The major product of 1,4-pentadiene NBS bromination Considering penta-1,4-diene, if allylic substitution is done with NBS, which product will be major, 3-bromo or 1-bromo?
In the first case I get stable free radical , in the second case I get conjugated diene.
Should I consider stability of free radical first as it is formed first , or the stability of the product which is formed in the end?

The Main reason for looking at the stability of the intermediate is Hammond's Postulate. Now the intermediate has the electron density spread all over it's body. SO it's reasonable to look at the activation energy of the transition state itself.

When the Br radical attacks at the end position, you have a much sterically favored transition state and the other two bonds are in conjugation and hence we would expect the activation energy would be much lower for this and not for the other transition state where the Br attacks at $3^{rd}$ position.

Hence we can conclude that conjugated diene is most probable product !.

Please correct me if I am wrong.


Having trouble understanding the iron-carbon phase diagram At room temperature, alpha and alpha+$\ce{Fe3C}$ are present (depending on carbon content), which is ferrite and pearlite. In the lab we polished and etched samples of carbon steel to observe the microstructure, and saw both ferrite grains AND pearlite grains. Not unlike the following:

However, from my understanding of the phase diagram, only one phase should be present. For any steel with less than 0.02% carbon we should see only ferrite, and for steel with >0.02% carbon we should see only pearlite. Though this was not the case, and rather there was a mix of pearlite and ferrite, with the higher carbon content steel having a higher proportion of pearlite.

It would be much appreciated if someone could explain to me what is happening here. Links to further reading would be helpful but not required.

For any steel with less than 0.02% carbon we should see only ferrite, and for steel with >0.02% carbon we should see only pearlite.

You have got a small misunderstanding here. The maximum solubility of carbon at $723\ \mathrm{^\circ C}$ is 0.02%. At room temperature it is very low ($0.005%$ at $0\ \mathrm{^\circ C}$). Also for carbon level between 0.02 and 0.8 you wont see only pearlite, you will see a combination of ferrite and pearlite for sure. Only at 0.8% you can observe pure pearlite(100%). Also note that the phase diagram is in such a way that if you cool from the ferrite+ austenite region at $800\ \mathrm{^\circ C}$ with carbon content < 0.02% we can still get a ferrite + pearlite + even austenite/martensite.
Phase diagrams give an ideal picture. In real situations, it is very tough to get very slow cooling rates to get the equilibrium structure.


Naming of hydroxides In naming of compound, for example in strong base; KOH and $\ce{Ca(OH)2}$ have different formula, in potassium hydroxide, one hydroxyl group is attached while in calcium hydroxide two hydroxyl group are present, but both named hydroxide, please explain.

A lot of chemical nomenclature is historic and not very systematic.

Common names for chemicals are often determined by history and don't always completely describe the detailed chemical structure or formula. The theory of chemical bonding and structure was not fully developed until the 20th century but we did a lot of chemistry before that. People got used to common names for things and those names persisted even when more systematic names were developed to introduce more clarity about the relationship between names and formula/structure.

For example, acetone is the commonest name for propan-2-one but the common name gives no indication at all about the chemical structure. Many organic compound have very unwieldy systematic names (reflecting the gargantuan complexity of the chemistry of carbon compounds). Fluoxetine (the active ingredient in prozac) is N-methyl-3-phenyl-3-[4-(trifluoromethyl)phenoxy]propan-1-amine and that is too big even to fit on the pill bottle and even the formula, C17H18F3NO, is a bit of a mouthful despite not giving away a lot of structural information about how the atoms are connected.

In the case of simple inorganic compounds like simple hydroxides the situation is similar. The names arose because they described the key chemical parts making up the compound, in this case a metal (as an ion) and a hydroxide ion (OH-). Rather than the unwieldy but more precise names describing the number of ether part of the compounds, chemists apply a little chemical knowledge and keep simpler names for many compounds. Most chemists know that alkali metals (Na, K etc.) form ions with a single positive charge and, therefore, if they form hydroxides they will only contain one hydroxide. Alkaline earth metals (Mg, Ca etc.) form ions with two positive charges and their compounds will contain two hydroxides. This is because all atoms want to reach the lowest possible energy state, which usually is the elecron structure of noble gases. Since most chemists know this, there is no point in using a more complex and longer name. This isn't always true: transition metals can form ions with several different "oxidation states" (which basically means the charge on the ion in simple ionic compounds: iron for example is commonly found both as Fe2+ and Fe3+) and, in these cases the name needs to contain extra information to help the chemist tell them apart.


How many decimal places should a volume measurement in a beaker include? I have a beaker which goes up in $0.5~\mathrm{mL}$ increments and my teacher told me that my measurement $(15.50~\mathrm{mL})$ was incorrect, and instead the answer was $15.5~\mathrm{mL}$. I don't understand. I know that the liquid is between $15.5~\mathrm{mL}$ and $16~\mathrm{mL}$. My estimated digit is $0$ (it looks like it lies on the line), so my answer is $15.50~\mathrm{mL}$. How is it $15.5~\mathrm{mL}$?

It may have to do with the fact that it's a beaker, rather than some other kind of glassware.

First, beakers are poor vessels for volumetric measurements due to their large diameter/height ratios.  You have to add a lot of liquid, relatively speaking, in order to increase the height of the fill level by (say) a millimeter, and that means the uncertainty in the fill volume is large.

As well, beakers are usually not graduated precisely. Thus, even if you very very carefully fill the beaker to exactly the $15.5~\mathrm{mL}$ line with liquid, there's a very real chance that the line doesn't actually correspond exactly to a $\mathit{15.5~mL}$ fill volume.

So, unlike with a piece of glassware specifically made for accurate volumetric measurements, such as the burette mentioned by MaxW in a comment, it is likely not appropriate to assign more significant figures to a beaker measurement than are "available" in the graduation increment.


Is pressure extensive or intensive? I've learnt that pressure is an intensive measure,
but I have also learnt that pressure can be added (partial pressures), which doesn't make sense since intensive variables are non-additive.
Where is the flaw in my understanding?

You are right that intensive properties are not additive like the properties which depend on the extent of the system. Pressure is indeed an intensive property because it is force divided by an extensive quantity (area). Now partial pressure is an interesting property. It is a hypothetical construct which cannot be directly or "experimentally" measured in a mixture of gases yet it can be calculated for individual gases in that mixture.
The basis of calculation of partial pressure is the number of particles in the container of each gas (hence an partial pressure originates from the number of particles of gas-A and gas-B).  Greater the number of particles of each gas in a container of fixed volume, greater is the calculated partial pressure of each component. The number of particles in a container determine the partial pressure which is an extensive property and hence partial pressures are additive.
Wikipedia explains the limitations of classifying quantities as extensive or intensive.
Consider reading the Journal of Chemical Education Article (free download from Google Scholar)
Redlich, "Intensive and Extensive Properties" J. Chem. Educ., 1970, 47 (2), 154–156. Redlich says that this classification is an incredible source of confusion in thermodynamics. I feel that he is not wrong.

The general validity of the division of physical properties into extensive and intensive kinds has been addressed in the course of science. Redlich noted that, although physical properties and especially thermodynamic properties are most conveniently defined as either intensive or extensive, these two categories are not all-inclusive and some well-defined physical properties conform to neither definition. Redlich also provides examples of mathematical functions that alter the strict additivity relationship for extensive systems, such as the square or square root of volume, which may occur in some contexts, albeit rarely used.

> Other systems, for which standard definitions do not provide a simple
> answer, are systems in which the subsystems interact when combined.
> Redlich pointed out that the assignment of some properties as
> intensive or extensive may depend on the way subsystems are arranged.
> For example, if two identical galvanic cells are connected in
> parallel, the voltage of the system is equal to the voltage of each
> cell, while the electric charge transferred (or the electric current)
> is extensive. However, if the same cells are connected in series, the
> charge becomes intensive and the voltage extensive. The IUPAC
> definitions do not consider such cases.
>
> Some intensive properties do not apply at very small sizes. For
> example, viscosity is a macroscopic quantity and is not relevant for
> extremely small systems. Likewise, at a very small scale color is not
> independent of size, as shown by quantum dots, whose color depends on
> the size of the "dot".


Home-grown diamond using CVD I want to make a diamond at home using chemical vapor deposition (CVD).

If I use a metal chamber filled with methane and hydrogen gas and heated with a microwave oven oscillator element, will the gas explode?
What is the ratio of methane and hydrogen for the diamond to be able grow?

The easiest way to grow diamond through chemical vapor deposition (CVD) is on a heated filament, which just takes a few watts.

You'd need a low-pressure chamber (~30 torr, or ~0.04 atm) with gas-tight fittings for electricity, feed-gas of ~1% $\ce{CH4}$ and ~99% $\ce{H2}$ and $\ce{Cu}$, $\ce{Ti}$, $\ce{W}$ or $\ce{Ta}$ filaments. Since this is a flow-though reactor, you'll also need to dispose of the flammable feed-gas without letting it accumulate near the vacuum pump.

N.B. This is not a simple project one could do at home, since you'd need vacuum equipment, pressure regulators, high-pressure bottles* of these highly flammable gases, and a vacuum-vessel which could implode. If you can get a knowledgeable instructor involved who has access to laboratory facilities and a thorough understanding of safety procedures, though, it's not impossible to accomplish.

* One of the greatest dangers would be a gas bottle falling, taking off like a rocket and leaving explosive gas in its wake!


Why And How are Unbranched chains of hydrocarbons easily biodegraded? My chemistry textbook says

  If detergents are highly branched then they can't be easily degraded by bacteria.So,nowadays the branching of hydrocarbon chain is controlled and kept to a minimum.

Upon searching this I got the answer that "side chains stop bacteria from attacking and breaking the chains".
I really don't understand how does this work.Why would having side chains prevent bacteria from attacking ? Won't they provide more sites for attack of bacteria ?

Here is a study1 showing this is the case, that generally speaking, branched hydrocarbons are more difficult to degrade, especially those exhibiting anteiso-terminal branching (branching  on the third carbon). Both the study and a book on oil spill forensics2 hypothesize that steric hindrance is the  mechanism responsible. They state, respectively:

  If alkyl branches are located near the terminus, it seems likely that decreased biodegradability could be the result of steric inhibition of terminal oxidizing enzymes

  Complex branching hinders both the initial oxidation and the subsequent lipid catabolism apparently because tertiary and quarternary carbon atoms interfere by steric hindrance with the oxidation enzymes.

However, this is not always the case as Boethling3 explains when discussing design objectives for molecule biodegradability:

  Similarly there is a commonly held view that any branching, for example, even a single methyl group on an otherwise linear alkyl group, is to be avoided. This is a gross oversimplification. All that can be said with any confidence is that quaternary carbon is usually to be avoided, as is extensive methyl chain branching, which has no strict definition.

1 Schaeffer, T. L., Cantwell, S. G., Brown, J. L., Watt, D. S., & Fall, R. R. (1979). Microbial Growth on Hydrocarbons: Terminal Branching Inhibits Biodegradation. Applied and Environmental Microbiology, 38(4), 742-746. Retrieved from https://aem.asm.org/

2 Wang, Z., & Stout, S. (2006). Oil Spill Environmental Forensics: Fingerprinting and Source Identification (1st ed.). Burlington, MA: Academic Press.

3 Boethling, R. S., Sommer, E., & DiFiore, D. (2007). Designing Small Molecules for Biodegradability. Chemical Reviews, 107(6), 2207–2227. https://doi.org/10.1021/cr050952t


Why is the probability for the signal at 160 in the bromine mass spectrum twice as high as for the signals at 158 and 162? The mass spectrum in Bromine, with the molecules $\ce{^{158}Br2+}$, $\ce{^{160}Br2+}$ and $\ce{^{162}Br2+}$:

As you can see, the $\ce{^{160}Br2+}$ is almost double in intensity compared to the $\ce{^{158}Br2+}$ and the $\ce{^{162}Br2+}$ peak.

The book I am reading simply states that this is because

  The probability of two different isotopes occurring in a $\ce{Br2}$ molecule are twice that of the same isotope appearing in a $\ce{Br2}$ molecule.

This is supported by the $\ce{^{160}Br2+}$ peak, formed from the $\ce{^{79}Br}$ and $\ce{^{81}Br}$ isotopes.
Likewise, $\ce{^{158}Br2+}$ peak is formed from two $\ce{^{79}Br}$ isotopes and $\ce{^{162}Br2+}$ is formed from two $\ce{^{81}Br}$ isotopes.

However, I am confused by the explanation given by the book above.
Why is the probability of two different isotopes occurring in a $\ce{Br2}$ molecule twice that of the same isotope appearing in a $\ce{Br2}$ molecule?

All possible arrangements of $\ce{Br2}$ molecule:

$\displaystyle 79 + 79 = 158$
$\displaystyle \color{red}{79 + 81} = 160$
$\displaystyle \color{red}{81 + 79} = 160$
$\displaystyle 81 + 81 = 162$

The amount of $\ce{^{79}Br}$ and $\ce{^{81}Br}$ in nature is roughly the same, thus each permutation is equally probable. There are two arrangements that lead to $160$. While $158$ and $162$ each have only one arrangement. Therefore $160$ is twice as likely to be found compared to other masses.


Why is CuI2 not stable in solution while other copper(II) halide salts are? Why is $\ce{CuI2}$ (copper (II) iodide) not stable in solution while other copper(II) halide salts are? And how can you tell this from the redox potentials?

It is common knowledge that copper (II) iodide ($\ce{CuI2}$) does not exist. Thus, your question "why is $\ce{CuI2}$ not stable in solution?" is clear on that, but unlike this previous question, you wasn't clear about what happens (or what you have observed). I have no knowledge of what would happen when some one add $\ce{KI}$ solution to $\ce{CuSO4}$ solution. Yet, I can speculate what would happen by considering few redox half reactions:

$$\ce{I2 + 2e- <=> 2I-} \qquad E^\circ = \pu{0.536 V} \tag{1}$$
$$\ce{Cu^2+ + 2e- <=> Cu} \qquad E^\circ = \pu{0.342 V} \tag{2}$$
$$\ce{Cu^2+ + e- <=> Cu+} \qquad E^\circ = \pu{0.153 V} \tag{3}$$

For convenience, let;s rewrite equation $(1)$ in oxidation format:

$$\ce{2I- <=> I2 + 2e-} \qquad E^\circ = \pu{-0.536 V} \tag{4}$$

Sum of $(2)$ and $(4)$ would gives the appropriate redox equation:

$$\ce{Cu^2+ + 2I- <=> I2 + Cu_{(s)}} \qquad E_\mathrm{EMF}^\circ = \pu{-0.194 V} \tag{5}$$

Similarly, sum of $(3)\times 2$ and $(4)$ would gives the appropriate redox equation:

$$\ce{2Cu^2+ + 2I- <=> I2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-0.383 V} \tag{6}$$

The excess $\ce{I-}$ would react with $\ce{Cu+}$ and $\ce{CuI}$ would precipitate as a result:

$$\ce{Cu+ + I- -> CuI_{(s)}} \qquad K_\mathrm{sp} = 1 \times 10^{-12} \tag{7}$$

Therefore, albeit forward reactions of equations $(5)$ and $(6)$ are not thermodynamically favorable (e.g., for equation $(5)$, $K_\mathrm{eq} \approx 2.7 \times 10^{-7}$), both forward reactions persist and they would be driven to completion because of one product separate as a solid (Le chatelier principle).

Let's now consider oxidation half reactions of other three halides:

$$\ce{2F- <=> F2 + 2e-} \qquad E^\circ = \pu{-2.866 V} \tag{8}$$
$$\ce{2Cl- <=> Cl2 + 2e-} \qquad E^\circ = \pu{-1.358 V} \tag{9}$$
$$\ce{2Br- <=> Br2 + 2e-} \qquad E^\circ = \pu{-1.087 V} \tag{10}$$

If $\ce{Cu^2+ -> Cu+}$ reduction happens, the respective three redox reactions are:

$$\ce{2Cu^2+ + 2F- <=> F2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-2.713 V} \tag{11}$$
$$\ce{2Cu^2+ + 2Cl- <=> Cl2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-1.205 V} \tag{12}$$
$$\ce{2Cu^2+ + 2Br- <=> Br2 + 2Cu+} \qquad E_\mathrm{EMF}^\circ = \pu{-0.934 V} \tag{13}$$

And if $\ce{Cu^2+ -> Cu}$ reduction happens, the respective three redox reactions are:

$$\ce{Cu^2+ + 2F- <=> F2 + Cu} \qquad E_\mathrm{EMF}^\circ = \pu{-2.524 V} \tag{14}$$
$$\ce{Cu^2+ + 2Cl- <=> Cl2 + Cu} \qquad E_\mathrm{EMF}^\circ = \pu{-1.016 V} \tag{15}$$
$$\ce{Cu^2+ + 2Br- <=> Br2 + Cu} \qquad E_\mathrm{EMF}^\circ = \pu{-0.745 V} \tag{16}$$

Since values of $E_\mathrm{EMF}^\circ$s are so high negatives, these reactions are practically not happen. For example, $K_\mathrm{eq}$s are in the range of $6.14 \times 10^{-26}$ (for eq. $(16)$) and $1.55 \times 10^{-92}$ (for eq. $(11)$). Also keep in mind that, contrary to $\ce{CuI}$, $\ce{CuBr}$ and $\ce{CuCl}$ are also water soluble and $\ce{CuCl}$ is water soluble as well as hygroscopic.


Is dehydrating copper(II) sulphate pentahydrate a chemical or physical change? Is removing the waters of hydration from $\ce{CuSO4 * 5H2O}$ a chemical or physical change? Sorry for the trivial question. I get varying answers across the net.

The traditional answer is that it's a chemical change when bond-making and bond-breaking is involved.  The material certainly changes appearance, and heating copper sulphate somehow releases bound water.

That said, bonding in metal-aquo complexes runs the gamut from very weak to exceedingly strong - exchange rates span 15 orders of magnitude (Fig. 1 in Jan Reedijk, Metal-Ligand Exchange Kinetics in Platinum and Ruthenium Complexes. Platinum Metals Rev. 2008, 52, 2–11. DOI: 10.1595/147106708X255987), and where do you draw the line between a "physical" electrostatic interaction (in, say, $\ce{Na(OH_2)_6^+}$) and a chemical bond (as seen in $\ce{Cr(OH_2)_6^3+}$?


Is sodium percarbonate the safest and by weight the most efficient oxygen generator? I have been reading about $\ce{Mn}$-catalyzed reaction of sodium percarbonate and water to produce gaseous oxygen in medical emergency situations.

This seems safe due to its reagents and products, however I am wondering if there are other reagents that would be suitable and safe enough to fulfill above use.

I did read about lithium perchlorate being used in so-called oxygen candles and capable of a high ratio of oxygen per weight output, however they hardly seem as safe as the above reaction.

Careful, as sodium percarbonate (basically $\ce{3Na2CO3.2H2O2}$) is usually marketed as a chlorine-free bleach. The latter usually has an additive, TAED, which acts as a key activator. See, for example, Wikipedia on OxiClean, where not even listed on the label is TetraAcetylEthyleneDiamine, which is apparently commonly employed as a bleach activator in many laundry products, see this, for example.

TAED reacts with H2O2 in alkaline conditions (referred to as perhydrolysis) creating, in part, Peracetic Acid. The latter PAA is actually preferred over sluggish H2O2 for bleaching. In other words, it is not the H2O2 as the active bleaching agent, but PAA created in situ.

So, other than possibly misleadingly implying that eco-friendly hydrogen peroxide is the active agent, it is, in reality, PAA, a probematic compound with associated health concerns (see Marquand, E. C.; et al. , 2007. "Asthma Caused by Peracetic Acid-Hydrogen Peroxide Mixture". J. Occup. Health. 49 (2): 155–158.). Here is an MSDS on an actual PAA mix, ( 76-61% water, PAA 20-35%, 3% Acetic acid and 1% H2O2), which is far from friendly but does eventually totally decompose into harmless O2, CO2, H2O,...

As such, anyone using a commercial percarbonate product should at least be aware of potential safety issues that one may likely be employing a Peracetic acid mix.

Note, there is an indicator that you do not have pure sodium percarbonate. It is the presence of some dire warning labels.


Thermodynamics of a spontaneous isothermal chemical reaction. Which of the following is/are correct for spontaneous isothermal
  chemical reaction?

  (A) $\Delta H = 0$,  because $\Delta T = 0$

  (B) $\Delta S = 0$

  (C) $\Delta U = 0$,  because $\Delta T = 0$

  (D) $\Delta G < 0$

My answer: $\mathrm{A, C, D}$

Reasoning:

$U$ is a state function and only depends on the initial and final states and their temperature. Since $\Delta T = 0$, $\Delta U = 0$

$\Delta H = \Delta U + \Delta(PV)$, but $PV = \text{constant}$, so $\Delta H = 0$

$\Delta G <0$ - the requirement for any reaction to be spontaneous.

But answer given is only option D.

I think the flaw in my $\Delta H = 0$  argument is that reactants can be solids and liquids as well and we cannot apply $PV = \text{constant}$ to them. Am I right?

But I  still can't figure out why $\mathrm C$ is wrong.

The internal energy and enthalpy of a pure ideal gas depends only on temperature.  But, the internal energy and enthalpy of an ideal gas mixture depends both on temperature and on the amount of each chemical species that is present.  To be more precise, it is equal to the internal energy and enthalpy per mole of each component as if it were a pure species times the number of moles of that species present, summed over all the species in the mixture:  $$U=\sum{n_iu_i}$$and$$H=\sum{n_ih_i}$$So, even at constant temperature, $$\Delta U=\sum{(\Delta n_i)u_i}$$and$$\Delta H=\sum{(\Delta n_i)h_i}$$You should have been taught this when they were covering the thermodynamics of mixtures and chemical reactions.


Method of allylic and benzylic oxidation I am told that $\ce{MnO_2}$ oxidizes allylic and benzylic alcohols. However, what is the mechanism of this oxidation? All the professor had to offer me was that the metal ion, manganese 2+, chelates the substrate; there is some sort of interaction between a pi bond, the metal ion, and the electrophilic carbon atom bearing the hydroxyl group. No further explanation was given, and a Google search hasn't unearthed anything of use. What, therefore, is the mechanism of this oxidation? I would like to know more about the mechanism to better understand why $\ce{MnO_2}$ only touches allylic and benzylic alcohols rather than any vanilla alcohol.

This is one of those heterogenous reactions occurring on a surface which are  mechanistically messy. As expected for a reaction that occurs on the surface of a reagent, factors such as surface area and reagent preparation (freshness, surface contamination) are significant variables.  A free radical mechanism is generally accepted and a reasonable mechanism is posted below.

image source

When an allylic alcohol with stereochemistry (cis, trans) around the double bond is oxidized, the stereochemistry around the double is preserved.

image source

This suggests that we are not really dealing with a free, long-lived radical intermediate, but rather a free radical held in close association with the manganese ion.


Are amines amphoteric? An amine can react with an acid just like any other base to form a salt:

$$\ce{RNH2 + HCl-> RNH3+ Cl-}$$

According to my textbook (unreliable), amines also react with bases just like any other acid:

$$\ce{RNH2 + NaOH -> RNH- Na+}$$

Do amines (if not all amines, at least the ones substituted with a strong withdrawing group) react with bases as shown in the above reaction?

In  other words: Are amines amphoteric?

Amines do not react with hydroxide to give amide ions - not significantly at least.

The typical $\mathrm pK_\mathrm a$ of an amine is $\sim 30$ (see Evans pKa table) and that of $\ce{H2O}$ is $\sim 15$. The equilibrium constant for the second equation you wrote is therefore on the order of $10^{-15}$.

In order to generate any amide ions you need a much stronger base than simply hydroxide. A typical choice is an alkyllithium (LDA is commonly made by reacting diisopropylamine with BuLi).

Back to the question whether amines are amphoteric: If your definition of an amphoteric compound is simply "can react either as an acid or a base", then yes, it technically is, since it reacts both with acids ($\ce{HCl}$) or bases ($\ce{RLi}$).

However, it may be useful to draw the line at reacting with aqueous acid and aqueous base. Otherwise, nearly everything with a hydrogen can be considered to be amphoteric (for example, $\ce{HF}$ can technically act as a base to form $\ce{H2F+}$). Under this definition amines are not amphoteric.


Is dew and water vapour the same? Is it possible to say water vapour and dew is the same? I know that the water vapour is the gaseous phase of water. But what is the dew? I just wonder how can I categorised it. Is it the liquid  or the gas phase of the water.

Figure below shows dew formed in grass. It seems as a liquid. But is it actually  exhibit in liquid phase or gas phase?

Let's have a look at the definitions of dew:

  dew (the free dictionary)

  Water droplets condensed from the air, usually at night, onto cool surfaces.
  Something moist, fresh, pure, or renewing: "The timely dew of sleep / ... inclines / Our eye-lids" (John Milton).
  Moisture, as in the form of tears or perspiration, that appears in small drops.

Therefore dew is water in the condensed phase, i.e. liquid. Water vapour on the other hand is the gaseous phase. They are definitely not the same.


Why is lithium bromide used in absorption refrigerators? Lithium bromide refrigerator

Why is lithium bromide used in absorption refrigerators?  It is more expensive than sodium chloride.  The wikipedia article on lithium bromide says it is extremely hygroscopic.  However, it is less soluble in water than NaCl.  I am thinking that a salt capable of generating a more concentrated solution would work better in the absorption refrigerator as it would increase the propensity of water to leave the gas phase and enter the solution.  Am I wrong about that?  Is the affinity of water for lithium bromide so high that it is the most important factor?

Your summary is correct; $\ce{NaCl}$ is more soluble than $\ce{LiCl}$ or $\ce{CaCl2}$, for that matter, but both $\ce{LiCl}$ or $\ce{CaCl2}$ are far  more hygroscopic. The explanation of hygroscopicity involves various mechanisms, including adsorption (surface forces) and absorption (requiring transport to the interior of the substance) due to hydrogen bonding, coordinate covalent bonding etc.

So these are distinct, though related, properties: solubility (limit on how much solute the solvent holds) and hygroscopicity (how avidly the solute attaches to the initial molecules of solvent).

For that reason, $\ce{CaCl2}$ is used to absorb water in damp closets, though it will eventually deliquesce into a sodden mess.

BTW, Isaac Asimov wrote of a compound so hygroscopic that it dissolved before it hit water: The Endochronic Properties of Resublimated Thiotimoline, often cited.


Why is Q/nF equivalent to the amount of substance in electrolysis? I am told that

$$Q = nF$$

where $Q$ is the quantity of current, $n$ is the amount of electrons and $F$ is Faraday's constant. Since Faraday's constant is the charge carried by 1 mol of electrons, multiplying it by the amount of electrons to give the total charge makes perfect sense. But why does dividing $Q$ by $nF$ gives the amount of substance? Won't the answer just be 1 for all possible values?

For a general electrolytic reaction,

$$\ce{M^{n+} + $n$ e^- -> M}$$

where $n$ moles of electrons are consumed in depositing 1 mole of substance ($n$ is also known as $n$-factor). So, for $x$ moles of substance, $nx$ moles of electrons would be consumed. Therefore,

$$Q = nxF,$$

dividing by $nF$ on both sides, we have

$$\frac{Q}{nF} = x,$$

which is the amount of substance.


Dipole moment of molecule (isolated vs in solution) Would you expect a dipole moment determined in solution to be different from that in an isolated molecule? If so, why?

I've actually been investigating this in my comp chem coursework. A large dielectric can support more charge separation and therefore stronger dipoles, but the effect diminishes. Going from an $\epsilon$ of 2 (cyclohexane) to 9 (DCM) gives a much bigger change in dipole than going from 9 to 80+ (water).


Solubility of potassium silicate in water It is somewhat surprising to me that there is almost no data about solubility of potassium silicate in water under normal conditions. The only thing I was able to find is the following:

<0.336 g/L @ 25ΕC, increasing  to 300 g/L @ 80ΕC. Once  dissolved,
material will not  precipitate out
Technical Document for Potassium silicate also referred to as a
BRAD

However, it is not clear what does the last statement mean. Does it mean that the silicate will not precipitate even if the solution cools down to 25°C?
The solubility of sodium silicate can easily be found on Wikipedia. It is mystery in case of potassium silicate.
So let us assume we want to dissovle potassium silicate in water with initial pH of 7. What amount of potassium silicate can be dissolved?
Edit:
Regarding the comment of @M. Farooq about being more specific. The second sentence on Wikipedia says:

The most common potassium silicate has the formula K2SiO3, samples of
which contain varying amounts of water.

We can consider the anhydrous form of K2SiO3 and couple of most common hydrates. Also, if the exact amount is not known then at least knowing the order of magnitude is fine as there is huge difference between 0.336g/L and 300g/L.

Potassium silicate is synthesized by melting $\ce{KOH}$ above $400$°C, and adding some solid $\ce{SiO2}$. Its composition depends on the amount of silica used. The dissolution process does not look so. But it is a slow chemical reaction with a rather viscous liquid. The result can be cooled down and gives a solid substance practically insoluble in water at room temperature.
However it is possible to dissolve it with boiling water under pressure. Surprisingly enough, the obtained potassium silicate becomes more easily dissolved at $T>80$°C under pressure. The concentrated solution so obtained can be cooled down. It yields an extremely viscous liquid, more viscous than honey. This liquid is usually not really transparent, but it can be diluted with pure water without producing a precipitate. Such solutions are probably colloidal solutions of $\ce{H2SiO3}$ or $\ce{H4SiO4}$ in a concentrated solution of $\ce{KOH}$. As a matter of fact, if such a solution is heavily diluted, neutralized to destroy KOH, washed and evaporated in a vacuum, hydrated silica is obtained and it is made of $\ce{H2SiO3}$ flakes which are extremely light : the density of the stuff is not much bigger that air, as if it was made of rather long chains $\ce{HO -[- Si(OH)2 - O -]_n-H}$ sometimes directed in all directions around a central Silicium atom.
The chemistry of the silicic acid and its salts is extremely complex. The ternary diagram $\ce{Na2O - SiO2 - H2O}$ contains $11$ different phases.
Ref.: J. G. Vial, Soluble silicates, Reinhold, New York, 1952.
N. Greenwood, A. Earnshaw, Chemistry of the Elements, Pergamon, 1986, § 9.3.4. p. 398


Is there any difference between getting dissolved and getting dissociated? In the context of electrolytes only, is there any difference between electrolyte getting dissolved and electrolyte getting dissociated?

According to my professor, they are different. He says a substance may dissolve completely but it may not dissociate completely and a substance may not dissolve completely but amount that dissolves can dissociate completely and, thus, substances which dissolve (either wholly or partially), amount that gets dissolved if dissociate completely are called strong electrolytes and amount of substances which gets dissolved if dissociates sparingly are called weak electrolytes (although it may dissolve completely).

How is it possible? I mean, how can solubility be different from dissociating?

If something dissociates it means it’s dissolving and in the context of electrolytes if something dissolves in water it has to dissociate into ions and even Arrhenius’ theory of electrolytic dissociation sys that substance which are water soluble, dissociates into ions and get dispersed in water.

And if someone says they are different to them, I ask the following two questions:

If solubility and dissociation are different then what would be the criteria for getting dissolved and what will be criteria for getting dissociated (for me the criteria for getting dissolved is dissociation).
how would you arrange the following in increasing order of 1.solubility
2.dissociation

NaCl, MgCl2, FeCl2, ZnCl2

This is a simple logical and semantic problem, which is not a problem at all. Your professor is right and wrong- both at the same time. He is creating a classification which does not exist and which is meaningless.

Look at the word origin of electrolyte:

  Etymology from OED: < electro- comb. form + ancient Greek λυτός that may be
  dissolved, soluble (see -lyte comb. form). Compare electrolysis n.,
  electrolyse v.

  and the portion -lyte "< ancient Greek -λυτος, combining form (in e.g.
  εὐδιάλυτος : see eudialyte n.) of λυτός that may be dissolved, soluble
  < λύειν to loose (see lysis n.) + -τός, suffix forming verbal
  adjectives. Compare French -lyte, German -lyt.

  A substance which gives rise to ions when dissolved (typically in
  water) or fused; a liquid or gel which contains ions and can be
  decomposed by electrolysis; spec. the conducting fluid used in a
  battery.

Does this solve the semantics problem?

By definition, an electrolyte is soluble in a given solvent. It is also understood among electrochemists that an electrolyte is added to a solvent to impart conductivity to that solvent. I will never call sugar an electrolyte because nobody would use sugar to increase conductivity of water. You should ask: Does a given substance produce ions in a solution? If yes, it can be called as an electrolyte.

Look at this logic: grapes is a fruit but not all fruits are grapes.

Electrolytes are soluble in the solvent, but not all soluble substances in that solvent behave as electrolytes. Anything which produces ions in that solvent, can be labelled as an electrolyte.

Q-1 If solubility and dissociation are different then what would be the criteria for getting dissolved and what will be criteria for getting dissociated (for me the criteria for getting dissolved is dissociation).

The measure of dissociation is conductivity. How much electrical resistance is offered by that dissolved (and dissociated) substance in that solvent

Q-2 how would you arrange the following in increasing order of 1.solubility 2.dissociation

NaCl, MgCl2, FeCl2, ZnCl2

There is no correlation between solubility and dissociation.  If I want to study their dissociation I will measure the conductivity of their solutions systematically.


What's wool made of? For example, silk is made of fibroin. But what's wool made of?

Unlike silk or cotton, wool does not have a homogeneous structure. Its core is considered to be made of keratin, a generic type of protein found in human hair and epidermis. A detailed PDF on Merino wool can be downloaded from here.

Thanks for encouraging me, through your question, to read that PDF, which explains both the chemical and physical properties in a straight-forward fashion.


Why don't the electrons move through the electrolyte (instead of the circuit) in a galvanic cell? I was learning about galvanic cells and I had a problem understanding why electrons do not travel through the electrolyte solutions themselves, instead preferring to travel through metals. Can electrons travel through an electrolytic solution?

Imagine a galvanic cell, without the wiring between the two electrodes and instead we only have the salt bridge. Won't the electrons (although it is very unfavorable to happen) travel through the salt bridge and the solutions? In other terms will the transformation of the electrodes occur?

Not in water. Free electron in water is really unfavorable, so no significant concentration of them can be generated chemically, and it almost immediately reduces water itself to hydrogen (but I heard rumors about generation of solvated electrons in water in very special experiment with short half-life)

In liquid $\ce{NH_3}$, however, solvated electrons can occur, so self-discharge of galvanic cells with $\ce{NH_3}$-based electrolyte may occur through travel of electrons via electrolyte.


Why is the concentration of protons equal to the concentration of a strong acid? First, I know that strong acids are assumed to dissociate completely in water, so that the concentration of $\ce{H+}$ which they provide equals their concentration.

What I am asking for is not the concentration of $\ce{H+}$ which they provide, but $\ce{[H+]}$ for the solution, since the $\ce{H+}$ ions that they provide will not stay aloof, but rather will react with water and with $\ce{OH-}$ ions that are produced in the self-ionization of water.

Won't the reaction of $\ce{H+}$ ions with water shift the self-ionization of water
$$\ce{H2O <=> H+ + OH-}$$
to the left and won't their reaction with $\ce{OH-}$ shift it to the right according to Le Châtelier’s principle?

So, what will $\ce{[H+]}$ be for the solution?

The statement $[\ce{H^+}] = [\ce{A^-}]$ is an approximation for not too diluted acids. However you can pretty fast tell that this won't hold if you add a strong acid to water so that the acid has a concentration of let's say $\pu{10^-8 mol//l}$. Via the formula $\mathrm{pH} = \log\left([\ce{H^+}]\right)$ this would result in $\mathrm{pH} = 8$ so if that was true, you just made the solution less acidic by adding an acid.

In fact the total concentration $[\ce{H^+}]$ is given by $[\ce{H^+}] = [\ce{A^-}] + \pu{10^-7 mol//l}$ where the last term stems from the autoionization of water. However in most use-cases $[\ce{A^-}]$ is far greater than $10^{-7}$. In those cases the autoionization of water can be neglected and thus the approximation $[\ce{H^+}] = [\ce{A^-}]$ gives sufficient results.
However, if you are working with really diluted solutions of acids this approximation is no longer applicable (as demonstrated above) which forces you to actually respect the autoionization of water.

If you want to be on the safe side you can always respect the autoionization. This will always give you the correct pH (even for not so diluted solutions) but you may neglect it if the acid-concentration is high enough (this of course depends on the accuracy of the results you need).

EDIT:
As @MaxW pointed out in the comment a general expression would be $[\ce{H+}] = [\ce{A-}] + [\ce{OH-}]$ which for a neutral solution at 25°C (thanks to @Martin for pointing this out) becomes the formula proved above.


Formation of toluene from (cyclobutane-1,2,3-triyl)trimethanol using conc. sulphuric acid In my book this reaction is given without any mechanism:

I thought that simple dehydration reaction is going to take place with rearrangement of the carbocation but I have no idea as to how toluene turned out to be the product.

Seeing the reaction given here, I assume that is an on-paper reaction that takes place and that all three alcoholic groups are dehydrated.

The mechanism (on paper) seems to be as follows
Dehydration $1$
For the first dehydration, the carbocation formed is stabilised via ring expansion and then forms a double bond as follows:

Dehydration $2$
Similar to the first dehydration, a ring expansion occurs to stabilize the carbocation by forming a 6-member ring:

Dehydration $3$
In the third dehydration, a hydride shift takes place concomitantly forming a tertiary carbocation stabilized by conjugation with the double bonds. The double bond is then formed giving the compound aromaticity.


Evaluating (∂U/∂T)P - (∂U/∂T)V for an ideal and van der Waals gas What is the value of $$\left(\frac{\partial U}{\partial T}\right)_{\!P} - \left(\frac{\partial U}{\partial T}\right)_{\!V}$$ for an ideal gas and for a van der Waals gas?

($U$ refers to the molar internal energy.)

Here I am going to use $U$ to refer to the total internal energy. If you want the molar internal energy ($U_\mathrm m \equiv U/n$) just divide your final result by $n$.

In another of my answers I have proven (equation $(14)$) that

$$\left(\frac{\partial U}{\partial T}\right)_{\!p} = \left(\frac{\partial U}{\partial V}\right)_{\!T}\left(\frac{\partial V}{\partial T}\right)_{\!p} + \left(\frac{\partial U}{\partial T}\right)_{\!V}$$

The proof up until there is completely general, so this applies for both an ideal and a van der Waals gas.

Furthermore here I have shown that (combining equations $(6)$ and $(8)$)

$$\left(\frac{\partial U}{\partial V}\right)_{\!T} = T\left(\frac{\partial p}{\partial T}\right)_{\!V} - p$$

Putting these together gives you

$$\begin{align}
\left(\frac{\partial U}{\partial T}\right)_{\!p} - \left(\frac{\partial U}{\partial T}\right)_{\!V} &= \left(\frac{\partial V}{\partial T}\right)_{\!p}\left[ T\left(\frac{\partial p}{\partial T}\right)_{\!V} - p\right]
\end{align}$$

All that's left is to evaluate the partial derivatives. The ideal gas case is very easy; since $p = nRT/V$, $(\partial p/\partial T)_V = nR/V$ and the term in square brackets goes to zero.

For the van der Waals gas you need to work at those partial derivatives which are a bit tedious, but should be doable, especially if you employ implicit differentiation. For example, if you take the van der Waals equation of state

$$\left(p + \frac{an^2}{V^2}\right)(V - nb) = nRT$$

and differentiate with respect to $T$ throughout, keeping $V$ constant, one gets

$$\begin{align}
\left[\left(\frac{\partial p}{\partial T}\right)_{\!V} + 0\right](V - nb) + \left(p + \frac{an^2}{V^2}\right)(0) &= nR \\
\left(\frac{\partial p}{\partial T}\right)_{\!V} &= \frac{nR}{V - nb}
\end{align}$$

and likewise, barring any careless mistakes, I obtained

$$\left(\frac{\partial V}{\partial T}\right)_{\!p} = nR\left(p - \frac{an^2}{V^2} + \frac{2abn^3}{V^3}\right)^{-1}$$

Maybe somebody has a less tedious way (and I hope so), but this is all I can offer you myself.


Why is there sulfur in black powder / gun powder? Black powder is composed of charcoal, sulfur and a nitrate salt. The charcoal and sulfur serve as fuel and the nitrate is the oxidizing agent.
However, wouldn't it be cheaper to leave out the sulfur, thus having only charcoal as fuel?
Wikipedia says about this (emphasis added):

Black powder is a granular mixture of

[...]
sulfur (S), which, while also serving as a fuel, lowers the temperature required to ignite the mixture, thereby increasing the rate of combustion.

I guess after this initial research my question boils down to: Why is sulfur required as an ingredient in gun powder, and how does it lower the ignition temperature of the mixture?

Black powder is a mixture of solids. As solids, they are not particularly inclined toward fast reactions between each other. Elevated temperature generally liven things up, because it allows diffusion of solids into each other. However, both carbon (charcoal) and potassium nitrate are relatively stable compounds, so even when potassium nitrate is melted, they still do not react immediately,  further heating is required.

Sulfur, on the other side, is quite easy to melt. Nitrates are not very powerful oxidizers, but potassium nitrate in particular is a low-melting compound. And at the moment both potassium nitrate and sulfur are molten, their interdiffusion becomes quite easy, so they ignite, and charcoal follows.

Summing it up: charcoal is almost impossible to melt and potassium nitrate is a relatively weak oxidizer and also hard to melt, so the mixture is hard to ignite. Sulfur is easily flammable, easy to melt, so it is much easier to react with potassium nitrate solid, and its presence makes mixture much more ignitable.


Have Ionic Liquids been used to refine aluminum on an industrial scale? Ionic liquids have been shown capable of reducing aluminum trichloride to solid aluminum at 3kw-hrs per kg of Al(s) produced instead of the current ~15kw-hr/kg. However, this would seem to neglect the fact that bauxite is aluminum trioxide not trichloride. This is no small detail and leads me to wonder if it is an insurmountable hurdle to the widescale adoption. Has this technology been implemented (or slated for implementation) on an industrial scale? If not, what holds it back?

I think you are misunderstanding the article linked in the question.

In the process described in the article, the starting material is A360 aluminum metal alloy.

The goal is to make very pure aluminum metal from impure aluminum metal.

The process is an electrolysis process.

$$\ce{Al(anode) + 7AlCl_4^{−} -> 4Al2Cl_7^{−} + 3e-}$$

$$\ce{4Al2Cl_7^{−} + 3e- -> Al(cathode) + 7AlCl_4^{−}}$$

The existing industry techniques referred to by the article as requiring ~15kw-hr/kg are also for refining impure aluminum metal to yield high purity aluminum.  This number is not referring to purification from bauxite.  The "three layer process" is one of the existing refining techniques, and it does involve ionic liquids, but the electrolyte is liquid only at high temperature.


How does radiation facilitate the formation of tetravalent iron? In reading the article abstract What Oxidation State of Iron Determines the Amethyst Colour?, the author states that in regards to the gemstone amethyst, from clear quartz:

  The crystal was transformed into violet amethyst by gamma-irradiation. The change in colour was accompanied by changes in the Mössbauer spectrum that can be interpreted as the conversion of trivalent iron into the tetravalent state: $\ce{Fe^{3+}→Fe^{4+}}$

How does gamma radiation facilitate the conversion of oxidation state of iron, from $\ce{Fe^{3+}}$ to $\ce{Fe^{4+}}$?

The ionization energy of $\ce{Fe^3+}$ to $\ce{Fe^4+}$ is 54.8 eV.
The energy of typical gamma radiation is much higher (in the order of roughly some keV to a few MeV). Depending on the energy, the interaction of average gamma radiation with the affected electron is mainly due to Compton scattering.


Can assumptions made during calculation of ka of buffer solutions be justified? We make many assumptions in sciences which are not always true but as they make little difference we ignore them for simplicity. But today I have decided that I can no longer be lured into using expression of $K_{a}$ and $K_{w}$ by assumptions which I deem to make a significant difference.

We say that $[H_{2}O]$ is constant. Why I think it is not justified?
$$\ce{H2O + HCL <=> H3O+ + Cl-}$$
$[H_{2}O]$ is almost $ 55.6 \:\mathrm{mol\:dm^-3}{}$.
At around $20^{o}C$ $20\:\mathrm{mol}$ of $HCl_{aq}$ can be dissolved. Let's say we dissolve $20 \:\mathrm{mol}$ of $HCl$ and try to find it's WE ASSUME that % disassociation of water is $100 \%$. So we must take $[HCl]\approx0$ that means that $20\:\mathrm{mol}$ of $H_{2}O$ has reacted and $[H_{2}O]$ is no longer a constant but has rather changed to $35.6\mathrm{mol}$.
So this expression can no longer be used. $$K_{a}=\frac{[H_{3}O^{+}].[Cl^{-}]} {[HCl]}=K_{c}.[H_{2}O]$$
Note: you might point out that $K_{a}$ is not used for strong acids but I just used $HCl$ as an example but make my point easier to explain.
Assumption made about concentrations when calculating $Ka$ of buffer solutions.
$$CH_{3}COOH_{aq} \rightleftharpoons CH_{3}COO^{-}_{aq} + H^{+}_{aq}$$
We have added $CH_{3}COONa$  to make a buffer solution.
$$CH_{3}COONa_{aq} \rightarrow CH_{3}COO^{-}_{aq} + Na^{+}_{aq}$$
So,
$$[CH_{3}COO^{-}]_{total}=[CH_{3}COO^{-}]_{CH_{3}COONa} + [CH_{3}COO^{-}]_{CH_{3}COOH}$$
$$[CH_{3}COO^{-}]_{CH_{3}COOH} = [H^{+}_{aq}]$$
BUT WE ASSUME that $[CH_{3}COO^{-}]_{total}=[CH_{3}COO^{-}]_{CH_{3}COONa}$ because we say $[CH_{3}COO^{-}]_{CH_{3}COOH}$ is negligible yet we include [$H_{aq}^{+}]$ in the expression which has same value as $[CH_{3}COO^{-}]_{CH_{3}COOH}$. Same value is considered negligible at one place and not negligible at another in the same expression. How is that justified?

Answering your first question, $[\ce {H2O}]$ is constant because the square brackets indicate a concentration. Concentration is the number of moles of the solute on the volume of the solvent.

Except in this case, water is the solvent. So the concentration will always be constant, since the volume is proportional to the number of moles.

Besides, while dealing with the law of mass action, for liquids we don't take concentration, we take activity. Which is approximately one for water.

As for your second question, we basically have the following product:

$$\ce{([CH3COO- ]_{CH3COOH} + [CH3COO- ]_{CH3COONa})[H+ ]}$$

which can be rewritten as $(a + b)b$, where $b << a$. Let's say $a$ was $1$ and $b$ was $0.0001$.

The actual value of the expression is $1.0001\times 10^{-4}$. Neglecting only the first $b$ will give $1\times 10^{-4}$, which is not too much different.

Neglecting only the second $b$ will give $1.0001$, and neglecting both will give $1$. Both of these answers are very wrong.

When a small number is added to a large one, the small number can be ignored. But not when it is multiplied.


Is CH₃CH₂CH₂Cl saturated or unsaturated? I believe the molecule $\ce{CH3CH2CH2Cl}$ to be saturated, since there are only single bonds between the carbon atoms, but my workbook says it's unsaturated. Why?

The workbook must be wrong, it is a saturated molecule.


The maximum no. of isomers (including stereoisomers) that are possible on mono-chlorination of the compound: 3,3-dimethylpentane is? The maximum no. of isomers (including stereoisomers) that are possible on mono-chlorination of the compound: 3,3-dimethylpentane is?
 The answer given is 5.

My working: The fourth carbon is chiral when chlorine is added to it, so it'd have an enantiomer, and there are two other carbons where chlorine can be substituted, so how is the answer 5?

$\ce{CH3-CH2-C(CH3)2-CH2-CH2Cl}$,
$\ce{CH3-CHCl-C(CH3)2-CH2-CH3}$ (this is chiral and would have an enantiomer),
$\ce{CH3-CH2-C(CH2Cl)(CH3)-CH2CH3}$

are the compounds I could make out.


Given advancements of computational power an machine learning, how is it still not possible to predict products from chemical reactions? Given that it seems every answer to this question is that it is "impossible to predict the outcome of a chemical reaction." Is chemistry just trial and error? Given how fundamental and revolutionary a simulator capable of predicting possible outcomes of a reaction would be, why has it not been done yet? Do you think it will ever happen?

It's not an all-or-nothing situation, where either we can reliably prediction reaction products, or we have not tried to.

For instance, drug companies, for many years, have been making extensive use of increasingly powerful, sophisticated, and refined in silico predictive models to choose likely candidates for in vitro testing.  And I would be surprised if they've not also begun to develop, evaluate, and incorporate ML.

At the same time, these models are not perfect.  So the answer is that it is being done, it's been done for a while, we're getting better and better at it, but we still have a long way to go.

It will be interesting to see what happens with quantum computing, which is currently in its infancy.


Rate of exothermic/endothermic reactions Is it necessary that all endothermic reactions be slow and all exothermic reactions fast? This is based on the fact that endothermic reactions require more activation energy than exothermic reactions to initiate a chemical reaction. Or  are there some endothermic reactions that can be fast without the presence of a catalyst?

A reaction being endothermic or exothermic depends on thermochemistry while the rate of a reaction depends on its chemical kinetics. Hence, there is no relationship between the activation energy of a reaction and its enthalpy change. Also there is no relationship between the change enthalpy and the rate of a reaction. A rate of reaction is dependent on the concentration of the reactants which are to a certain power which is multiplied by a certain constant, k, which is called the reaction rate constant. For example, for the general reaction: $$\ce{A + B -> C}$$ the rate of the reaction is given by the following equation: $$ rate = k[A]^a[B]^b$$
The reaction rate constant is not dependent on whether a reaction is exothermic or endothermic. However is does depend on the temperature and and also the activation energy of the reaction, given by the equation: $$k = Ae^\frac{-E_a}{RT}$$ So you are right to say that a reaction will be slower if it has a higher activation energy, but you are wrong to say that endothermic reactions have a higher activation energy than exothermic reactions.

There are actually several endothermic reactions that are fast, for example, dissolving common table salt (NaCl) occurs pretty quickly, but it is actually a endothermic reaction. Meanwhile, the spontaneously decomposition of diamond to graphite is actually a exothermic reaction, however it is extremely slow. It is so slow that it not even observed after several decades.


What does the "h" "k" and "(5.2.2.02,6)" mean in Dibenzo(h,k)-4-oxatricyclo(5.2.2.02,6)undec-8,10- diene-3, 5-dione I am preparing to do an experiment that involves synthesizing Dibenzo(h,k)-4-oxatricyclo(5.2.2.02,6)undec-8,10- diene-3, 5-dione via a Diels-Alder reaction.

Could someone please explain the nomenclature for this molecule? Specifically, what the "h", "k" and (5.2.2.02,6) mean?

Below is the reaction:

The systematic (but not necessarily preferred) name tricyclo[5.2.2.02,6]undecane describes the structure of octahydro-1H-4,7-ethanoindene (see this answer on how to name tricyclo compounds):

Expanding the name from tricyclo[5.2.2.02,6]undecane to 4-oxatricyclo[5.2.2.02,6]undec-8,10-diene-3,5-dione adds the functional groups, which leads to the structure of 3a,4,7,7a-tetrahydro-4,7-etheno-2-benzofuran-1,3-dione:

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), isomers of fused polycyclic ring systems are distinguished by lettering each peripheral side of the parent component using the italic letters a, b, c, etc., beginning with a for the side numbered ‘1,2’, b for ‘2,3’, etc.
For example:

Accordingly, fusion of two benzene rings to the above-mentioned parent structure 4-oxatricyclo[5.2.2.02,6]undec-8,10-diene-3,5-dione at the relevant sides yields the name dibenzo[h,k]-4-oxatricyclo[5.2.2.02,6]undec-8,10-diene-3,5-dione.
(Note that this name does not necessarily correspond to the preferred IUPAC nomenclature and notation.)


Is steric inhibition of resonance or steric inhibition of protonation dominant in o-toluidine? I wish to know which effect out of steric inhibition of resonance (SIR) or steric inhibition of protonation (SIP) is dominant over the other when comparing basicities of o-toluidine and aniline:

Case 1, using only SIR effect
We see that due the bulky methyl present on o-toluidine there is inhibition of resonance which increases the availability of nitrogen's lone pair. Hence, we can say o-toluidine is more basic than aniline (as nitrogen's lone pairs are in resonance with benzene).

Case 2, using only SIP effect
If we consider that methyl's bulkiness shields the lone pair of nitrogen from protonation in o-toluidine and no such inhibition or shielding happens in aniline, we can conclude that aniline is more basic than o-toluidine.

Both these cases are contradictory. Can anyone please resolve the discrepancy?

The reason for the decreased basicity is essentially the ortho-effect as outlined here: Ortho-effect in substituted aromatic acids and bases.

Your question on the other hand asks about a seeming contradiction, but I believe there is a small misconception on your part.

  We see that due the bulky methyl present on o-toluidine there is inhibition of resonance which increases the availability of nitrogen's lone pair. [...]

I don't see how resonance could be inhibited, as that essentially requires that the lone pair orbital of nitrogen would have to rotate significantly towards an in-plane arrangement. The methyl group is simply not big enough for that to happen, and if it were, the lone pair would more likely rotate toward the more shielded side, further reducing acidity.

The highest occupied molecular orbitals (HOMO) of the molecules show this quite nicely, as they are very, very similar.

On the DF-B97D3/def2-SVP level of theory the RMSD of the core structure (excluding the hydrogen and the methyl group) is negligible with only 0.03 Å.

From this I would conclude that there is no steric inhibition of resonance.

On the other hand, steric inhibition of protonation is given in 2-methyl aniline, where it is obviously missing in aniline.
Additionally there is a stabilising non-covalent interaction between the methyl and amine group, which can be analysed with NCIPLOT, and then displayed. The green blob is weakly attractive, red is strongly repulsive, blue would indicate strongly attractive interactions.

TL;DR: The dominant effect is steric inhibition of protonation, as steric inhibition of resonance is negligible (if possible at all).


How to convert sodium aluminate back to sodium hydroxide? I had dissolved a bunch of aluminium foil in an aqeous solution of sodium hydroxide. The resulting solution consists mainly of sodium aluminate with very little excess sodium hydroxide.
What I want to do is convert this back to sodium hydroxide. I am thinking electrolysis should do the trick. But, I am not completely sure if this is right. If it is, what kind of electrodes should I use? If not, why not? (What will be the product?)
Thank you for your time.

I don't think you will have much luck recovering the hydroxide.  To do so you would need to reverse the reaction that happened when the aluminum dissolved, which was

$$\ce{2 Al + 2 NaOH + 2 H2O -> 2 NaAlO2 + 3 H2}$$

Your idea is essentially to substitute reduction at an electrode for reduction with hydrogen gas.  Unfortunately electrochemical reduction of $\ce{Al(III)}$ is very difficult.  If your idea was feasible, it would revolutionize aluminum manufacture!  Instead of doing electrolysis of molten cryolite at ~1000 degress with sacraficial carbon electrodes, aluminum manufacturers could use your process.  So if you could get it to work, you'd be rich...but I am not hopeful.

Note however that you don't necessarily need to do electrolysis to recover sodium hydroxide.    There is hope!  You could just boil down your solution so that it is supersatured with respect to aluminum hydroxide, and then wait for the aluminum hydroxide to precipitate, which would happen faster if you add seed crystals of aluminum hydroxide $\ce{Al(OH)3}$) to initiate precipitation.

$$\ce{2H2O + NaAlO2_{(aq)} -> Al(OH)3_{(s)} + NaOH_{(aq)}}$$

This step is part of the Bayer process for commercial refining of aluminum ore.


Major product of the reaction of m-hydroxyphenol in boiling aq NaHCO3 The reaction: (a) m-hydroxyphenol + aq. $\ce{NaHCO3}$/boil; (b) workup with $\ce{H3O+/H2O}$

I know that $\ce{CO2}$ will be formed by boiling aq. $\ce{NaHCO3}$, and so the reaction will be like Kolbe-Schmitt type, i.e. $\ce{COOH}$ group will be attached on the benzene ring.

But I have a doubt that where will the addition of $\ce{COOH}$ group take place,

  (1) ortho-position w.r.t. both hydroxy groups
  $$\text{or}$$

  (2) ortho-position w.r.t. one hydroxy group and para-position w.r.t. the other group.

So, Which will be the major product and why?

The major product (proportions dependent on conditions) is option 2 (beta-resorcylic acid) according to these papers$\ce{^{[1]} ^{[2]}}$. The relatively low proportion of option 1 (gamma-resorcylic acid) is attributed to steric hindrance$\ce{^{[3]}}$.

References

The carboxylation of resorcinol and the separation of β-and γ-resorcylic acid by ion-exchange chromatography by
D. K. Hale,  Audrey R. Hawdon, J. Idris Jones  and  D. I. Packham. DOI: https://pubs.rsc.org/en/content/articlelanding/1952/JR/JR9520003503#!divCitation
The chemistry of resorcinol carboxylation and its possible application to the CO2 removal from exhaust gases by Vincenzo Barbarossa, Francesco Barzagli, Fabrizio Mani, Sarah LaicGiuseppina Vangaa
DOI:https://doi.org/10.1016/j.jcou.2015.04.004 (picture source)
https://www.studocu.com/en-us/document/university-of-north-carolina-at-charlotte/organic-chemistry-2/assignments/lab-report-4-carboxylation-of-resorcinol/2141490/view


Is this online fluorophore database reliable? I need fluorescence spectra for various fluorophores in certain dispersion media for some calculations. There is an online database of University of Arizona: http://www.spectra.arizona.edu/
Is using such online resources reliable (not for formal purposes)?
I'm sorry for not asking a technical chemistry question.

That is the correct domain name for the respected and accredited University of Arizona. I would feel confident using their data.

In general, that's the key: being certain that the source is who they say they are, and that they are a source that you have some reason to trust.

Regarding security, it's unlikely someone would spend the time, money and effort required to falsify some fluorescence spectra database.  If it were a database of controversial climate data, or clinical trials for a new medication, etc., you might want to investigate what the site does to secure their data.

I would also be cautious of using open wiki data like Wikipedia. Although I've personally gotten into the habit of trusting those sites more than I should, anybody can edit them. And while they try to police the sites for poorly documented data and such, it's a lot to keep up with. A good wiki article will have sources for their data, and reaching out directly to the source, if possible, is the better way of getting reliable data from a wiki site.


Kinetics and Half Lives Using Pressure For a certain first-order gas-hase reaction
$$\ce{A \rightarrow B + C}$$
the time required for half of an initial amount of A to decompose is 8.8 min. If the initial pressure of A is 400 mmHg, the time required to reduce the partial pressure of A to 50 mmHg is:
(a) 1.1 min
(b) 17.6 min
(c) 26.4 min
(d) 35.2 min
(e) 70.4 min

The correct answer is (c).
When I approached this problem, I started off thinking about it as a zeroth order reaction for simplicity.
$$\mathrm{400~mmHg \rightarrow 200~mmHg \rightarrow 100~mmHg \rightarrow 50~mmHg}$$
So, if this was a zeroth order reaction (not dependent on the amount of A remaining), then time would be $\mathrm{3 \cdot 8.8~min = 26.4~min}$. But, since this is a first order reaction, then [A] does factor into the half life and thus the time to reach $\mathrm{50~mmHg}$ should be slightly higher than $\mathrm{26.4~min}$.
However, this doesn't seem to be the case. Perhaps it's because we're using pressures instead of concentrations? But isn't pressure synonymous with concentration in the gas phase?

Your approach isn't bad: it seems to me that you're considering 3 half-lives to get from 400 mm $\ce{Hg}$ to 50 mm $\ce{Hg}$, and that works for the purpose of answering the question. That said, I think the following steps are going to serve you better when approaching problems like this one, as we need to explicitly solve for $k$ and then find $t$ where some concentration of reactants is specified.

What you've been given is simply the half-life ($t_{1/2}$) of reactant $\ce{A}$ and the change in pressure of the reactant at some time after the reaction begins. That change in pressure is a fine proxy for concentration, since we are in the gas phase and are considering the change of one reactant only. I'm using the concentrations in the expressions here and not the partial pressures, given that the conversion factors of one to the other cancel: we're interested in ratios.

For a first-order reaction, the integrated expression describing the disappearance of $\ce{A}$ is

$$\ln\left({\ce{[A]}\over\ce{[A]_0}}\right) = -kt$$

and the usual expression for half-life (where I've rearranged to solve for $k$) is

$$k = -{\ln(1/2)\over 8.8\;\mathrm{min}} = 0.0787\;\mathrm{min}^{-1}$$

So now, we have $k$. We also have $\ce{[A]_{0}} = 400$ mm $\ce{Hg}$ and $\ce{[A]} = 50$ mm $\ce{Hg}$ (from the problem statement). Rearranging the first equation I've written above, we solve for $t$:

$$-\left({1\over 0.0787\;\mathrm{min}^{-1}}\right)\cdot\ln\left({50\over 400}\right) = t = 26.42\;\mathrm{min}$$

where I have omitted the units of mm $\ce{Hg}$ in the natural logarithm argument because they cancel.

As you properly note, the time is slightly longer than the 26.4 minutes given in your choice of answers. I believe that the answers are just rounded to the tenths place, which obviously has created a situation where one can calculate the "right" answer using the wrong methodology.


Why elements with high atomic mass (like tungsten) are used in the production of X-rays, in the Cathode-Ray Tube? I was studying about the production of X-rays, and was struck at how elements having high atomic mass (like tungsten) are used for the production of high energy X-rays. I am wondering why low atomic mass elements are not used for the production of high energy X-rays.

I have read that "tungsten has the highest melting point of any metal, and because over 99 percent of the energy put into the X-ray tube to produce the beam is turned into heat energy, the ability to dissapate the heat without loosing structural integrity is critical".

If a sample containing a particular element is irradiated (e.g. with photons or, in case of an X-ray tube, with electrons) with an energy high enough to excite inner electron orbitals, X-rays are emitted in the de-excitation. In contrast to the continuous bremsstrahlung energy spectrum, the energy liberated in the transition from the excited to the ground state takes the form of a characteristic X-ray photon whose energy is given by the energy difference between the initial and final states. If the temporary vacancy has been created in the K shell of an atom, then a characteristic K X-ray is liberated when that vacancy is subsequently filled. In particular, if the concerned electron comes from the L shell, than a Kα photon is generated whose energy is equal to the difference in binding energies between the K and L shells.

The K-series X-rays are generally of most practical importance because their energy is greatest. Their energy increases regularly with atomic number $Z$ of the element, for example, only about $1\ \mathrm{keV}$ for sodium ($Z=11$) and about $100\ \mathrm{keV}$ for uranium ($Z=92$). An empirical law concerning the characteristic X-rays that are emitted by atoms is expressed by Moseley’s law.

Hence, in order to produce characteristic X-ray photons with a high energy, K-series X-rays of a heavy element are preferred.

Since only a small fraction of about $1\ \%$ of the energy generated in an X-ray tube is actually emitted as X-rays and the rest of the energy is released as heat, the temperature of the anode target can be very high during the operation of the X-ray tube. Therefore, the anode has to be made of high-temperature materials. You already mentioned the high melting point of tungsten ($Z=74$); whereas for example the heavier elements lead and bismuth cannot be used because of their low melting points.


Confusion regarding DFT basics, i.e. the connection between Hohenberg–Kohn theorems, external potentials and the Kohn-Sham method I have a question on the basics of DFT, which bases on the Hohenberg-Kohn theorem: there exist an unique mapping between the external potential, the electronic wave function and thus also the electronic density n(r). In other words, from wikipedia, where ${\displaystyle v(\mathbf {r} )}$ is defined as an external potential,
"If two systems of electrons, one trapped in a potential ${\displaystyle v_{1}(\mathbf {r} )}$ and the other in ${\displaystyle v_{2}(\mathbf {r} )}$, have the same ground-state density ${\displaystyle n(\mathbf {r} )}$, then ${\displaystyle v_{1}(\mathbf {r} )-v_{2}(\mathbf {r} )}$ is necessarily a constant."
My confusion comes from the Kohn-Sham equations, i.e.
${\displaystyle v_{\text{eff}}(\mathbf {r} )=v_{\text{ext}}(\mathbf {r} )+e^{2}\int {\frac {\rho (\mathbf {r} ')}{|\mathbf {r} -\mathbf {r} '|}}\,d\mathbf {r} '+{\frac {\delta E_{\text{xc}}[\rho ]}{\delta \rho (\mathbf {r} )}},}$
${\displaystyle \left(-{\frac {\hbar ^{2}}{2m}}\nabla ^{2}+v_{\text{eff}}(\mathbf {r} )\right)\varphi _{i}(\mathbf {r} )=\varepsilon _{i}\varphi _{i}(\mathbf {r} ).}$
which was derived by decomposing the energy functional of the interacting-electron system into solvable energy functionals of a non-interacting electron system affected by an alternative external potential ${\displaystyle v_\mathrm{eff}(\mathbf {r} )}$, plus an exchange-correlation functional that attempts to account for the difference between the two systems. The optimal orbitals are obtained through this iterative scheme.
My question is, since the HK theorem states the unique mapping between between an external potential and the electron density / wavefunction, why would we obtain a common electronic state when ${\displaystyle v_\mathrm{eff}(\mathbf {r} )}$ - ${\displaystyle v_\mathrm{ext}(\mathbf {r} )}$ is not necessarily a constant?
${\displaystyle v_\mathrm{eff}(\mathbf {r} )}$ in this case is the external potential experienced by a single non-interacting electron defined by the first Kohn-Sham equation, while ${\displaystyle v_\mathrm{ext}(\mathbf {r} )}$ is the external potential exerted by the stationary nuclei to the electrons, in the electron-interacting picture.

I guess more specifically, I assumed that the KS method meant that we can solve a simpler system of non-interacting electrons affected by a specific ${\displaystyle v_\mathrm{eff}(\mathbf {r} )}$ that would give the same electron density of the same system but with interacting electrons.
However, the "external potential" governed by the interacting and non-interacting systems are different, so how come they will yield the same ground state, when we minimize their respective energy expressions?

Its a tricky question but anyway when you go through the HK theorem you'll see that there is a part where you have $v_{\mathrm{ext}} - v_{\mathrm{ext}}' = H - H'$ as the electron-electron and other operators are the same for both systems and will therefore cancel. The only important part is the external (nuclear-electron) potential and you get that mapping.
Now with KS essentially what you do is take the DFT functional and switch out the kinetic energy functional to the kinetic energy operator from QM and move some other stuff around. This functional now depends on orbitals and we now want to optimise these for the lowest energies. Using some mathematical methods you can get a set of equations to do this known as the KS equations.
With the KS equations you get a set of orbital equations which you solve collectively. If you run the HK theorems with one of the KS equations the mapping between that orbital densities would be with $v_{\mathrm{eff}}$. All this is now saying is that the total electron density which maps with $v_{\mathrm{ext}}$ will be different to your orbital densities which maps to $v_{\mathrm{eff}}$ which is reasonable. You only get the total electron density when you sum all the orbital densities up.


What is the hybridization of terminal fluorine atoms in molecules like boron trifluoride? What is the hybridisation of $\ce{F}$ in $\ce{BF3}$ or $\ce{CH3F}$ or $\ce{PF5}$ or $\ce{SF6}$? My book states that the p orbitals of fluorine overlaps, but I feel that sp3 hybridisation occurs in $\ce{F}$, so its sp3 hybrid orbital overlaps with the empty p orbital of boron.

According to Ron's answer,  all atoms with p or d orbitals can hybridise.

TL;DR: As a rule of thumb, terminal (heavy) atoms are almost always best described as having sp hybrid orbitals (at the most).

For more on this, I refer you to my answer on What is the hybridization of chlorine in vinyl chloride?

Here are the relevant visualisations for $\ce{BF3}$ at the DF=BP86/def2-SVP level of theory. First I'll start with the canonical orbitals obtained from a molecular orbital point of view. The molecule has $D_\mathrm{3h}$ symmetry, therefore some orbitals are degenerated.

With appropriate transformations, we can interpret these orbitals in a localised fashion, which will give us hybrid orbitals. In this view the σ bond orbitals, and the lone pairs are triply degenerated. Note that there is a significant amount of delocalisation of the fluorine lone pairs into the empty boron p orbital.

Here is the representative section from the natural bond orbital analysis:

     (Occupancy)   Bond orbital / Coefficients / Hybrids
 ------------------ Lewis ------------------------------------------------------
(core and symmetry equivalent orbitals skipped)
   5. (1.99345) LP ( 1) F  2            s( 63.79%)p 0.57( 36.19%)d 0.00(  0.01%)
   6. (1.94124) LP ( 2) F  2            s(  0.00%)p 1.00( 99.95%)d 0.00(  0.05%)
   7. (1.86896) LP ( 3) F  2            s(  0.00%)p 1.00( 99.94%)d 0.00(  0.06%)
  14. (1.99778) BD ( 1) B  1- F  2
               ( 17.29%)   0.4158* B  1 s( 33.25%)p 1.98( 65.98%)d 0.02(  0.76%)
               ( 82.71%)   0.9094* F  2 s( 36.24%)p 1.76( 63.62%)d 0.00(  0.14%)
 ---------------- non-Lewis ----------------------------------------------------
  17. (0.38551) LV ( 1) B  1            s(  0.00%)p 1.00(100.00%)d 0.00(  0.00%)
  18. (0.05374) BD*( 1) B  1- F  2
               ( 82.71%)   0.9094* B  1 s( 33.25%)p 1.98( 65.98%)d 0.02(  0.76%)
               ( 17.29%)  -0.4158* F  2 s( 36.24%)p 1.76( 63.62%)d 0.00(  0.14%)

Here you can clearly see the approximate two sp orbitals and the remaining p lone pair orbitals.

The above remains true for (almost) all terminal atoms, because of the local linear symmetry (only one bonding partner, negligible external field), although with lesser extend for heavier atoms (like bromine). This is because hybridisation in general becomes a less reliable description due to the increase in the s-p gap.


Equilibrium of more than two inter reacting species We know that equilibrium in a chemical system is attained when forward and backward reaction rates are equal. What if the reaction mixtures involve more than one reaction?

For example, consider three inter-reacting species A, B and C. The rate constants for the forward and backward reactions of $\ce{A -> B}$ are $k_1$ and $k'_1$, that of $\ce{B -> C}$ are $k_2$ and $k'_2$ and that of $\ce{C -> A}$ are $k_3$ and $k'_3$. What is the relation between the six rate constants so that the system remains in equilibrium? Assume first order kinetics for all the reactions. Thanks in advance.


Effect of pressure on viscosity of liquids and gases My textbook says the following on viscosity and pressure:

  increase in pressure decreases the viscosity of water but for other liquids it increases. Viscosity of gases does not change much

No further explanation is provided.

I have two doubts regarding this:

How does the viscosity of water decrease? Does it have something to do with the anomalous expansion of water?
How does the viscosity of gases not change much on increasing the pressure? The gas molecules should have an increase in kinetic energy, contributing to more collisions and hence more viscosity

One of the best websites for all advanced water properties is a website maintained by Dr. Martin Chaplin http://www1.lsbu.ac.uk/water/physical_anomalies.html#Pvisc

Keep in mind that viscosity is a really complicated subject and water is among the unique solvents. In general viscosity of liquids increases with pressure. At pressure extremes, let us say, 100,000 psi, many liquids become solids. Water also becomes a solid at really really high pressures. So when you talk about viscosity, you should talk about what pressure and what temperature.

From the website:

  Water's pressure-viscosity behavior [534, 2890] can be explained by
  the increased pressure (up to about 100-200 MPa) causing deformation,
  so reducing the strength of the hydrogen-bonded network, which is also
  partially responsible for the viscosity. This reduction in
  cohesiveness more than compensates for the reduced void volume. It is
  thus a direct consequence of the balance between hydrogen-bonding
  effects and the van der Waals dispersion forces [558] in water; with
  hydrogen-bonding prevailing at lower temperatures and pressures. At
  higher pressures (and densities), the balance between hydrogen-bonding
  effects and the van der Waals dispersion forces is tipped in favor of
  the dispersion forces and the remaining hydrogen bonds are stronger
  due to the close proximity of the contributing oxygen atoms [655].
  Viscosity, then, increases with pressure.

Second question on the viscosity of gases

The textbooks seems to be talking about a limited range of P. Viscosity of gases is dependent on very low or very high pressures. However, for ideal gases and normal pressures, it is only dependent on the temperature. Yes, it is quite counterintuitive. Also viscosity of gases increases with temperature. This is another surprise. https://en.wikipedia.org/wiki/Viscosity#Pure_gases

You can try posting this second query in Physics forum. I don't know the derivation of that Wikipedia expression.


Do metals react with ammonia gas (not liquid phase)? In my chemistry notes I have two chemical equations as follows:

$$
\begin{align}
\ce{2 NH3(g) + 6 Li(s) &-> 2 Li3N(s) + 3 H2(g)}\tag{R1} \\
\ce{2 NH3(g) + 2 Na(s) &-> 2 NaNH2 + H2(g)}\tag{R2}
\end{align}
$$

Before digging into the reasons for different products in two equations, when I searched in the internet to find the clear picture of what are the metals (like alkali metals, alkaline earth metals, transition metals) that can react with ammonia in gaseous form, I could only find metals dissolving in liquid ammonia.

I would be grateful if someone could tell that metals don't react with gaseous ammonia, or, if it reacts, what are the metals those react with gaseous ammonia? And are there any conditions needed for the reaction (like heat)?

Wikipedia reports that magnesium nitride can be prepared:

By passing dry nitrogen over heated magnesium:
${\displaystyle{\begin{matrix}{}\\{\ce {{3Mg} + N2->[{\ce{800^{\circ }C}}]Mg3N2}}\\{}\end{matrix}}}$
or ammonia:
${\displaystyle{\begin{matrix}{}\\{\ce{{3Mg} + 2NH3->[{\ce{700^{\circ }C}}]{Mg3N2} + 3H2}}\\{}\end{matrix}}}$

No references are cited.


Why isn't aluminium involved in biological processes? There are so many biological processes which are dependent upon ions of lighter metals (upper part of periodic table) such as $\ce{K+}$, $\ce{Na+}$, $\ce{Mg^{2+}}$ and even early transition elements ($\ce{Fe}$, $\ce{Mn}$, $\ce{Cu}$, $\ce{Ni}$), but I haven't yet come across dependence of biological phenomena on aluminium. Is this because there is less use of trivalence or is there some other reason?

One argument put forward has been that aluminum is very poorly bioavailable, moreso than many other elements.  Aluminum oxide is very insoluble in water.  In addition, any dissolved aluminum that does form in seawater is likely to be precipitated by silicic acid, forming hydroxyaluminosilicates.

From Chris Exeter's 2009 article in Trends in Biochemical Sciences:

  But how has the by far most abundant metal in the Earth's crust remained hidden from biochemical evolution? There are powerful arguments, many of which influenced Darwin's own thinking [15], which identify natural selection as acting upon geochemistry as it acts upon biochemistry. I have argued previously that the lithospheric cycling of aluminium, from the rain-fuelled dissolution of mountains through to the subduction of sedimentary aluminium and its re-emergence in mountain building, depends upon the ‘natural selection’ of increasingly insoluble mineral phases of the metal [7]. The success of this abiotic cycle is reflected in the observation that less than 0.001% of cycled aluminium enters and passes through the biotic cycle. In addition, only an insignificant fraction of the aluminium entering the biotic cycle, living things, is biologically reactive. However, my own understanding of such an explanation of how life on Earth evolved in the absence of biologically available aluminium was arrived at by a somewhat serendipitous route! In studying the acute toxicity of aluminium in Atlantic salmon I discovered that the aqueous form of silicon, silicic acid, protected against the toxicity of aluminium [16]. Subsequent work showed that protection was afforded through the formation of hydroxyaluminosilicates (HAS) [17] which, intriguingly, are one of the sparingly soluble secondary mineral phases of the abiotic cycling of aluminium! The discovery that silicic acid was a geochemical control of the biological availability of aluminium, though now seemingly obvious in hindsight, was a seminal moment in my understanding of the bioinorganic chemistry of aluminium, and although it helped me to understand the non-selection of aluminium in biochemical evolution, it also provided me with a missing link in the wider understanding of the biological essentiality of silicon.

Dr. Exeter is one of the few scholars who appears to have written in depth about this issue.  Thus, perhaps it is fair to say that (a) your question doesn't have a definitive answer, but (b) the poorly accessible nature of aluminum over geological time due to its interaction with and precipitation by silicic acid is the leading hypothesis.

It's worth noting that when aluminum is artificially introduced into metalloenzymes in place of naturally occuring metals, the resulting alumino-enzymes can retain activity, as a 1999 article in JACS by Merkx & Averill shows.


Reaction mechanism of imine reduction to amine using an aluminium amalgam I can not find any good resources on the Internet except for some shady drug syntheses. I was hoping that someone could provide me with some more educational information.

I'm interested in imine-amine reduction methods, but it was hardly discussed in my class. It would be very nice if someone could explain the exact reaction mechanism of the imine-amine reduction in the presence of an aluminium amalgam.

I also found that the very dangerous compound Mercury(II) chloride was used a lot in this method and I was wondering why for example Mercury sulfide is not used, as I thought it would be a lot safer since it is hardly soluble in water and is therefore less likely to do any physical harm.

The mechanism is electron transfer. The imine will take one electron from metal to form a radical anion ($\ce{.CR2-NR'^{-}}$) intermediate; the $\ce{-NR'^{-}}$ will grab one proton from solvent to give you the neutral radical ($\ce{.CR2-NR'H}$); the carbon radical grab one hydrogen atom ($\ce{.H}$) from solvent to form your amine product and a solvent radical ($\ce{.OH}$ if water used as solvent); and the resulting solvent radical will take another electron from aluminum to finish the whole process. The reason to add mercury is to destroy the relatively less reactive aluminum oxide layer. Mercury also serve as the medium for electron transfer.

$\ce{HgCl2}$ is not that dangerous, since it is not volatile at all like metal mercury or alkylmercury. After reaction, simply add some sulfide or sulfur to transfer all of them to $\ce{HgS}$ which is not soluble and not reactive in non-oxidative condition. This is also the reason that you can not use $\ce{HgS}$ instead, because it is neither soluble nor reactive.


Super-saturated solutions If a solution is super-saturated is it possible to add a substance to it without allowing it to crystallise? Does a change in temperature have to happen and reduce it's saturation if I want to add another substance?

Yes, adding additional solute to a super saturated solution can be done without crystallization.

Adding additional solute to a super saturated solution and preventing crystallization is dependent on temperature, degree of agitation, and rate of solute addition. Changing the temperature is usually the easiest way to control the degree of super saturation, but you can experiment with the other variables as well. The degree to which these three variables affect any given system is dependent on the identity of the system and the amount of impurities (generally you do a lot of work to minimize impurities before you create a supersaturated solution).


How does a halogen react in an electrophilic substitution reaction From what I understand, electrophiles are electron deficient and therefore would be attracted to the electron-rich benzene ring. So would the correct answer be A?

A is correct because the question specifically says an electrophilic substitution reaction (D will also react, but it's an electrophilic addition reaction). However, you also need a Lewis acid catalyst like FeCl$_3$. This is because, while the delocalised system of electrons in benzene is indeed electron rich, the fact that it's aromatic (there is a system of pi electrons delocalised throughout the ring) makes it very stable.
The mechanism of the bromination of benzene is shown below (image from Wikipedia as I don't have ChemDraw on this computer). The mechanism is the same for Cl$_2$.


Why does fluorine stabilise a carbocation? I have read that halogens have a stonger -I (inductive effect) than +M (mesomeric effect),then why do they stabilise a carbocation instead of destabilising it?

I asked this doubt to my teacher and he said that halogens like fluorine do not exert -I on a carbocation and hence they stabilise a carbocation through +M,is this true?

Fluorine belongs to the same period as carbon. Carbocation is electron deficient which has its two $\mathrm{2p}$ orbitals vacant. One electron of carbon in $\mathrm{2p}$ orbital goes into the formation of $\ce{C-F}$ σ bond. $\ce{F}$ has $\mathrm{2p}$ orbital containing a lone pair which can be donated to the $\mathrm{2p}$ vacant orbitals of carbocation. Same size $\mathrm{2p}$ orbitals of carbon and fluorine overlap effectively to form a π bond.


How to properly work with potassium HMDS What is the proper procedure for working with KHMDS?  I would like to use it to deprotonate a compound and have a bottle of KHMDS in toluene.  Does it need to be handled under inert atmosphere?  And is it soluble in halogenated solvents?

If you would look up the properties of KHMDS from a chemical supplier, you will find it will react violently with with water. So do keep it under inert atmosphere.

As for solubility, it is a non-nucleophilic base, so probable no substitution and addition like reactions will occur. I wont suspect it will be a problem there.

A comment on the use though. If you plan to use this, explore the dangers and possibilities of a substance like this a bit further than asking it here. For safety considerations. Look up the MSDS sheets from your supplier and be sure to only use this in a properly equipped lab.

Hope this helped.


Is it possible to chemically treat grass to make it digestible? Is it possible to chemically treat grass after "harvesting" it to make it digestible by humans?  I know we can "eat" grass, but we can't break it down in our stomachs. If there is some chemical treatment to partially break grass down into something we can digest, I see it possibly being used to end world hunger!  I know we already treat bone to make it into gelatin, and I am wondering if something similar can be done for grass.

The indigestible component of grasses is cellulose, which is polymerized glucose.

Yes, it is possible to hydrolyze the cellulose to glucose, which is of course digestible.

The hydrolysis can be catalyzed enzymatically with a cellulase, or using acid hydrolysis.

The process of hydrolyzing cellulose is often discussed in the context of so called cellulosic ethanol, as the glucose is then fermented to ethanol subsequent to hydrolysis.


How to run electrochemical tests on Microorganism (bacteria in a medium)? I am trying to run some electrochemical tests(e.g.CV (Cyclic Voltammetry) & EIS (electrical impedance spectroscopy)) using half cell configuration on micro-organisms (e.g.bacteria). The bacteria are cultured in a medium (liquid). The half cell configuration is working, counter, and a reference (AgCl) electrodes.

Usually, it is easy to clip a metal or any thin solid material. However, I don't know how to deal with a solution. Since the beaker will be filled with electrolyte, I am afraid it mixes with bacteria solution causing it to be removed from the electrode. I am avoiding add another material like tape or something as it might alter the measurement.
I am looking for a way to attach the Micro-organism to the working electrode.

From what I understand, your main issue is connecting a solution in an electrical circuit without changing the composition of the solution due to electrochemical redox reactions.
So, there are two problems here.  Firstly, passing direct current
(DC) changes the composition of the solution. Secondly, a solution cannot
be connected to the bridge like a metallic wire or any other solid conductor.
The first difficulty is resolved by using an alternating current (AC) source
of power. The second problem is solved by using a specially designed
vessel called conductivity cell.


Are specific heat capacity and boiling point of a substance related or proportional to each other? That is, if the substance has a high specific heat, will it also have a high boiling point? ( and vice versa)

According to the Ehrenfest classification, during a first-order transition (such as boiling) the heat capacity $C$ goes to infinity. This makes sense, since at the transition point the temperature does not change with input of heat and $C = dQ/dT $. This and other considerations add a little complexity to the problem, as it requires you to state which heat capacity (at which T) you are talking about.

More importantly, however, the boiling point depends on the magnitude of attractive intermolecular forces compared to kT, whereas the heat capacity depends on all the available degrees of freedom (including intramolecular). Therefore you can envision large and small molecules with similar overall solution composition (for instance hydrocarbons) - and thus similar heat capacities - but very different BPs. See for instance this
table of specific heats, where C for hexane through dodecane are very similar, whereas their BPs differ quite a bit.


Does excessive sulfuric acid as catalyst affect the synthesis of aspirin? Recently had a lab experiment to synthesize aspirin.
I was instructed to add 4 drops of concentrated sulfuric acid but misread the "4 drops" as "4 ml".
Ended up with a yield of just 10 %. But the crystals that I obtained upon recrystallisaion weren't of aspirin but of salicylic acid (as confirmed by a melting point of 160 °C).

Did the addition of excessive sulfuric acid affect this in any way? What could have caused salicylic acid crystals to be produced instead of aspirin?

Some technical details of the steps I took:

mixing 2 g of salicylic acid with 5ml of acetic anhydride
4 ml of sulfuric acid added as catalyst (should've been 4 drops)
Heated the mixture in a water bath for 15 min.
Added 1 ml of cold water slowly then added 20 ml of cold water again quickly and stirred for 5 min.
Allowed solution to cool at room temp.
Crude product was dissolved in 5 ml of ethanol. (I noticed that my crude product was able to dissolve immediately whereas my peers' had to heat theirs and took awhile to dissolve, any reason for this?)
30 ml of hot distilled water was added to the mixture and heated on a hotplate.
Allowed solution to cool to room temperature then placed in an ice-bath for 15 min.
Crystals collection by suction filtration.
Crystals were dried in the oven at 100 °C for 15 min.
Crystals placed in desiccator for 10 min.
Melting point of crystals were found to be 159–161 °C

Did the addition of excessive sulfuric acid affect this in any way? What could have caused salicylic acid crystals to be produced instead of aspirin?

The intended reaction was the formation of an ester from a a phenol and an anhydride.

However, with an excess of sulfuric acid, acidic hydrolysis of the product according to an $\mathrm{A_{Ac}2}$ mechanism takes place.


Typesetting oxidation state and charge simultaneously I'm currently trying to typeset the ion $\ce{Cr2O7^{2-}}$ with oxidation numbers on. My first thought was to put the oxidation state above the atom, but according to IUPAC, the oxidation state should be in the right superscript. Combined with charge, it looks ugly and wrong:

My current ways of doing this are below

The oxidation number in right superscript. Looks weird with both numbers so close.
The oxidation number above. Looks better, but is against IUPAC.

My question therefore is: Which is the preffered way to typeset this?

Also, I would like to excuse beforehand if this should rather be in tex.stackexchange.

My personal preference in this situation would be to put square brackets around the whole ion and put the charge outside.

$$\ce{[Cr^{VI}2O^{-II}7]^{2-}}$$ gives:
$$\ce{[Cr^{VI}2O^{-II}7]^{2-}}$$

This is sometimes done with polyatomic species anyway (particularly metal-ligand complexes) to emphasise that the charge is on the whole ion, not on any one particular atom.


Boric Acid - Mechanism of Acidity I was wondering how one could test the hypothesis that boric acid behaves not as a Brønsted acid but as a Lewis acid in that boric acid cleaves water molecules as to make hydrogen protons.

rad
Could one run an experiment involving boric acid in heavy water ($\ce{D_2O}$)?

Thus, if boric acid behaves as a Brønsted acid, the concentration of deuterium cation should stay the same, right? Boric acid might release hydrogen protons but those wouldn't affect the deuterium protons, right? Wouldn't heavy water still undergo auto-ionization and create a certain equilibrium concentration of the heavy equivalents of hydronium and hydroxide ions?

And thus, if boric acid behaves as a Lewis acid, couldn't we somehow precipitate the $\ce{B(OH)_3(DO)^-}$, and test whether the precipitate has D rather than H? I.e. introduce sodium chloride into the system and evaporate to dryness. Wouldn't that leave behind a precipitate consisting partly of sodium ion and partly of the borate? On the other hand what would happen to the excess chloride ion?

As for acidity of boric acid you can find on https://en.wikipedia.org/wiki/Boric_acid that it dissociates in both ways, but is very weak. However it's acidic enough that it's only a matter of time until it would become almost completely deuterised if dissolved in D2O. It happens because in mixture there would be much more deuterium then normal hydrogen. The acid is too weak to capture or detect its anions. Maybe kinetics of deuterisation would give us some info about properties of its anions.


Is there a convention for breaking chemical formulae across lines? I found a chat where a user asked for chemical formulae to be broken across lines (with LaTeX/mhchem). I could imagine offering a switch to allow this. But only, if line-breaking within formulae is not completely uncommon.

According to him/her, possible breaking points in $\ce{[nBu4N]2[(UO2)2(\mu-\eta^2:\eta^2-O2)(NO3)2(\mu-Au(CN)2)2]}$ would be
$\ce{[nBu4N]2~\color{red}{|}~[(UO2)2~\color{red}{|}~(\mu-\eta^2:\eta^2-O2)~\color{red}{|}~(NO3)2~\color{red}{|}~(\mu-Au(CN)2)2]}$.
So a rule could be "allow a line-break before (, [ or \{ (which would introduce another possible breakpoint into the example).

Is it common to break long formulae across lines and is doing so sensible?

I remember a couple of books from 1970-80s, where long formulas were split with what would be a $\rm \LaTeX$ \cdot "$\cdot$" notation, I believe. This is an example from Pope, M. T.; Jeannin, Y.; Fournier, M. "Heteropoly and isopoly oxometalates" in Inorganic Chemistry Concepts; Springer: Berlin, 1983; Vol. 8.:

The reason for that, it seems, is that the book was printed in A5 format. Personally, I find an option for mhchem to be able to split formulas across pages very useful.


How can change in entropy be the same for all processes if the entropy production $\sigma$ is present for irreversible processes? From the definition of entropy change,
$$S_2-S_1=\left ( \int_{1}^{2} \frac{\delta Q}{T}\right )_{int.rev}$$
From the closed system entropy balance, we have
$$S_2-S_1=\left ( \int_{1}^{2} \frac{\delta Q}{T}\right )_{b}+\sigma $$
where $\sigma$ is the entropy produced within the system, vanishing to zero in the absence of irreversibilities. I don't quite understand how the entropy change between two states is the same for all processes. Is it the case that the entropy transfer in the case of internal irreversibilities present is lower than that of the entropy transfer in an internally reversible process between these same two states, and the entropy production makes up for this difference?

When a system changes from state 1 to state 2, the change in entropy of the system is the same for all processes, because entropy is a state function.
The difference between a reversible and an irreversible process is that, in the latter case, there is net entropy production for the universe, i.e., for the system+surroundings.
Since the entropy change for the system is independent of the process, the difference in entropy production between rev. and irrev. processes is seen in the entropy change of the surroundings.
More broadly, since path differences (which in turn means differences in heat flow and/or work flow) have no effect on the final state of the system, they instead always manifest themselves as differences in the final state of the surroundings.