Does NaOH react with O3 to produce NaO3? I read two reactions which I want to specify first:

$\ce{Na2O2 + O3 + H2O -> 2NaOH + 2O2}$
$\ce{2KOH +  5O3 -> 2KO3 + 5O2 + H2O}$

Does $\ce{NaOH}$ also undergo a further reaction to produce $\ce{NaO3}$ or not? Since both are alkali metals and with relatively similar sizes(?), I thought it could, but Google doesn't show me anything apart from a reaction balancer site which just shows me the reverse equilibrium of the above reaction.

My own Google search does show that sodium ozonide can form [1].  But it requires low temperature:

Sodium ozonide was produced by the ozonization of sodium hydroxide at -80 to -100°[C] followed by extraction with liquid ammonia and removal of the latter under vacuum at-50° [2]. Composition of the product obtained: $\ce{NaO3}=80.1, \ce{NaOH•H2O}=11.2, \ce{H2O}=8.6$ [%].

Upon warming it decomposes to give the superoxide, and then in the presence of water back to sodium hydroxide:

The process of thermal decomposition of sodium ozonide to $\ce{NaO2}$ occurs at −20 and −10°.

At temperatures above 0°, the process of thermal decomposition of sodium ozonide is accompanied by the process of interaction of the $\ce{NaO2}$ formed with $\ce{H2O}$, and decomposition proceeds to the formation of $\ce{NaOH}$.

References
1.
Tokareva, S.A., Pilipenko, G.P. Thermal decomposition of sodium ozonide. Russ Chem Bull 13, 686–688 (1964). https://doi.org/10.1007/BF00845322.
2.
S. A. Tokareva and M. S. Dobrolyubova, Izv. AN SSSR, Ser. Khim. 1964, No. 4.


What is the proper way to prepare a solution aiming to minimize errors? I am a undergraduate student in chemistry, and my university is very weak in teaching solution preparation. I need to prepare a 10 mg / L methylene blue solution and now the way I do is, using an analytical balance of 5 decimal places, I tare a 50 mL beaker and simply add the 10 mg of methylene blue, trying be as accurate as possible (the amount is very small due to the large molecular weight). Afterwards, I add deionized water to a certain volume and transfer to a 1 L volumetric flask using a funnel (I wash the small beaker several times to guarantee that I transfer as much as possible of the compound) and complete with deionized water until the meniscus line. And so, I shake it many times.

Is this procedure correct? If not, how can I prepare this solution avoiding errors?

Methylene blue isn't a primary standard so I'm not sure how much accuracy/precision you need.

Typically when wanting the best accuracy the solid would be dried in
a hot oven, then cooled in a desiccator.
Do not to handle the beaker with your hands. Even if it has been
"left out" in humid air that amount of water would be stable. So use
tongs to pull in and out of balance.
Another option is to weigh more and do a serial dilution. So 1 gram
in liter, then 10 ml of that solution into 1 liter.


Help to kick out the anomaly in proving $Q$ = $K_\text{eq}$ Consider a hypothetical bidirectional reaction ,  $\ce{aA + bB <=> cC + dD}$
(A , B are reactant(s) and C , D are reaction product(s)) ,
As we know the following expressions $Q=\frac{P_C^c.P_D^d}{P_A^a.P_B^b}$ (Q=Reaction Quotient , $P_x$ = Partial Pressure od $x_\text{th}$ reactant/reaction product) and while the reaction is in equilibrium , $K_\text{eq}=\frac{[C]^c.[D]^d}{[A]^a.[B]^b}$ ($K_\text{eq}$= Equilibrium constant of the reaction , [ ] = Concentration for Ideal Solution or Activity for Non-Ideal Solution) please help me to understand why at chemical equilibrium $Q=K_\text{eq}$ holds ,
although $Q$ and $K_\text{eq}$ both are dimensionless quantity , "The expression of $Q$ contains Partial Pressure and The expression of $K_\text{eq}$ contains Concentration for Ideal Solution or Activity for Non-Ideal Solution"

In order to compare $K$ and $Q$, you have to define them in terms of the same measures of concentration and the same standard state. The only difference in the definition should be that in order to determine $K$, you are required to use the concentrations or activities of an equilibrium state, while $Q$ may be determined for states that are not necessarily at equilibrium.

please help me to understand why at chemical equilibrium $Q=K_{eq}$ holds

Once you have the definitions match, it becomes fairly trivial. The equilibrium constant is what you observe whenever the reaction has had enough time to go to equilibrium. The individual equilibrium concentrations might be distinct for different starting concentrations, but $K$ comes out the same (if it does not, it means the reaction is not reaching equilibrium). If $Q$ is not equal to $K$, you are not at equilibrium.
I realize this is a bit of a tautology. The explanation why the particular form of the equilibrium expression leads to a value that is a constant requires a knowledge of how chemical potentials (or $\Delta_r G$) vary with concentration. The reaction will proceed in the direction that minimizes $G$, so at equilibrium $\Delta_r G = \frac{dG}{d \xi}$ will be zero.


Clarification on the $\mathrm{S_N2}$ mechanism The question is about the specifics of the $\mathrm{S_N2}$ mechanism (nucleophilic bi-molecular substitution).

(Source: Wikipedia)

As we know that in this reaction the nucelophile ($\ce{Nu-}$) approaches the molecule. The main step (According to the molecular orbital theory) of the reaction is that the the nucleophile bonds with its lone pair in the anti-bonding molecular orbital (ABMO) of the central carbon atom. As this happens the leaving group ($\ce{Lg}$) is kicked off by gradual weakening of the $\ce{C-Lg}$ bond. Now the thing that bothers me here is:

Why does carbon prefer to add electrons in its ABMO? As the $\ce{Lg}$ has to bond with the carbon atom there has to be an increase in bond order whereas, while electrons are pushed into the ABMO it clearly decreases the bond order, which directly indicates the instability of bond formation.
The main question being, what makes the reaction favourable? What is the 'driving force'? Why does it take place when there is a decrease in bond order?

Why does carbon prefer to add electrons in its ABMO?

Well, preferring is a bit of a strong word. Orbital interactions that are productive towards forming and breaking bonds can only happen between a filled and an empty orbital. Therefore, you should always identify a HOMO (highest occupied molecular orbital) and a LUMO (lowest unoccupied molecular orbital), compare their energies and thereby decide whether they are able to react.

Here, the lowest unoccupied orbital of the carbon compound is typically the $\ce{C-Lg}$ antibonding $\unicode{x3c3}^*$ orbital — because that is the antibonding orbital belonging to the leaving group (I am aware that this is circular reasoning). Its energy is also somewhat favourable with respect to the energy of the filled lone pair orbital of the nucleophile meaning the two can react well.

  As the $\ce{Lg}$ has to bond with the carbon atom there has to be an increase in bond order …

No. Why?

  electrons are pushed into the ABMO it clearly decreases the bond order …

To specify: decreases the bond order *of the $\ce{C-Lg}$ bond.

  which directly indicates the instability of bond formation.

No, it doesn’t. It only indicates that the $\ce{C-Lg}$ bond is being broken and the leaving group is getting ready to leave.

  Why does it take place when there is a decrease in bond order?

You fail to see that while the bond order of the $\ce{C-Lg}$ bond is steadily decreasing the bond order of the $\ce{C-Nu}$ bond is steadily increasing. At any point in time on the reaction coordinate, the sum of both bond orders is $1$.

  The main question being, what makes the reaction favourable? What is the 'driving force'?

Two things: the formation of a more stable $\ce{C-X}$ bond (meaning that $\ce{C-Nu}$ is a stronger bond than $\ce{C-Lg}$) and the liberation of a better stabilised anionic species $\ce{Lg-}$ as opposed to $\ce{Nu-}$.

Actually, in many $\mathrm{S_N2}$ reactions only one of those two factors is the driving force. And while we tend to write $\ce{Nu-}$ and $\ce{Lg-}$, we should always remember that these compounds need not be anions. For example, $\ce{H2O}$ is a very good leaving group (after alcohols have been protonated) but not an anion. The same is true for $\ce{N2}$ (from a diazonium salt). And equally, water or ammonia can be nucleophiles although they are neutral compounds.

So it can well be that the reaction is only driven by the liberation of a neutral compound and thereby the removal of charges (as would be the case in the $\mathrm{S_N2}$ reaction of iodide and a protonated alcohol when $\ce{HI}$ is used to generate iodoalkanes). It might also be that the reaction is driven only by the stronger $\ce{C-Nu}$ bond — or by the breaking of the weaker $\ce{C-Lg}$ bond as would be the case in nucleophilic substitutions with tosylate ($\ce{Me-C6H4-SO3-}$) or mesylate ($\ce{MeSO3-}$) as leaving groups.

There is not one same truth for all $\mathrm{S_N2}$ reactions.


What is the stereochemistry resulting from hydroboration of 1-methylcyclopentene? What is the major product of the following reaction sequence?

I'm supposed to work out the structures of both 2 and 3, but I already know 2 to be 1-methylcyclopentene (as the most substituted alkene is formed during elimination).

For the hydroboration step, boron should add to the less substituted carbon (in an anti-Markovnikov fashion). However, I'm not sure which of the two choices B and C is correct. Why should one diastereomer be formed preferentially over the other?

The second step is a syn addition of the boron hydride. As depicted below, it will result in an anti-Markovnikov product with the boron on the less-substituted carbon. Both hydrogen and boron are on the same side of the former double bond and these two are the ones we refer to when we say syn addition.

The final step, the oxidation by hydrogen peroxide, is stereospecific with retention. So the newly-formed $\ce{C-O}$ bond is on the same side as the former $\ce{C-B}$ bond. Thus, an anti relationship is established between the methyl group and the hydroxy group.

Since 1-methylcyclopentene is planar, a racemic mixture of enantionmers is formed; neither re nor si side are favoured. The scheme below shows the attack from the si side of the $\mathrm{C2}$ atom.


Organic solutions that corrode metals Most of the common organic solvents are regarded as noncorrosive, but  stainless steel can be attacked by formic, acetic, and propanoic acids.
Corrosion of stainless steel by organic solvent mixtures, Alice Y. Ku , David H. Freeman, Anal. Chem., 1977, 49 (11), pp 1637–1638
DOI: 10.1021/ac50019a044
Are there some aromatic hydrocarbon that attack stainless steel? Which organic solvents can corrode copper and aluminium?

Picric acid (2,4,6-trinitrophenol) is an aromatic compound has been used as an etchant in the past. Picral is an ethanol solution of picric acid and could etch steel. Look up www.metallographic.com/etchants/etchants.htm


Help with absolute configuration of a substituted cyclohexane? How is the absolute configuration decided when it is a cyclohexane and several stereocenters? Up to now we have only worked with simple molecules which are only 1. I believe the stereocenters here are the 2nd, 4th and 5th carbon but what are their configurations?

Thanks!

The most helpful way that I was taught to identify the configuration of stereocenters is a right-hand/left-hand rule. First, assign priority numbers to each of the substituents according to atomic mass, or the atomic mass of their first point of differentiation. Next, identify the direction that the substituent with the lowest priority is pointing in (into the plane/paper, out of the plane/paper, or parallel with the plane/paper). Then imagine that your thumb is the lowest priority substituent. If the fingers of your right hand can curl in the direction of the 1-2-3 substituents, then the configuration is R. If the fingers of your left hand curl in the direction of the 1-2-3 substituents, then the configuration is S.

I found this method to be extremely helpful because no matter the orientation of the stereocenter, this method will work because you can adjust your hands to match the orientation of the stereocenter and identify it as R or S.


How do you find the bond enthalpy from an equation? Which equation represents the $\ce{N-H}$ bond enthalpy in $\ce{NH3}?$
$$
\begin{align}
&\textbf{A.} &\ce{NH3(g) &-> N(g) + 3 H(g)}\\
&\textbf{B.} &\ce{1/3 NH3(g) &-> 1/3 N2(g) + H(g)}\\
&\textbf{C.} &\ce{NH3(g) &-> 1/2 N2(g) + 3/2 H(g)}\\
&\textbf{D.} &\ce{NH3(g) &-> .NH2(g) + .H(g)}
\end{align}
$$

From what I've learnt, I understand that the bond enthalpy is defined as the energy required to break one mole of a specific bond.
In the question above, I opted for answer C as it was the only one with the products in the form of $\ce{N2}$ and $\ce{H2}.$ But since there are three $\ce{N-H}$ bonds in $\ce{NH3},$ I am unsure about answer B.
Can you clarify this?

Option D is the only one corresponding to the definition "energy required to break one mole of a specific bond" and nothing more.


What does the 129Xe NMR spectrum of XeOF4 look like? What will the $\ce{^129Xe}$ NMR spectrum of $\ce{XeOF4}$ look like?

I know that its electron pair geometry is octahedral, and the molecular pair geometry is square pyramidal.

I believe that the lone pair on $\ce{Xe}$ will take the equatorial position (less sterically crowded), alongside the oxygen and 2 other equatorial fluorines in the same plane, to minimize electron pair repulsion.

So there will be 2 equatorial $\ce{F}$ and 2 axial $\ce{F}$. Therefore, am I right to say that the $\ce{^129Xe}$ NMR will be a triplet of triplets?

The $\ce{^{129}Xe}$ nmr spectrum of $\ce{XeOF4}$ is a sharp, quintuplet of peaks (reference, p. 445).  As you note, $\ce{XeOF4}$ adopts a square pyramidal geometry (reference), but the lone pair occupies an axial position.

Therefore, the 4 equatorial fluorines are equivalent and since they are spin 1/2 nuclei they will couple with the $\ce{^{129}Xe}$ nucleus (also spin 1/2) producing the observed quintuplet. $\ce{XeOF4}$ is an IUPAC suggested standard for $\ce{^{129}Xe}$ nmr spectroscopy (but xenon gas is more commonly used because $\ce{XeOF4}$ can decompose in the presence of moisture to produce $\ce{HF}$).


Exchange of electrons in f orbital If an f orbital has 6 unpaired electrons then how many electrons can be exchanged.
According to me if 2 electrons are added then a half filled f orbital would be formed which is stable (not as much as a completely filled orbital but still quite stable) however, the answer is supposedly 15.

If one knows the reason please do tell.

There are six unpaired electrons, and we can swap any two of them, so we have:
$${6\choose2} = 15$$


How to calculate n factor for salts in this reaction? My source says : For  salts which react such that no atom undergoes change in oxidation state ,
n-factor for such salt is the total moles of cationic /anionic species replaced in 1 mole of the salt .

For Example $$\ce{Na3PO4 + BaCl2  -> NaCl + Ba3(PO4)3}$$

N factor of $$\ce { Na3PO4} $$ is 3.

Please can anybody give more examples and explain that to me ?

$$\ce{2 Na3PO4 + 3 BaCl2 -> 6 NaCl + Ba3(PO4)2}$$
Now this is clearly evident that dividing whole equation by 2 we get 1 mole of $\ce{Na3PO4}$ and at the right hand side 3 moles of $\ce{NaCl}.$ Therefore, 1 mole of phosphate ion is replaced by 3 moles of $\ce{Cl-}$ ion.
Hence $n\text{-factor} = 3.$
In reactions like this where there's no change in oxidation number, the $n\text{-factor}$ will always be the net positive charge in 1 molecule of that compound, in this case since there are 3 $\ce{Na}$ ions in sodium phosphate, the $n\text{-factor} = 3 × 1 = 3.$


Oxidation of cycloalkenes Non-cyclic alkenes can be oxidized to carboxylic acids in presence of hot alkaline $\ce{KMnO4}$ and quenched with dilute $\ce{H2SO4}$ to form carboxylic acids. I suddenly got a thought. Can cycloalkenes be oxidized to form carboxylic acids? For example: can cyclohexene be oxidized to cyclohexanoic acid using the same conditions?

Both ends of the alkene would be converted into a carboxylic acid functional group.

This would be a 'ring-opening' or 'permanganate cleavage' reaction and form adipic acid (hexanedioic).


Mechanisms involved in iodination of salicylamide protocol In the laboratory, I performed an eletrophilic aromatic subsitution by reacting $\ce{I+}$ with salicylamide. The mechanism for this reaction is pretty straightforward, but it's the mechanisms for the supplementary reactions used in this lab that I can't seem work out.

In order to prepare the electrophile  $\ce{I+}$, sodium iodide was reacted with sodium hypochlorite:

$$\ce{ ClO- (aq) + I- (aq) + 2H3O+(aq) -> I+ (s) + Cl- (aq) + 3H2O (l)}$$

The equation (I think) is right, but can anyone provide some mechanistic insight or literature on how the hypochlorite manages to oxidize the Iodide?

Does it first have to go into an iodine ($\ce{I2}$) form? The reason I propose this is because as the hypochlorite was added to the salicylamide/sodium iodide absolute ethanol solution, color change was observed from dark red-brown to increasingly lighter shades of yellow.

It seems a similar question was asked about a year ago, but I was hoping to take it a step further.
Once the reaction had finished, a 10% (w/v) sodium thiosulfate solution was added. I assume this was because thiosulfate is a well known bleach (hypochlorite) neutralization agent: $$\ce{HOCl + 2 S2O3^2- -> Cl- + S4O6^2- + OH-}$$

Assuming the above equation is correct, how exactly (mechanistically) does thiosulfate neutralize hypochlorite/hypochlorous acid? The $\ce{S-S}$ bond of the tetrathionate is whats tripping me up.

The order of events is probably:

  Protonation of the hypochlorite ion

$$\ce{ClO- + H3O+ <=> ClOH + H2O}$$

  Nucleophilic displacement of chloride by iodide

$$\ce{I- + ClOH \phantom{0}<=> Cl- + IOH}$$

  Protonation of hypoiodous acid

$$\phantom{000}\ce{IOH + H3O+ <=> IOH2+ + H2O}$$

  Fragmentation of the $\ce{IOH2+}$ cation

$$\phantom{00000}\ce{IOH2+ <=> I+ + H2O}$$

Alternatively, the second step could have been protonation of hyochlorous acid, followed by displacement of water by iodide, and then fragmentation of iodine monochloride:

$$
\begin{align}
\ce{ClOH + H3O+ &<=> ClOH2+ + H2O}\\
\ce{I- + ClOH2+ &<=> ICl + H2O}\\
\ce{ICl &<=> I+ + Cl-}
\end{align}$$


Molarity of pure water The molarity of pure water is 55.5 M at which temprature is obtained this value? Further, you will come across that 1000 mL = 1000 g (for water only, but at which temprature?)

Water is very unique liquid, because it has a higher density in the liquid state than the solid state. The maximum density of water is found at $\pu{4 ^\circ C}$, which is reported as $\pu{999.9720 kg\:m^{-3}}$ (Temperature Effects on Density). Therefore, molarity of water at $\pu{4 ^\circ C}$ can be calculated as $\frac{\pu{999.9720 g\:L^{-1}}}{\pu{18.015 g\:mol^{-1}}} = \pu{55.51 mol\:L^{-1}}$.

Following chart shows accepted densities of pure water at range of temperatures:

$$\text{Table of Densities of Pure Water at Various temperatures under }\pu{1.0 atm}$$
$$
\begin{array}{ccc}
\hline
\text{Temperature }(\pu{^\circ C }) &\text{Density }(\pu{g\:cm^{-3} }) & \text{Temperature }(\pu{^\circ C }) &\text{Density }(\pu{g\:cm^{-3} }) \\\hline
-30 & 0.98385 & +30 & 0.99565 \\
-20 & 0.99355 & +35 & 0.99406 \\
-10 & 0.99812 & +40 & 0.99225 \\
-5 & 0.99930 & +45 & 0.99025 \\
0 & 0.999840 & +50 & 0.98807 \\
+4 & 0.99997 & +60 & 0.98313 \\
+10 & 0.99970 & +70 & 0.97763 \\
+15 & 0.99910 & +80 & 0.97180 \\
+20 & 0.99821 & +90 & 0.96506 \\
+25 & 0.99705 & +100 & 0.95838 \\\hline
\end{array}
$$

So you may calculate the molarity of pure water at each temperature using the chart. An example for step by step calculations to find the molarity of pure water at particular temperature ($\pu{20 ^\circ C}$):

$$\text{Density of pure water at }\pu{20 ^\circ C } = \pu{0.99821 g\:cm^{-3}}$$
Since molarity $= \pu{mol\:L^{-1} }$, let's consider $\pu{1.0 L}$ of water.
$$\text{Mass of } \pu{1.0 L} \text{ of pure water at }\pu{20 ^\circ C} = \pu{1.0 L}\times \frac{\pu{1000 cm^{-3}}}{\pu{1.0 L}} \times \pu{0.99821 g\:cm^{-3}}= \pu{998.21 g}$$
$$\text{Molar mass of pure water at any temperature} = \pu{18.015 g\:mol^{-1}}$$
$$\therefore \text{ # of moles in } \pu{1.0 L} \text{ of pure water at }\pu{20 ^\circ C} = \frac{\pu{998.21 g}}{\pu{18.015 g\:mol^{-1}}} = \pu{55.41 mol}$$
$$\therefore \text{Molarity of pure water at } \pu{20 ^\circ C} = \frac{\pu{55.41 mol}}{\pu{1.0 L}} = \pu{55.41 mol\:L^{-1}}$$


Correctness of writing H3C as opposed to CH3 in a carbon chain I have been studying alkanes, alkenes & alkynes for a while, and I see in the examples that all the basic structural formulas start with $\ce{CH3}$, then $\ce{CH2}$, and the last carbon atom is $\ce{CH3}$, but the one that is attached to the single/double/triple bond is $\ce{CH}$. (Please correct me if I'm explaining it with the right terms.)

Anyways, there's a question about drawing the structural formula for compounds.

  Draw a structural formula for hex-3-ene.

What my answer is:

$$\ce{CH3CH2CH=CHCH2CH3}$$

I thought this was right, then I looked at the solutions and it was

$$\ce{H3CCH2CH=CHCH2CH3}$$

So now my question is, why does it start with $\ce{H3C}$ not $\ce{CH3}$ like some other examples?

Your answer is the same as the book's. The book wrote the first carbon as $\ce{H3C}$ to stress that the chain continues via a carbon to carbon bond, not bonded a carbon to hydrogen to carbon bond.

As for why it's not like this in other examples, technically it is not necessary to write it this way, but it's strange your book lacks that consistency.


Sugar solutions have a neutral pH in themselves, but it makes your body acidic. Why? I'm not an expert, but as far as I understood a sugar solution is completely neutral since sugar can't take hydrogen ions out of the water or donate them in. Sugar is a non ionic compound, so it does not release H and OH ions in the water so it will not make the solution acidic or alkaline.
I keep on reading and seeing charts of how sugars make your body acidic, like this one:

What process makes a neutral pH solution into an acidic one?
I'm not into chemistry at all and therefore the simpler the answer the better.

It is not proven that "sugar makes your body acidic"!

Your body's pH is very tightly regulated by the body's internal systems; it is also different in different parts of the body - the stomach is acidic (1.0-2.5), the intestine are mildly basic (jejunem 7-9) terminal ileum 7.5 reference here. Blood pH is 7.35, and any deviation from this is indicative of serious illness.


Is it possible to calculate specific heat capacity of a compound based on its elements' heat capacities? What the question says. For example, if I knew the specific heat capacities of lithium and oxygen, could I work out the specific heat capacity of lithium oxide with no further information?

There is not much connection between the heat-capacity of a compound and the heat-capacity of the elements it is made of. Heat capacity is primarily coming from the low frequency vibrations of the lattice (if it is solid), and vibration+rotation+translation if it is liquid or gas. These have no connection with the properties of the lattice etc properties of the given elements.

So short answer: no


what would happen if you got a full package of wet silica gel on your hand after it tore open? I accidently opened a bag of silica gel and got it on my hand what will happen to my hand because now it kinda feels weird.

Like user137 has already written in the comment section it should not be harmful unless exposed for a long time. The main charateristic is its hygroscopy, so skin may feel dry.

To cite from a safety data sheet (source):

4 FIRST AID MEASURES
4.1 Description of first aid measures
If inhaled
    If breathed in, move person into fresh air. If not breathing, give artificial respiration.
In case of skin contact
    Wash off with soap and plenty of water.
In case of eye contact
    Flush eyes with water as a precaution.
If swallowed
    Never give anything by mouth to an unconscious person. Rinse mouth with water.

Nevertheless: Always use the safety data sheets your distributor provided and read them before dealing with the chemicals, if the safety hazards are not clear.


Are there known compounds of caesium with oxidation number >1? Given that the left neighbor of caesium, xenon, does have fluorides and oxides, it is not inconceivable that caesium can have oxides or fluorides with an oxidation number higher than 1. Are such compounds known and synthesized? Or, alternatively, are there computations that rule out such compounds as unstable?

As noted in the answer to the other question, alkali metal can exist in higher oxidation state when bonded with polycylic multidendate ligands like cryptands etc. This is an excerpt from an eBook:

  The chemistry of group 1 elements have been dominated by +1 oxidation
  states. However, there have been indications that caesium might form
  higher oxidation species. Thus, electrochemical oxidation of
  $\ce{[CsL]PF6}$ L=18-crown-6 or cryptand-[222] gives evidence for
  $\ce{Cs^2+}$ and $\ce{Cs^3+}$. Compounds containing caesium in higher
  oxidation state is yet to be isolated.

Also, compounds of form $\ce{CsF_n}$ is known to contain stoichiometric amount of caesium in higher oxidation state($\ce{Cs^2+,Cs^3+,Cs^4+,Cs^5+}$). Experimental calculations have confirmed the presence.(See here and here).

Furthermore, it has also been observed by Moock and Seepelt, Angew, Chem. Intl. Ed. Engl. 28, 1676(1989) that $\ce{Cs+}$ can be oxidised in acetonitrile solution to $\ce{Cs^3+}$ at a potential of +3.0 V($\ce{E°}$). - (See this eBook)


What is the difference between a configuration state function and a Slater determinant? I do not understand the distinction between configuration state functions and Slater determinants. Is not every Slater determinant a CSF ?

Please explain the difference and the use/need of configuration state functions.

Slater determinants are not eigenfunctions of the $\hat S^2$ operator, but CSFs are.

The Hamiltonian commutes with the operators for total and projected spin
\begin{align}
  [\hat H, \hat S^2] &= 0 \\
  [\hat H, \hat S_z] &= 0
\end{align}
Therefore, a set of common eigenfunctions to all three operators exists. As Slater determinants are eigenfunctions of $\hat S_z$, but not $\hat S^2$, using them as a basis for the electronic wave function (=eigenstate of $\hat H$), will not guarantee the (approximate) solution to be an eigenfunction of $\hat S^2$. The spin multiplicity of the found solution may not be a pure Singlet (or Doublet, Triplet, etc.).

As an example, consider a 2 electron system, with both electrons occupying different spatial orbitals. The situation with same spin for both electrons can be represented as the Slater determinant $|\alpha\alpha\rangle=\frac{1}{\sqrt{2}}[\alpha(1)\beta(2)-\alpha(2)\beta(1)]$. The spatial part is omitted here, to keep the notation short. The corresponding eigenvalue equations are then
\begin{align}
    \hat S_z |\alpha\alpha\rangle &= 1 |\alpha\alpha\rangle \\
    \hat S^2 |\alpha\alpha\rangle &= 2 |\alpha\alpha\rangle
\end{align}
Thus we have $S=1$, from $S(S+1)=2$, and the spin multiplicity is $2S+1=3$, i.e. a Triplet. Here the Slater determinant directly corresponds to a CSF.

For the case of opposite spin, we have
\begin{align}
    \hat S_z |\alpha\beta\rangle &= 0 |\alpha\beta\rangle \\
    \hat S^2 |\alpha\beta\rangle &= |\alpha\beta\rangle + |\beta\alpha\rangle
\end{align}
and therefore $|\alpha\beta\rangle$ is not an eigenfunction of $\hat S^2$. The expectation value for total spin would yield $\langle\hat S^2\rangle=1$, which is neither a Singlet nor a Triplet (not even a Doublet where $S(S+1)=0.75$).

This can be fixed by taking suitable linear combinations of Slater determinants. In this example we have 2 options:
\begin{align}
    |^1\Psi\rangle &= \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle - |\beta\alpha\rangle \right) \\
    |^3\Psi\rangle &= \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle + |\beta\alpha\rangle \right)
\end{align}
This yields the eigenvalue equations
\begin{align}
    \hat S^2 |^1\Psi\rangle &= 0 |^1\Psi\rangle \\
    \hat S^2 |^3\Psi\rangle &= 2 |^3\Psi\rangle
\end{align}
which correspond to a Singlet and a Triplet state respectively.

Overall, this yields 1 Singlet component $|^1\Psi\rangle = \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle - |\beta\alpha\rangle \right)$, and the 3 Triplet components $|^3\Psi\rangle = \frac{1}{\sqrt{2}} \left( |\alpha\beta\rangle + |\beta\alpha\rangle \right)$, $|\alpha\alpha\rangle$ and $|\beta\beta\rangle$.

For the arithmetic on how to apply the $\hat S_z$ and $\hat S^2$ operators to multi-electron systems, see Chapter 2.5 in Modern Quantum Chemistry by A. Szabo and N. Ostlund.


How do IR spectrometers produce light spanning the entire infrared spectrum? An IR spectrum covers thousands of wavenumbers, typically $\sim 4000$ to $\pu{400 cm-1}$.

How are the spectrometers able to generate different wavelengths of light covering this entire range of wavelengths?

This could be accomplished by a scanning instrument. A broad-band light source could output IR radiation over the whole range. This incident radiation is passed through a monochromator to individually select wavelengths to send to the sample. This monochromator can work many ways, using mirrors, gratings, prisms, etc.

Alternatively, you could expose the sample to the broad-band radiation and use a detector that captures all of the transmitted radiation at once. UV/Visible spectroscopy can be done with a photodiode array detector capable of this, but I know of no IR spectrometers that use this method.

Most modern IR spectrometers use Fourier transform. An interferometer, converts an IR laser beam into broad-band IR pulses through interference generated with moving mirrors or by other methods. This action converts the coherent beam into an interference pattern in time. The interferogram below shows what this might look like (where this mirror position is also the time axis). The image has been released to the public domain by Wikipedia user Sbyrnes321.

Absorption of IR radiation by the sample perturbs the interference pattern before it hits the detector. Electronics or software perform the Fourier transform, converting time data into frequency data.


Differences between Fe3+ and Fe2+ I would like to know the differences between Fe3+ and Fe2+ ions. I'm guessing they must differ in reactivity but I'm not sure how. Also, is there a way that we can observed the differences

The only thing that's different about ${Fe^2}^+$ and ${Fe^3}^+$ is the number of electrons, which in turn results in different properties of these species.

${Fe^2}^+$, aka ferrous, is pale green and turns violet when added to water. ${Fe^3}^+$, aka ferric, is yellow-brown in solution. You can use several methods to distinguish these two ions. I will mention some of them.

${Fe^2}^+$ can oxidised to ${Fe^3}^+$ with the help of oxidizing agents like $KMnO_4,\space HNO_3$, etc.. $KMnO_4$ is better because its pink color decolorizes on addition of ferrous ions.

Another method to distinguish them is by using colored complexes. For eg., ${Fe^2}^+$ forms a red-orange compound with amine lignads. ${Fe^3}^+$ forms blood red solution with thiocyanate ions, which is used as fake blood in movies, dramas and so on.

Ferrous ions normally have paramagnetic properties, but can become diamagnetic due to the formation of low-spin complexes. A ferric ion is only paramagnetic due to the presence of only one lone electron.


What happens if you make a gas of magnetic particles I was reading todays SMBC which asked what happens if you make a gas of magnetic particles

So what does happen if you make a gas of magnetic particles?

If you heat up the magnet above Curie temperature (or more drastically just evaporates it), the magnetic spins will be randomly aligned and the magnetic field  is not produced anymore (at higher temperature).


What is the heat energy required to perform electrolysis equivalent? The question may be off-topic here. If my concept is flawed, please poke me in the rib (+:

Sources on the internet indicate a minimum 1.2Volts is adequate for electrolysis of water; nominally 6V. Applying a DC source to water provides adequate energy to break the hydrogen bond. Apparently some heat is also released in the process..

Say, one were to collect 100ml (an arbitrary figure) oxygen by electrolysis applying 6V DC to 1 Litre of water.

Can the energy to break the hydrogen bond be provided by applying heat to water? All other things being identical, how many calories would be required to collect the same amount of oxygen from the same volume of water?

I won't do any rib poking, but there is some flawed thinking here.  Certainly not off topic though.

Nomenclature: First, when you pass a current through water, you are not breaking the hydrogen bonds, which are intermolecular forces, but (a) breaking the bond between hydrogen and oxygen and (b) forming molecular oxygen and molecular hydrogen, $\ce{2H2O -> 2H2 + O2}$, and this reaction can be broken down into two half reactions: $$\ce{4H2O + 4 e- -> 2H2 + 4OH-}$$ $$\ce{2H2O -> O2 + 4H+ + 4e-}$$

Potentials: The 1.2 V you refer to likely arises from the standard reduction potential of oxygen, which is the reverse of the second reaction above.  I won't go in to the thermodynamics here, but suffice it to say that a brief look at a standard reduction potential table indicates that the reduction of water (first equation) has a potential of -0.83 V and the oxidation of water has a potential of -1.23 V for a grand total of ~ -2 V. (Note, these values are pH dependent and standard reduction potential tables report values at a pH of 0.)

Electrochemistry When you apply an overpotential to a system (the potential beyond what is needed to get the oxidation/reduction reactions to go forward) then two things will happen (a) the reaction will go faster and (b) heat will be generated.  This heat would be analogous to heat due to friction in a physical system and is considered wasted energy.

Thermochemistry Finally, we get to the answer to your question, which is "probably not, but the way you asked it, yes"  (hunh?)  Recall the nomenclature issue above.  Adding heat to water will break the hydrogen bonds in water eventually leading to the water changing from a liquid to a gaseous state.  However, adding heat to water would not result in the oxidation/reduction reactions outlined above because an important ingredient is missing: electrons.


Is table salt sticking to my laptop? I was reviewing first posts here, and eating a juicy and salty (table salt) tomato at the same time. Suddenly, I realized that a few table salt crystals have been poured down on my mousepad.

I got curious and I rotated the laptop $90°$. I witnessed that the salt is somehow resisting to fall! Not sticked (very firmly attached) though, and the mousepad wasn't wet and didn't contain any possible chemicals that could cause the "attaching";

except that apparently the mousepad is "soaked" with the sebum produced by the tip of my fingers. Sebum, however, is a mixture of non-polar lipids such as fatty acids, triglycerides, sterols, sterol esters, wax esters and squalene. (See here also)

There can be another reason which lies in the electrical field that my laptop produces, and this could affect a compound like table salt that is primarily ionic. (No clue how, am I missing an easy chemistry concept?)

So, is there any possible "mild" (as in invisible) reaction, either chemical or physical, as a reason of the phenomenon? Or is this behavior because of the composition of table salt?

Note that I'm certain about the existence of such thing, though research showed nothing relevant in the Net.

More than likely the case of your laptop became charged and was attracting the salt. Yes salt crystals can be coerced by an electrostatic field.

Here is a post on a teaching site that actually uses salt (and pepper) to demonstrate electrostatic attraction.

https://www.teachervision.com/electricity/lesson-plan/5787.html


What units do I use for energy in the Boltzmann distribution? I am a mathematics student doing a course in thermodynamics and am working on the basics now. The question I am asked is:

  Suppose two states differ in energy by one electronvolt. What can be said about the ratio of the population at $T = \pu{300 K}$?

Now it is clear that I will use the formula:

$$\frac{N_i}{N_j} = \exp\left(-\frac{E_i-E_j}{kT}\right)$$

where $k = \pu{1.381 \times 10^−23 J K-1}$. Now if let $E_i$ be the higher energy state, then am I right in saying that $E_i-E_j = \pu{1 eV}$, or should I convert this to some other unit?

Any quantity that is used as a power has to be dimensionless (that is, a pure number with no units). This means that $kT$ has to have the same units as $E_i-E_j$, so the terms in the exponent cancel. If you're measuring temperature in $\pu{K}$ and Boltzmann's constant in $\pu{J/K}$, then the units of energy you have to use are joules, $\pu{J}$.


Thermionic Emission Conservation of Energy I am learning about thermionic emission and how it led to the vacuum tube. It is said that thermionic emission was a way to increase the flow of electrons from a signal (e.g. in a radio receiving a weak signal, boosted by thermionic emission). What I don't understand is how it "generates" electricity; even if the coil is heated to it's work threshold, wouldn't a coil only be able to emit as many electrons as are supplied?
Thank you for your time

I was mistaken on the principle that governs vacuum tubes. It is not that the signal is being amplified by thermionic emission; rather, the radio wave is being used as a gate to allow or block electrons from being received. Thermions are being emitted from a coil using electricity, emitting at a higher current, modulated by the radio wave as a control grid (preventing or allowing the thermions from being collected at the anode).


Why do weak acids like ethanol not act as a strong base in water? In water, ethanol has a $\mathrm{p}K_\mathrm{a} = 15.9$ which means that its $\mathrm{p}K_\mathrm{b}$ is $-1.9$. Which implies that all weak acids in water are in fact strong bases. But this is not true, weak acids in fact do act like weak acids. That would mean that this equation isn't true always.

$$14 = \mathrm{p}K_\mathrm{a} + \mathrm{p}K_\mathrm{b}$$

Considering this is just the ionic product of water it should be valid always?

There seems to be a misunderstanding: The conjugate of a strong base like EtONa can't be a weak acid, because weak acids, like acetic acid for example, are still quite acidic.

Strong acid (HCl) - very weak conjugate base (chloride anion)
strong base (EtONa) - very weak conj. acid (Ethanol)

A strong acid or base reacts strongly (with water), the resulting conjugate is rather inert, with the equillibrium fully on the side of the latter.

weak acid (acetic acid) - weak conj. base (acetate anion)
weak base (ammonia) - weak conj. base (ammonium)

A weak acid or base just reacts partly towards its conjugate, so does the pure conjugate: They are both rather weak. The equillibrium in an equimolar mixture is rather stable: A buffer solution.

The ideal weak base or acid has a $pKa$ of 7 (in water), that is $1/2$ of the -log ion product of the given solvent. If it's even weaker, the conjugate becomes stronger again.


Calculation of empirical formula from elemental analysis A compound contains 40.0% Carbon, 6.7% Hydrogen, and 53.3% Nitrogen (by mass). Calculate the empirical formula.

I'm getting $\ce{C7H14N8}$, but my professor says it is $\ce{CH2N}$.

Which is the correct answer?

When brought to moles, 40 g C is 3.33 moles of C, 6.7 g of H is 6.63 moles of H, and 53.3 g of N is 3.80 moles of N.

Divided by the lowest mole count (3.33), you get one carbon, two hydrogen, and 1.14 nitrogen. I am multiplying all by 7, to get $\ce{C7H14N8}$.

He is saying 1.14 is 1, basically, and that is can all be simplified as it is, no need to multiply, and your final answer is $\ce{CH2N}$, marking my answer half off. It's not worth going to battle for, but I want to know who is really right, because I feel my answer is better.

So the real question is, with N, should you get it closer to a whole number, or leave "as is"? Thanks.

A sad Ph.D. student had been working on the synthesis of a natural product, with empirical formula $\ce{CH2N}$, for several years. He finally succeeded in synthesising it - or so he thought - because his NMR data seemed to fit the target structure. Excited about his breakthrough (in fact, he was so excited that he didn't even bother taking a mass spectrum!), he sent a sample for elemental analysis, expecting to get some results along the lines of 42.8% carbon, 7.2% hydrogen, and 50.0% nitrogen.

Not long after this, the results came back, and he discovered that his compound inexplicably had 40.0% carbon, 6.7% hydrogen, and 53.3% nitrogen. "That's not what I expected," he thought to himself - "I wonder if I can publish it and try to gloss over the data."

Upon checking the Author Guidelines for the Journal of Organic Chemistry (section 2.2.2), he found that they were actually pretty strict.

  Found values for carbon, hydrogen, and nitrogen should be within 0.4% of the [calculated] values for the proposed formula.

Thus was the story of the sad Ph.D. student who had to go back to the drawing board. (Thankfully, he had synthesised enough of the immediate precursor and by varying the conditions, he eventually succeeded in making his target.)

TL;DR Rounding down 1.14 to 1 is a gross error and won't ever get published in a reputable journal. Considering that this is not even a real-life scenario, there's absolutely no reason why standards should be lowered.


Removing Iron Oxide from a Floppy Disk? Don't ask me why, but I did an experiment where I tried to remove the magnetic iron oxide layer from a floppy disk by soaking it in HCl for a week. I put pieces of floppy disk in 0 - 1 M HCl, then measured their absorbance at 700 nm. HCl concentrations at 400 mM or more were able to remove about half of the absorbance, but not more than that.
We can even visibly see the change in color intensity.

This suggests that something is preventing the acid from dissolving the remaining iron oxide, and that simply increasing the concentration beyond 1 M won't help. However, I can still scratch the surface and remove some of the color, suggesting that it isn't buried throughout the plastic film, but still accessible on the surface. But I don't want to spend forever scratching the iron oxide off floppy disks, so I would still like a simple chemical method to remove the metal. Does anyone have any suggestions for removing it?

I have to say thank you to Maurice for his suggestion to use more concentrated HCl. I took the same pieces of floppy disk that I used before and increased the acid concentration by a factor of 10 and let them sit for another week.
At first I was disappointed, because there was no visible change in the plastic or the solutions. But when I took the pieces out of the acid and cleaned them off with an alcohol soaked kimwipe, the remaining Iron Oxide came right off. It's not perfectly clean, most likely because I had taped the pieces into my notebook after the first treatment, and I didn't clean the tape residue off before the second treatment. This is also probably why the remaining absorbance at 700 nm is so variable. But I think this method works as long as I don't get the plastic covered in tape goo. The acid treated plastic also might feel a little less stiff than the 0 M control, but this is subjective.


How does Hofmann elimination work in this reaction? In my Chemistry class, we were given this question to solve, and I'm not sure I fully understood the solution.

I understand that since the substrate is bulky, Saytzeff elimination can't happen, so it has to be Hofmann. What I don't understand is how it happens here. Where does the base attack the substrate? Essentially, I'd like to know the mechanism of the reaction.

Also, our teacher made us write that had all three substituents on the nitrogen atom been methyls, Saytzeff elimination would have been followed as there wouldn't have been any other choice. However, even if all three substituents were methyls, wouldn't the substrate still be bulky? Why would it follow Saytzeff elimination?

Wow, this is an amazing question. The expected reactivity is strongly dependent on the exact structure.

For starters, trimethylammonium is about the size of a tert-butyl group. So, expect an A value around 4.9, which is very large. So, put that group equatorial on the ring.

Once you do this, you see that there are only 2 hydrogen atoms that are anti-periplanar for an E2 elimination. One is on the axial methyl group (red). The other is on the N-ethyl group (blue). Only these two can easily be deprotonated for elimination. Of these two hydrogen atoms, the one on the axial methyl is very sterically crowded. Specifically, any base has to get around the ring structure and the encroaching flag pole hydrogen atoms. On the other hand, the hydrogen atom on the ethyl group sees no such steric encumberance.

Of the three elimination products, I expect to see:

Almost exclusively ethylene
Maybe a small amount of the methylidenecyclohexane
No methylcyclohexene

EDIT:
@Martin-マーチン was kind enough to compute the structure below. Notice the structural features we've already discussed: the ring conformation and the position of the anti-periplanar hydrogen atoms.


Lab safety - Long vs. Short Sleeves on Lab Coats I have three lab coats, two with long sleeves, and one with short sleeves.
I saw that the concepts about lab safety vary from country to country; in some places it looks like the short-sleeve coat aren't allowed or they tell you to avoid it.

I'm a little confused about it: When should I use long-sleeves and when should  I use the short-sleeve coat?

Are the short-sleeves really wrong in a chem lab?

The short sleeve has no place in a chemistry lab.

Considering the coats purpose as personal protective equipment (PPE) is to reduce or eliminate the potential or risk of exposure to materials then the long sleeve coat would be the correct PPE to wear in a chemistry laboratory along with extra PPE which should have been assessed prior to performing the procedure.

Even in a microbiology or molecular biology lab, where a lot of the solutions might not seem harmful, the cellular material in suspension is classed as a bio hazard, which the long sleeve will provide better protection.

Hazardous chemicals are still commonly used in these labs, too, and will most likely be present whether you using them or not. It's not just for chemical exposure either also for general protection.


Which has the highest melting point among NaF, NaCl, NaBr, NaI? I've been trying to understand why $\ce{NaF}$ has the highest melting point among all of the given solids. I thought it would be $\ce{NaI}$ as it will have the biggest molar mass among them which would mean it has the strongest intermolecular forces. Is it electronegativity difference or molar mass that you look at when analysing intermolecular forces? Am I missing something or misunderstanding something? Thank you very much.

All those are ionic compounds, so it is just Coulomb forces that hold them together. Two charges are attracted with the same force regardless of their mass. The distance, however, does matter a great deal. And the distance depends on the ionic radii, and I'm sure you know how those change down the group.

Things could have turned out otherwise if the compounds had different crystal structures. Luckily for us, this is not the case.

So if you are confronted with an unknown compound, you first look at the electronegativity difference, and if it tells you the compound is ionic, you use the same line of thought as I just did. Otherwise it is probably a molecular compound held together with dispersion forces, and they tend to grow with molar mass. Oh, and there might be a covalent solid...

See, chemistry is complicated, and it is not always possible to get by with a few simple rules of thumb. But at least I think I've answered your original question in the narrow sense, so let's leave it at that.


Solutions, solvents and solutes In all the chemistry books and eBooks it says that "like dissolves like"... but when it comes to some molecules like ether, cellulose, vitamin A and $\ce{CO_2}$, it confuses me.

If we take ether, it is true that it will not form hydrogen bonds with water when it dissolves, but they have polar bonds between $\ce{O}$ and $\ce{C}$ and it's not in a linear form like $\ce{CO_2}$ where the dipoles cancel out.

Vitamin A also has $\ce{-OH}$ group but it still is hydrophobic. In some book, I read that this is because it has more hydrophobic molecules or that they have more $\ce{C-H}$ bonds than $\ce{-OH}$. If it is so, octadecanol and nonadecanol also have way more $\ce{C-H}$ bonds than $\ce{-OH}$ groups, so  why is it still soluble? $\ce{CO_2}$ is dipolar — the dipoles cancel out, so why is it still soluble? Cellulose also has a bunch of $\ce{-OH}$ but still it is soluble. I am very confused by this.

When you make assumption for guessing the solubility of the molecule in some solvent you should consider the whole molecule, his structure and his size.

We can start with the ether, in the below figure (from wikipedia the one at right ) is shown a diethylether molecule.

You can see that is quite a non-polar molecule (charge 0,green, in the most part of the surface) so this explains why has a low solubility in a polar solvent like water. Same thing for Vitamin A:

The most part of the molecule is not polar. Only one $-OH$ can't make the difference. And so is for 1-Octadecanol that has a solubility in water of $1.1  \times 10^{-3} mg/L$ and Nonadecyl alcohol (1-nonadecanol) seems something like $4.8  \times 10^{-3} mg/L$ both up to my references have very low solubility in water (where did you find that were not?). For the $CO2$ is true what you have stated it has not a very strong polarity (but in fact it has a little bit!) because is a linear molecule.

This last example let me introduce to an important factor that influence the solubility: molecule size.
Little molecule are more easily solvatated by solvent molecule, this bring us to the case of cellulose:

Cellulose is very hydrophilic and so should be soluble in water, but this doesn't happen because cellulose is a polymer (in figure is show his monomer) that can have a molecular weight very hight. Polymers may swell but can't be solubilize if their chains are bonded with chemicals bonds until the bonds are break.


what is the mass of chromium in this sample? If the total number of atoms in a sample of potassium dichromate is $2.13\cdot10^{24}$ , then the amount of chromium in this sample is?
the answer that I've come up with is $367.64\ \mathrm g$
but in fact the solution in my book is $33.4\ \mathrm g$.

I like to solve such assignments using Python as a scientific calculator.
import scipy.constants

# number of atoms in sample
N_ATOMS_TOTAL = 2.13E+24

# atomic mass of chromium
M_CR = 51.9961 # g/mol

# ratio of chromium atoms to total atoms in K2Cr2O7
CR_RATIO = 2/11

# moles of Cr in sample
n_cr = N_ATOMS_TOTAL * CR_RATIO / scipy.constants.Avogadro

mass_cr = n_cr * M_CR # mass of chromium in g

mass_cr

Let me explain:

You have an unknown mass (in gramm) of potassium dichromate, $\ce{K2Cr2O7}$.

The total number of atoms in this sample is $2.13\cdot10^{24}$.

How can you calculate the number of chromium atoms in the sample?

You need to determine the ratio of $\ce{Cr}$ to all atoms in $\ce{K2Cr2O7}$, and that's $\frac{2}{11}$ atoms.

I know the number of $\ce{Cr}$ atoms now, but how many moles is that?

You know that 1 mol of an element has $6.02214129\cdot10^{23}$ atoms of that element (Avogadro's number).
Divide the number of $\ce{Cr}$ atoms determined before by Avogadro's number to determine the number of moles of $\ce{Cr}$.

I know the number of moles of $\ce{Cr}$ now, but what is the weight?

You know that the standard atomic weight of $\ce{Cr}$ is 51.9961.
Multiply  the standard atomic weight with the number of moles to get the weight.


Anion-cation balance in particle phase I'm now analysising the water-soluble ions in atmopsheric aerosol. With ultrasonic concussion and filtering impurity, the anion and cation from the particle sample are examined  in Ion Chromatography System(IC) separately.

We test $\ce{Cl-, SO4^2-, NO3-}$ and $\ce{Ca^2+, Mg^2+, Na+, NH4+, K+}$ as 8 major ion species to quantify the concentration level and seasonal variation in the sampling area.

But when I combine the anion and cation data together for same particle sample, the ratio of [ Total charge of anion]/[Total charge of cation] are much larger than  1.

In my opinion, the cation and anion inside particle should keep the balance of charge. The 8 ion species I choose should take up the majority of the total water-soluble ions (added with $\ce{F-}$, organic acid radical, etc).

I have also searched some paper, some results  of their ion-sample are also not keep the balance, but some observed.

Are the ratio of (+-)charges differed by sampling sites? The anion especially $\ce{SO4^2-}$ are larger comparing with the cation species. Are there any missing cation that I have not deteced?

Yes, you have left out one very important cation from your cation/anion balance: $\ce{H+}$

The primary pathway for new particle formation in the atmosphere is the photo-oxidation, via hydroxyl radicals, of volatile reduced sulfur species to form very non-volatile sulfuric acid.  For the most part then, new particles start out as puny droplets of sulfuric acid and grow via accretion, etc. from there.

I recommend measuring the pH of the filtered aerosols, and I strongly suspect that will account for your missing cations.

If I have missed the point of the question, please feel free to ask for any clarifications in the comments.


What is the purpose of this glassware? I have seen these small test tube looking doohickeys being supplied from many chinese suppliers, but never with any description as to what they are for. They almost invariably have the description "Automatic Filling Tube Liquid Test Tube With Spout" which didn't lead me anywhere.
With most glassware, even relatively complex pieces, I can usually discern what it is for and why it is designed in that way but this has me stumped. My best guess so far is that it is meant to collect liquid from a sealed flask where the long stem passes through a bung to the bottom of the flask and gas pressure pushes some liquid up and into the tube. So maybe it is for volumetric reagent dispensing? If you had a squeeze bulb on the bottle it would push liquid up into the tube, then suck back and reduce the level to a consistent volume, but you would have to tip the whole bottle to pour it out so idk?
If anyone is familiar with what these are for and how they are used it would be much appreciated because this has been bothering me for a while.

It is an apparatus for taking a sample of a liquid from a big container. Let's suppose you have at your disposal a big bottle (at least one liter) which has two holes on a level with one another at middle height. Now imagine that a flow of a small river is entering the bottle from the upper neck, at such a rate that the level of the water inside is always on a level with the holes. The water gets out through the lateral holes. Now you put a punched rubber stopper in one of these holes. And the long thin pipe of your glassware is horizontally inserted into the stopper hole. Your glassware can then be turned and placed in such a way that the broader tube is vertical with its neck above. With this equipment, the broad tube is all the time half-filled with water. If the composition of the water slowly changes, a sample of its composition may be taken by turning the broad tube so as to empty it upside down, then putting it vertically again.


Mixing liquids when applying a voltage I'm thinking in the following experiment:

A) Two (non mixable (like water and oil)) liquids are going to lie in a container, in the extremes of the container there are two electrodes, those electrodes will have a potential difference $\Phi$. Just as the figure $1.a)$.

I'm looking for  some interesting liquids that mix, change of color when a voltage difference is present.

B) The next one is really similar, now there are some solid particle present, analogously I look for a mixture that has different properties when a voltage difference is present.

I'm a physics student that want to study different properties of mixtures when applying different voltage differences, and maybe study the diffusion of the solid particles when they mix, if the diffusion coefficient is grater, etc.

Any help with the liquids that has this property is really  appreciated.

Cross posted Physics stack Exchange

I won't comment on B for the moment, but liquid-liquid interface electrochemistry is quite a vibrant field. There's a review of it here, if you're interested. The simplest interaction involves ions: if one applies a potential across the interface of sufficient strength and polarity, the energy barrier of transferring a given ion from one phase to the other can be overcome. If you do this with something like a solvatochromic dye, it will change colour as it moves from one phase to the other. (It would have to be something that doesn't partition into the second phase easily on its own, of course) One could also have some reagent in one phase that reacts with the ion being pushed across the interface that produces some coloured product. (iodide-starch, perhaps?) There are also voltage-sensitive dyes, though I don't have any experience with them.

One thing to note is that neither water nor oils and other immiscible solvents are generally very conductive. To keep the cell resistance down to a reasonable level, it will be necessary to dissolve electrolytes in both. Things like Tetrabutylammonium hexafluorophosphate are commonly used as electrolyte in non-aqueous electrochemistry.


What does "lit." mean in boiling point or melting point specification? For example, this Xanthene 99% specification from Sigma Aldrich states:

\begin{array} \\\hline
\text{bp}  & 310-312 ~^\circ\mathrm{C~ (lit.)} \\\hline
\text{mp}  & 101-102 ~^\circ\mathrm{C~ (lit.)} \\\hline
\end{array}

My guess is that it's from literature.

Yes, you are correct: “lit.” refers to the literature value.
Usually one would also expect the corresponding reference, but it looks like Merck doesn't attach any to the provided data directly in place.
From The ACS Style Guide, section Reporting Analytical Data [1, p. 274] (emphasis mine):

Melting and Boiling Points
mp 175.5 °C (lit.25 mp 175–176 °C)
mp 225 °C dec
bp 127 °C
Abbreviations: mp, melting point; bp, boiling point; lit., literature value; and dec, decomposition. A full space is used between the number and the unit °C; the
degree symbol is closed up to the C. A superscript number after “lit.” denotes the
number of the reference.

References

The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. ISBN 978-0-8412-3999-9. https://doi.org/10.1021/jm068034q


1H NMR splitting of 3-methyl-1-butanol I have a question about 1H NMR splitting of 3-methyl-1-butanol.

The spectrum looks as the following.

The assignment of the NMR spectrum is the following.

Assign.     Shift(ppm)

  A             3.673
  B             1.66
  C             1.57
  D             1.49
  E             0.922

So based on the results, A hydrogen is splitted into triplet. Is this splitted by D hydrogen? If so, why not C hydrogen on the alcohol group?

The results are retrieved from the following link: https://www.chemicalbook.com/SpectrumEN_123-51-3_1HNMR.htm

In this example, HA is indeed split by HD, but you have quite rightly identified that we don't see coupling to the alcohol proton, HC. This is pretty normal that we don't observe couplings to protons that are termed labile; -OH, -NH2, -SH, for example. Whenever samples contain trace amounts of an acid or base, there will be very fast exchange between the alcohol proton and that other species (water is often the culprit). In solvents like chloroform (which this spectrum was run in), the exchange is very fast, and you just observe a single peak. In other solvents, such as dmso, you might observe two broadened peaks. But, the exchange means that any coupling to that nucleus is lost

We do sometimes observe coupling to alcohol protons. If you can eliminate exchange, or slow it sufficiently, you will observe 3J couplings as you would for any other nuclear coupling pathways. 3J couplings to alcohol protons are in the order of 3-6 Hz, commonly.


Dichromate ion: Tetrahedral or bent? I am currently studying about the vespr theory and I'm stuck on an ion which is dichromate (VI) ion $\ce{Cr2O7^2-}$. So this is what I am thinking about:

1) If I make the oxygen atom at the middle the centre atom, it is a bent.
2) If I make the chromium atom thecentre atom, it is a tetrahedral.

So, my problem is that whether dichromate (VI) ion bent or tetrahedral?

Both, Bent about Oxygen and tetrahedral about chromium:

$\hskip2.5in$


How can change in Enthalpy for a reaction be written like this? If a gas expands in a container with constant external pressure then:
$W=-P_{ext}\Delta V$
Instead of $P_{ext}$ if we use $P_{int}$ then we will not get the same value because  $P_{int}$ is constantly decreasing from a certain value until equilibrium is attained, while the change in volume remains the same.
This is the reason why we use $P_{ext}$ here:
$\Delta H= \Delta U +P_{ext} \Delta V$
But then, $\Delta H$ for a reaction is:
$\Delta H= \Delta U +\Delta n_{g}RT$ ($n_{g}$ is for change in number of gaseous moles as the solids and liquids are ignored)
where $PV=nRT$ is used. However, the $P$ in the ideal gas equation is $P_{int}$.
So my question is how can $P_{int}$ be substituted in the place of $P_{ext}$ if the values of work done change when we do so?

Your claim that $\Delta H = \Delta U + P_{ext} \Delta V$ is incorrect. It is only correct when the system is always in mechanical equilibrium with the external pressure, or in other words irreversible expansion against constant pressure. The more general statement is $\Delta H = \Delta U + \Delta (PV) $. Here as you can see even when you compare it with $\Delta H = \Delta U + \Delta n_gRT$ there is no problem as we don't have to replace $P_{in} $ with $P_{ext} $.


Why is (tri-)ammonium phosphate ((NH4)3PO4) so unstable? Why is triammonium phosphate, $\ce{(NH4)3PO4}$, so much more unstable than diammonium phosphate, $\ce{(NH4)2HPO4}$, or monoammonium phosphate, $\ce{NH4H2PO4})$?

Ammonium phosphates are solids between 0 and 75°C. Because ammonium
  hydroxide $(\ce{NH4OH})$ is a much weaker base than the common metal
  hydroxides, ammonium phosphates are comparatively unstable.  Both
  triammonium phosphate ($\ce{(NH4)3PO4}$) and the double salt
  ($\ce{(NH4)3PO4.2(NH4)2HPO4}$) are unstable at room temperature and
  evolve ammonia to form diammonium phosphate. Even the commercial mono-
  and diammonium phosphates exhibit an ammonium vapor pressure in the
  solid form and in solution. Although MAP is more stable than DAP, it
  decomposes at high temperature to give ammonia and polyphosphoric
  acid. Diammonium phosphate decomposes to give ammonia and monoammonium
  phosphate at around 70°C.

Source: (2005) Ammonium Phosphates and Ammonium Sulfate. In: Synthetic Nitrogen Products. Springer, Boston, MA. DOI: 10.1007/0-306-48639-3_12.


Materials used in metallic glass I've been looking over the lists of materials used together to make metallic glass.

Is it purely coincidental that if you round up/down any decimals from the atomic weight of elements used to construct metallic glass, you're always very close to +-1 (if not at) a prime number for the atomic weight of each element?

Yes, it’s coincidental. There is no expectation that prime (or near-prime) atomic weight means anything for the element itself. In fact, your affirmation is not even true: on the list given by Wikipedia, the first two examples are:

$\ce{Au_75 Si_25}$, your assertion is true
“an alloy of 77.5% palladium, 6% copper, and 16.5% silicon”, and your assertion isn’t met for copper

Let's go further. For a given element, how likely is it that its atomic weight, rounded, is within ±1 of a prime number? Well, atomic weights are relatively low integers, and prime numbers are plenty numerous among small integers, so it is relatively frequent. The answer can be computed with Mathematica:

weights = {#, Round@ElementData[#, "AtomicWeight"]} & /@ ElementData[All];
f[i_] := PrimeQ[i] || PrimeQ[i - 1] || PrimeQ[i + 1]
res = {f[#[[2]]], #[[1]]} & /@ weights
Select[res, ! #[[1]] &][[All, 2]]

74 out of the first 118 elements turn out to satisfy your “near-prime atomic weight” condition, which is 63%. This seems high, but it's not: if you do the same check, not only for atomic weights, but for all integers below 200, 61% of them are near-prime:

Total@Boole@Select[f /@ Range[1, 200], # &]


Nitrogen chain nomenclature I know that nitrogen is versatile enough that it can form branched chains if it is all single bonds.

For example:

$\ce{H2N-N(NH2)-NH2}$

Is this compound a nitrane (nitrogen alkane)?

If so how would I name this branched nitrogen compound?

According to the IUPAC Gold book, these are known as azanes, from the information:

  Saturated acyclic nitrogen hydrides having the general formula $\ce{N_{n}H_{n + 2}}$

For the example you are looking at, according to the IUPAC document P-67 Mononuclear and polynuclear noncarbon acids and their functional replacement
analogues as parents for naming organic compounds (p. 15) is named as 2-aminotriazane.


What is the pKaH of pyrrole? I am supposed to say what is more basic and I ran into a problem. I would like to ask if anyone has any idea what is the correct $\mathrm{p}K_\mathrm{a}\ce{H}$ value of pyrrole. I know it is easy to just google it or open a book to find the answer. I have done that, but got pretty interesting (confusing?) results:
Wikipedia: "Pyrrole is weakly basic, with a conjugate acid $\mathrm{p}K_\mathrm{a}$ of (−3.8)"
Human metabolome database (HMDB): "Pyrrole is a very weak base with a $\mathrm{p}K_\mathrm{a}\ce{H}$ of about (4)"
PubChem: "Pyrrole is a very weak base with a $\mathrm{p}K_\mathrm{a}\ce{H}$ of about (4)"
McMurry 9th edition: "pyrrole - $\mathrm{p}K_\mathrm{a}$ of ammonium ion: (0,4)"
University of Calgary: "pyrrole conjugated acid $\mathrm{p}K_\mathrm{a}$:  (-3,8)"

Op's question is what is the correct $\mathrm{p}K_\mathrm{a}\ce{H}$ value of pyrrole. Thus, I'm not going to elaborate OP's findings, but would try to give a reasonable answer to the question. The most reasonable answer I found for $\mathrm{p}K_\mathrm{a}\ce{H}$ value for the pyrrole is $0.4$ (for the novices, this is the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid of pyrrole).

It is well known that the best way to quantify the basicity of an amine is to examine the $\mathrm{p}K_\mathrm{a}$ of its conjugate acid (Master Organic Chemistry). The concept is the higher the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid, the stronger the base.

Let's compare two cyclic amine bases under this concept: Pyridine and Pyrrole. The following diagram illustrate the concept:

Accordingly, pyrrolium ion is about $10^5$ times stronger acid than pyridinium ion. Thus, pyrrole is about $10^5$ times weaker base than pyridine. Why? Read the note on the bottom of the box in the diagram. This is also well explained in chemistry.msu.edu with the diagram attached:

  In each case the heteroatom has at least one pair of non-bonding electrons that may combine with the four π-electrons of the double bonds to produce an annulene having an aromatic sextet of electrons. This is illustrated by the resonance description at the top of the following diagram. The heteroatom Y becomes $\mathrm{sp^2}$-hybridized and acquires a positive charge as its electron pair is delocalized around the ring. An easily observed consequence of this delocalization is a change in dipole moment compared with the analogous saturated heterocycles, which all have strong dipoles with the heteroatom at the negative end. As expected, the aromatic heterocycles have much smaller dipole moments, or in the case of pyrrole a large dipole in the opposite direction. An important characteristic of aromaticity is enhanced thermodynamic stability, and this is usually demonstrated by relative heats of hydrogenation or heats of combustion measurements. By this standard, the three aromatic heterocycles under examination are stabilized, but to a lesser degree than benzene. Additional evidence for the aromatic character of pyrrole is found in its exceptionally weak basicity ($\mathrm{p}K_\mathrm{a}$ ca. $0$) and strong acidity ($\mathrm{p}K_\mathrm{a} = 15$) for a $2^\circ$-amine. The corresponding values for the saturated amine pyrrolidine are: basicity $11.2$ and acidity $32$.

Furthermore, for everybody's convenience, I attached a orbital representation of pyrrole and furan to show their resistance to protonation.

Comment on different $\mathrm{p}K_\mathrm{a}$ for pyrrole in literature: The $\mathrm{p}K_\mathrm{a}$ value of $3.8$ (and $\approx 4$) is probably for the protonation of pyrrole at $\ce{C2}$ position. For protonation at $\ce{N}$, $\mathrm{p}K_\mathrm{a}$ is most probably $0.4$, as described above. We can trust the value given in McMurray's Organic Chemistry textbook, since it is widely used in sophomore organic chemistry university course in USA.


Copper chloride color in electrolysis of salty water I tried a simple water electrolysis experiment at home, with $\ce{NaCl}$ as electrolyte, and a $12 \,\rm{V}$ battery.

My wires were made of copper and $\ce{H2}$ bubbles were only forming at negative terminal (cathode). On the positive side (anode), nothing was happening.

After $3$-$4$ minutes, water started to change into a greenish-blue color and I realized that it must be copper reacting with chloride ions. After $15$ minutes, my solution was light blue with exceed $\ce{NaCl}$ at the bottom, after taking the wires out the solution started to turn into a yellow color in about $10$ minutes.

What was the yellow solution and why didn't cathode produce any oxygen?

I'm also worried that the yellow solution might be chlorine because I later boiled the solution.

The color of the solution is probably due to the presence of blue copper(II) ions ($\ce{Cu^2+}$) and greenish tetrachlorocuprate(II) ions ($\ce{[CuCl4]^2-})$.

The change in color comes from the equilibrium

$$\ce{Cu^2+ + 4 Cl- <=> [CuCl4]^2-}$$

Low concentrations of chloride favor the formation of the blue copper ions. High concentrations of chloride favor the formation of the greenish tetrachlorocuprate ions.

If a significant amount of chlorine had been created you should have observed bubbles at the anode and the typical chlorine smell.


Do we need to standardize 10M of NaOH? I made $10~\mathrm{M}$ of $\ce{NaOH}$ solution by diluting $40~\mathrm{g}$ of its pellets in distilled water. I'm not sure if it's really $10~\mathrm{M}$ or at least close enough, but I don't think standardize can be done in such concentration. I've been careful enough in making the solution such as measuring the mass in low humidity room and closing it with plastics/glass.

This is a highly concentrated base, which is very corrosive. I would assume, that its concentration is quite unstable.

It will, for example, quite rapidly react with carbon dioxide from the air:
$$\ce{NaOH + CO2 -> NaHCO3}\tag{1}$$

Also it will corrode glass containers, dissolving minerals like magnesium and calcium from it:
$$\ce{Mg^{2+} + 2NaOH -> Mg(OH)2 v + 2Na+ }$$

A PE bottle should be used to avoid that.

Even if you worked in a very clean and low humidity environment, the surface of your pallets will be corroded since sodium hydroxide is hygroscopic and will further react according to $(1)$, even when stored carefully.

Therefore the concentration you obtain by weighing the pallets can only be quite approximate. If you need to know the exact concentration of your base, you have to calibrate it versus a primary standard, like dried succinic acid, before and after the intended use.
In your case standardisation will indeed be not easy to perform, since your concentration is very high. Most likely there are uncertainties to the procedure that will likely leave you with similar error bars than not doing the calibration.


Difference between Side Chains and Functional Groups From what I've read, a functional group seems to be a specific case of a side chain, but this group must be directly responsible for some of the chemical characteristics of the molecule. However, in some cases, methyl groups (such as R-CH3) are listed as functional groups as well. For example, I have the following question:

The correct answer is listed as C - but isn't the methyl a site of reactivity as well?

So the example you present is of a poor question. It's a classic case of a "best answer" MCQ that requires you pick the answer that is most accurate, rather than an answer that is unambiguously right. (I hate these.) Alkanes are listed in most textbooks and guides to organic chemistry as a "functional group." The reality is that it's hard to do chemistry on an alkane in most circumstances, so it's probably the least "functional" of the functional groups.

That said a side chain is an imprecise term. It's most consistently used in amino acid chemistry to refer to the differentiating moiety. But it's really used to indicate any moiety that hangs off of a common structure. What constitutes the side chain versus the primary structure is a function of context. A large polymeric structure can have side chains that consist of aromatic rings, while with a smaller structure consisting of a single benzene ring with some alkanes hanging off of it, the alkanes will be considered the "side chain." There's no substantive difference between a side chain and a functional group except for emphasis in a discussion.

So in brief: Functional groups are the things you learned about like amines, amides, carboxylic acids, etc. Side chains are parts of a molecule that are considered to be "hanging off" the "main" part of the molecule, and are highly dependent on the context of the discussion. Sometimes these side chains are single functional groups, and sometimes they're much bigger structures. It really depends. But functional groups are always functional groups. Though in general, chemists tend to "forget" alkanes as functional. But as I indicated earlier, the question isn't all that great, so don't feel too bad if it confused you.


What are the ideal pair of (everyday) elements to pair in a thermocouple I want to create a thermocouple (to carry out experiments on thermoelectricity).

I want to use the pair of everyday* elements (e.g $\ce{CuO}$ and $\ce{Al}$ wires) to create the thermocouple. The temperature range of the heat source will be between 300° C - 500° C

My (little remaining) knowledge of physical chemistry informs me that the periodic table would provide useful clues as to which elements to use (that's why I posted this question here, rather than in physics or engineering) - but unfortunately, it's not clear to me how I may use the periodic table in this case, to select the two elements that meet my twin requirements of:

Maximum voltage and current generated (via the Seebeck Effect) within the operating temperature indicated above
Both elements are "everyday" elements

Note: By everyday elements, I mean elements that I can buy off the shelf (preferably, in wire or sheet form), and are relatively inexpensive.

My book has a few metals/alloys listed in a Seebeck Thermoelectric Series:

  Antimony, Nichrome, Iron, Zinc, Copper, Gold, Silver, Lead, Aluminum, Mercury, Platinum-Rhodium, Platinum, Nickel, Constantan, Bismuth

There's also the Seebeck formula for emf of a thermocouple:

$$E_{\text{AB}} = a_{\text{AB}}\theta + \frac{1}{2}b_{\text{AB}}{\theta}^2$$

Here, $a_{\text{AB}}$ and $b_{\text{AB}}$ are constants that depend on the pair of metals $\text{A}$ and $\text{B}$ chosen. $\theta$ is the temperature difference between the junctions of the thermocouple.

There's another defined quantity called thermoelectric Power, which is defined as $\frac{dE_{\text{AB}}}{d\theta}$.

The graph of $E_{\text{AB}}$ with respect to $\theta$ would be a parabola. If $a_{\text{AB}}$ and $b_{\text{AB}}$ are of opposite signs, then the parabola crosses the $x$-axis twice, and thus, the thermo-emf becomes zero for two values of $\theta$ and maximum for one value of $\theta$.

From the above thermoelectric series, a iron-nickel might seem like a good option to consider building a thermocouple with, considering they're quite far away in the series and are easily accessible. For this couple, we get the values of thermoelectric constants as $a_{\ce{Fe-Ni}}=\pu{2.5 \mu V/°C}$ and $b_{\ce{Fe-Ni}}≈0$.Thus, you get:

$$E_{\ce{Fe-Ni}}=2.5\theta$$

If you're using your temperature range of 300°C-500°C for the hot junctions and room temperature (25°C) for the cold junction, you would get your thermo-emf in the range $\pu{0.687-1.1875mV}$.


Pseudo chiral centre I was reading about stereoisomers and I came across a term pseudo chiral centre.  I looked up on the net for its definition and found a weird formula for calculating no. of isomers if the compound has a pseudo chiral centre.
I can't understand how to deduce the formula and how to do analytically instead of using the formula in case of complex molecules say of more than 3 chiral carbon atoms.

Carbon number 4 is a pseudo chiral center because its chirality depends on the chirality of the second and sixth carbons. For example, if the second carbon is S and the sixth carbon is R, then carbon number 4 can be r or s, but if the chirality of the second and sixth carbons are the same, then carbon number 4 is not the chiral center, because it doesn't have 4 different groups around itself.


What's the pH of vinegar containg 5% acetic acid? Vinegar generally contains 5% acetic acid. We would expect the pH of vinegar to be approximately:
  a. 0
  b. 3
  c. 7
  d. 9
  e. 12

I don't have the key for this question, so I just want to make sure that I got it right and what I got doesn't seem to be one of the answer choices.

My answer was 1.3, so I rounded it to 1 and there wasn't an answer choice. Here is what I did:

$$\log(.05/1) = -1.3010$$

How else would you do it because you were already given the final concentration from the original concentration?

Am I missing something here?

Edit: The professor said to use this formula for calculating it:

$$K_\mathrm{a} = \frac{(.05x)^2}{x}$$

and got 3, which made no sense, any way this is possible?

Consider the choices. Since it's acetic acid, you can rule out any pH values that indicate a basic or neutral solution. Knowing that acetic acid is a weak acid and that vinegar is a fairly dilute solution, you can also rule out any pH values that would indicate either a strong acid or a highly concentrated solution. Eliminating those impossible values leaves only one choice as the answer.

The trick with a question like this is to realize that it's multiple choice for a reason. It's intended to be a conceptual problem, not a calculation problem. In general, a good rule for multiple-choice questions like this is to first see if you can solve it using general concepts; only if you can't should you reach for you calculator.


Conformation of six-membered ring constituting a bridged compound What is the conformation of the cyclohexane ring constituting [2.2.1]-bicycloheptane?
Is it boat or twisted boat? Is that conformation abundant in other bicyclic systems containing a cyclohexane ring?

EDIT: From Modern Organic Synthesis by Zweifel I took the following image:

which is in contradiction with current answers. Can anyone account for this observation?

According to structure determined by electron diffraction, norbornane is present in bridged boat conformation. http://pubs.acs.org/doi/abs/10.1021/ja01014a032, see Figure.

Norbornane and its derivatives are often used for studying effect of equatorial and axial substitution of cyclohexane and is considered rigid.
Therefore, I don't expect significant population of the second conformer you sketched.

Edit:
For the [2.2.2]bicyclooctane (not the lactone mentioned in question), both conformers are known in crystalographic database.

Bear in mind, that bicyclooctane constist of three hypothetical cyclohexanes, possibly equivalent by symmetry.

For any compound of interest, you should find enough data in crystallographic database, but I'd expect all reasonable conformations to exist.


Are electrostatic effects temperature dependent? A charged particle like an ion, does its repulsive or attractive effect on other particles change with temperature? Are electrostatic effects temperature dependent?

The strength of the electric charge does not change with temperature (unless approaching that of the Big Bang). However, effects of the charges do change with temperature. For example, at low temperatures (near 0 K), ions can form arrays in space, though their charges repel. The history of Bose-Einstein condensates begins with trapped ions, extending to neutral gases.


What does "spin degeneracy" mean in this context? I am currently studying Diode Lasers and Photonic Integrated Circuits, second edition, by Coldren, Corzine, and Mashanovitch. In chapter 1.2 ENERGY LEVELS AND BANDS IN SOLIDS, the authors say the following:

On the other hand, in a covalently bonded solid like the semiconductor materials we use to make diode lasers, the uppermost energy levels of individual constituent atoms each broaden into bands of levels as the bonds are formed to make the solid. This phenomenon is illustrated in Fig. 1.4. The reason for the splitting can be realized most easily by first considering a single covalent bond. When two atoms are in close proximity, the outer valence electron of one atom can arrange itself into a low-energy bonding (symmetric) charge distribution concentrated between the two nuclei, or into a high-energy antibonding (antisymmetric) distribution devoid of charge between the two nuclei. In other words, the isolated energy level of the electron is now split into two levels due to the two ways the electron can arrange itself around the two atoms.$^1$ In a covalent bond, the electrons of the two atoms both occupy the lower energy bonding level (provided they have opposite spin), whereas the higher energy antibonding level remains empty.
$^1$The energy level splitting is often incorrectly attributed to the Pauli exclusion principle, which forbids electrons from occupying the same energy state (and thus forces the split, as the argument goes). In actuality, the splitting is a fundamental phenomenon associated with solutions to the wave equation involving two coupled systems and applies equally to probability, electromagnetic, or any other kind of waves. It has nothing to do with the Pauli exclusion principle.

If another atom is brought in line with the first two, a new charge distribution becomes possible that is neither completely bonding nor antibonding. Hence, a third energy level is formed between the two extremes. When $N$ atoms are covalently bonded into a linear chain, $N$ energy levels distributed between the lowest-energy bonding state and the highest-energy antibonding state appear, forming a band of energies. In our linear chain of atoms, spin degeneracy allows all $N$ electrons to fall into the lower half of the energy band, leaving the upper half of the band empty.

What does "spin degeneracy" mean in this context?

I'm not a solid-state chemist, but I think the meaning seems reasonably clear from context to me. "Spin degeneracy" here means that each energy level is capable of holding a spin-up electron as well as  a spin-down electron, i.e. in each orbital there are two different spin states which are degenerate (some would call these spin orbitals). Thus, $N$ atoms form $N$ different energy levels which $N$ electrons need to be distributed between. Because of this degeneracy, only the lowest-energy $N/2$ energy levels are actually populated: each energy level has $2$ electrons.


Electrolysis of Strongly Basic Water What happens when strongly basic water is electrolyzed? I know that the electrolysis of water in strongly acidic solution involves hydrogen protons; will these protons still be involved in strongly basic solution?

In a basic solution $\ce{H2O}$ will act as the $\ce{H}$-source for the hydrogen evolution reaction. The reaction will look something like:

\begin{equation}
 \ce{2H2O + 2e- -> 2OH- + H2}
\end{equation}


Why do some ions form tetrahedral complexes more readily than others? Why does Co$^{2+}$ readily form tetrahedral complexes such as [CoCl$_4$]$^{2-}$, whereas Fe$^{2+}$ usually forms octahedral complexes?

Ligands use lone pairs of electrons to form coordinate bonds with the metal ion of the complex. The coordination number is the number of coordinate bonds from the ligands to the central ion. For instance, if a central ion has two coordinate bonds formed to it, one from each ligand, then the coordination number would be 2.(1)

According to Lumen Learning, for certain metals the maximum coordination number is related to the "electronic configuration of the metal ion (specifically, the number of empty orbitals) and to the ratio of the size of the ligands and the metal ion". Furthermore, it states that large metals with small ligands result in high coordination numbers, presumably due to there being more attraction from the positively charged central ion. Conversely, small metals with large ligands will have low coordination numbers, due to there being less electrostatic attraction towards the lone pairs on the ligands as its surface area is smaller.(2) Extrapolating this knowledge towards your question, we can parse the question to look at its components:

Co2+ is has an ionic radius of 74 pm while Fe2+ has an ionic radius of 76 pm. If we examine the example complex you gave, [CoCl4]2-, Chlorine has a large atomic radius of 175 pm relative to Co2+ ionic radius.(3) Using what we learned from Lumen Learning, this is a small central ion with relatively large ligands. Hence, the coordinate number must be smaller, such that it is 4, meaning it forms a complex with a tetrahedral geometry. Fe2+ on the other hand has an ionic radius of 76 pm and for a complex with an octahedral geometry, let us say [Fe(H2O)6]2+, it has 6 water molecules that function as ligands. The hydrogen and oxygen atoms have atomic radii of 25 and 60 pm respectively.(4) Consequently, the size of the water atoms are considerably smaller than the Chlorine atoms and thus, according to Lumen Learning, since we have smaller ligands, more of them can form coordinate bonds with the Fe2+ central ion. If we are to look at a different example, e.g. [Fe(CO)4]2−, we can see that it has ligands with large atomic radii (i.e. Carbon and Oxygen) and thus fewer ligands can bond with the central ion, resulting in a tetrahedral geometry.

References:

Brown, Catrin. "Chapter 3 – Periodicity." Higher Level Chemistry. Ed. Mike Ford. 2nd ed. N.p.: n.p., n.d. 123-33. Print.
Mott, Vallerie. "Introduction to Chemistry." Lumen. N.p., n.d. Web. 16 Apr. 2019.
"Cobalt Chemistry." Cobalt Transition Metal Chemistry Cobalt(II)Co2+ Complex Ions Stabilised Ligand Substitution Cobalt(III) Co3+ Complexes Redox Chemical Reactions +2 +3 Principal Oxidation States GCE AS A2 IB A Level Inorganic Chemistry Revision Notes. N.p., n.d. Web. 16 Apr. 2019.
Kwilkis. "Helium Atom, Radon Atom and Water Molecule Sizes and Concrete?" Helium Atom, Radon Atom and Water Molecule Size... | ACS Network. N.p., 03 Oct. 2018. Web. 16 Apr. 2019.


Helmholtz Free Energy Minimization Proof In the proof that Helmholtz free energy is minimized at equilibrium, we use the fact that $$\mathrm dS_\text{total}=\mathrm dS_\text{system}+\mathrm dS_\text{bath}\geq 0$$
But we also use the fact that
$$\mathrm dS_\text{bath}=\frac{\mathrm dU_\text{bath}}T=-\frac{\mathrm dU_\text{system}}T$$
Why does this not imply that $\mathrm dS_\text{system}=\frac{\mathrm dU_\text{system}}T$ and therefore $\mathrm dS_\text{total}=0$?
Is this because only the bath is always at temperature $T$, so we can imagine a reversible path, whereas the system may have some temperature fluctuations during the process of heat exchange with the bath?

Who says that $dS_{bath}=\frac{dU_{bath}}{T}=-\frac{dU_{system}}{T}$?  The equation should correctly read $\Delta S_{bath}=\frac{\Delta U_{bath}}{T_{bath}}=\frac{Q_{bath}}{T_{bath}}=-\frac{Q_{syst}}{T_{bath}}$, where the Q's are for the irreversible path.  The Q of the system for the reversible path is different from that for the irreversible path because of both temperature gradients and viscous dissipation of mechanical energy to internal energy during the irreversible path (both of which generate entropy).  For the reversible path of the system, these entropy generation effects are not present.


Gibbs energy: the term −TdS. Where does the reversible heat go? When we have an exothermic reaction as well as a decrease of the entropy in the system after the reaction, why does the decrease in the entropy of the system diminishes the amount of useful work we can extract from the system?

In my opinion, if we have at the final state less accessible microstates than at the  initial state, the kinetic energy (at STP most of them as translational energy) of the system must be more concentrated and therefore have a higher capability of doing useful work. Thus, it gains reversible heat from its system, "stores" it as kinetic energy and could do work.

Where is my mistake? Could you please help me, maybe with a depiction?

The way to think about microstates is as alternative configurations of a system which satisfy constraints consistent with the observed macroscopic properties of the system. The collection of such microstates forms an ensemble.

The misconception lies in the idea of "concentrating" kinetic energy, and in thinking that reducing the number of available microstates changes the available energy. It doesn't necessarily do so. It is necessary to be careful, since the idea that "concentrating kinetic energy reduces the entropy" is not generally correct. In general you should consider all available degrees of freedom. An ideal gas only has translational degrees of freedom (kinetic energy) so in this case this restriction to kinetic energy applies.

In addition, during an isothermal process performed on an ideal gas, the internal energy does not change, and neither does the average kinetic energy ($K=\frac32 N \mathrm{k_B}T$), but the number of available microstates may change (for instance with an expansion), as reflected in the dependence of entropy on volume. The change in entropy reflects a change in the probability distribution over the ensemble.

To illustrate what is meant with "dispersion" (and by extension entropy) envision a maze in which the system wanders. As the maze gets larger the system gets more lost in that maze, it has more choices where to go.

If you use a particle in a box model, assuming quantum like behavior, then the gaps between energy levels become smaller as you increase the size of the container, that is, with expansion. This results in spreading probability of occupation (dispersion) over more states. Now, irrespective of how the isothermal expansion is performed - reversibly or irreversibly (assume a free expansion) - the entropy cost for the system is the same. However, in the case of the spontaneous expansion the surroundings remains at constant entropy, whereas its entropy is reduced in the reversible case. In addition (and this is a key point), no work was performed in the free expansion.


Please explain why dioxygen difluoride is so dangerous I just read this article that mentioned that dioxygen difluoride is very dangerous. The terms it uses are "awful", "violently hideous" and "deeply alarming".

But I couldn't get a handle on exactly what it would do that warrants those descriptions.

XKCD's source material is an article from the blog of one of the funnier chemists I've read, Dr. Derek Lowe. The chemical in question made his "Things I Won't Work With" list, and the article is found here.

Dioxygen difluoride, $\ce{O_2F_2}$, sometimes evocatively labeled according to its atomic arrangement $\ce{FOOF}$, is first and foremost a vigorous oxidizer; in plain English, it starts roaring fires with almost anything it touches, even with things that simply will not burn in any other circumstances, like sand, concrete, brick, iron, copper, even asbestos and fiberglass. Both the oxygen and the fluorine are pretty highly electronegative (fluorine is the most electronegative element in the table, and oxygen's #2), and so they will readily react with anything even slightly more willing to donate an electron; carbon, silicon, metals, alkalis, most other gases, etc. When that happens, the high-energy bonds of this molecule (it can only be formed at 700*C, and that thermal energy remains "locked" in the structure of the molecule) are rearranged into structures that require less energy, and the excess energy is released. The resulting rapid, aggressive, highly exothermic reaction of dioxygen difluoride with just about any other material would not be described as anything less than an explosion.

Second, even in the absence of something to oxidize, the oxygen would much rather exist in its "native" diatomic gas state. Again, there's a lot of energy tied up in the molecular bonds of this chemical, and the lower the energy required to form a bond, the more stable the molecule. Therefore, even at extremely cold temperatures (-160*C) in complete isolation, dioxygen difluoride decomposes to oxygen difluoride ($\ce{OF_2}$) and $\ce{O_2}$ gas. At room temperature, this decomposition is uncontrollable (no-one's gotten it anywhere close to RT without it coming apart), and even at extremely cold temperatures, the heat produced by slow decomposition can cause what's called a "thermal runaway"; the heat produced by one molecule decomposing is absorbed by nearby molecules causing them to decompose, releasing more heat, and so on. When things rapidly decompose into more stable gaseous components, the event is normally accompanied by a loud noise and shrapnel, formed from the remains of whatever had been holding the substance and anything else nearby.

Of the products formed by the self-decomposition of FOOF, oxygen gas is a pretty strong oxidizer, but it's the least of our worries here. The other product, oxygen difluoride, is almost as strong an oxidizer as dioxygen difluoride, which is only stronger because it's oxygen difluoride plus more oxygen. The "shared" electron pairs forming the bonds between the oxygen and each fluorine atom are much more closely attracted to the fluorine, as opposed to a structure like water where the oxygen is the more attractive. This gives the oxygen a formally positive oxidation state of +2.

By its very nature, oxygen yearns to flip that sign and achieve its "natural" -2 ox state by bonding with something less electronegative (which, as previously mentioned, is almost everything else in the periodic table besides fluorine). However, unlike the other oxygen atom in the FOOF molecule which flew the coop, this second oxygen can't escape, because the two fluorine atoms won't let it go. The fluorines prefer to bond to it rather than to each other... until something else, anything else, is introduced that's even more attractive.

$\ce{OF_2}$, like its FOOF parent, reacts very strongly and exothermically with almost anything, especially water. It will literally burn ice, rapidly; that's one of the reactions guaranteed to produce a powerful explosion. The most stable ultimate products of that reaction are hydrogen fluoride gas and more oxygen, and when you call HF gas a "stable" product of any reaction you are speaking in very relative terms; a release of HF gas into the air is one of those "drop everything and run" types of industrial accidents. HF gas dissolves in water, including the mucus of your lungs, to produce hydrofluoric acid, a persistent, toxic, corrosive acid that will shred just about any organic chemical it comes across from hemoglobin to proteins to DNA, and you really want to keep those kinds of things intact.

Basically, there are a lot of things in an inorganic chemistry lab that have a Medusa-like ability to kill at a glance, but fluorine deserves special notice, in part because of the number of notable chemists who were poisoned or blown up in the early heady days of isolating it. When you combine it with other elements on the right hand side of the periodic table in "lightweight" combinations (between two and about ten atoms per molecule), your day is about to get reeeeealy interesting; see the work of A.G. Streng, who was tasked not only with producing large quantities of dioxygen difluoride, but also with reacting it with things that normal chemists would keep very far away from it (and in fact "normal chemists" stay very far away from FOOF itself for the obvious reasons).

Fluorine only starts becoming more stable and easier to work with as an ionic salt, such as sodium or calcium fluoride (in which form it naturally occurs as the mineral fluorite; still poisonous in sufficient doses but much less of an explosion/reactivity hazard), or when you start throwing in elements of the carbon group (the carbon-fluorine bond is among the strongest known to chemistry, and fluorocarbons have a variety of uses as propellants, refrigerants and polymers, including Teflon), or heavier chalcogens like sulfur (sulfur hexafluoride, $\ce{SF_6}$, has virtually no reaction chemistry and is favored as a dielectric and to blanket air-sensitive materials, though other sulfur-fluoride compounds like sulfur tetrafluoride and disulfur decafluoride are highly toxic).