide. But why does sodium hydroxide break it down? Also since aluminum oxide is amphoteric can you do this reaction with a strong acid instead of a strong base?

If you use a "strong acid" to break down the passive oxide you need to be careful which acid you choose.  Sulfuric or nitric acid is oxidizing and thus could maintain the oxide.  Hydrochloric acid would be best, among common mineral acids, for actually getting to the metal while being relatively easy to handle.


Why does shampoo lather much more the second time it is applied? When I shampoo my dirty hair, the first time it does not lather very much. The second time it lathers a lot. Why is this?

Quoting a reddit post from chemist nallen:

The short answer is that the dirt and oils from your hair compete for the surfactants making them less available to form lather, which is small bubbles.

To better understand the mode of action, you have to know a bit about the formulation of shampoos and the nature of dirt and oils.

Dirts and oils deposit on hair and fibers because they are at least partially hydrophobic (not soluble in water.) The surface of hair, skin, etc... is more hydrophobic than water (well of course!), so these dirts have greater adhesion for the surface than for water (the water actually pushes them out to minimize surface area.) This is why water is ineffective for removal, it won't pull the dirt off the surface by itself. Surfactants associate with the hydrophobic surfaces to make them more dispersible or soluble in water, allowing them to be rinsed off.
So what happens when there is more surfactant than there is hydrophobic dirt surface?

The answer is that the surfactant orients to the surface of the water, because air is hydrophobic, and the water wants to push the hydrophobic portion of the surfactant out of the water molecule matrix. This orientation on the air-water interface is how skin of bubbles is formed. When the water/surfactant solution undergoes sufficient shear mixing, bubbles are formed, and that is lather.
Lather typically forms when there is surfactant in excess of the hydrophobic surface. More dirt means less free surfactant, and therefore less lather.

Now, it's not purely that simple, because the amount of lather that a surfactant develops varies with the type of surfactant, some are better than others.

Lather does not mean that the shampoo is cleaning better, it's purely cosmetic. Typical shampoos are primarily SLES/Betaine, that is sodium laureth sulfate and cocamidopropyl betaine. The SLES is there for cleaning, and the betaine is there for lather, it's specifically added to produce lather because, well, people like it. There are some other benefits, but that's the primary one. If you used SLES by itself is would clean just as well, but people just don't like poor lather, so it would not sell.


What is the CO2 content in the air in a compartment of air and water, as a function of temperature? Update 31.12.2015: In a closed system of water and air, how does the concentration of CO2 change as a function of temperature, in the gas phase?

What is the graph of carbon dioxide ($\ce{CO2}$) content in the air in a compartment consisting of 80 percent air and 20 percent ocean sea water as a function of temperature, in the temperature range 0 to 60 degree Celsius?

Example but not the complete answer:

As can be seen from the graph above, increasing the temperature of the water will cause the water to release more carbon dioxide ($\ce{CO2}$) in to the gas phase. This happens for most gases as can be seen from this page:

http://www.engineeringtoolbox.com/gases-solubility-water-d_1148.html

What I am interested in is to see a graph of the percentage, or how to put it, of carbon dioxide ($\ce{CO2}$) in air as a function of temperature.

In other words, what does the plot of:

$$\%\ce{CO_2} \text{ in air (temperature)} = \frac{\text{kilogram } \ce{CO_2} \cdot 100 %}{\text{kilogram  air}}$$

look like in the gas phase in a compartment consisting of 20 % sea water and 80 % air as a function of temperature?

Ideally the compartment should consist of a column with the height upwards of average atmosphere thickness on planet earth filled with air, and a depth downwards with the average depth of sea water also at planet earth filled with sea water. But this is probably not feasible. Therefore the simplifcation 80 percent air, 20 percent sea water. If sea water is too complicated I am also satisfied with an answer with pure water.

The definition of air without water steam in volume percentages:

Nitrogen $\ce{N_2}$ = 78.08
Oxygen $\ce{O_2}$ = 20.95
Argon $\ce{Ar}$  = 0.934
Carbon dioxide $\ce{CO_2}$ = 0.036
Neon $\ce{Ne}$ = 0.00182
Helium $\ce{He}$ = 0.00052
Krypton $\ce{Kr}$ = 0.00011
Hydrogen $\ce{H_2}$ = 0.000050
Xenon $\ce{Xe}$ = 0.000009
Ozone $\ce{O_3}$ = 0.000001
Radon $\ce{Rn}$ = 0.000 000 000 000 000 006

Edit 18.4.2015:
I will try to clarify the question:

  Does an increase in temperature cause an increase or decrease in the carbon dioxide (CO2) content in the atmosphere?

Edit 22.4.2015: According to engineering toolbox dot com, the four most common gases in the atmosphere have the following solubilities in water:

Related question: https://mathematica.stackexchange.com/questions/80586/how-can-i-get-the-solubility-value-of-nitrogen-in-water-at-10-degree-celsius-wit

Simple, partial pressure of $\ce{CO2}$ is misleading.

$\ce{CO2}$ actually reacts with sea water:
$$\ce{H2O + CO2 -> H2CO3}$$
which then ionizes:
$$\ce{H2CO3 -> H+ + HCO3-}$$
which lowers the pH of the sea water (making it more acid).

In previous high $\ce{CO2}$ periods, $\ce{CaCO3}$ came out of sea water
making limestone and marble rock formations geologically. $\ce{CO2}$ at high pressure is responsible for other side effects, also killing
fish, and microbes, damaging coral reefs and generally messing up ocean life.


How to measure concentration of arsenite and arsenate in a water sample A colleague and I are developing a transgenic bacterial strain that contains an arsenite oxidase gene. The bacteria can oxidize arsenite into arsenate autotrophically. We need a method to measure the concentration of arsenite and arsenate over time in a water sample. First we thought of using a pH meter, but we need to measure the arsenite and arsenate concentration over a 48 hour period, and we are worried that the pH meter wouldn't hold up. Could there possibly be a chemical that reacts with arsenite to change the color, and we could use a spectrophotometer or a colorimeter. If there are any ideas for measuring concentrations of arsenite (around 5 mM), please let me know.

Speciation analysis is very challenging because you would like to separate the species first before detection. In order times electrochemical detection was done by now ion chromatography is the way to go for arsenic speciation. You will need ion chromatography system with inductively coupled plasma with mass spectrometry. Have a look at this paper: Determination of Anionic, Neutral, and Cationic Species of Arsenic by Ion Chromatography with ICPMS Detection in Environmental Samples, Anal. Chem., 1998, 70 (17), pp 3649–3655 The good thing about Analytical Chemistry papers is that most of the methods are reliable.


Acid chlorides and reduction by hydrogenation An acid chloride being reduced through catalytic hydrogenation is an "unobserved" event in the context of our class, according to my professor. Nonetheless, we are expected to predict that the hydrogenation occurs because acid chlorides are very reactive in general.

Now, this sounds like a case of the right answer with the wrong reason.

Yes, acid chlorides are reactive. We've covered their hydrolysis with water. We've covered their role in Friedel-Crafts acylation. We realize that the $\ce{C-Cl}$ bond in an acid chloride is very weak and that the carbonyl carbon is highly electrophilic. This gives rise to most of the reaction pathways with acid chlorides.

But what does catalytic hydrogenation have to do with the electrophilicity of the carbonyl carbon in an acid chloride? Nothing, in my mind. Hydrogen gas and platinum metal simply work here because the $\ce{C=O}$ part of the acid chloride is flat. I don't think it has anything to do with how reactive the acid chloride is. Right? Hell, even plain jane alkenes can be hydrogenated and they aren't exactly reactive.

Hydrogen gas and platinum metal simply work here because the C=O part
  of the acid chloride is flat.

It's more than just that.  Acid chlorides are significantly more reactive towards catalytic hydrogenation than other carbonyl compounds.  They are reactive enough that a poisoned catalyst must be used along with mild reaction conditions to achieve reduction to the corresponding aldehyde, and have the reaction stop there (Rosenmund reaction).

So the question becomes, "why are acid chlorides so reactive towards catalytic hydrogenation."  It is thought that because the acid chloride $\ce{C-Cl}$ bond is so reactive that the palladium inserts into this bond to form intermediates such the one pictured below.

Other typical carbonyl compounds can't form such an intermediate; consequently acid chlorides are noticeably more reactive towards catalytic hydrogenation. Anhydrides, where the carbonyl is roughly as reactive as an acid chloride's carbonyl towards nucleophilic attack, have not been reported (to my knowledge) to produce aldehydes under Rosenmund conditions supporting this mechanism involving a palladium intermediate.

Historical Note: Prior to the application of metal hydrides to the reduction of carbonyl groups, the Rosenmund reduction was a major synthetic tool for chemists.  It was one of the few methods available for the straightforward transformation of acids, esters, amides, etc, to aldehydes.


How to understand the concept of moles? I have trouble relating to the mole concept. For example in a question like the following:

  How many moles of $\ce{K}$ are in 1 mole of $\ce{KOH}$?

How is it that in 1 mol of $\ce{KOH}$ there is 1 mol of $\ce{K}$ and x moles of $\ce{OH}$? How is it that 1 mol of $\ce{KOH}$ has 6.02×1023 atoms, and then $\ce{K}$ has 6.02×1023 atoms as well. In addition, $\ce{OH}$ will have 6.02×1023 atoms, but if we add these two it's greater than the original number of atoms in 1 mol of $\ce{KOH}$?

I had the same problem as well, until I sketched out the actual molecules.

When you have one mole of $\ce{KOH}$, you have one mole of the collective "molecule".

If you were to break it down into $\ce{K}$ and $\ce{OH}$, you would get 1 mole of $\ce{K}$, and 1 mole of $\ce{OH}$. This is because you need 1 mole of each to make 1 mole of $\ce{KOH}$.

It's kind of like if you have one hundred keyboards, you would get one hundred 'Enter' keys, one hundred 'Up arrow key' and so on...


What are the oxidation states of sulfur in the tetrathionate ion? What are the oxidation states of the sulfur in the tetrathionate ion $\ce{S4O6^{2-}}$?

Consider the structure of tetrathionate. The two central sulfurs each have two lone pairs and are assigned half of the electrons from the two bonds they make, since the electrons of bonds between atoms of the same element must be distributed evenly (due to there being, by definition, an electronegativity difference of zero between two atoms of the same element). A neutral sulfur atom has six valence electrons, so the oxidation state of the central sulfurs can be calculated as follows:

$$6 - 4 - \frac{1}{2}(4) = 0$$

That is, six electrons in neutral sulfur, minus four from the lone pairs, minus half of the four sulfur-sulfur bonding electrons, gives zero.

The terminal sulfurs, on the other hand, have no lone pairs, and all the electrons from the sulfur-oxygen bonds are assigned to oxygen, since it's the more electronegative element. The only electrons assigned to those sulfurs are half of those from the sulfur-sulfur single bonds. Hence, their oxidation state is:

$$6 - \frac{1}{2}(2) = +5$$

That is, six electrons in neutral sulfur, minus half of the two sulfur-sulfur bonding electrons, giving an oxidation state of +5.


Can the oxidation of hydrogen peroxide be reversed? Hydrogen peroxide is said to be unstable, for it undergoes auto-oxidation on standing/heating:

$$\ce{2H_2O_2 -> 2H_2O + O_2}$$

where $\Delta S=\pu{70.5 J {mol}^{-1}K^{-1}}$ and $\Delta H^{\Theta} = \pu{-98.2 kJ {mol}^{-1}}$.

I speculate if the decomposition can be reversed in some way under suitable conditions, particularly at infinite pressure and near absolute zero temperature. At once this seems to be a problem of entropy reversal. But I seek a more rigorous treatment of the problem with the help of quantum theory. Consider the molecular wavefunctions $\Psi_{1A}$ , $\Psi_{1B}$ of water (2 molecules) and
$\Psi_2$ of dioxygen respectively. By studying the interaction potentials of these three wavefunctions as a function of different thermodynamic coordinates, it can be possible to find the desired suitable conditions.

However, the problem begins here – this is not a case of just two molecules where we have a number of methods of analysing the system, I am talking about three molecules, so the calculation is not simple. Moreover, I doubt if I need to study the wavefunction of hydrogen peroxide too in some respect. Overall, I am lacking some theoretical and computational knowledge. Any help is greatly appreciated.

Yes, if you use some quantum chemistry tool like density functional theory to find the enthalpy of the compounds at many different conditions then you can determine where hydrogen peroxide is the ground state.

You would probably find this at low temperatures and high pressures since it is denser than the other two phases and oxygen is condensed as a liquid or solid so the entropic contribution is smaller.


Can anions be reduced at the cathode in an electrochemical cell? Two related questions:

(1) It's my understanding that for a galvanic cell, as electrons flow from the anode to the cathode, the cathode (+) becomes negatively charged due to the accumulation of electrons. If I have a sodium chloride electrolyte, the sodium ions will move toward the cathode. However, does that mean chloride will not be reduced at the cathode?

(2) Is the same true for an electrolytic cell? The cathode is (-), so I assume it stays negative. But, what if I wanted to force the reduction of an anion like perchlorate - is it possible or will there be too great of an electrostatic barrier?

I see people reporting on the electrocatalytic reduction of anions, but I don't understand how this is possible if the ionic current is the opposite direction.

Theoretically, it is possible for anions to get reduced more. According to wikipedia's reduction potential table page, this reaction occurs at +0.08V vs SHE:

$\ce{S4O6^2− + 2 e− ⇌ 2 S2O3^2−}$

In your chloride example, though, you're trying to reduce $\ce{Cl^-}$ to (at least) $\ce{Cl^2-}$ and the potential for that reaction would be extremely negative. In most real-world scenarios, you would start reducing other things way before you hit that potential. Most likely you'd start reducing water at -0.8V.

$\ce{2H2O + 2 e− ⇌ H2(g) + 2 OH−}$

As an aside, be careful with your terminology. Reduction always happens at the cathode, that's what "cathode" means. Sometimes we get sloppy and don't switch the terms when we go from galvanic to electrolytic mode, but that doesn't change the chemistry. If electrons are flowing into an electrode, it's the cathode. I find the (+)/(-) labels on terminals more confusing than anything else.

Also, you said that

  for a galvanic cell, as electrons flow from the anode to the cathode, the cathode (+) becomes negatively charged due to the accumulation of electrons

This is backwards. The cathode does not become negatively charged because electrons have flowed into it. Rather, reduction occurs spontaneously at the cathode, leaving it electron-deficient. Electrons then flow from the anode to the cathode to compensate this charge imbalance and make both sides neutral again.


Why do atoms generally become smaller as one moves left to right across a period? It seems to me that the addition of electrons and protons as you move across a period would cause an atom to become larger. However, I'm told it gets smaller. Why is this?

As you move from left to right across a period, the number of protons in the nucleus increases.  The electrons are thus attracted to the nucleus more strongly, and the atomic radius is smaller (this attraction is much stronger than the relatively weak repulsion between electrons).

As you move down a column, there are more protons, but there are also more complete energy levels below the valence electrons.  These lower energy levels shield the valence electrons from the attractive effects of the atom's nucleus, so the atomic radius gets larger.


Would graphite stick to basalt cooled from its molten state in a graphite mold? Molten metal that is poured into a graphite mold pops out easily once it has cooled. Would basalt poured the same way pop out just as easily?

I am particularly interested in lunar basalt, if that makes any difference. The image shows the composition used to make a lunar regolith simulant, taken from chapter 13 of Resources of Near Earth Space, by B. J. Pletka.

No, it would not stick. In our lab we commonly use graphite dishes/cups/etc to contain small amounts of liquid rock. All you need to do then is to hold it upside down and the solidified rock (now glass mostly) falls out. Sometimes there's a bit of graphite dust on it, but it can be easily scraped off.

Doing it in a reducing (or oxygen-free) environment will make it even easier, because graphite slowly burns to carbon dioxide with time, on the scale of hours, from my own personal experience.


Calculating the pH given number of moles of HI and 500. ml of water Shouldn't we also add the volume of the HI to the volume of the water (500mL) when we calculate the molarity? My book simply divided the number of mol by 500 mL.

Simple questions are often interesting.

(1) Most importantly, volumes are rarely additive. So the notion of adding the volume of the HI and the volume of water just doesn't work.

(2) In an answer to a question about hydroiodic acid,  Curt F. showed a chart of concentration and density. For the $\pu{20 ^\circ C}$ values the liters of water in a liter of the acid solution can easily be calculated. So for the OP to get an error of 1% or less, the molarity would have to be about 0.2 molar or less.

\begin{array}{|c|c|c|c|}
\hline
\text{Concentration, %}(w/w) & \text{Density, }\pu{kg/L}(\text{@ }\pu{20 ^\circ C}) &  \text{Molarity} & \pu{L(water)/L(acid)}  \\ \hline
5.2 & 1.0342 & 0.420 & 0.980 \\ \hline
10.8 & 1.0812 & 0.922 & 0.964\\ \hline
16.4 & 1.1226 & 1.44 & 0.938 \\ \hline
22.4 & 1.1765 & 2.06 & 0.913 \\ \hline
27.2 & 1.2333 & 2.62 & 0.898 \\ \hline
33.1 & 1.2918 & 3.34 & 0.864 \\ \hline
38.7 & 1.3605 & 4.12  & 0.834\\ \hline
42.9 & 1.4208 & 4.77 & 0.811 \\ \hline
48.7 & 1.5072 & 5.74 & 0.773 \\ \hline
53.0 & 1.5913 & 6.59 & 0.748 \\ \hline
57.0 & 1.6933 & 7.55 & 0.728 \\ \hline
\hline
\end{array}

You can see from the red line that the data is not linear.


How to differentiate a metal from its ion? I am working on metal and its ion. The problem is, I would like to know whether the metal has been converted to its ionic form or not.

The hypothesis for the analysis is that metals are good conductors and ions possess redox potential. Therefore, by measuring the conductance and red-ox potential is it possible to  differentiate whether I have metal or metal ion.

The questions are

Based on the hypothesis, can I expect a positive result?
Apart from this, is there any other technique where I can qualitatively / quantitatively measure  metal and its ion?

Other than the excellent answer by F'x, there is also actual spectroscopy, specifically UV/Vis spectroscopy. This'll only work if the metal has transitions in the right range, though -- transition metals, lanthanides and actinides are usually fine, s or p-block metals and metalloids, probably not.

Then you're looking for the absorption/transmission spectrum of the ion and the metal separately, and how much of each there is in a combined spectrum. It's not strictly quantitative, but you can get a good idea of the ratios.


induction of van der Waals forces Is there a way to maximize van der Waals forces between two materials, by electrifying one material, thus creating dipoles/instantaneous dipoles and thus creating stronger van der Waals forces? has this been documented?

Perhaps an example of what you mean can be seen with the noble gases. As you go from helium to neon to argon, etc. you are adding inner electron shells and making the electron cloud around each atom more polarizable. That in turn increases the van der Waals attraction between atoms, resulting in a dramatic increase in boiling points down the series.

You might be able to make an atom more polarizable (and increase van der Waals forces) by electronically exciting it to a higher energy level. Can anyone else elaborate on that possibility?


Aluminium in homemade liquid castile soap? I made some liquid castile soap in a slow cooker.  There was a slight scratch in the non-stick surface of the container.  I read that potassium hydroxide when coming into contact with aluminium creates hydrogen gas and hoped my container is stainless steel.  However, the container has corroded along the scratch by about a millimetre deep and wide.   The soap is good, but I am concerned that there is a small amount of alum in the product and this can be absorbed into the body.  Is the soap safe to use?

The question is actually is skin absorption of Al3+ a health risk. The answer from a legal perspective varies depending on where you live. In the USA, for example, the answer is no, however, there appears to be a growth industry in selling Aluminum-free products, but not in Europe, where apparently stick deodorants containing the Aluminum ions are banned.
But what is the suspected chemistry possibly suggesting health problems with the Aluminum ion? My prior exploration of this topic points to some recent research. Based upon a review of two articles noting the pro-oxidant property of the aluminum ion, a potential health risk pose by the ability of aqueous Al3+ to recycle transition metals (like Fe, Cu, Mn,..) to lower valence states (via the superoxide radical anion) in Fenton and Fenton-like reactions, contributing to so-called oxidative stress related diseases (including Alzheimer's disease, various cancers,...). More explicitly, with iron ions, for example, recycling between ferric and ferrous ions feeding a fenton reaction via superoxide radical anion (produced in our bodies):
$\ce{ Fe^{3+} + O2^{•−} <=> Fe^{2+} + O2}$
Source: See Eq (1.xxxiii) which can feed a fenton reaction forming the hydroxyl radical:
$\ce{ Fe^{2+} + H2O2  -> Fe^{3+} + •OH + OH-}$
The first alluded to article is Oxidative Stress Gated by Fenton and Haber Weiss Reactions and Its Association with Alzheimer’s Disease by Tushar Kanti Das, et al, published in Archives of  Neuroscience, July 2014 2(3): e20078, DOI: 10.5812/archneurosci.20078.
The authors cite, in Figure 4, “Formation of Aluminum Superoxide Semi reduced Radical Ion and Aluminum Superoxide Complex (43)”, with described reactions proceeding as follows (also adopting notation and water complexing from the second article):
$\ce{ [Al(H2O)4]^{3+} + O2^{•−}   <=> [Al(O2^{•−})(H2O)4]^{3+} }$
$\ce{ [Al(O2^{•−})(H2O)4]^{3+} + Fe^{3+} -> O2 + [Al(H2O)4]^{3+}  + Fe^{2+} }$
And, in the presence of H+ :
$\ce{ [Al(H2O)4]^{3+} + O2^{•−} + H+  <-> [Al(O2^{•−})(H+)H2O)4]^{3+} }$
$\ce{ [Al(O2^{•−})(H+)H2O)4]^{3+} + [Al(O2^{•−})(H+)H2O)4]^{3+} -> 2 [Al(H2O)4]^{3+} + H2O2 + O2 }$
Note per below, the action of hydrogen peroxide on ferric also leads to ferrous:
$\ce{ H2O2 <=> H+ + HO2- }$
$\ce{ Fe^{3+} + HO2- -> Fe^{2+} + HO2• }$
or, pH> 4.88, we have:
$\ce{ H+ + O2^{•−} -> HO2•}$
Another work: Pro-oxidant Activity of Aluminum: Stabilization of the Aluminum Superoxide Radical Ion by J. I. Mujika, F. Ruiperez, I. Infante, J. M. Ugalde, C. Exley, and X. Lopez in J., published in Phys. Chem. A 2011, 115, 6717–6723, American Chemical Society, dx.doi.org/10.1021/jp203290b .
In the Mujika article to quote “In addition, the presence of LMM ligands such as citrate could also have an indirect effect in the oxidation capacity of aluminum by augmenting the bioavailability of Al3+ species, shifting the formation of Al(OH)4- to higher pH’s. However, one should also take into account the effect of citrate chelation itself in the thermodynamic equilibrium of [AlO2•]2+ formation.”
So, my short takeaway, our body produces the short-lived superoxide radical anion capable of recycling problematic transition metals which can engage in fenton (and fenton-like) reactions forming powerful radicals. The latter can attack your DNA leading to associated diseases. The role of Al3+ is that it can form a complex with the superoxide, likely increasing its effective half-life and augmenting destructive radical activity in one's body.
To answer the question: "Is the soap safe to use?", likely that it is a personal decision based on one's understanding of the proposed underlying chemistry, as jurisdictions apparently vary in opinions.


When our hard water is boiled, it leaves a blue-green residue in our kettle. The inside of our kettle is lined in a blue-green deposit. Our water is very hard, but what minerals could cause the coloration?

If your building's water pipes are made of copper, that could explain the blue-green coloration. Hopefully the inside of your pipe doesn't look like this, though:


What is the order of nomenclature priority among an alkyl side group, a double bond, and a triple bond? I saw the below structure in a video on YouTube and was confused by the fact that while they were numbering it, higher priority was assigned to the methyl than the triple bond:

According to the priority table on Master Organic Chemistry, the opposite is true, and the triple bond should have priority.

Also I saw this structure:

According to the same priority table, shouldn't a double bond have higher priority than a triple bond, and hence the numbering should start from the left?

Your so-called priority table (the “table of functional group priorities for nomenclature” shown here) is relevant for the choice of a characteristic group for citation at the end of a name by means of a suffix or a class name. It does not apply to the numbering of locants. This particular table could also be misleading since it mingles the concepts of classes, suffixes, endings, and prefixes.

The most important simplified criteria for the numbering of simple acyclic compounds are:

lower locants for suffixes
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

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

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Since both compounds that are given in the question do not have any principal characteristic group (a characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name), the first criterion (greater number of suffixes) given in P-14.4 (c) is not relevant in this case.

The systematic name of the first compound is 6-methylhept-2-yne rather than 2-methylhept-5-yne since low locants are given first to the ‘yne’ ending and the locant ‘2’ is lower than ‘5’.

The systematic name of the second example is (4E)-hex-4-en-1-yne rather than (2E)-hex-2-en-5-yne since low locants are given to multiple bonds as a set (even though this may give ‘yne’ endings lower locants than ‘ene’ endings), and the locant set ‘1,4’ is lower than ‘2,5’. Only if a choice remains, preference for low locants is given to the double bonds.


What happened when the cherry juice turned blue? When I was eating cherries today, some of its juice spilled on my shirt. Attempting to clean it, I poured a solution (baking soda, vinegar, toothpaste, and salt) onto the fabric. Though it did not remove the stain, the colour turned from a wine red to a subtle blue.

So the Question is: Why did it happen?

(This is not that technical so I'm not sure how to classify it)

Cherries, like red cabbage, hibiscus and many red or purple colored fruits, contain a class of compounds called anthocyanins of general formula:

where $\ce{R_x}$ is either $\ce{OH}$, $\ce{OR}$ or $\ce{H}$.
In basic medium it loses a proton and it converts to a non charged chinone like structure, which has a blue color.


Intuitive understanding of standard electrode potential I have read many books on standard electrode potential and they define the term in various ways but I just can’t seem to get an intuitive idea of it.

$$\ce{Cu^2+ + 2e- <=> Cu}$$

Standard electrode potential, $E= 0.34\ \mathrm{V}$

Does this mean that when $1\ \mathrm{C}$ of charge passes from the hydrogen half cell to copper half cell, $0.34\ \mathrm{J}$ of energy is released?

$$\ce{Zn^2+ + 2e- <=> Zn}$$

Standard electrode potential, $E= -0.76\ \mathrm{V}$

Does this mean that when $1\ \mathrm{C}$ of charge passes from the hydrogen half cell to zinc half cell, $0.76\ \mathrm{J}$ is absorbed?

The absolute electrode potential, i.e. the actual volts between electrode and electrolyte, is incredibly difficult to measure accurately.    Instead we simplify: declare the hydrogen electrode potential to be zero, when actually it's roughly +4.4V @25C deg.    Then measure other electrode potentials using a standard hydrogen electrode to contact the electrolyte.


Safely lowering bleach pH without chlorine gas? According to Eurekalert, by lowering the pH of household bleach (e.g., by adding vinegar or citric acid), we can increase $\text{[HOCl]}$ and consequently greatly improve the disinfecting capabilities of the bleach itself.

However, doing so could technically liberate chlorine gas, either through hypochlorous decomposition
$$ \ce{4H+(aq) + 4OCl- (aq) -> 2Cl2(g) + O2(g) + 2H2O (\ell)} $$
or mere comproportionation
$$ \ce{2H+(aq) + OCl- (aq) + Cl- (aq) ->  Cl2(g) + H2O (\ell)} $$
(depending on how the bleach was originally made) and so, mixing bleach and acid seems to be contraindicated everywhere else on the internet.

However, the authors of that link also suggest

  ... first diluting one cup of household bleach in one gallon of water and then adding one cup of white vinegar.

So then does that make it safe? Wouldn't chloroacetic acid formation also be a concern?

The oxidizing effect of chlorine is quite powerful even at concentrations of 1% or less. Mixing bleach and vinegar gives a much more concentrated solution. Although chlorine is soluble to the extent of 1.46 grams per 100 mL cold H2O (CRC Handbook), its vapor pressure above the water may be significant. A solution diluted by a factor of 7 or 8 would be in contact with a surface long enough to disinfect with a much reduced odor. Safer for people, but still deadly to bacteria.


In a substance do atoms actually touch each other? In a substance do atoms actually touch each other?

Quantum mechanics fuzzes out the electrons into clouds of probability distribution (the motion of little electrons orbiting the nucleus is from an old outdated model).  In molecules these electronic clouds do not so much touch, rather they overlap.  See, for instance, https://en.wikipedia.org/wiki/Covalent_bond.


How is hybridisation of a carbon atom determined by looking at the molecule's structure and observing the lone pair and/or sigma bond present? How can I judge the hybridization of a carbon atom of any carbon containing compound just by looking at its structure? How can I say this by the help of lone pairs and sigma bonds present?

The answer you are looking for is probably this:

Count the number of $\sigma$ bonds made by that atom, and add to that the number of lone pairs present on that atom. This number is called the steric number.

If the steric number is $2$, hybridisation is $sp$.
Similarly, $3 \Rightarrow sp^{2}$, and $4 \Rightarrow sp^{3}$ hybridization.


What will form when Dihalides or Trihalides or Polyhalides react with Magnesium? I know that
"R-X + Mg" gives RMgX.

I want to ask if the Halide is Poly, then How's the formation of Grignard take place?

For example.
If 1,4 - Dibromobutane react with Mg, then what will be the product?

1,4-dibromobutane reacts with magnesium to form cyclobutane as main product. The  the side products are buta-1,3-diene, ethene, butane, (E) and (Z) but-2-ene. The reaction has been performed in xylene solvent to give a mixture of hydrocarbon(81%) and cyclobutane(13%). The best yield for cyclobutane from 1,4-dibromobutane can be obtained by using lithium amalgam in dioxane for 3 hours.

In general, for any dibromoalkane reacting with magnesium will always lead to formation of alkene or cycloalkane and magnesium bromide as side product. For e.g.- 1,2 dibromoethane reacts with magnesium to form ethene and magnesium bromide. In fact, it is being used as activating agents as its action can be monitored by the observation of bubbles of ethylene and also the side-products($\ce{MgBr2}$) is almost negligible and does not affect the reaction. (Wikipedia)
$$\ce{Mg + BrC2H4Br → C2H4 + MgBr2}$$
The reaction mechanism is that any dibromoalkane reacting with magnesium will first lead to magnesium bromine adduct of the alkane chain which will then decompose to form alkene or cycloalkane and magnesium bromide as side product.
$$\ce{Br-(CH2)_n-Br + 2 Mg ->[ether] BrMg-(CH2)_n-MgBr}$$
1,3-dibrompropane reacting with magnesium will form cyclopropane and magnesium bromide as side product.
$$\ce{BrCH2CH2CH2Br + Mg -> (\Delta) + MgBr2}$$

Scientists have discovered more exotic species called magnesacycloalkanes which is said to maintain a equilibrium between magnesium bromine adduct of the alkane chain and the final product. Some of the species include:

Now, the species formation depends on position of bromine atoms in alkane chain:

1,3 DiGrignard reagent species: formed by reacting 1,3-Dibromopropane and magnesium to form 3-Bis(bromomagnesio)propane (4a) and 1,6—bis(bromomagnesio)hexane (15%) as side product.

1,2-DiGrignard reagent species: Formed when aromatic 1,2 Dihalides are used

1,1-DiGrignard reagent species: Also called methylene—di—Grignard reagent It is prepared from dibromo- or diiodomethane.

$$\ce{CH2Br2 + Mg ->[(i-Pr)2O] CH2(MgBr)2 }$$
References

Bickelhaupt, F.. “Di-Grignard reagents and metallacycles.” Pure and Applied Chemistry 58 (1986): 537 -542, DOI: 10.1351/pac198658040537
https://www2.chemistry.msu.edu/faculty/reusch/virttxtjml/alhalrx4.htm
Science of Synthesis: Houben-Weyl Methods of Molecular Transformations Vol. 48: Alkanes, Georg Thieme Verlag, May 2014.


Oxidation number of hydrogen in compounds In a given compound that contains three elements (e.g. a metal, a non-metal, and hydrogen), how can it be determined whether hydrogen is connected to the metal or the non-metal so that its oxidation number can be assigned accordingly?

You can look at the electronegativity of the atom Hydrogen is bonded too.

For instance, in NaH the electronegativity of H is greater than that of Na (~2.1 as opposed ~.9) so H two would have an oxidation state of -1)

In HCl, we observe that Cl is more electronegative so it will have an oxidation state of -1 and H will have an oxidation state of +1

In borane (even though B is a non-metal) each hydrogen displays an oxidation state of -1 (again electronegativity of H is 2.1 as opposed 2.0 of B)

Hope this makes sense.


Does CHT=CHT have a geometric isomerism because of tritium atoms? Is the radioisotope of hydrogen considered as a new atom? Can you detect it with an H NMR?

Yes isotopes can induce geometric isomerism.  You can induce chirality by using different isotopes of atoms; the molecule $\ce{CHDClF}$ is chiral with the deuterium atom (you do not need radioactivity) while ordinarily $\ce{CH_2ClF}$ is not.  Chirality won't work for ethylene (that molecule is still planar and thus mirror-symmetric about its own plane no matter what iostopes are used for any atoms), but cis-trans isomerism with two deuterium atoms still can work.


All roads lead to S4N4 Why does thionyl chloride $\ce{(SOCl2)}$ when it reacts with liquid ammonia form a trace of $\ce{S4N4}$. What is the mechanism of the reaction forming $\ce{S4N4}$.

See discussion of sulfur chemistry for details.

Yes, you are indeed right in the comment section that the reaction between thionyl chloride and ammonia will give trace amount of tetrasulfur tetranitride(1) and the intended product is thionyl imide ($\ce{HNSO}$)(2).

From the 'thionyl chloride' entry from the same site you used
in the question:

  Thionyl chloride reacts with ammonia giving rise to various complex
  products, the nature of which varies with the conditions.

As said, the main product from the reaction in gas phase is thionyl imide and trace amount of thiazyl chloride($\ce{NSCl}$) and sulfur dioxide. The reaction are as follows:

$$\ce{SOCl2(g) + 3NH3(g) -> HNSO(g) + 2NH4Cl(s)}$$
$$\ce{HNSO + H2O -> SO2 + NH3}$$

Ref. : Inorg. Chem., 1977, 16 (1), pp 216–217
DOI: 10.1021/ic50167a050


Procedure vs. synthetic method? I am genuinely confused on the difference between a procedure/flow chart and the synthetic method for a lab report and what exactly a synthetic method includes, and I can't find any information online. I would really appreciate the help if anyone has knowledge about it.

This is what the directions say, but it almost sounds the same as what a procedure would cover.

Thank you very much!

Procedure/flow chart is a short, often diagrammatic, summary of a reaction sequence with reagents, solvents, temperature & yields. It is expected that a competent synthetic chemist would be familiar with the reactions shown and would be able to understand how the product was made.

Synthetic method contains all the details of the reaction. Sources and quantities of reagents and solvents, times and temperatures of reaction, workup and purification procedures, characterisation of product.


Advice on forming a 10% solution of Maltol I’m having great trouble forming a 10% solution of Maltol. Specifically, I’m trying to dissolve 3 grams of Maltol in 27 grams of solvent.

I’ve tried everything from Propylene Glycol, to Deionized Water, to Ethanol, including some mixtures of these few solvents, but none of them can work to fully dissolve all the Maltol.

The best one so far appears to be hot Deionized Water, which appears to dissolve the most Maltol. Adding a small amount of Ethanol helps dissolve the Maltol further, but not completely.

Here I’ve attached a photo of the result. The entire mixture was immersed in a hot water bath. As you can see, the Maltol appears to form fluffy clumps that float to the surface, and are really stubborn to dissolve.

If anyone could advise me on how to achieve a complete dissolution of the Maltol, that would be very much appreciated!

Thank you!

Maltol (3-Hydroxy-2-methyl-4-pyrone) appears to be most soluble in ethanol.
https://pubchem.ncbi.nlm.nih.gov/compound/maltol#section=Melting-Point gives solubility as
1 to 10 mg/mL at 72° F in water, 10900 mg/L at 15 deg C (about 1%); one gram dissolves in 82 mL water, 80 mL glycerin, 21 mL alcohol, 28 mL propylene glycol; Freely soluble in hot water, chloroform; sparingly soluble in benzene, ether, petroleum ether; soluble in alkali hydroxides giving yellow solutions

Your fluffy floating clumps suggest something like water, which would reduce the solubility of maltol and could clump it up. or maybe there is some other impurity that complicates things.

You could try is to make a 5% solution in warm dry ethanol (apparently no problem...) and then evaporate off excess solvent. If you can get 10% when hot, but the solution drops crystals on cooling, you will have a saturated solution. But maybe it won't precipitate...


Can rate of reaction depend on both the reactants and the product? In introductory chemical kinetics, an elementary reaction is taken for instance
$$\ce{A -> B}$$
Then, we go on to derive integrated rate kinetics equations for them by assuming
$$\text{rate} = k [\ce{A}]^n$$

Could it possible in a more generic case that :
$\text{rate}=k[\ce{A}]^n [\ce{B}]^m$ and the total order of the reaction be $m+n$? In other words, can the rate of reaction depend on both the molar concentrations of reactants and products independently?

There exist autocatalytic reactions than can show such a behaviour. E.g. a reaction $\ce{A + B \rightarrow 2\ B}$ might show the rate $k[\ce{A}] [\ce{B}]$.

The reaction of permanganate with oxalic acid is a well known autocatalytic reaction.
$$
\mathrm{ 5\ (COOH)_2 + 2\ MnO_4^- + 6\ H^+ \rightarrow 10\ CO_2 + 2\ Mn^{2+} + 8\ H_2O}
$$
Here the generated manganese(II) ions catalyze the reaction. The reaction starts slowly if no initial manganese(II) ions are present, and then speeds up.


Work function and ionization energy I've seen that the first ionization energy of a particular metal is always greater than its work function. This implies that the valence electrons in an isolated atom are more strongly bound than the de-localized conduction electrons in the solid metal.

  For example, the first ionization energy of sodium is $5.14\mathrm {eV}$/atom.

  (source: here)

  while its work function is just $2.28 \mathrm {eV}$.

  (source: here)

I'm thinking that in a metal lattice, the de-localized electrons are attracted by many positive kernels at the same time. Also, they suffer repulsive forces from the electrons around. So the concerned electron is shielded from the attractive force from the positive centers  by the other nearby de-localized electrons and the the case isn't much different from that of an isolated atom. But that doesn't seem to be the case because the work function is (very roughly), just half the value of the ionization energy.

In metallic solids, after the formation of bands, are the de-localized electrons (which now exist in bands) attracted less strongly by the positive cores which gives rise to such a difference?

Is it the case always? If so, why?

Thank you very much for your valuable suggestions.

They differ by definition, which states that work function is difference in energy between Fermi level (work required to add one electron to the body) and vacuum level (energy of an electron that drifts in vacuum, free from any material), while ionization energy is difference in energy between HOMO (highest occupied molecular orbital) and our old friend, vacuum level.

It's clear that electrons that occupy orbitals (HOMO energy level) are lower in energy than Fermi level (the boundary between occupied and unoccupied states in a continuum of states). Below is the picture which you might find useful. Also check for the source of picture: it has good and detailed explanation about your next possible questions on topic.

[source]: Antoine Kahn. Fermi level, work function and vacuum level. DOI: 10.1039/C5MH00160A (Focus) Mater. Horiz., 2016, 3, 7-10.


Do Slater's rules apply to neutral lithium (at least approximately)? For lithium, the electronic configuration is $\ce{(1s)^2(2s)^1}$. I tried to calculate $Z_{eff}$ for the $\ce{2s}$ orbital of the neutral atom as follows:

$$Z_{eff} = Z-\sigma$$
$$Z_{eff} = 3-(2\times0.3)$$
$$Z_{eff} = 2.4$$

But the Wikipedia article's table of effective nuclear charges (linked below) states that $Z_{eff}$ is 1.279 for the $\ce{2s}$ lithium orbital, which is not even close to my answer (above). So do Slater's rules not apply to lithium or I have done something wrong?

Wikipedia article: https://en.wikipedia.org/wiki/Effective_nuclear_charge

You need to use the factor of 0.85 for each of the $\ce{1s}$ electrons:
$$Z_{eff} = 3 - (2\times 0.85) = 1.3$$
as per rule #2 in the Wikipedia article:

  If the group is of the [s p] type, an amount of 0.85 from each electron with principal quantum number n one less than that of the group, and an amount of 1.00 for each electron with principal quantum number two or more less.

The $\ce{1s}$ electrons both qualify as having "a principal quantum number $n$ one less than that of the group" (here, the principal quantum number of the group of the $\ce{2s}$ electron is 2). Here is a nicely worked example for further edification.


Why isn't thallium triiodide stable? We know that stability of higher oxidation state (+3) decreases and stability of lower oxidation state (+1) increases, so thallium is most stable in its +1 oxidation state due to the inert pair effect and paired s electrons.

In the compound $\ce{TlI3}$, since $\ce{I3}$ exists as a linear molecule $\ce{I3-}$  so the oxidation number of $\ce{Tl}$ is +1 which is stable and the compound should exist but I have read in many of the books that it does not exists. Why is it so?

Thallium triiodide does exist  and it is formulated as $\ce{Tl+I3-}$ and not $\ce{Tl^3+(I^{-})3}$. Thus $\ce{TlI3}$ is a thallium(I) compound and contains the triiodide ion, $\ce{I3^{−}}$. This is confirmed by considering the standard potential which indicate that $\ce{Tl(III)}$ is rapidly reduced to $\ce{Tl(I)}$ by the iodide.

$$\begin{align}
\ce{Tl^3+ + 2e− &-> Tl+} & E^\circ &= +1.252~\mathrm{V} \\
\ce{I2 + 2e- &-> 2I-} & E^\circ &= +0.5355~\mathrm{V} \\ \hline
\ce{Tl^3+ + 2 I- &-> Tl+ + I2} & E^\circ_\mathrm{cell} &= +0.717~\mathrm{V} > 0 \\ \hline
\end{align}$$

(data from Wikipedia).

$\ce{TlI3}$ can be prepared by the evaporation of stoichiometric quantities of $\ce{TlI}$ and $\ce{I2}$ in concentrated aqueous $\ce{HI}$, or by reacting $\ce{TlI}$ with $\ce{I2}$ in ethanol.


Why does some coffee smell like urine? I can smell a urine like odour from some coffees. I don't seem to be alone in this as there are multiple reports on the internet of people having the same experience.

I have read from a few posts that results of the Maillard reaction create ketoamines that could be causing it.

What's the science behind this?

References:

Reddit
Yahoo Answers

This answer is purely speculative in nature fueled by high school science and a research paper

From medicalnewstoday

  Urine does not usually have a strong odor to. However, occasionally, it will have a pungent smell of ammonia.

Googling coffee + ammonia led me to this little article Filter coffee nanoparticles to turn ammonia sensors.

The article  says that the researchers at IISC Bangalore were using coffee powder for detecting ammonia because coffee powder contains carbon and nitrogen in abundance and they've used that to make carbon nanotubes whatnot.

Then I hit gold when I found the research paper itself
- Understanding the ammonia sensing behavior of filter coffee powder derived N-doped carbon nanoparticles using the Freundlich-like isotherm.

The detecting technology is based upon adsorption which is a surface phenomenon basically the adsorbate(ammonia) gets concentrated only at the surface of the adsorbant(coffee powder).

Now, there are two types of adsorbtion.

Chemisorption
Physisorption

Now comes the speculation I'm assuming that the process over here is physisorption because physisorption is inversely proportional to temperature.

How it works?.

Filter coffee is essentially a powder this provides a large surface area for adsorbtion.

Intially when the coffee in your kitchen at room temperature some amount of ammonia gets adsorbed on it. Now, when you make the coffee it is exposed to high temperatures and as you can recall I just discussed that in physisorption the amount of gas adsorped on the surface is inversely proportional to temperature so the hot coffee can hold less amounts of ammonia on its surface as compared to the room temperature coffee.

So that's how ammonia is released on making coffee. Presumably.


Which measurement should I use? In lab, I gathered everything and my partner weighed everything, I didn't realize til now that there is a problem. She measured the copper and the crucible separately, she got 0.89 g for the copper and 11.63 g for the crucible. Those two numbers combined equal 12.52 g. However, when we put the copper in the crucible it weighed 12.54 g.

Which number should I use for the calculation, 12.52 g or 12.54 g?

Given that a 2-place balance is no more accurate than $\pu{+/- 0.01 g}$, you can say that the two measurements were within $\pu{0.01 g}$ of the average, $\pu{12.53g}$, then just use this average for your calculations.


Calculating the ionization energy for hydrogen What energy would be needed to remove the electron from the $n = 4$ level of the hydrogen atom?

  $\pu{−3.49 * 10^{−17} J}$
  $\pu{−1.36 * 10^{−19} J}$
  $\pu{+2.18 * 10^{−18} J}$
  $\pu{+1.36 * 10^{−19} J}$

I assume that the way to do this is to start from the Rydberg formula,

$$
E = \mathcal{R}Z^2 \left( \frac{1}{n_i^2}-\frac{1}{n_f^2} \right),
$$

set the initial level of the electron as $n_i = 4$, and the final level corresponding to removing the electron (ionization) as $n_f = \infty \implies \frac{1}{\infty^2} = 0$, leading to

$$
E = \pu{13.6*\frac{1}{4^2} eV = 0.85 eV = 1.36*10^{-19} J}.
$$

I assume you refer to Bohr's atomic model.

The energy of an electron in Bohr’s orbit of Hydrogen atom is given by the expression:

$$
\begin{align}
E_{n} &= \frac{2\pi^{2}me^{4}Z^{2}}{n^{2}h^{2}\left(4\pi\epsilon_{0}\right)^{2}} \\
&= \color{\navy}{-13.6\frac{Z^2}{n^2}\pu{ eV}}
\end{align}
$$

Since $Z = 1$ for hydrogen, the above equation can be further simplified. Now just plug in the required values.


Color of precipitate formed on reaction of acetylene with Tollen's Reagent I am asked to find the color of the precipitate formed by the passage of acetylene gas into an aqueous solution of ammoniacal silver nitrate.

I thought ammoniacal silver nitrate was Tollen's reagent, but acetylene is neither a ketone nor an aldehyde so I don't know how to figure out the color of the precipitate.

You are correct that ammoniacal silver nitrate is Tollen's reagent. Tollen's reagent is most notable for its reaction with aldehydes to produce a silver mirror, but it also reacts with several other types of compounds, notably alpha-hydroxy ketones and terminal alkynes.

With terminal alkynes, the alkyne reacts as a acid to form a silver acetylide which precipitates. In the case of acetylene, both hydrogens can react: $$\ce{HC#CH + 2AgNO3 + 2NH4OH -> AgC#CAg + 2NH4NO3 + 2H2O}$$

The colour of the precipitate is variously described as yellow, gray-white, or white. The actual colour you observe is probably strongly dependent on the purity of the sample and the size of the precipitated particles.


Why don't effervescent vitamin C tablets react with the vitamin C? I have a packet of effervescent Vitamin C tablets. When placed in a glass of water, a tablet will fizz and dissolve. They have the following ingredients (all included for completeness):

citric acid
ascorbic acid
sodium hydrogen carbonate
sorbitol
sodium carbonate
inulin
potato starch
zinc citrate
tricalcium phosphate
maltodextrin
riboflavin phosphate sodium
sweetener & flavours
tocopherol

I'm guessing that the citric acid is intended to react with the sodium hydrogen carbonate to effervesce.

Given that vitamin C (ascorbic acid) is also an acid, surely it could also react with the sodium hydrogen carbonate, removing the vitamin C from the finished drink?

Does this happen to any extent? If so, in what proportions? If not, why not?

The ingredients state how much vitamin C is present in the tablet, not in the finished product. This is also specified as a proportion of government recommended recommended daily amounts, which suggests that one would receive that amount if one took a dose as directed. If the vitamin C does react, does this mean that the label doesn't indicate how much vitamin C is available?

(I last studied chemistry at GCSE level in [high] school, so please excuse the lay question. A simple answer would be most useful!)

Acid-base reactions are fast and reversible (barring any additional reaction that occurs). That is, once you move the conjugate base of ascorbic acid to a low-pH environment (e.g. your stomach), it rapidly reverts to the protonated form. (Ascorbic acid does not undergo any pH-dependent further reactions, as opposed to carbonate, which when protonated to carbonic acid can decompose to carbon dioxide and water.)

Actually, as the pKa (roughly the "halfway pH" of acid/base reactions) of ascorbic acid is 4.2, the moment the protonated ascorbic acid enters your bloodstream (roughly pH 7.3), it immediately loses that proton again, reverting to the deprotonated ascorbate ion.

So in a practical sense, it really doesn't matter which form you take your Vitamin C in - either (protonated) ascorbic acid or (deprotonated) ascorbate acts in you body the same way, because they can rapidly interconvert based on the pH of the solution they happen to be in at the time.

(Because they typically work in buffered aqueous environments, biochemists typically play fast-and-loose with the difference between the protonation states of molecules. Unless you're explicitly dealing with pH-dependance, they're typically lumped all together, with the implicit understanding that at a given pH they'll adjust to the appropriate relative concentration. This can get confusing if you're a stickler for nomenclature.)


Negative charge of methoxide anion Why is the methoxide ($\ce{CH_{3}O^{-}}$) anion negatively charged?

It has 13 valence electrons and has 13 free electrons. In spite of that the methoxide anion is negatively charged.

The negative charge is due to the unpaired electron in the oxygen atom, remained after all bonds(molecular orbitals) are filled. Oxygen has 2 unpaired electron in its valence shell and is singly bonded to carbon in this molecule. Hence being un-bonded, the other electron exhibits the charge.

Check this link, hover the cursor on the $\ce{O}$-atom, you will get a graphical representation of the same.


Stoichiometry of Gaseous Reactions The Question:

How many liters of $H_2S$ would be required to react with excess oxygen to produce $12.0$ $L$ of $SO_2$? The reaction yield is $88.5$%. Assume constant temperature and pressure throughout the reaction.

My Approach:

I used dimensional analysis to convert from liters of $SO_2$ to $moles$ of $H_2S$ and then used Avogadro`s Law ($\frac{V_1}{n_1} = \frac{V_2}{n_2}$) to find the volume of $H_2S$ in liters.

Dimensional Analysis:
$\frac{1}{.885} * \frac{12.0L}{1} * \frac{1 g}{1 mL} * \frac{1 mol SO_2}{64g} * \frac{2 mol SO_2}{2 mol H_2S} * \frac{1 mol H_2S}{34 g} = 6.23$ $mol$ $of$ $H_2S$.

Avogadro`s Law:
$\frac{12.0L}{166 mol} * \frac{ L }{6.23 mol}$

End Result = $.028 L$

^ The end result does not seem right and any help in remedying my approach or shedding light on another approach would be much appreciated. Thanks in advance!

I never really learned the method of solving this stuff with Avogadro's Law, but here's an almost identical way.

Your balanced equation is as follows...

$\ce{2H2S + 3O2 -> 2SO2 + 2H2O}$, and I think you agree on that by looking at your calculations.

At a given temperature and pressure, $1 mol = x L$. For example, at STP (standard temperature and pressure) this is $1 mol = 22.4 L$. It wasn't exactly defined what temperature and pressure are, but they are constant, so we can use a variable $x$ to denote some conversion (which cancels out anyway, I assume giving rise to Avogadro's Law).

$12.0L_{\ce{SO2}} * \frac{1mol_{\ce{SO2}}}{xL_{\ce{SO2}}} * \frac{2mol_{\ce{H2S}}}{2mol_{\ce{SO2}}} * \frac{xL_{\ce{H2S}}}{1mol_{\ce{H2S}}} = 12.0L_{\ce{H2S}}$

But this is assuming $100\%$ yield. I see you divided by $0.885$ above, which is correct, but I've gotten a metric ton of these wrong before, so I prefer the proportions way.

Regardless, $\frac{12L_{\ce{H2S}}}{0.885} = 13.6L_{\ce{H2S}}$

I think you definitely overcomplicated things by assuming a density of $1g/mL$ among other things. Your dimensional analysis also seems strange near the end of the left hand side where you somehow cancel $g$ with $mol_{\ce{SO2}}$.

If I were to do it with Avogadro's Law (taking a guess here), I'd still get $12.0L$ because the mole ratio is the same. And then I'd divide like you did.


Optical Density numbers If it's OD of 600nm I know that's the wavelength used, but what are the other numbers in something like  "OD 600nm of 0.6-0.7". The 0.6-0.7?

That is the optical density (absorbance). It's just a range of between 0.6 and 0.7. Without the context, it's hard to say why they don't have a precise value, but in something like a protocol, it's common to see that kind of thing when there is some natural variation or the exact number isn't important.


Is there an isotope effect for diffusion of an ion through a crystal lattice? Consider the diffusion of natural abundance $\ce{Li^{+}}$ and $\ce{{}^{6}Li^{+}}$ through a crystalline matrix, say a spinel such as $\ce{LiMn2O4}$.  The two lithium ions will have the same ionic radius and electric charge.  They will differ only in atomic weight.  Is there any reason to believe that the lighter isotope, $\ce{{}^{6}Li^{+}}$, will diffuse through the crystal any faster than the natural abundance isotope?  When I consider the forces acting upon the ions in the crystal, I come up with coloumbic, dispersion and hard shell repulsive forces.  I cannot see how any of those would change by simply changing the mass of the ion.  Am I missing anything?

$F = ma$, so a less massive isotope should move faster, depending on the forces moving it.


How are intermolecular forces between non-metallic substances affected when the substance is heated to its melting point? I have 3 main questions that I am unsure of:

Are intermolecular forces acting simultaneously causing attraction between particles in a sample of a substance?
Example: Hydrogen bonding and dipole-dipole attraction present in a substance both influencing the attraction between particles.

If there are multiple intermolecular forces present, would they be overcome in an order?

What actually happens to the intermolecular forces between particles as a non-metallic molecule is heated?

First of all, multiple intermolecular forces could be present between molecules, and, yes, all forces influence attraction, binding and etc. However, if you study any chemical system, you will see that is usually dynamic.
It all depends on conditions like temperature, pH and especially compound properties, etc. For example, the drug captopril (an antihypertensive drug) could form multiple bonds with the protein angiotensin-converting enzyme (ACE).:

How about one specific atom of a molecule? In theory, one atom could form multiple intermolecular forces at a specific time (let's say picosecond), however, it is not practical - energetically unfavorable. Imagine that atom should be in such a position, an electronic configuration that reduces its potential energy and also form multiple intermolecular bonds. It is possible that such systems exist, but they are not common.
Second answer:
Yes, intermolecular forces have specific values of bonding energy:

For example, it takes less energy to break London dispersion forces than to break hydrogen bonds.
Third answer:
It all depends on the level you want to know... Basically, when you heat molecules (give them more energy) they absorb it and move faster.  Vigorous movements make bonds break. For example, water molecules at 100 °C start to move so fast that almost all its intermolecular forces are broken. Fun fact: you can smell lemons at room temperature because it has volatile molecules (molecules that form very few intermolecular bonds) called Limonene.


Why don't gases escape Earth's atmosphere? Some gases are lighter than others and rise. Why don't they continue going up, leave the atmosphere, and then enter outer space?

The atmosphere actually loses gases to outer space.

The average velocity $\bar v$ of gas molecules is determined by temperature $T$.
However, not all the molecules travel with the same velocity.
The probability of finding a molecule with a velocity near $v$ is described by the Maxwell distribution of speeds
$$\begin{align}
f{\left(v\right)}&=4\pi\sqrt{{\left(\frac m{2\pi kT}\right)}^3}v^2\exp\left(-\frac{m{v^2}}{2kT}\right)\\[6pt]
&=4\pi\sqrt{{\left(\frac M{2\pi RT}\right)}^3}v^2\exp\left(-\frac{M{v^2}}{2RT}\right)
\end{align}$$
where $m$ is the mass of the molecule, $k$ is the Boltzmann constant, $M$ is the molar mass of the gas, and $R$ is the molar gas constant.

Individual molecules may reach escape velocity $v_\mathrm e$ and thus be able to leave the atmosphere.

Escape velocity is the minimum velocity that is sufficient for an object to escape from the gravitational attraction of a massive body. For a planet, the escape velocity may be estimated by using the formula
$${v_\mathrm e}=\sqrt{\frac{2Gm_\text{planet}}r}$$
where $G$ is the gravitational constant, $m_\text{planet}$ is the mass of the planet, and $r$ is the distance from the centre of mass of the planet.

Therefore, atmospheric escape depends on the mass of the planet, the temperature of the atmosphere, and the molar mass of the gas.


Resemblance of molecular orbital formed by mixing of s and p orbitals to hybridised sp orbital See the following two images.

I think that the two orbitals should not resemble each other because s and p orbitals overlap in different fashions in the two cases.

In the case of sp hybrid orbital, both s and p orbitals are centred at the same point. As a result, s orbital does constructive interference with one lobe and destructive interference with another lobe of p orbital.

But in the case of molecular orbital, s and p orbitals are centred at different nuclei (and those nuclei can't be at the same point). Then how come the resultant molecular orbital resembles the sp hybrid orbital?

The right lobe of the molecular orbital should get enlarged due to constructive interference of s and p orbitals, but why does the left lobe get diminished? Shouldn't it remain as it is in the p-orbital?

It's important to realize that both of these drawings are highly stylized renderings of the electron density that are intended to provide a qualitative understanding. They don't accurately represent the actual shape of the orbitals. For example, $p$ orbitals are much rounder and taller relative to their length than we usually draw them. They are drawn elongated so that they don't cover each other up if you are drawing more than one on an atom. So the fact that the drawings of the $sp$ hybrid and the $p+s$ $\sigma$ bond look similar should only be taken to mean that they both have a bias of density to one side. You can't really conclude much else from the drawings.

The Orbitron website is a great resource for seeing the actual shape of the electron density of different types of atomic orbitals, which you can compare to the drawings.

MolCalc produces similar images of molecular-level orbitals. The bonding orbital of HF is probably the closest you can get to the $p+s$ $\sigma$ bond orbital in your drawing.


Why is formic acid a stronger acid than acetic acid? I am told that because the methyl group is electron donating in the conjugate base of acetic acid, this destabilizes the conjugate base by exacerbating the existing negative formal charge on the deprotonated oxygen, while in formic acid the electron donating methyl is absent in lieu of a hydrogen, which is neither withdrawing nor donating.

Is this reasoning correct, and are there any other reasons why formic acid might be stronger than acetic acid?

We are discussing the following equilibrium

We can make the acid a stronger acid by pushing the equilibrium to the right.  To push the equilibrium to the right we can

destabilize the starting acid pictured on the left side of the
equation, and \ or
stabilize the carboxylate anion pictured on the right side of the
equation.

Comparing acetic acid ($\ce{R~ =~ CH3}$) to formic acid ($\ce{R~ =~ H}$), the methyl group is electron releasing compared to hydrogen.  Therefore the methyl group will stabilize the dipolar resonance form of the starting acid where there is a partial positive charge on the carbonyl carbon.  This should stabilize the starting acid. Further, this electron releasing ability of the methyl group will tend to destabilize the resultant carboxylate anion which already has a full unit of negative charge.

Therefore, because the methyl group 1) stabilizes the starting acid and 2) destabilizes the carboxylate anion product, the methyl group will push the equilibrium to the left, compared to the case where the methyl group is replaced by a hydrogen.  Consequently, acetic acid is a weaker acid than formic acid.


Making elemental sulfur from pyrite FeS2 Has anyone done this or know of a good experiment that would allow someone to make elemental sulfur from $\ce{FeS2}$ (iron pyrite)?

You may just need to heat it.

$$\ce{FeS2 ->[\Delta] FeS + S}$$

In the wikipedia article of pyrite, it is written that:-

  Thermal decomposition of pyrite into $\ce{FeS}$ (iron(II) sulfide) and elemental sulfur starts at 540 °C.

A good experiment for recovering sulfur from iron pyrite is called ' steam-pyrite interactions'.  The following is an excerpt from the original journal:-

  The reactions of natural iron pyrite with steam, hydrogen, and carbon monoxide have been studied in a fixed
  bed tubular reactor at 900-1 100°C. A new process has been developed for the recovery of elemental sulfur
  by the reaction of pyrite with steam. In this process, the yields of elemental sulfur and sulfur evolved as hydrogen
  sulfide and sulfur dioxide were, respectively, 84.5 and 13.7% of total sulfur under the following
  conditions: particle size, below 100 ASTM mesh; temperature, 1000°C; reaction period, 1 5 min; space velocity,
  33.2 X lo4 cc/cc/hr; and partial pressure of steam, 1 atm. Under these conditions, the effluent gas also
  contained an appreciable amount of hydrogen. The calcine contained ferrous oxide, ferrosoferric oxide,
  silica, and 1.3% sulfur. The residual sulfur could be further brought down to a minimum of 0.2% in the calcine
  at 1 100°C using a 30-min reaction period. The incorporation of hydrogen in the steam-iron sulfides reactions
  at 1 100°C increased considerably the yield of hydrogen sulfide (59.5%), but only a small amount of metallic
  iron (1.6%) could be obtained.


Calculating ΔH with non-integer coefficients in the balanced equation I am working a practice problem out Ebbing's General Chemistry and for part of the problem I needed to calculate $ΔH.$ The problem considers the burning of ethane in oxygen to yield carbon dioxide and water vapor.

The coefficients used in the solution manual were 1 for ethane 7/2 for oxygen 2 for carbon dioxide and 3 for the water vapor. The coefficients I used were 2, 7, 4, and 6, respectively.

Naturally, it follows the $ΔH$ value I got is twice as large. Can anyone provide clarification here?

There is not much to clarify: you determined dimolar enthalpy of combustion, whereas standard enthalpy of combustion is determined for a complete oxidation of one mole of the substance:

$$
\begin{align}
\ce{C2H6(g) + 7/2 O2(g) &-> 2 CO2(g) + 3 H2O(l)} &\quad &Δ_\mathrm{c}H^\circ\\
\ce{2 C2H6(g) + 7 O2(g) &-> 4 CO2(g) + 6 H2O(l)} &\quad &Δ_\mathrm{c}H^\circ ×2
\end{align}
$$

When composing the corresponding reaction equation, make sure to balance it in such a way that the stoichiometric number $ν$ of the substance the enthalpy is determined for is always $-1.$


Reduction of amides with lithium aluminium hydride Since $\mathrm{p}K_\mathrm{a}$ of primary amides are close to alcohols, will there be an acid-base reaction between amides and $\ce{LiAlH4}$ besides reduction?

Not according to the commonly accepted mechanism here

The first step is

  The nucleophilic H from the hydride reagent adds to the electrophilic C in the polar carbonyl group of the ester.

Note: the pKa of primary amides is in the 22-25 range so not close to alcohols.


If dS ≥ dq/T for a spontaneous change is dS ≤ dq/T for a non-spontaneous change and is that even possible? I am struggling with the concept of entropy but might be on the verge of understanding it. The Clausius inequality, $$\mathrm{d}S \geq \mathrm{d}q/T,$$ is true for all spontaneous changes with the equality for reversible changes. But what if a change is not spontaneous or is that impossible i.e a change cannot be forced. Any help with answering this question or to help conceptualize and understand the entropy would be greatly appreciated.

As you perfectly say, Clausius inequality states that $$\mathrm{d}s\geq \mathrm{d}q/T.$$ This has to be true for any process.

For reversible processes, what happens is that $\mathrm{d}s= \mathrm{d}q/T$, and for non-reversible processes, $\mathrm{d}s > \mathrm{d}q/T$. If for a given process $\mathrm{d}s< \mathrm{d}q/T$, it would not fulfill Clausius inequality.

Consider the situation where the system is isolated from its surroundings. In this case, $\mathrm{d}q_\mathrm{(sys)} = 0$ , and thus the Clausius relation implies $\mathrm{d}S\geq0$, so this tells us that the entropy of an isolated system cannot decrease in the course of a spontaneous change.

Since the universe is itself an isolated system, this result shows that we can use entropy as the signpost of spontaneous change. Processes are only spontaneous if they cause an increase in the total entropy of the universe.


What happened to this 10 year old unopened plastic bottle? A colleague of mine found this unopened Aquarius bottle (non-carbonated drink) that was 10 years past its expiration date.

The bottle was collapsed inward which would suggest that the pressure in the bottle was lower then outside the bottle.

Can someone explain what happened here? Also, what happened to the air inside the bottle?

Since the drink is not carbonated, no liquid has been escaping the bottle, and the components don't seem to be able to participate in any kind of chemical reactions involving gaseous products, I assume the gas that's been leaked is air.

I also cannot imagine any kind of chemical reaction involving consumption of this amount of air leading to such a severe bottle deformation. My theory is that air simply escaped over time through the bottleneck. There are 2 parts made of 2 materials with different thermal expansion coefficients:

the bottle itself (polyethylene #1 (PET), or polyethylene terephthalate (PETE));
the screw cap (polypropylene (PP)).

Due to the construction of the joint, both parts basically act as a pressure equalization valve. I suspect that the coefficients of thermal expansion are chosen in such a way that when heated, excess gas would slowly escape in order to minimize the risk of explosion when the storing conditions for the finished product is not met. So, when the bottle is heated, the air is slowly leaking out. Once the temperature drops, the "valve" closes and the process stops. Over the course of 10 years, there surely were many temperature fluctuations making air leave the bottle little-by-little.


How do we determine molecule chirality in these examples? I understand that in the first example, the molecule is chiral as there is no symmetry. For the second and third example, I see no symmetry in the 3d conformation after twisting the right hand side. In the third example, I do see symmetry if we twist the right hand side so that the carbon molecules are in one plane. What am I missing?

Edit: typo corrected. Also I was not clear in that the picture I attached is the answers the professor provided. I see why the first one is right as there is no symmetry. But I did not see why the second and third ones were considered chiral and achiral, as I saw symmetry in the third example and not in the second. From the answers though, I am understanding that we consider if the molecule is super imposable with it's mirror image?

Although you have arrived at the correct answers and seem to have followed a reasonable process for getting there, it's important to be precise with the use of terms like symmetry and chirality. For a molecule to be chiral, it must be non-superposable with its mirror image. It's important to really embed this definition before talking about symmetry and things like chiral centres. For example, it is possible for molecules to be chiral without any chiral centres, and it's possible for molecules with multiple chiral centres to be achiral.

the guaranteed way to determine if a molecule is chiral: draw (or build a model) of the mirror image and determine whether they are superposable

Now, symmetry is a different, but related, topic. A molecule with a point or plane of symmetry will be achiral. However, molecules with a rotational axis of symmetry can still be chiral, eg. many often-used chiral ligands are C2 symmetric.

So, in answer to your question, "What am I missing? - I would just say: the rigor in use of the terminology. Your approach of drawing different conformers of the molecules with two chiral centres and looking for a plane of symmetry (that shows the molecule is a meso compound, and therefore achiral) is sound and led you to the correct answers.


What makes an atom favor another one when bonding, and should there be enough atoms for both parts? I'm learning chemistry through Khan Academy and I have 2 pretty basic questions.

Oxygen (O) has 6 electrons on it's last shell, so it needs 2 more to become stable. Hydrogen (H) has 1 electron, so it needs 1 more. O might combine with 2 H's, creating H2O, and it might combine with another O, creating O2. It can also create H2O2.

QUESTION 1: So, given a hypothetical solution (or environment, or whatever), with only O and H atoms, what will make O to favor a bond with another O instead of a bond with 2 H's (or vice-versa)? Both situations will make it stable.

QUESTION 2: Also, if we assume that, in that solution, there's a limited number of both O and H atoms, and after all bonds are made, there's only one O atom and one H atom left, will that H atom bond with the one O atom left, since that will make the H stable (albeit not the O atom), creating a HO molecule ? So, I guess what I'm asking is, will bonds only be made when there's enough atoms for both parts to become stable?

Thanks.

The answer to your question boils down to one thing, bond energies. When a bond is formed, energy is released. When a bond is broken, energy must be absorbed. Elements prefer to be in a low energy state, so if you just added H and O into a container, the reaction with the most negative net bond energy will take place. To calculate the overall bond energy of the reaction, the equation is:

$E_{total} $= (Energy of bonds broken) - (Energy of bonds formed).

(As a side note, the answer is in joules of energy per mole)
For naturally occurring hydrogen and oxygen the bond energy is:

$2H_2 + O_2 $ ->$2H_2O$

= $(436*2 + 493*1)-(463*4)= -487 \frac{kj}{mol}$

(2 H-H bonds broken + 1 O=O bond broken) - (4 O-H bonds formed)

This reaction is favored and would occur under typical conditions if oxygen was in the presence of hydrogen and would release -487 joules of energy per mole. The bond energies for each bond can be found by simply looking them up online. The bond energy for the formation of $H_2O_2$ I calculated turned out to be -149kj/mol. Therefore the formation of $H_2O$ would be favored over $H_2O_2$. In your question, you asked about single O and H atoms, not $H_2$ and $O_2$, which would not occur naturally, but is still very possible. This result however would be the same. The atoms would first bond to form $O_2$ and $H_2$, then would proceed to form water as noted above.

Finally, to answer your last question, if there is one hydrogen and one oxygen left, they will indeed bond to make an $OH$ molecule.


Does the nitrogen atom move with in the molecule in an ammonia maser, or does the molecule flip? This intriguing answer to Can reactions produce microwave or radio wave radiation? says (in part):

An ammonia maser qualifies as an inorganic RF-emitting reaction. As the nitrogen changes position in the $\ce{NH3}$ molecule, ~24 GHz radiation is emitted.
Similarly, organic methanol, $\ce{CH3OH}$ makes a maser found in nature, emitting at ~36.2 GHz and also at ~6.7 GHz

The first link is to The Ammonia Maser in Feynman's Lectures on Physics and it includes a picture of a triangular pyramid with the nitrogen pointing "up" in one case and "down" in the other as an illustration of the transition.
But how does it get there? Does it "tunnel through" the nitrogen triangle, or does the whole molecule flip around? Can we even ask that question? Could we, if for example we labeled the corners with $\ce{{}^{1}H}$ $\ce{{}^{2}H}$ and $\ce{{}^{3}H}$ as a gedankenexperiment to make the two distinguishable?

Let me expand a little on my earlier comment. If we only consider the inversion of the ammonia molecule (we ignore rotation and vibrations other than the inversion mode), the Schrödinger equation is given by
$$
-\frac{\hbar^2}{2\mu_\text{red}}\frac{\text{d}^2\psi{z}}{\text{d}z^2}+V(z)\psi(z)=E\psi(z),
$$
with $\mu_\text{red}$ the reduced mass and $z=d_\text{NH}\sin{\alpha}$, where $d_\text{NH}$ is the NH bond distance and $\alpha$ the so-called umbrella angle. The potential V(z) is shown in the link given by porphyrin in one of his comments. Although the reduced mass does, in principle, depend on $\alpha$ (see Eq. 4 of this paper), the dependance is weak and we may consider that the hydrogen atoms move simultaneously with respect to the nitrogen atom. In this case
$$
\mu_\text{red}=\frac{3m_\text{H}m_\text{N}}{3m_\text{H}+m_\text{N}}\approx2.5\text{ u}.
$$
The tunneling does not change the center of mass (c.o.m.) of the molecule, so while the nitrogen moves up, the plane of hydrogen atoms moves down to ensure the stationary c.o.m. (and vice versa). I made the following animated gif at some point during my PhD (sorry for the watermark, but I've  had some negative experiences in the past):

The first paper - to my knowledge - that treats the tunneling of ammonia appeared already in 1932. The first application of QM tunneling was done by Gamow appeared 4 years prior to the ammonia work and solved the mystery of the huge range of observed lifetimes of alpha emitters (20 orders of magnitude).

In the 1932 paper, Dennison and Uhlenbeck also derive the magnitude of the tunneling splitting as a function of the reduced mass
$$
E_\text{inv}=\frac{h\nu_0}{\pi}\exp\left \{-\frac{2\pi}{h}\int_\limits{-z_0}^{z_0}\left [2\mu_\text{red}\left (V(z)-E \right ) \right ]^{1/2} \right \},
$$
where $\pm z_0$ are the equilibrium positions corresponding to the potential minima, $\nu_0$ is the vibration frequency in one of the two wells and $E$ is the energy of the system with respect to the potential minimum. This expression is derived from the WKB approximation and it basically is the product of the tunneling probability with the frequency with which the reduced mass 'hits' the barrier. As can be seen from the formula, the tunneling frequency scales exponentially with the square root of the reduced mass, which is also observed experimentally. The following picture is taken from my thesis and basically a copy of Fig. 2 of this paper.

If the molecule would simple rotate instead of tunnel, the change in the energy splitting would be inversely proportional to the reduced mass (via the moment of inertia) and this is clearly not observed.


Is there any book where specific heat capacities are given? I want to find the value of the specific heat capacity of aluminium nitrate nonahydrate, in order to calculate mixing effects in a chemical reaction. I am looking for a source, such as a book or link to an online PDF, which would be appreciated.

There are several databases of thermochemical properties, a number of them being listed on this Wikipedia page (at the bottom). Most of them are not freely available online, but require subscription. I've searched online but couldn't find that information freely available.

Another possible way to get the information would be through a search of the published literature (scientific papers). You can, for example, find this information on magnesium nitrate in a 1944 paper in J. Am. Chem. Soc.. However, I had no such luck with Al(NO3)3 · 9 H2O .

Clarification: the answer above is for the crystalline compound aluminium nitrate nonahydrate, which as @Ivan stated, may not be what you actually need for your calculation if the reaction is in water.