How to determine the equivalent weight of CO2 We were estimating dissolved $\ce{CO2}$ in water by American Public Health Association method. It was a titrimetric method using phenolpthalein indicator.

Titrant used was $\ce{NaOH}$ and analyte was sample water.

After the end point was reached and the volume of titrant estimated the ultimate calculation was done to find out the amount of $\ce{CO2}$ present in the sample per litre. The formula mentioned in the book was

$$\ce{CO2} \mathrm{(mg/l)} =\frac{\text{[Volume of NaOH} \times \text{Conc. of NaOH (in Normality)}\times 22 \times 1000]}{\text{Volume of sample water}}$$

$22$ here is the equivalent weight of $\ce{CO2}$. My question is how's it so?

NOTE:

Our teacher said that since 2 molecules of $\ce{NaOH}$ react with 1 molecule of $\ce{CO2}$ to form $\ce{Na2CO3}$ so $n=2$ here.

Is it right? I doubt it because I have always known $n =$ the number of $\ce{H+}$ or $\ce{OH-}$ released by an acid /base or the number of electrons released or accepted in a redox reaction.

The reaction here is
$$\ce{2NaOH + CO2 -> Na2CO3 + H2O}.$$

But as $\ce{CO2}$ is not an acid or base and even doesn't undergo redox so how do we determine its equivalent weight?

P.S. I'm a Zoology student and have learned chemistry for XII + 2 years.

This is the danger of using formulae from books without fully understanding where the formulae came from.

So let's take it step by step.

We have a titration of an unknown concentration solution with a known (standardised) solution.

The reaction $$\ce{2 NaOH + CO2 -> Na2CO3 + H2O}$$ gives us the stoichiometric relationship between $\ce{NaOH}$ and $\ce{CO2}$, i.e. 2 moles of $\ce{NaOH}$ react with 1 mole of $\ce{CO2}$.

How much $\ce{CO2}$ Was used? Half the amount of $\ce{NaOH}$!

How much $\ce{NaOH}$ was used?

$$\begin{align}
c(\ce{NaOH})  &= \frac{\text{amount of substance}}{volume}\\
              &= \frac{n_{\ce{NaOH}}}{V}\\
n_{\ce{NaOH}} &= c(\ce{NaOH})\times V\\
n_{\ce{CO2}}  &= \frac{1}{2} n_{\ce{NaOH}} \\
              &= \frac{1}{2} c(\ce{NaOH})\times V\\
\end{align}$$

What is the concentration of $\ce{CO2}$ in sample?

$$c(\ce{CO2}) = \frac{n_{\ce{CO2}}}{\text{Volume of sample}} = \frac{1}{2}\frac{c(\ce{NaOH})\times V}{\text{Volume of sample}}$$

Concentration is also expressed as $\mathrm{mg/L}$. 1 mole of $\ce{CO2}$ weighs $44~\mathrm{g}$ or $44\times 1000$ mg so sticking this into the above

$$c(\ce{CO2})\mathrm{(mg/L)} = \frac{1}{2}\frac{c(\ce{NaOH})\times V}{\text{Volume of sample}} \times 44 \times 1000 = \frac{c(\ce{NaOH})\times V\times 22\times 1000}{\text{Volume of sample}}$$

That's where the $22$ came from.
This may not address exactly where the equivalent weight of $\ce{CO2}$ comes from but at least it shows you the derivation of how the $22$ got into the formula.


Are compounds which are used to seed clouds like silver(I) iodide hygroscopic? I have read a teensy bit of the subject matter and the explanation they provided for silver iodide's ability to being a good site for ice to form is because of its size resembling an ice nucleus.
How exactly does silver iodide adsorb water onto its surface? Is it hygroscopic?
Ice crystals growing larger due to vapour deposition or rapid freezing of supercooled water upon contact with ice makes sense. But how does silver iodide do it?
If the hygroscopic properties of silver iodide are what enables it to seed clouds, can the common salt also be used to seed clouds? I do apologize as the question wasn't aptly structured. Please do help in understanding this!
P.S. I found an article which I find is relevant to this question [1], but unfortunately, I'm not experienced enough to go about reading scientific articles.
Reference

Pena, R. G. D.; Caimi, E. A. Hygroscopicity and Chemical Composition of Silver Iodide Smoke Used in Cloud Seeding Experiments. Journal of the Atmospheric Sciences 1967, 24 (4), 383–386. DOI: 10.1175/1520-0469(1967)024<0383:HACCOS>2.0.CO;2.

Silver iodide is not hygroscopic. It is used to seed clouds, because when the atmosphere is saturated with water vapor, liquid water should start condensing. But to do this, it needs a seed, a place when the first molecules will start to condense. No water molecule feels that it should be the first one to show other molecules where to be condensed. But, as silver iodide has the same hexagonal shape and same dimensions as snow, it may make believe it is an ice crystal. And the neighboring water molecules will be mistaken, follow the example and deposit on it. The common salt cannot do this. It does not look like ice.


Mechanism of dehydrogenation with nosylate What is the mechanism of the attached reaction? How could nosylate do the dehydrogenate the species?

The reaction is from this synthesis (46th IChO preparatory problem 26):

This is the reaction mechanism that I propose for this reaction. Though it does not show the complete reduction of the 3-nitrobenzene sulfonate, it could be reduced further through the nitroso-hydroxylamine pathway. This means that in total, a single nitrosyl anion could oxidize three molecules of the starting material.

Extraction of a hydrogen by the nitro group. The hydrogen is extremely acidic compared to an ordinary aliphatic $\ce{C-H}$ bond due to the massive conjugation in the resulting ion:

Amide-imidic acid tautomerization:

Resonance structures of the molecule, showing its extensive conjugation, and hence stability:

Radical reaction pathway and formation of the product. Every atom in product is $\mathrm{sp^2}$ hybridized and is conjugated over the entire system. It is much more stable than the starting material, and is the driving force of the entire reaction:


Recovery and Disposal of Unknown Copper Compound About a month ago, I ran a small electrolysis experiment in an old fish bowl with a $\ce{MgSO4}$ electrolyte and copper electrodes. It ran for about 3-4 hours.

Now, a month later, I am left with about 40% of the initial volume of water and a blue copper compound on the bottom of the bowl. I believe this is $\ce{Cu(OH)2}$. I've provided a picture.

This has been sitting outside for the entire month under a little greenhouse type thing (but air can still get at it). As winter is coming, I was hoping to collect what I've formed and dispose of it properly. So

1) Could it be $\ce{Cu(OH)2}$?

2) Is there a way I can get the solids out? Or should I just pour everything into a plastic container for disposal?

Unlikely. Copper hydroxide adsorbs carbon dioxide from air rather
fast, forming basic carbonates. if you want to test, try to boil it.
Copper hydroxyde looses water under mild heating, forming black
copper oxide.

Most likely it is either basic copper carbonate or one of basic
copper sulfates or a mix of both. I highly doubt that it is a pure
copper sulfate, as Mg had to go somewhere.
If you want to be a responsible citizen (which is not actually much
a point with relatively safe copper salts), add some soda, leave
overnight and then filter off the precipitate with filter paper.
If no filter paper is present, cotton white, toilet paper, and
similar should work, just be careful to not tear them. The
precipitate then may be heated until black and disposed any way
suitable.


How does pH depend on temperature, when taking into account volume expansion? Since autodissociation of water is endothermic, increasing the temperature will drive the following equilibrium to the right, which has the effect of increasing $[\ce{H+}]$:

$$\ce{H2O <=> H+ + OH-}$$

However, increasing the temperature of liquid water will also increase its volume, which has the effect of reducing $[\ce{H+}]$ (cf. this table on Wikipedia showing how density of water varies with temperature).

So as we increase the temperature, which of the two effects dominates? Does $[\ce{H+}]$ experience a net increase or decrease?

Your qualitative understanding is correct.  As you increase the temperature, you have two effects:

The dissociation reaction, which is endothermic, shifts to the right.  This effect increases the concentration of hydrogen ions.
The concentration of hydrogen ions is expressed as molarity, which is number per unit volume.  As you increase $T$, water expands, resulting (everything else being equal) in all concentraitons decreasing.

And you're asking which one wins.

The answer is the first, overwhelmingly.  For instance, as you go from $\pu{25 ^\circ C}$ to $\pu{50 ^\circ C}$, $K_\mathrm w$ increases by five times.  By contrast, the increase in volume over this temperature range is only $1\%$.


Serious introduction to the chemistry of explosives recommendation I'm looking for a serious introduction to the chemistry (and perhaps to a lesser extend the physics) of explosives. Presupposition of basic university courses in Organic Chemistry and Thermodynamics is no problem, though it would be helpful if some of the specifics needed to understand subsequent material would be briefly outlined as a reminder (e.g. in an introductory chapter). Sadly this search is quite hard because of all the Uncle Fester and like trash clogging up search results. To be clear, that is absolutely not what I'm after. I want an academically oriented introduction, not some "recipe" book.

A book that more or less resembles (I guess) what I'm after is The Chemistry of Powder and Explosives by Tenney L. Davis. However, it is over 70 years old and might be a little to advanced for me (might, I just don't know without some guidance).

Every help is appreciated!

PS. I actually found an online copy of the book I mentioned. It does not look really theoretical?

To be clear, that is absolutely not what I'm after. I want an academically oriented introduction, not some "recipe" book.

I'm glad to read that!

I'm aware of Chemistry of Energetic Materials (Eds. George A. Olah and David R. Squire, Academic Press, 1991) which seem to cover

  The Structural Investigation of Energetic Materials.
  Studies of Initial Dissociation Processes in 1,3,3-Trinitroazetidine By Photofragmentation Translational Spectroscopy.
  Studies of Molecular Dissociation By Means of Ultrafast Absorption and Emission Spectroscopy and Picosecond X-Ray Diffraction.
  Computer-Aided Design of Monopropellants.
  Polycyclic Amine Chemistry.
  Metallacarboranes of the Lanthanides and Alkaline-Earth Metals: Potential High-Energy Fuel Additives.
  Methods for Preparing Energetic Nitro-Compounds: Nitration with Superacidic Systems, Nitronium Salts, And Related Complexes.

On a side note, if you are interested in current research on high-energy materials free of nitro groups, have a look at the work of Thomas M. Klapötke.


How to obtain hydrophobic Doxorubicin from its hydrochloride salt? I have read literatures on dissolving Dox hydrochloride in DMSO and adding Triethylamine to the resulting solution. The purification is done using dialysis (incase of polymers). But I cannot do dialysis.
Another method I tried was neutralizing the salt solution of Doxorubicin with PBS (pH 8.04), and centrifugation of the resulting mixture. But centrifugation yielded no pellet.

Add a base to an aqueous solution of the compound.  The base could be ammonia water or ammonium hydroxide.  Make sure you add enough base to raise the pH to 12 or higher.
Add a water-immiscible organic solvent with a low boiling point to the solution and mix well.  A good solvent might be dichloromethane.
Let the phases separate and carefully remove the organic phase.  (If you use dichloromethane, the organic phase will be the bottom phase, but this is not true of all solvents.)
Repeat step 2 several times, each time combining the newest organic extract phase with the extract from the previous steps.
Evaporate the solvent under air, or, if you have it, an inert gas such as nitrogen.


Palladium hydride forming water and heat at room temperature I just read that saturated palladium hydride, $\ce{PdH}$0.7, when exposed to oxygen will generate heat and water on it's surface, $\ce{4 H}$(Pd) $\ce{+O2}$(g) $\ce{\to 2 H2O}$(g, l).
What is going on?

I'm not an expert, but typically there are a series of surface reactions where the reactive species are all adsorbed on the surface. So atomic hydrogen wouldn't "leave" the palladium before it reacts, it just diffuses from an internal crystal site to a surface site.

Likewise, the oxygen would adsorb onto the Pd surface, I'll assume dissociatively. So now there is atomic O and H diffusing around the surface sites of the Pd. These can react and form adsorbed OH, which can further react and form adsorbed H2O, which can then desorb and give you your water.

Your enthalpy calculation is off, the equation for water already includes the fact that an oxygen molecule must break apart to form it.

If you include the hydrogen gas reaction with Pd in your picture, all you're doing is turning hydrogen gas and oxygen gas into water, using Pd as a catalyst. So the total enthalpy change is just -285.83kJ/mol. You don't have enough information to determine how much heat is released (or absorbed possibly) in the hydrogen absorption reaction and how much is released in the water production reaction, but the total of the two will be -285.83kJ/mol.


Why doesn't the enthalpy change of a reaction affect the rate of reaction? For example, an exothermic reaction will release energy to provide energy to overcome the activation energy. So, why doesn't the enthalpy change affect the rate of reaction?

The rate of a chemical reaction is directly dependant to the activation energy of that particular reaction. These two values are inversely proportional. The higher the activation energy, the lower the rate. As it has been mentioned though, the rate can be increased, in such cases, by increasing the temperature. All of the above is described as kinetics, which shows how fast the reaction will reach its end phase.

Enthalpy is linked with thermodynamics, which shows what kind of entities will be present at the end of the reaction. At what degree the reactants will turn to products.

I don't know your chemical background, so I tried to make this as simple as possible.


Can NaOH and NaH2PO2 exist together in aqueous solution? Can $\ce{NaOH}$ and $\ce{NaH2PO2}$ exist together in aqueous solution?

My textbook denies it, but I feel that since $\ce{NaH2PO2}$ is acidic, it should react with $\ce{NaOH}$, because in my textbook it is written that salts of even weak acids can react with a strong Base (e.g., $\ce{NaHCO3}$ is mentioned to react with $\ce{NaOH}$).

Your question: Can $\ce{NaOH}$ and $\ce{NaH2PO2}$ exist together in aqueous solution?

I assume what you meant is your textbook is saying when two separate solutions of $\ce{NaOH}$ and $\ce{NaH2PO2}$ are combined, there should be no reaction. If that is what you meant "exist together," then your textbook is correct because there won't be an acid-base reaction since $\ce{NaH2PO2}$ has no "acidic hydrogen" contrast to $\ce{NaH2PO4}$ and $\ce{Na2HPO4}$ as Poutnik mentioned in his comment (more descriptive presentation would be $\ce{Na^+^-OP(=O)H2}$). The following diagram summerizes this point:

As diagram shows $\ce{NaH2PO2}$ (sodium hypophosphite) does not have any acidic protons to react with strong bases like $\ce{NaOH}$. The two hydrogens in $\ce{NaH2PO2}$  have attach ed to phosphorous atom in covalent bonds. And also showing is sodium bicarbonate, which has one acidic proton to react with $\ce{NaOH}$.


Number of spatial arrangements of tetrasubstituted biphenyl According to my book 2-bromo-2'-(1-chloroethyl)-6'-iodo-6-(prop-1-en-1-yl)-1,1'-biphenyl has eight possible spatial arrangements. How can this be so? I can think of only four – left two groups being planar with right two groups up-down and down-up, and vice versa.
For this configuration of I and $\ce{CHClCH3}$ I can think of four, for I and $\ce{CHClCH3}$ interchanged I can think of four more, but that is not the same molecule and is just a configurational isomer.

There are three independent stereogenic centers or elements in this molecule,

chiral carbon atom of the chloroethyl group,
double bond of the propenyl group,
axis of the substituted biphenyl with bulky substituents at 2,2′,6,6′ positions.

Each of them has two configurations. It means that the total number of stereoisomers is 23 = 8. (No one of them is symmetrical and cannot be superimposed with its mirror image, no one can be superimposed with any other one.)

(1M)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
(1M)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
(1M)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl
(1M)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl
(1P)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
(1P)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
(1P)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl
(1P)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl


Alternative PCB etching solutions - Is KMnO4 or NaClO suitable oxidizer? I need to etch small PCB for my project, since I do this once in a blue moon I would like to try some easy approach, preferably with household items. I'm aiming at the use of acetic acid with hydrogen peroxide to aid the reaction. But this got me wondering what are some other household items usable in the process?

I was able to find potassium permanganate ($\ce{KMnO4}$) which is a strong oxidizer, but I'm not sure if this still holds for water soltuion mixed with acetic acid.

My other idea for strong oxidizer was sodium hypochlorite ($\ce{NaClO}$) found in bleach, but I'm not really sure how safe this is. Even though I will etch only small PCB, 1 by 2 inches, I would rather not endanger myself or anyone else.

Does any of the proposed ideas make sense? Is $\ce{KMnO4}$ or $\ce{NaClO}$ suitable oxidizer for etching PCB using acid solution?

If you don't want to use the classical solid etchants, such as

Iron(III)chloride ($\ce{FeCl3}$) or
Ammonium persulfate ($\ce{(NH4)2S2O8}$) or
Sodium persulfate ($\ce{Na2S2O8}$)

there's the mixture of hydrogen peroxide ($\ce{H2O2}$) and hydrochloric acid ($\ce{HCl}$) in various ratios.

Typical recipes use

water
30% $\ce{H2O2}$
37% $\ce{HCl}$

in a volume ratio of 1:1:1.

Hydrochloric acid can be considered a "household chemical", it was/is available at DIY stores as muriatic acid and typically used to clean brickwork.

I would advise against mixing concentrated (glacial) acetic acid and hydrogen peroxide solutions of higher concentrations. You might end up forming peracetic acid, which is unhealthy and tends to decompose spontaneously (explode) at higher concentrations.

There's another method worth to be mentioned, although the etchant is usually not a household item.

Cupric chloride $\ce{CuCl2}$ in aqueous solution can be used at room temperature to remove metallic copper from the PCB too:

$$\ce{CuCl2 + Cu -> CuCl}$$

Thinking in terms of transition metal waste management, this seems to be pretty nice, since the resulting copper(I) solution can be regenerated (oxidized) to copper(II) (= the etchant) using hydrochlorid acid and hydrogen peroxide. This conference proceeding provides further information.

As far as the permanganate is concerned, it seems that it has been used to desmear PCBs in industry, but it has been superseeded by plasma treatment to avoid damage to the boards.


How do you calculate the concentrations of multiple reactants in a certain point in time? For $\ce{A + B -> C + D}$, given that $v = k [\ce{A}] [\ce{B}]$. I know the derivation for $\ce{A->B + C}$ , being $\ln[\ce{A}]=\ln[\ce{A}]_0−νkt$, but I'd like to know how you can, given a point in time, calculate the concentrations of A and B in the first equation, given the initial concentrations.

$$\frac{dA}{dt}=-kAB$$
$$\frac{dB}{dt}=-kAB$$subject to initial conditions $A=A_0$ and $B=B_0$ at time t = 0.  Let x = number of moles of A that react = number of moles of B that react.  Then,$$\frac{d(A_0-x)}{dt}=-k(A_0-x)(B_0-x)$$and$$\frac{d(B_0-x)}{dt}=-k(A_0-x)(B_0-x)$$Therefore, from both equations  $$\frac{dx}{dt}=k(A_0-x)(B_0-x)$$ subject to the initial condition x = 0 at t = 0.  This differential equation can readily be integrated using separation of variables and partial fractions.


Clemmensen reduction vs. Wolff-Kishner reduction In which cases is the Clemmensen reduction is preferred over the Wolff-Kishner reduction?

One thing is that the Clemmensen reduction is done under acidic conditions whereas the Wolff-Kishner reaction is done under basic condition, and that I think decides somewhat which to choose depending of the nature and other groups present in the reactant.

But are there any other points to keep in mind while choosing the best suitable reaction?

A key consideration for the Wolff-Kishner reduction is thermal stability of the substrate. High temperature is required for the classical reaction using hydrazine and KOH, sometimes up to 200 °C. This can rule out plenty of compounds.
The Clemmensen reduction may be fine for the laboratory,  but fundamentally many industries wouldn't use it due to scaling up the use of mercury.
Scalable alternatives are molecular hydrogen with a catalyst under acidic conditions or, the trifluoroacetic acid-silane combination, such as triethylsilane or tetramethyldisiloxane.


Intramolecular hydrazide condensation Say I have acetohydrazide ($\ce{CH3CONHNH2}$).

Is it possible, that when I add an acid, an intramolecular condensation reaction occurs, forming the cyclic compound 3-methyl-1H-diazirine:

Furthermore, in general, the mechanism of such a reaction is nucleophilic substitution, right?

Your reaction product is a double bond isomer of a diazirine.

Halogene-substituted derivatives of the latter are typically prepared via oxidation of amidines with hypochlorites (Graham oxidation).

Heating your hydrazide in an aqueous solution of a mineral acid will only turn it to the corresponding carboxylic acid.


Heterogeneous Equilibrium Without Gas We know that for homogeneous mixtures the equilibrium constant is determined using both concentration of reactants and of products. For heterogeneous equilibrium involving a gas and a liquid or a solid, we only use the concentration of the gas because the concentration of liquids and gases are constants, and can thus be omitted from the equation. But what of the heterogeneous equilibrium between a solid and a liquid? Both are incompressible, so what do you do?

Thanks!

You're right, they don't really have concentrations or pressure to work with, which would lead to 0/0 craziness.

I think we can go back a step and look at how the equilibrium constant is defined. When you derive the equilibrium constant you don't use pressure or concentrations. You actually use the substances' activities which we can relate to change in Standard Gibbs Free Energy for the reaction (which is easy to find with tabulated data):

For a full derivation refer to Wikipedia's article on the equilibrium constant.


Is Internal Energy = (3/2)nRT for a ideal monoatomic gas? Internal Energy is a state variable and its value at a particular state cannot be measured; only the change in internal energy can be measured. So how come we write that
$$U=\frac32nRT$$
where
$U$ = internal energy of the system,
$n$ = no of moles of the gas,
$R$ = Universal gas constant,
$T$ = Temperature of the system.

Till now, I knew that it was 'Avg Kinetic energy' that followed the above relation that is  $\mathrm{K.E.} = (3/2)nRT$, but today while looking at a video in Khan Academy, I came across the relation $U=(3/2)nRT$ and got entirely confused.

Molecules have internal energy due to intermolecular interactions, as well as translational kinetic energy, rotational energy, vibrational energy, electronic energy (and if you care to include them, nuclear energy and the mass-energy of the protons, neutrons and electrons themselves).

By saying "ideal", energy due to intermolecular interactions is eliminated.

By saying "monoatomic", energy due to rotation and vibration is eliminated.

So only translational kinetic energy and electronic energy remains.

(3/2)nRT is the translational kinetic energy, and since almost all atoms are in the ground electronic state at low temperature, it is a good expression for internal energy as long as the temperature is low enough that essentially all atoms are in the electronic ground state.


The "Ostwald Isolation Method": How does this work? Recently started reading up on Chemical Kinetics, and I came across a certain "Ostwald Isolation Method" that figures in my school-issued workbook. Apparently, it's used to determine (approximately) the order of a given reaction.
However, the painfully short passage that constitutes that section of the book, coupled with my teacher's reluctance to take up the said section in class, means I still don't know how this "method" works. Heck, I don't even know if the said "method" even exists (at least not under that name anyways...the book is notorious for its numerous errors that have gone unchecked over the years. We fondly view it as the Mater Omnium Errorum)

My book, on the "Ostwald Isolation Method":

This method is particularly useful in case of reactions where two or more reactants are involved.
Consider the reaction between $a$ moles of species $A$ and $b$ moles of species $B$ to give product(s) $P$: $$\ce{aA + bB -> P}$$
The reaction is first carried out with $B$ present in great excess. So the reaction must only depend on the concentration of $A$: $$\ce{r = k[A]^{x}}$$
Similarly, when the reaction is now carried out in a large excess of $A$, the rate of reaction must only depend on the concentration of $B$: $$\ce{r = j[B]^{y}}$$
The order, $n$, of the overall reaction is therefore: $$\ce{n = x + y}$$

Okay, I understand how the first two bits work (i.e- Concentrations of whatever species present in excess can safely be considered constant/invariant).
But I don't get how directly adding $x$ and $y$, both obtained from different rate equations, having different rate constants , gives us the overall order of the original (without excess) reaction.
Could someone explain the logic behind this to me? That is, if there is any logic behind it...

The powers $x$ and $y$ are intrinsic to the reaction (at least if the reaction pathway remains the same) and will not vary with reaction parameters, so that adding them then poses no difficulty.

More specifically, it is presumed that the reaction rate is given by an equation of the form $$r = c\ce{[A]}^x\ce{[B]}^y;$$ the Ostwald isolation method finds the powers $x$ and $y$ by flooding the system with $\ce{B}$ and $\ce{A}$ respectively in excess in two separate experiments, so that in the former case we have $$r = (c\ce{[B]}^y)\ce{[A]}^x \approx k\ce{[A]}^x$$ and in the latter $$r = (c\ce{[A]}^x)\ce{[B]}^y \approx j\ce{[B]}^y.$$

Adding the powers $x$ and $y$ gives the order of reaction by definition.

Incidentally, Wikipedia lists this method as the 'method of flooding', here.


What is the general phenomenon taking place here? For many cases while calculating the activity of a particular species I find them calculating like this:
$$a=\frac{p_{Mn}}{p^{\circ}_{Mn}}$$
What I really do not understand is what is evaporating into what. A detailed explanation of the topic and quantification of the parameter as activity is most welcome.
This link very well explains how thermodynamic activities are defined. However what I really do not understand is how the "estimating activity" technique is to be applied. A detailed answer to that is most welcome.

At low pressures the activity of a gas can usually be related in a straightforward way to its partial pressure as
$$a=p/p^{\circ}$$
Since the activity of a substance present in multiple phases is the same in all of the phases, for a volatile compound it suffices to determine the partial pressure in the gas phase to know its activity in condensed phases. This is how the activity of water in different materials including foodstuffs is usually determined.
For the case of a non-volatile solute in a condensed phase things are not so simple. For electrolytes, theories such as Debye-Hückel do a reasonable job of predicting activities at low concentrations, and other theories do better at higher ionic strengths. Electrochemical measurements can provide experimental values of activities. Because ions usually exhibit strong deviations from ideality this field has been studied extensively and even introductory physical chemistry or analytical chemistry textbooks will address it at some length.
One path to determine the activity of a non-volatile non-electrolyte - for which a vapor-pressure measurement is impractical - is to determine the activity of the solvent and to invoke the Gibbs-Duhem equation relating the activities of the solute and solvent. This is what is meant in the Libretext:

The activity coefficient for a nonvolatile, neutral solute is often estimated by non-linear curve fitting, taking into account the molality of the solute and the activity of the solvent (usually its vapor pressure).


Finding 1/V from absorbance in Lineweaver-burk plot I’ve got six test-tubes with $0.5\ \mathrm{ml}+0.5\ \mathrm{ml}$ of 0.005 %, 0.0025 %, 0.00125 %, 0.000625 %, 0.0003125 % and 0.0003125 % starch and 4.2 pH phosphate buffer solution (2, 4, 8, 16, and 32-fold dilution of 1 % starch solution).
0.5 ml of beta-amylase was added to each test-tube, and heated (without sixth test-tube). Then the absorbance of each specimen was measured (sixth test-tube used as reference) in 530 nm wavelength.

I have to find the $K_\text{M}$ of β-amylase from Lineweaver–Burk graph.
There is no problem with plotting $x$ points on graph (it’s just a reciprocal of substrate concentration).
But how can I use absorbance data to find $V$ that I need to plot on $y$-axis?

First, let me tell you about the pitfalls of Lineweaver-Burke plots.  The tranformations of your data (i.e. taking the reciprocal) distort the error structure of your data and can lead to erroneous fits.  It is better to directly use nonlinear fitting techniques to find kinetic parameters.  These techniques do not require transformation of the data and thus do not distort error structure.

Second, let me answer your question.  The classical Lineweaver-Burke plot puts inverse reaction rate on the $y$ axis, i.e. $1/V$.  Your question is how to find $V$ from absorbance data.  The answer is, you don't need to.  Let $A_i(t)$ be the absorbance data for tube $i$ as a function of time.  What is $V_i(t)$ for that tube?  We won't know unless you know the molar absorption coefficient of the substrate (or product -- I'm a little unclear from your question what you are hoping to detect by measuring $A_{\textrm{530 nm}}$).  Assume that it is the product that is absorbing, and that it has a molar absorption coefficient of $\epsilon$ and that you are using a photometer with path length $L$.  Then the reaction rate for the $i$th tube will be

$V_i(t)= \frac{1}{\epsilon L}\frac{dA_i(t)}{dt}$

This means take the slope of the absorbance measurement and divide it by $\epsilon$ and $L$ -- numbers you don't know -- to get $V$.  But the key assumption is simply that $V \propto \frac{dA}{dt}$.  Try making up any numbers you like for $\epsilon$ and $L$.  Find $K_m$.  Then make up new numbers and find $K_m$ again.  Did the result change?


Is this endo or exothermic? $$\ce{4Fe + 3O2  -> 2Fe2O3(s) + 1625~$\mathrm{kJ}$}$$

This should be endothermic because the energy is in the product. However, if the equation is viewed from the perspective of the reactant, it is endothermic.

Is this endo- or exothermic?

The reaction is exothermic left to right and endothermic right to left. Yes, as you noted, it's wise to note which way you are looking at the reaction, but generally, we look at reactions from left to right (because the arrow generally points left to right). For example, the following reaction is endothermic as written, i.e. very hot water (steam) is very corrosive and adding heat to water encourages further ionization of water.

$$\ce{2H2O ⟶ H3O+ + HO-}$$


Global reactivity parameters from open shell DFT calculations using Koopman's theorem Koopmans' theorem is a useful approach to calculate the global reactivity parameters from the HOMO-LUMO energies. My question is, does it apply to the open-shell systems where we get two sets of singly occupied (alpha and beta) HOMO-LUMO energies?

I am working on rare earth systems. Rare earth (RE) elements generally show +3 oxidation states which result in open shell configurations for $\ce{Ce^3+}$, $\ce{Eu^3+}$, $\ce{Gd^3+}$ etc. I am performing unrestricted open shell DFT calculations for these systems, as restricted open shell calculations for RE systems require larger computational time and cost. In unrestricted open shell calculations, we get two sets of alpha and beta HOMO and LUMO. My question is, can I apply Koopmans' theorem here to calculate the global reactivity parameters?

Could anyone help with some literature on the reactivity parameters for open shell systems? Thank you!

You (probably) cannot apply Koopmans' theorem. (Please also note that it is named after Tjalling Charles Koopmans, the s at the end is part of the name.)
Open shell systems are usually hard to describe and unrestricted density functional approximations might not even describe the system correct qualitatively. The problem usually comes down to the multireference character of the partially occupied orbitals. In these cases you need a multireference wave function to correctly describe the ground state, e.g CASSCF methods. In a first order approximation, unrestricted methods can probably yield reasonable results, because (in a practical sense) it is using more than one determinante. However, when you use these methods, there will be fractional occupancy of multiple orbitals, which make the approximations of Koopmans' theorem break down. I would suggest a thorough literature review on the best practises for such systems. In short, HOMO and LUMO are not defined in these systems.


E and Z configuration In the above we need to find the configuration of the two double bonds whether (E) or (Z) according to me the one with the cyanides should be (E) and the other (E) however my book say that the correct answers are (E,Z) (in the same order as above) . Please explain how ?

You are correct; the answer in your book is wrong. I get (E) for both alkenes.

For the alkene on the upper right:

$\ce{COOH}$ has precedence over $\ce{CH2OH}$ because the first carbon atom is bonded to $\mathrm{ \{O,O,O\}}$ and the second is bonded to $\mathrm{\{O,H,H\}}$.
Clockwise around the ring has precedence over counterclockwise. While both initially are $\mathrm{\{C,H,H\}}$, the next carbon atom clockwise is $\mathrm{\{C,C,H\}}$ while the next carbon atom counterclockwise is $\mathrm{\{C,H,H\}}$.

For the alkene on the lower left, the nitrile groups win in both cases:

On the left: $\mathrm{\{N,N,N\}}$ versus $\mathrm{\{N,H,H\}}$
On the right: $\mathrm{\{N,N,N\}}$ versus $\mathrm{\{C,C,H\}}$


Different formulas for copper pyrites and bauxite In my book in the chapter on General Principles and Processes of Isolation of Elements, I found that the formula for copper pyrites was stated as $\ce{Cu2S.Fe2S3}$ at one place, $\ce{CuFeS2}$ at another place and $\ce{Cu2FeS2}$ at another place.

Are all these taken to be the formula for copper pyrites or are these printing mistakes. I have searched the internet but could not find any useful information regarding this.

Both $\ce{Cu2S.Fe2S3}$ and $\ce{CuFeS2}$ are the equivalent means to denote chalcopyrite. The first notation, $\ce{Cu2S.Fe2S3}$, commonly used a few decades ago, shows that two sulfides are not just a mechanical mix, but form a chemical compound (same as for crystallohydrates, e.g. $\ce{CuSO4 * 5 H2O}$). The second one, $\ce{CuFeS2}$, is a formula unit, a more universal and modern representation. Chalcopyrite is a mineral of $\ce{ABX2}$ type, crystallizes in $I\bar{4}2d$ space group.

Figure 1. Unit cell of chalcopyrite $\ce{CuFeS2}$. Color code: $\color{#FFFF30}{\Large\bullet}~\ce{S}$; $\color{#E06633}{\Large\bullet}~\ce{Fe}$; $\color{#C88033}{\Large\bullet}~\ce{Cu}$.

On the other hand, $\ce{Cu2FeS2}$ is a reduced formula of $\ce{Cu8Fe4S8}$, a superstructured bornite [1]. A compound of $\ce{AB2X2}$ type, crystallizes in $F\bar{4}3m$ space group.

Figure 2. Unit cell of superstructured bornite $\ce{Cu8Fe4S8}$. Color code: $\color{#FFFF30}{\Large\bullet}~\ce{S}$; $\color{#E06633}{\Large\bullet}~\ce{Fe}$; $\color{#C88033}{\Large\bullet}~\ce{Cu}$.

Structurally, chalcopyrite and superstructured bornite have very little in common. Unless there is a specific context given, I'd rather say that $\ce{Cu2FeS2}$ is an outlier among the three and is probably a typographic issue. Also, it's not a good practice to mix dot-notated formulas with formula unit representations unless one wants to underline some structural aspects (e.g. molecular assemblies/coordination polyhedra/domains etc.)

References

Ding, Y.; Veblen, D. R.; Prewitt, C. T. Possible $\ce{Fe/Cu}$ Ordering Schemes in the 2a Superstructure of Bornite ($\ce{Cu5FeS4}$). American Mineralogist 2005, 90 (8–9), 1265–1269. https://doi.org/10.2138/am.2005.1518.


In sp² hybrid orbital, why does the p orbitals only have 3 sides instead of 4? Usually two separate p orbitals would have '4 lobes' while in a $\ce{sp^2}$ hybrid those 2 p orbitals would only have '3 lobes'. why?

I'm referring to diagrams such as the one shown below:

First, an $sp^2$ orbital does have two lobes. One is large, and one is tiny. Secondly, $sp^2$ means "a 'mixture' of one $s$ and two $p$ orbitals". Like the following:

So, you plug in three orbitals, and you get out three more orbitals. When drawn in place, you get this:

There are a total of six lobes here, but they are distributed amongst three $sp^2$ orbitals.

The number of lobes need not stay constant--it's the number of orbitals that does, since each orbital corresponds to two electrons, and there's no way for new electrons to be added/removed (yet). What's happening is that the wavefunctions(orbitals) of the electrons are "mixing up" to distribute the electrons in a way that stimulates bonding (one per orbital). The lobes have a significance, but it's much deeper.


What would be the mechanism for this condensation reaction? Below is the reaction in question. Will it go like an aldol consensation would, where the base would deprotonate the methyl group of the benzoxazole moiety, the negative charge would be transferred over to the nitrogen and this anion would then bond with the aldehyde?
In any way, a drawing of the mechanism would be really helpful.
A bonus question: is this a specific type of condensation? (i.e. like aldol)

Your 2-methyl-1,3-benzoxazole is a masked acetic acid.

Remember that one way to synthesize oxazoles actually starts from aldehydes (and uses TosMIC).

So, yes, the reaction is an aldol condensation where the $\ce{CH}$-active component is carboxylic acid in disguise.


Bainite at room temperature? I read that if, after a tempering from T>723°C to T*=[250°C,550°C], I do an isothermal transformation of an eutectoid steel, I'll get bainitic microstructure.

Does this microstructure exist only if temperature remains at T=T* also after microstructure was formed?

In other words, if I do a cooling until room temperature (after bainite was formed), will it exist at this temperature?

Thank you very much.

The microstructure is retained upon cooling. After bainite forms, you can even quench it and it will remain bainite. The only way to get martensite is to cool it so fast that bainite has no time to form. The only way to get bainite is to cool it so fast that bainite can form, but not so fast you get martensite, and not so slow that you get pearlite. After the microstructures have formed, they remain, unless you reheat it.


Shouldn't it take more energy to break CO2 compared to CO? I read in a book that bond dissociation energy of carbon monoxide $(\ce{CO})$ is $\pu{749 kJ/mol}$ and carbon dioxide $(\ce{CO2})$ is $\pu{532 kJ/mol}.$ Shouldn't it take twice as much energy to break $\ce{CO2}?$

The bonds are not equivalent, therefore they have different bond energy.

  Carbon monoxide consists of one carbon atom and one oxygen atom, connected by a triple bond that consists of two covalent bonds as well as one dative covalent bond.It is the simplest oxocarbon and is isoelectronic with other triply-bonded diatomic molecules having ten valence electrons, including the cyanide anion, the nitrosonium cation and molecular nitrogen. In coordination complexes the carbon monoxide ligand is called carbonyl.

Source: Wikipedia page on Carbon Monoxide


Why are 'sec-' and 'tert-' prefixes printed in italics? While reading about nomenclature of branched alkyl groups, I read that certain common names are also approved by IUPAC. Among those names, only few prefixes are italicised (namely sec- and tert-) and others like iso and neo are printed normally. These names are documented in the IUPAC blue book. What can be the reason for italicising sec- and tert- and leaving out the others?

From the ACD Labs page about IUPAC italicization,it is stated that

  Italicizing mainly serves to mark letters which are not involved in the primary stage of alphabetical ordering.

According to the American Chemical Society Style Guide (p. 235), the use of italics for

  positional, stereochemical, configurational, and descriptive
  structural prefixes when they appear with the chemical name or formula.

(There are several prefixes in the list).

These are separated from the chemical name with a hyphen and are not capitalised


Name for HxOy compounds? Is there a name for the class of stand alone molecules that contains oxygen, water, hydrogen peroxide, and ozone? ($\ce{O2}$, $\ce{H2O}$, $\ce{H2O2}$, $\ce{O3}$) I would not want to include hydroxyls $\ce{OH}$ because they aren't neutral molecules that I can buy.

I think oxide and oxygenate is too generic.

Thanks for your help!

Not only I've never heard of a name for such a class, but the very choice of compounds seems quite artificial to me. What feature do they have in common?

Consisting of H and O? That would exclude oxygen and ozone.
Consisting of H and/or O? That would include $\mathrm H_2$ as
well.
Containing "active oxygen", whatever that might mean? That
would exclude water, but include a bunch of inorganic and organic
peroxides.
Oxide is indeed too generic; also, it does not include oxygen and ozone. Oxygenate is even farther off.

Why would you want to have such a class, really?


Do dishwashing detergent and stain remover powder/stick have the similar ability to remove collar rings? For yellow rings on shirts caused by skin grease or sweat:

Some recommended to use an "Oxyclean" stain remover or stain remover stick
for cleaning.
Some recommended to use dishwashing detergent to brush and wash the
stains, saying "dish washing liquid is designed to remove grease, and
why buy another product (that doesn't work very well -- I've tried)
when you already have one that does?"

From a chemistry perspective, do dishwashing detergent and stain remover powder/stick have similar chemical composition and a similar ability to remove collar yellow rings?

Original dirty collar

After dishwashing detergent and laundry detergent

Detergents seem to consist of both polar and non-polar compounds. Probably because they need the non-polar part of the mixture to dissolve the grease, and the polar compound part of the mixture to get washed off with the water so there is not much left over on your shirt. The ideal solution would probably depend on just how bad the stain is (how much grease needs to be dissolved). If your home solutions did not work the first time, try to look for signs if it did anything at all, and it might not be something that will remove all the grease the first go around.

I have no experience using detergents or removing stains, but I am going off a lot of speculation here as to how the process might work.

"Like dissolves like"

According to this article if there are lots of Mg, Ca, or Iron ions in your water (likely) then it could reduce the effectiveness of soaps. A good question to ask yourself is whether or not your water contains many ions such as these.


If some particles of liquid in a solution convert to vapour then why can't some liquid particles convert to solid? I was reading about the vaporisation of liquid when I came across the following statement.

  The average energy of the particles in a liquid is governed by the
  temperature. The higher the temperature, the higher the average
  energy. But within that average, some particles have energies higher
  than the average, and others have energies lower than the average.Some
  of the more energetic particles on the surface of the liquid can be
  moving fast enough to escape from the attractive forces holding the
  liquid together. They evaporate.

Now if some particles have energies lower than the average, then why don't they convert into solids in the same way the particles with higher energies convert to gases.

If we cover a vessel filled with water with a plate, after some time, we observe droplets of water accumulated at the plate because of vaporisation of water. Why don't we observe any changes that show liquid particles converting to solid?

If water molecules go into the gas phase from the liquid then they are in a different phase. But water molecules go into the gas phase one at a time. If there are the same number of water molecules leaving the liquid as coming into it then the atmosphere above the liquid would have 100% humidity and be in equilibrium with the liquid.

Now if some small number molecules of water happen to line up to form an ice crystal within the water itself what happens? Such an ice crystal is thermodynamically unstable and then "melts" within the liquid phase. Since the whole process takes place in the liquid phase it is virtually undetectable.

Now you also have to realize that form ice crystals big enough to detect visually requires more than a handful of molecules. You'd need billions of billions of molecules. That is just a statistical impossibility.


Why does a gas' ability to absorb infrared radiation affect its potency as a greenhouse gas? My book says that the ability of a gas to absorb infrared radiation affects its potency as a greenhouse gas, but I don't understand why. So what if a gas is able to absorb a lot of IR? How does that affect its potential as a greenhouse gas?

There is only one difference between a greenhouse gas and a non-greenhouse gas: a greenhouse gas has a greater ability to absorb radiation in the infrared (IR) region of the electromagnetic spectrum than does a non-greenhouse gas.

Once a greenhouse gas absorbs IR radiation, that energy must be converted to another type of energy such that the the energy is conserved. The only thing that can happen at IR energy levels is that the absorbed IR energy gets transformed into thermal energy, resulting in an increase in heat in the atmosphere.

From there it follows that the more strongly a particular gas absorbs IR radiation, the more heat is retained in the atmosphere, which might have otherwise just radiated off into space as IR radiation. This means a gas absorbing strongly in the IR is a more "potent" greenhouse gas than one that only absorbs weakly or not at all in the IR.


Difference between tendency of benzene and thiophene to undergo sulfonation Why does thiophene possess a greater tendency to undergo sulphonation than benzene? I came across this reaction which employs the use of sulphuric acid to separate thiophene from commercially prepared benzene.

In the chapter Aromatic five-membered ring heterocycles with one heteroatom in Organic Chemistry by  J. William Suggs (2002) the following explanation can be found (pp. 403-404) which summarizes the thoughts already pointed out in the comments:

  The resonance stabilization energy of benzene is greater than that of these heteroaromatic compounds.
  The order of aromaticity is benzene > thiophene > pyrrole > furan.

  […]

  All three of these ring systems undergo electrophilic aromatic substitution and are much more reactive than benzene.
  In part, this reactivity difference arises because the rate-determining step in electrophilic aromatic substitution is the first step, which breaks up the aromatic $\pi$ system.
  Since thiophene, pyrrole, and furan have less stabilization to lose than benzene, the intermediate is lower in energy and the overall reaction proceeds more rapidly.

  […]

  All three of these heteroaromatic rings undergo electrophilic aromatic substitution, preferentially at C-2.
  The reactivity order is pyrrole > furan > thiophene because of several factors, including the electronegativity of the heteroatom and the resonance stabilization of the aromatic ring.


Excited-states properties with traditional density functional theory Recently I was shown the book "Density Functional Theory II, Relativistic and Time Dependent Extensions" by Nalewajski in which a chapter from E. K. U. Gross et al. is printed. [1] This chapter contains an interesting statement in it's introduction:

  Excited-state properties, however, are notoriously difficult to calculate within the traditional density functional framework and time-dependent phenomena are not accessible at all.

What are those excited-state properties that are "notoriously difficult to calculate" and what makes it hard to calculate them without the TDDFT framework?

Literature:
[1] Gross, E. K. U.; Dobson, J. F.; Petersilka, M. Density Functional Theory of Time-Dependent Phenomena. In Density Functional Theory II; Nalewajski, R. F., Ed.; Topics in Current Chemistry 181; Springer Berlin Heidelberg, 1996; pp 81–172.

Conventional density functional theory (DFT) is strictly limited to describing the electronic density of ground electronic states. This is because the Hohenberg–Kohn theorems, on which it is based, are restricted to non-degenerate ground states (with no magnetic field). Also, because DFT aims at solving the time-independent Schrödinger equation (i.e. it explicitly finds stationary solutions), time-dependent phenomena are not accessible.

From the ground state density, any property of the system including excited state properties could in theory be derived given an appropriate functional[1] $X[\rho]$ that maps the ground state density $\rho$ onto the property $X$. However, in practice, no known functional exists that maps the ground state density to excitation energies or other excited-state properties. Thus, excited states and their properties are in practice unreachable by DFT: I suppose this is what is meant by Gross when stating them as “notoriously difficult to calculate”. I myself would have said impossible.

Finally, to give a short list of properties that can be described with TDDFT but not with DFT — are thus all excited state properties. In particular: bandplots/bandgaps, excitation energies, transition moments, photo absorption spectra, frequency-dependent response properties (optical & dielectric properties), etc.

Specifically, one has to be careful about band plots and band gaps, which are routinely reported from time-independent DFT calculations: they are a representation of the Kohn–Sham energies, i.e., the energies of the fictive non-interacting Kohn–Sham system, which has no physical interpretation at all.

[1] Note here that we are talking about a functional in the generic, mathematical meaning of the word. It is not the exchange–correlation functional.


If A reacts with B (aq), does that mean it is soluble in it? For example, ethanol is soluble in water, but doesn't react with it. On the other hand, H2SO4 is said to be soluble because it dissociates (reacts) in water.

Would ethanol be soluble in HCl (aq)? It won't react with the HCl, but it can form hydrogen bonds with the water, right?

How about ethylamine or ethyl methanoate? Would they be soluble in HCl (aq)? They react with the HCl, so I guess they will be. Is this correct?

Summary, TL;DR:
No, A reacting with B does not mean that A is soluble in B. Solvation and chemical reaction are two different processes, although the distinction can in some cases seem less than perfectly clear. All of the example compounds you discussed are soluble or miscible with both water and hydrochloric acid. The IUPAC definition of a chemical reaction is "A process that results in the interconversion of chemical species", which is fundamentally different from solvation, and does not apply to any of the examples in the question.

As you suggest, solvation and chemical reactions are two fundamentally different concepts. For example, if compound A reacts with water to form a homogeneous liquid solution, that is a different processes than if compound A were to simply be dissolved by water.

Solvation is defined by the IUPAC Gold Book as follows:

  Any stabilizing interaction of a solute (or solute moiety) and the solvent or a similar interaction of solvent with groups of an insoluble material (i.e. the ionic groups of an ion-exchange resin). Such interactions generally involve electrostatic forces and van der Waals forces, as well as chemically more specific effects such as hydrogen bond formation.

Whereas the IUPAC Gold Book defines a chemical reaction as:

  A process that results in the interconversion of chemical species.

One aspect of solvation vs. reaction that may seem confusing is the solubility of ionic species. Even though $\ce{NaCl}$ becomes $\ce{Na+}$ and $\ce{Cl-}$ in an aqueous solution, this does not constitute a chemical reaction as defined above, and we say that $\ce{NaCl}$ is soluble in water. The same is true of your example of $\ce{H2SO4}$; even though it dissociates in water, it is not converted to a new chemical compound. One way to think of this is if you remove the solvent, the solute should typically resume to it's original form. From our previous example, if we evaporate the water from our aqueous solution of $\ce{Na+}$ and $\ce{Cl-}$, we just get the solid $\ce{NaCl}$ back.

Your examples of solubility in hydrochloric acid are a bit complex because that is a two-component system of water and $\ce{HCl}$. All of the compounds you discuss are water soluble and it doesn't matter if the $\ce{HCl}$ is there or not. One slight exception in your examples is ethylamine. Ethylamine itself is miscible with water, but many higher molecular weight amines are not water soluble, but are soluble in hydrochloric acid. In this case, the $\ce{H+}$ from $\ce{HCl}$ protonates the amine, making the hydrochloride salt. This is still an example of solubility however, as once you remove the solvent, you are left with the original compound, in this case the amine.


What is the product when ferric chloride reacts with propan-1-ol? I tried to dissolve ferric chloride in propan-1-ol and I saw ferric chloride partially dissolve in it, but I couldn't identify which reaction had occurred. If any one knows about this, please give me your suggestions.

There is no reaction between propan-1-ol and iron(III) chloride because propan-1-ol isn't acidic enough to react with it. Instead, phenol reacts with iron(III) chloride because it is acidic enough to form either the phenoxide or iron-phenol complex. On this basis, aliphatic and aromatic alcohol are distinguished. The mechanism of reaction of phenol and $\ce{FeCl3}$ can be found here.

According to this quora post:

  Ferric chloride test is specific for determination of phenol presence.
  So as mentioned above, phenol is more acidic and can form phenoxide
  anions. This can form complex with Fe(III) in $\ce{FeCl3}$, which has a
  blue/green/red color depending on the nature of the phenol. Alcohol is
  too weakly acidic to even form the alkoxide or the complex.

So what's happening?

It actually forms a sol-gel. It is a part of iron-oxide based gels.

One noble application of this is non noble metals electrocatalysts for high temperature PEM fuel cells. For this purpose, Fe-containing carbon aerogels were prepared in resorcinol-furfuraldehyde resins. The resins are made in premixed solution of $\ce{FeCl3}$ and propanol.


Why is difluoride a stronger oxidizing agent than dichloride? I am preparing for a chemistry olympiad and I am stuck with the question that why $\ce{F_{2}}$ is stronger oxidizing agent than $\ce{Cl_{2}}$.

If you think about it, strong oxidizing agents must have a high "desire" or propensity to be reduced themselves (as the nature of redox reactions is such). We'll build up on the explanation from the ground up.

Comparing both fluorine and chlorine gas, both are halogens and both only need two electrons each to be reduced to fluoride and chloride ions respectively:

$$\ce{X2 + 2e- <=> 2 X-}$$

Since it appears that the two are similar in most respects, we should look at the clearest differences between Fluorine and Chlorine instead: the number of protons and the number of electrons.

Fluorine is a row 2 element with 9 protons an electronic configuration 1s2 2s2 2p5, whereas chlorine is a row 3 element with 17 protons and electronic configuration 1s2 2s2 2p6 3s2 3p5. Although chlorine has more protons to attract electrons (and therefore be reduced / act as an oxidizing agent), it also has two inner shells worth of electrons that create a shielding effect to reduce the attraction. Fluorine, however, only has one inner shell to do this.

From a simplistic physics point of view, electronic attraction is governed by the charge and the inverse square of the distance. You can approximate that the shielding effect will make the effective charge of the atomic nucleus the same in both halogens, so distance becomes the key deciding factor. Since chlorine is larger, and therefore the attracted electron further than it would be if it was instead bound to fluorine, the attraction (and therefore oxidizing power) is greater in fluorine than in chlorine.

However, looking at the electron affinity (EA) values you'd find that fluorine doesn't follow the group trend of decreasing EA and there's a very good discussion on that over at ChemLibre. Long story short, it's because fluorine is so small that you have to account for the electronic repulsion as well which reduces the bond strength of fluorine vs chlorine (36.6 vs 58 kcal/mol)

There is another measurement which seems to be consistent with this whole fluorine being more oxidizing than chlorine though - and that is reduction potential. The reason that there is a difference between the values for reduction potential and electron affinity even though they are both supposed to be measuring the same thing (that is the energy change related to accepting an electron), the key is the environment in which they are measured. Electron affinities are measured in the gas phase where it is purely the energy related to the electron transfer whereas reduction potentials are measured in aqueous solution and so factors such as hydration energy start to play a role.

As your textbook correctly points out, the high hydration energy is due to fluorine's higher charge density compared to that of chlorine.


Can a precipitate dissolve upon addition of a diluted acid? Is it true that precipitates dissolve upon addition of diluted acids?
If a precipitates is dissolve , does it disappear from solution , or  any thing else  happened out there?

I  am just curious about this. Because I have heard that precipitates can be dissolved if you  add a diluted acid. But why it would happen? And is it valid for any diluted or strong acid ?.Do all the precipitates dissolve upon addition of any acid ?.I would like to know more about this.

Another example of a precipitate dissolving in acid is the reaction $\ce{AgNO3 + HCl -> AgCl (v) + HNO3}$. At first, the silver chloride precipitates out as a white powder (if kept away from direct light). Adding more acid, though, forms ligands with the additional chlorine ions:  $\ce{AgCl(s) + Cl- (aq) -> AgCl2- (aq)}$. See also Physics Forums.


Mechanism for the reaction between trisilylamine and hydrogen chloride In Concise Inorganic Chemistry by J. D. Lee (adapted by Sudarsan Guha), in the chapter "Chemical Bonding", under the topic "Back Bonding with Nitrogen as a donor atom" the following reaction is given:

  $\ce{(SiH3)3N +4HCl \to NH4Cl + 3SiH3Cl}$ (bond cleavage through $\mathrm{S_N2}$ mechanism)

I have learnt about $\mathrm{S_N2}$ in organic chemistry. I know its mechanism there and tried to do the same here (there is no reaction mechanism given for this in my book).

As suggested by this comment, $\ce{(SiH3)3N}$ is still basic so the following set of reactions takes place:

$$\ce{\color{orange}{(SiH3)3N} + \color{blue}{HCl} \to (SiH3)3NH+ +Cl-}\tag{Protonation}$$
$$\ce{(SiH3)3NH+ +Cl- \to (SiH3)2NH + \color{red}{SiH3Cl}}\tag{Nucleophilic Attack}$$
$$\ce{(SiH3)2NH + \color{blue}{HCl} \to (SiH3)2NH2+ + Cl-}\tag{Protonation}$$
$$\ce{(SiH3)2NH2+ +Cl- \to (SiH3)NH2 + \color{red}{SiH3Cl}}\tag{Nucleophilic Attack}$$
$$\ce{(SiH3)NH2 + \color{blue}{HCl} \to (SiH3)NH3+ + Cl-}\tag{Protonation}$$
$$\ce{(SiH3)NH3+ + Cl- \to NH3 + \color{red}{SiH3Cl}}\tag{Nucleophilic Attack}$$
$$\ce{NH3 + \color{blue}{HCl} \to \color{green}{NH4+Cl-}}\tag{Protonation}$$

Summing up all the above reactions we get the overall reaction as follows:

$$\ce{\color{orange}{(SiH3)3N} + 4\color{blue}{HCl} \to \color{green}{NH4+Cl-} + 3\color{red}{SiH3Cl} }$$

I got the mechanism I was looking for! But the main problem is the first step - protonation. In my book, it's given that in $\ce{(SiH3)3N}$, due to the presence of back bonding, the lone pair on nitrogen atom is not available for donation. Or in other words, $\ce{(SiH3)3N}$ cannot act as a lewis base (electron-pair donor). Thus the first step of protonation (acid-base reaction) is ruled out.

So, how does the above reaction take place? Do we have any other reaction mechanism for the above reaction? If yes, could you please specify the mechanism? Or is the above one is itself correct?

I am unable to find relevant information regarding this on the internet.

I am not sure what back bonding you are referring to. It may be related to the outdated concept of silicon using its vacant, high-energy d orbitals in any meaningful way which is not actually the case.

There is a second effect which consists of overlap of nitrogen’s p orbitals with the σ* orbitals of the $\ce{Si-H}$ bonds (which are silicon-centred due to the low electronegativity of silicon). This indeed reduces nitrogen’s basicity slightly compared to other nitrogen compounds but the effect is not that important overall. If you check page 2 of the Evans $\mathrm pK_\mathrm a$ table, you can see that the difference between $\ce{(Me3Si)2NH}$ and $\ce{(iPr)2NH}$ is about 10 logarithmic units which is noticeable (i.e. the HMDS salts are milder bases than LDA) but it does nothing to counteract their overall behaviour. (I am aware that I am arguing with the wrong $\mathrm pK_\mathrm a$ values but sadly the table does not contain data for $\ce{(Me3Si)2NH2+}$ so this is the closest equivalent.)

It is also important to point out that you are using a very strong acid $(\mathrm pK_\mathrm a \approx 8)$ which will protonate practically any nitrogen lone pair that it can come across.


Assigning locants in halo alkanes/alkenes In the first example, if you add up the number of carbon which has a substituent, you will get 2+4+4 = 10. However, if you start from left to right, then you'd get 2+2+4=8. Why don't we follow the latter one when it gives us a lower number? (Lowest set of locant rule.)

The same applies for the second problem: we could've gotten 2+3+4, but we chose 2+4+5, why so?

Indeed, the first compound in your sketch is numbered incorrectly according to IUPAC. IUPAC rules for branched hydrocarbons say: "When series of locants containing the same number of terms are compared term by term, that series is "lowest" which contains the lowest number on the occasion of the first difference. This principle is applied irrespective of the nature of the substituents. " (emphasis mine). In your top case, this means that you have two numbering choices: either 2,2,4 or 2,4,4. The 2,2,4 choice has a lower locant at the first point of difference, so it is preferred. When you construct the name, you indicate the substituents in alphabetical order, so the full name of the top compound is 4-chloro-2,2-dimethylpentane.

If unsaturations are present, then these have a higher priority than alkyl and halogen substituents. From the rules for unsaturated hydrocarbons: "The chain is so numbered as to give the lowest possible numbers to the double bonds." This rule is sufficient to impose the numbering you report.

As Loong correctly pointed out, the links provided above refer to the text of the now obsolete 1979 recommendations. Both the 1979 and 1993 recommendations are available on the same site, maintained by ACD Labs with IUPAC's permission, and are in accordance with what shown above.

Unfortunately, the text of the most recent (2013) recommendations is not freely accessible (that I know of). This is a link to the electronic version of the book, published by The Royal Society of Chemistry. The relevant section is P-14.3.5 in Chapter P-1 (General Principles, Rules, and Conventions): "The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set '2,3,5,8' is lower than '3,4,6,8' and '2,4,5,7'."


Calculating the total uncertainty of the final product I have performed a large number of calculations in order to determine the ethanol content of wine. These included titrations using burettes, mass measurements using analytical balances, and others. I know the uncertainties of the apparatus.

I am given a task to find the ethanol content of wine (done) but also to calculate the total uncertainty by propagation of errors.
As I understand it, the total uncertainty is given by

$$\delta F=\sqrt{\left(\delta q\right)^2+\left(\delta w\right)^2+...}$$

But - does this include the standard error? That is, to calculate the total uncertainty of the final value, will I need to find the standard error for all the sets of titrations (eg - for titration 1, a set of 3 values, find the standard error taking into account all results, even non-concordant ones, and do this for titration 2), and treat those values as fractional uncertainties? Should I add this value to the fractional uncertainties of the burette measurements (ie its own measurement uncertainty, not the standard errors from the readings), and to the measurement uncertainties of the other apparatus?

Finally - does the operation which I apply to the values that the uncertainties effect change the equation for the total uncertainty? Or do I just use the same equation as above?

General approach

\begin{equation}
    \sigma_{F}^2 = \sum \left( \frac{\partial F}{\partial x_i} \right)^2 \mathrm{d}x_i^2
\end{equation}

Here we neglect all correlation terms formed by squaring what is essentially the first order Taylor expansion with respect to the variables about their uncertainty. $x_i$ is a variable or parameter which contains uncertainty and $F$ is the function which depends on the variables and parameters $x_i$. $\mathrm{d}x_i$ is the uncertainty in $x_i,$ i.e., you measure the mass of your solution 10 times and it is 1 kg ± 0.1 kg. Your $\mathrm{d}x_i$ for mass is 0.1 kg. If you want you can think of $\mathrm{d}x_i$ as $\Delta x_i$ instead.

For a given calculation $f$ you take its derivative $F$ wrt each variable/parameter, hence you know its rate of change, and $\mathrm{d}x$ is the known/measured uncertainty of that variable/parameters, i.e., how much the function might be out for a given error in a specific parameter.

So, in your original expression you only accounted for the uncertainty in the parameters, but not how their error would propagate into each and every function/equation/calculation they are used. If a function is linear in a given $x_i$ then at most it will be out by the uncertainties magnitude multiplies by some constant in the equation. However, it may be that the function contains $x_i^{10}$ in which case the error in $x_i$ will be to the 10th power which could be huge! So, use the above equation, with the appropriate derivatives to "propagate" error.

Some background derivation

The change in a function due to a variable/parameter can follow from its differentiation

\begin{equation}
\mathrm{d}f = \left( \frac{\partial f}{\partial x} \right) \mathrm{d}x + \left( \frac{\partial f}{\partial y} \right) \mathrm{d}y
\end{equation}

It is possible that the error due to variable $x$ is equal but opposite to $y,$ and the errors cancel. However, when we square it, which gives us

\begin{equation}
\sigma_f^2 = \left( \frac{\partial f}{\partial x} \right)^2 \mathrm{d}x^2 + \left( \frac{\partial f}{\partial y} \right)^2 \mathrm{d}y^2 + 2\left( \frac{\partial f}{\partial x} \right)\left( \frac{\partial f}{\partial y} \right) \mathrm{d}xy,
\end{equation}

this way errors don't cancel. Also, we almost always assume the correlation term $\mathrm{d}xy$ is negligible. Neglecting the correlation terms, we find ourselves left with the top equation.

More info can be found here: Wikipedia uncertainty propagation. Also, don't forget to donate to Wikipedia. :)


Why does electronegativity increase across a period? What explains why electronegativity increases as you move across a period? Does it have something to do with the shielding effect of added electrons?

The electronegativity is the tendency of an atom or a functional group to attract electrons (or electron density) towards itself.

This is because the attraction of bonding electrons by an atom increases with nuclear charge (Atomic Number) and decrease of atomic radius. Both these factors operate as we move to the right in period.

  Does it have something to do with the shielding effect of added electrons?

The electron cloud in the inner orbits act as a shield and reduces the nuclear attraction to the outer orbits. Because of the shielding effect, the tendency of nuclear attraction reduces and thus electronegativity reduces.


How can water with a very low hardness have substantial buffer capacity? My freshwater fish tank has a pH of 5.8 and a KH value in excess of 40 or 720+ppm while still maintaining a water hardness of 20 ppm or GH< 1 now I am a math/phsyics student but I have taken both first year chem class at university so I expect to follow most basic ideas pertaining to chemistry but this doesn't make a lot of sense. Most water with a high alkalinity also has a basic pH around 8-8.5 normally I'd say that a high KH value is simply saying I have an acidic solution that is very resistant to a pH change i.e. a large buffering capacity but my water is very soft to the point of almost not having anything in it ( GH<1) or less than 20ppm of minerals how can water be so acidic have such a large buffering capacity and have nothing in it?

(the reason why I am asking here is all the aquarium forums i have tried can't seem to figure out what's going on in my tank)

EDIT:

This answer poseing an intresting possibility my tank has Turtles in who obviously are advanced enough to excreet urea instead of amonium i Titrated out a reaction and Got KH value of 28 and GH value of 4.5 i also added several air stones ( my buddie has a bio chem degree and figured it could of been too much $CO_{2}$ we managed to get the Ph up to ~6.4 and drop the KH value ot be more closely in line with accpetable parameters ( a value of 8 still oddly high compared to the GH value of 4.5)

In an aquarium, there could be several non-mineral buffers. The most likely one is a phosphate buffer. There would also be carboxylic acid (amino acids) and probably urea, as well as carbon dioxide.


Why is ZnFe2O4 a normal spinel? Considering that $\ce{Zn^2+}$ is $\mathrm{d}^{10}$, $\ce{Fe^3+}$ is $\mathrm{d^5}$, and $\ce{O^2-}$ is a weak field ligand, the CFSE values of both the normal and inverse spinel structures should be $0$.

Why is it then that the normal spinel, with $\ce{Zn^2+}$ in tetrahedral holes and $\ce{Fe^3+}$ in octahedral holes, is formed?

The reason stated in the literature is fairly straightforward. However, actually tracking down the original source is harder than it looks.

Going back to Phys. Rev. B22, 115 (1980) one finds in the opening paragraph the statement

  Mixed Zn ferrites of the type Zn$_{x}$M$_{3-x}$O$_{4}$ where M
  is a magnetic ion have been the subject of numerous
  investigations in the past, especially with regard to
  their magnetic properties.$^{1}$ The metallic ions occupy
  the tetrahedral A or octahedral B sites in the spinel
  crystal structure. The Zn ions preferentially occupy
  the A sites because of their tendency to form covalent
  bonds involving sp' orbitals.

Reference number 1, on the first sentence, is a book [J. Smit and H.P.J. Wihn, Ferrites (Wiley, New York, 1959)]. The statement on $sp3$ bonding is not backed up by a reference. I can only assume that it was considered common knowledge in the spinel community in 1980.


Does the mixture of a ligand exchange reaction contain all the complexes obtained in the part reactions? A ligand exchange reaction involves many part reactions. So, does the mixture contain all the complexes from those part reactions?

Quoting an example from ChemGuide:

  This can be written as an equilibrium reaction to show the overall effect:

  $$\ce{[Cu(H2O)6]^2+ + 4NH3 <=> [Cu(NH3)4(H2O)2]^2+ + 4H2O}$$

  In fact, the water molecules get replaced one at a time, and so this is made up of a series of part-reactions:

  $$
\begin{align}
\ce{[Cu(H2O)6]^2+ + \color{blue}{NH3} &<=> [Cu\color{blue}{(NH3)}(H2O)5]^2+ + H2O}\\
\ce{[Cu\color{blue}{(NH3)}(H2O)5]^2+ + \color{blue}{NH3} &<=> [Cu\color{blue}{(NH3)2}(H2O)4]^2+ + H2O}\\
\ce{[Cu\color{blue}{(NH3)2}(H2O)4]^2+ + \color{blue}{NH3} &<=> [Cu\color{blue}{(NH3)3}(H2O)3]^2+ + H2O}\\
\ce{[Cu\color{blue}{(NH3)3}(H2O)3]^2+ + \color{blue}{NH3} &<=> [Cu\color{blue}{(NH3)4}(H2O)2]^2+ + H2O}\\
\end{align}$$

And if yes, then which complex has the highest concentration?

Yes, all those complexes from all those reactions will be present, and some more. Essentially Loong did all the mathematics in his answer to: Why is ligand substitution only partial with copper(II) ions and ammonia?

Let's shorten this a little bit and write the general reaction equation:
\begin{align}
  \ce{[Cu(H2O)_{(7-n)}]^2+ + n NH3 &<=> [Cu(NH3)_n(H2O)_{(6-n)}]^2+ + n H2O};&
  n&\in\{\mathbb{N}; 1,\dots,6\}
\end{align}
Therefore we can write the general equilibrium constant:
\begin{align}
  K_n &=
    \frac{{\left[ \ce{[Cu(NH3)_n(H2O)_{(6-n)}]^2+} \right]}}
   {{\left[ \ce{[Cu(H2O)_{(7-n)}]^2+} \right]\left[ \ce{NH3} \right]^n}}\\
%  K_\mathrm{B} &=
%    \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}
%    {{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]^6}}
%  &&= \prod_{n=1}^{6} K_n
\end{align}

Since all reactions are in equilibrium, all possible species will exist, although they might not be quantifiable. The overall equilibrium depends on the concentration of the involved species, especially ammonia.

If $[\ce{NH3}]$ is low, then a complex with smaller $n$ will be more stable, if you go to higher concentrations, larger $n$ will be preferred.

In all of this it should be mentioned that there are also complexes with vacant coordination sites $\square$, like anything in between
$$\ce{[Cu(H2O)5\square_1]^2+},\dots,
  \ce{[Cu(H2O)3(NH3)2\square_1]^2+},\dots,
  \ce{[Cu(NH3)5\square_1]^2+},\\
  \left(\ce{[Cu(H2O)4\square_2]^2+},\dots,\ce{[Cu(NH3)4\square_2]^2+}\right),$$
and possibly others.


Variation of equilibrium constant In an exothermic reaction, the value of equilibrium constant increases with the rise of temperature because,

$\Delta H < 0$ ( exothermic reactions)

$\Delta G = \Delta H - T\Delta S$

Now if $\Delta H $ and $T\Delta S$ are negative and positive respectively, the overall value of $\Delta G$ is also negative. (For positive entropy)

$\Delta G = -RT\ln K$   (where $K$ is equilibrium constant)

Since $\Delta G$ is negative, the value of $RT\ln K$ becomes positive and as $\Delta G$ increases equilibrium constant increases.

Is this logic right?

First of all, How can you assume that, $\Delta S$ is positive in any exothermic reaction? Secondly, the relation you have written $\Delta G = -RT\ln K_\text{eq}$ is not at all correct. The correct relation is $\Delta G = \Delta G^\circ + RT\ln Q_c$ (or $Q_p$ if the reactants are gaseous). Now at equilibrium $\Delta G= 0$ and $Q = K$. So, the correct relation is

$$\Delta G^\circ = -RT\ln K_\text{eq}$$

And, there are obviously differences between $\Delta G$ and $\Delta G^\circ$. So, the reasoning you have given is not totally correct.

The correct reasoning comes from Van't Hoff Equation, which states $$\frac{\mathrm d}{\mathrm dT}(\ln K_{\text{eq}}) = \frac{\Delta H}{RT^2}$$

Integrating both sides, we have

$$\int_{K_1}^{K_2}\mathrm d(\ln K_{\text{eq}} ) = \int_{T_1}^{T_2}\frac{\Delta H}{RT^2}\,\mathrm dT$$

So, we have $$\ln\frac{K_2}{K_1}= \frac{\Delta H}{R}\left(\frac{1}{T_1} -\frac{1}{T_2}\right)$$ Thus we see that increasing the temperature from $T_1$ to $T_2$, the equilibrium constant changes. Here $\Delta H < 0$ (exothermic reaction) and $\displaystyle\frac{1}{T_1}- \frac{1}{T_2} > 0$ so, $\displaystyle \ln\left(\frac{K_2}{K_1}\right) < 0$, which means $\displaystyle {\frac{K_2}{K_1}< 1}$, which implies $K_2 < K_1$. So, the equilibrium constant decreases.

Note: Without any mathematical equation also, the same result can be deduced. In an exothermic reaction, the products are formed along with heat production, which is the definition of exothermic reaction. That means, temperature of the system will increase due to this heat production. Now if you increase the temperature, the system will obviously want to go in the direction where this temperature minimisation is possible, which is actually the backward reaction. So, the reaction shifts backwards, and thus the equilibrium constant decreases as now the product concentration has become lesser and reactant concentration has become larger.


Why zinc sulphate solubility is higher in ethylene glycol than water? I'm trying to find saturation point for dissolved zinc sulphate in water and ethylene glycol.

Based on my knowledge, water is more polar than ethylene glycol as reflected in its higher dielectric constant value. However, max solubity of this salt in water is roughly limited to 3M and for ethylene glycol is much higher (experiment in progress to determine the right number).

What causes this opposite behaviour ?

Zinc is an amphoteric metal which means it can dissolve both in acidic and basic conditions. However, neutral pH is the transition point. Please see this article Recovery of Rare Earths from Waste Cathode Ray Tube (CRT) Phosphor Powder by Selective Sulfation Roasting and Water Leaching on its behavior in water and eg. I believe the reason is that in eg there is no mechanism for zinc to form insoluble hydroxide complexes which will decrease the solubility of zinc. But in ethylene glycol you simply cannot form hydroxides..


Gold Plating on Glass? I am considering labware options to handle NaOH solutions at ~150 °C. One appealing idea would be to coat my current glassware with gold. I've read different sources mentioning the possibility without further detail.
The most detailed description of a process for this is INSTRUCTIONS FOR GOLD PLATING ON GLASS by Peacock Laboratories, Inc who sells (or used to sell) products for doing so. Unfortunately, this says little as to the chemical mechanisms involved.
Would anyone guess what chemicals are being used in this process and know how to explain the mechanisms at hand, and perhaps describe properties to expect from the result?
Can I expect impermeability from the coating or at least a significantly reduced glass surface exposure to hydroxides? What is the best resistance to expect vs routine cleaning of glassware that would be coated with such a method? Can I expect it to not dissociate upon heat shocks that borosilicate glass can handle? Would there be any chemicals to avoid with such a coating, other than the few that would be expected to react with / dissolve gold metal?

Gold plating glass could probably be 99.99% complete. It would make a nice mirror and would probably stay unoxidized for a long time.

However, if I wanted to remove the gold, I think I would rinse the glass with HF solution, or perhaps NaOH solution, either of which would find those teeny tiny gaps and undercut the whole gold plate.

I don't think there would be any strong chemical bonding, and no mechanical strength to a gold film. Glass is just a reactive substrate, and protecting it would require a finite thickness, with mechanical strength.

An example of plating not working well is chromium plating (although it can be done very well, it requires much attention to detail). Automobile bumpers used to be chrome plated, but developed rust spots after a few years because of porosity in the chromium film. I believe nickel plating to reduce corrosion tendency (but Ni is not shiny enough) followed by chrome plating works reasonably well. But then there is hexavalent chromium toxicity... So now we just paint the bumpers or make them out of plastic.

Perhaps you could coat the glass with teflon or polypropylene.


How to recognise the significant figures in given input values for the calculated output value? If the transfer of $1~\mathrm{kJ}$ of energy to a quantity of water causes its temperature to rise $10~^\circ\mathrm{C}$, what is the mass of the water?  I know how to do the math and everything but I'm so confused with significant figures. If $1~\mathrm{kJ}$ or $100~\mathrm{J}$ only has 1 significant figure, then would I write $\pu{23.90 g}$ as $\pu{20 g}$?

You have a reasonable question: if the data supplied specified 1.00 $\mathrm{kJ}$, and the temperature change was given as $10.00~^{\circ}\mathrm{C}$, then "23.90 g" would be a justifiable answer.

However, since the data precision was given to only one digit for energy, then the mass would be "20 g".

Personally, given enough space, I'd write an explanation, of the sort you've give, e.g. "Though I calculated 23.90 g, it could be rounded to 20 g due to lack of significant digits in the data given."

BTW, you've shown good effort in thinking through the homework assignment, in my opinion, which is as important as a numerically correct answer.


What exact color does ozone gas have? This answer to a now closed question says that ozone gas has a "slight blue" color. But there are lots of blue colors: sky blue, ultramarine, phthalo blue, many others — what is closer to the color of ozone?

This page shows photos of ozone solutions in water, like the one below, but this may not be representative of the color of the gas.

 (source)

So, what exactly color does ozone gas have? Is it a single hue, or does it change depending on concentration/amount of ozone?

Due to Chappuis absorption, ozone does have a bluish color. To determine exactly what kind of blue it is, let's first look at the spectrum of absorption in the Chappuis band. The following plot was done using these data for 293K.

This is spectral cross-section of absorption. To determine color from this spectrum, we need to choose some parameters:

Number density of ozone molecules,
Thickness of ozone layer we're trying to visualize,
Illuminant.

If we denote spectral radiance of our illuminant as $L(\lambda)$, thickness of ozone layer as $d$, ozone molecule number density as $\rho$, and absorption cross-section as $\sigma(\lambda)$, then we'll get the following expression for spectral radiance transmitted through the layer:

$$L_T(\lambda)=L(\lambda)\exp\big(-\sigma(\lambda)\rho d\big).$$

The most sensible illuminant to choose for showing color of a material on the web is the CIE illuminant D65, whose color is the white point of the sRGB color space. Its spectrum can be found e.g. here.

We can find the color in XYZ space using CIE 1931 color matching functions (can be found e.g. here). The expression is

$$c_X=\int_{300}^{830}L_T(\lambda)\bar x(\lambda)\,\mathrm d\lambda,$$

and similarly for $Y$ and $Z$ coordinates. Then these can be transformed to sRGB using linear transformation matrix $\mathrm{XYZ}\to\mathrm{sRGB}$ given e.g. here and gamma-correcting to $\gamma=1/2.2$ to yield final sRGB values.

Then, for ozone molecule number density $\rho=10^{25} \frac{\mathrm{molecule}}{\mathrm{m}^3}$ we'll get the following colors for different layer thicknesses:

For comparison, typical ozone column in the atmospheric ozone layer is about 300 Dobson units, which is equivalent to $2.687\times10^{20}\frac{\mathrm{molecule}}{\mathrm m^2}$; with our $\rho$ chosen above this corresponds to $d=8\,\mathrm{mm}$. So for daylight, ozone column has negligible effect on the sky color (unlike the evening — see the history of Chappuis absorption!).

As can be seen in the above plot, hue does change with increasing layer thickness. If we normalize the RGB values to see the hues of the thick layers (this would correspond to increasing illuminant power to compensate for absorption), we'll get the following hues:

Note that the violet hues in the above plot aren't accurate: they can't be accurately represented on sRGB monitors, so the plot only approximates them. They should be more saturated. Here's how the chromaticity changes from the white point to the most violet with increasing layer thickness (dashed triangle denotes the sRGB gamut):


Splitting of $d$ orbitals when ligands approach central metal ion In my high school chemistry book, it is written that when ligands approach the central metal ion (transition metal ion) to form dative bonds, the $3d$ orbitals split into two: two which are in higher energy level and the other 3 in lower energy orbital...

One thing that doesn't make intuitive sense to me is that why 3 orbitals move to a lower energy level... if the ligands repel the electrons in the $3d$ orbitals, how can some of the orbitals possibly lose energy and be at a lower enrgy state than they were currently at before...

The picture you pose is only a part. The full picture is posted below.

This solves your doubt. The splitting does not occur about the original level, i.e. not about the energy level the d-orbital has in complete absence of the ligand's electrostatic field. But, the splitting occurs about a hypothetical Barycentre. This is the energy level of the d-orbitals (still degenerate) in a hypothetical spherical electric field (with spherical symmetry). Due to the symmetry of the hypothetical barycentre field, the d-orbitals remain degenerate but at an energy higher than in the absence of the ligands. Now, the $t_{2g}$ group which is lowered in case of an octahedral field, is still above the original free ion energy level in the complete absence of the ligands.

EDIT
By conservation of energy, we mean that the energy is conserved during splitting. The very process of splitting occurs symmetrically about the barycentre, so that the net amount by which $t_{2g}$ is lowered ($3\times 0.4\Delta_o$) is equal to the amount by which $e_g$ is raised ($2\times 0.6\Delta_o$). But the net energy in absence of field and in presence of field need not be same for energy conservation to hold. Whatever the rise in the net energy (from free ionic to barycentre) will be taken from the kinetic energy (or potential) of the approaching ligands.


Why can alkoxides leave in the E1cb mechanism? Why can alkoxides leave in the $\text{E}_{\text{1cb}}$ mechanism? I refer to the mechanism below. I would think that the anion formed from the first step would be more stable than the alkoxide ion, because the charge on the alkoxide ion is intensified by the electron-donating alkyl group, making it less stable. Correct me if I am wrong.

image source: https://en.wikipedia.org/wiki/E1cB-elimination_reaction

The fact that the elimination of $\ce{EtO-}$ leads to the generation of a carbonyl bond makes this a feasible reaction. Also, the $\ce{EtO-}$ ion generated can then take $\ce{H+}$ from water taking the equilibrium further in the direction of the forward reaction ( the net constant would be ~ $10^2$).


Free refined value in Rietveld I have a question about Rietveld refinement

The author of this paper claimed that the free refined CN ligand is 1.081.
However, they does not show how they got that number. They stated it was based from this table however I am not sure how you can get 1.081 from this table. What is the calculation?

Thank you.

Source:

Reference

Wang, W.; Gang, Y.; Hu, Z.; Yan, Z.; Li, W.; Li, Y.; Gu, Q.-F.; Wang, Z.; Chou, S.-L.; Liu, H.-K.; Dou, S.-X. Reversible Structural Evolution of Sodium-Rich Rhombohedral Prussian Blue for Sodium-Ion Batteries. Nature Communications 2020, 11 (1), 980. DOI: 10.1038/s41467-020-14444-4

I guess that 1.081 refers to the distance length (in Angstrom) separating the C and N atoms. You obtain this value by applying by using the values listed in the table; this is quite tedious, takes some time and require quite good knowledge about crystallography. Otherwise (and much more easy), you can use the crystallographic data listed in the table to build the structure using a proper crystallographic software (such as Vesta); then the software will calculate bond distances for you.


How to remove Ethanethiol from household natural gas? Household natural gas is injected with a little bit of Ethanethiol to make it a distinctive odor for safety purposes.

But, if it is used to drive cars (*), its odor is sensible and it is annoying.

Is there a way to remove this from the household gas (which is mainly the mix of 90% methane and 10% ethane) on a home budget scale?

My first idea is to have some type of chemicaly which could be inserted into the empty bottle.

(*) In most countries, this can be done legally, although it has to follow relatively strict safety and tax regulations.

Just an idea: You can pass the gas through a bleach solution that can oxidize the thiol. I have no idea how effective this can be since the human smell is extremely sensitive to ethanethiol (to the scale of parts per billion). One possible set-up (very common in organic labs) is shown in this picture. The end Buchner flask should contain bleach and not HCl. The other conical is a safety measure in case you have suck back of the bleach solution.


Consequences of ortho effect on resonance Consider any carboxylic acid with any group at ortho position. According to ortho effect, the resonance of the $\ce{C=O}$  with the ring is inhibited.

Is this resonance entirely lost? Or some little amount still exists?

Can resonance be present when the participating orbitals are not in the same plane?

If they are off the plane due to steric effect,  then resonance wouldn't occur, hence the (-)R effect of - COOH group vanishes and hence "generally" acidity of the respective acid increases.


Why is dU an exact differential and dq an inexact differential? dU, dG, dH etc are all exact differentials and the variables themselves are known as state functions because they only depend on the state of the system. However, dq and dw for example, are inexact differentials. My questions is, what does this actually mean? I've been told that exact differentials are not dependent on the path (what does 'path' mean?) but inexact differentials are. I have been told this is related to line integrals but I am not sure how. Also is this relevant?

As opposed to an exact differential, an inexact differential cannot be expressed as the differential of a function, i.e. while there exist a function $U$ such that $U = \int \mathrm{d} U$, there is no such functions for $\text{đ} q$ and $\text{đ} w$.
And the same is, of course, true for any state function $a$ and any path function $b$ respectively: an infinitesimal change in a state function is represented by an exact differential $\mathrm{d} a$ and there is a function $a$ such that $a = \int \mathrm{d} a$, while an infinitesimal change in a path function $b$ is represented by an inexact differential $\text{đ} b$ and there is no function $b$ such that $b = \int \text{đ} b$.

Consequently, for a process in which a system goes from state $1$ to state $2$ a change in a state function $a$ can be evaluated simply as
$$\int_{1}^{2} \mathrm{d} a = a_{2} - a_{1} \, ,$$
while a change in a path function $b$ can not be evaluated in such a simple way,
$$\int_{1}^{2} \text{đ} b \neq b_{2} - b_{1} \, .$$
And for a state function $a$ in a thermodynamic cycle
$$\oint \mathrm{d} a = 0 \, ,$$
while for a path function $b$
$$\oint \text{đ} b \neq 0 \, .$$
The last mathematical relations are important, for instance, for the first law of thermodynamics, because while $\oint \text{đ} q \neq 0$ and $\oint \text{đ} w \neq 0$ it was experimentally found that $\oint (\text{đ} q + \text{đ} w) = 0$ for a closed system, which implies that there exist a state function $U$ such that $\mathrm{d} U = \text{đ} q + \text{đ} w$.


Why does Cobalt have different coordination numbers in its complexes with water and thiocyanate? I'd like to know why does cobalt(II),  $\ce{Co^2+}$, form $\ce{[Co(H2O)6]^2+}$ complex with water while  it forms $\ce{[Co(SCN)4]^2-}$ with the thiocyanate? Why is there $\ce{6H2O}$ but only $\ce{4 SCN-}$? And is the cobalt's  coordination number 6 in these cases?
Thanks.

Firstly, in water Co(II) is not exclusively $\ce{[Co(H2O)6]^2+}$, there is also a small amount of tetrahedral $\ce{[Co(H2O)4]^2+}$, 0.08% at room temperature increasing substantially at high temperature. Canadian Journal of Chemistry, 1980, 58(14): 1418-1426.

According to Greenwood's "Chemistry of the elements" 2nd edition page 1131, the more polarizable the ligand, the more the tetrahedral form of Cobalt (II) is preferred since fewer ligands are required to neutralize the metal ion's positive charge.


Are all volatile liquids combustible? I realized that liquids like petrol, kerosene and even perfumes are combustible when they are volatile.

Could that be the reason why water is not combustible is because it is not volatile?

Are volatile liquids all combustible?

Many volatile liquids are not combustible

Dichloromethane (DCM) is a widely used solvent by chemists. It boils at around 40°C (the same as diethyl ether) but is not remotely combustible or flammable. Ether is both very volatile and very flammable, so much so that most labs would prefer not to have it used anywhere where flames or sparks could be present. HFC-134a is a refrigerant gas at room temperature and used to be preferred to alternatives like propane (also a gas) because it didn't risk creating an explosive atmosphere when it leaked.

Combustibility has almost noting to do with volatility (other than in the trivial sense that things that are hard to vaporise are hard to set alight even if they are inherently flammable–like bitumen). Combustibility is determined by the chemical properties of a substance, usually whether the reaction of the substance with oxygen releases energy easily. Ether is both volatile and very combustible as that reaction releases a lot of energy and is easy to start. DCM can be burned but releases far less energy and the reaction is very hard to start or maintain as the chlorine in the molecule inhibits the combustion reactions.

Other compounds, like water, are certainly volatile enough to combust (how would we get rain if water were not volatile?) but there are no possible reactions that release any energy from reacting water with oxygen.

So two factors matter in combustibility: is there an energy releasing reaction from burning the compound?; is there a reaction mechanism that makes it easy to cause that reaction to happen? Neither are true for water; many other compounds either don't release enough energy from reacting with oxygen or have chemical components that inhibit the reaction.

Volatility has little or nothing to do with either.


Convert benzene to isopropylbenzene Which of the following reagents will convert benzene to isopropylbenzene? Choose all that are appropriate.

I know for certain that (b) can be used, but I also think the following mechanism with (d) is also possible:

Is my line of thinking correct with this one?

You are right, both abstracting $\ce{Cl-}$ from the alkyl chloride with Lewis acid as well as dehydration of isopropanol with $\ce{H2SO4}$ are suitable ways to generate the 2-propyl carbocation, your electrophile in this reaction. Option (c) will not work because $\ce{AlCl3}$ reacts with alcohols to form aluminium alkoxides.

Source: Master Organic Chemisry


Why does aqueous NaCl conduct electricity? Could someone please explain why an aqueous solution of NaCl conducts electricity? I've Googled this up but couldn't get a satisfactory answer. This has an answer on Physics.SE but that's too complicated. My textbook doesn't answer my question.

Original Question:
Why an aqueous solution of $\ce{NaCl}$ conducts electricity
Because $\ce{NaCl}$ is an electrolyte. This means it yields ions in solution.
Simply put, solid $\ce{NaCl}$ consists of $\ce{Na+}$ cations and $\ce{Cl-}$  anions bound together in a rigid crystal lattice. When it melts or is dissolved in water, the crystal lattice breaks. The ions are now able to move around. Similar to charged particles in a metal conductor (in this case electrons), in liquid form or aqueous solution the ions are the charged particles that can move, allowing the solution to conduct electricity.
Follow-up question:
Why $\ce{NaCl}$ dissociates in water
The (rather oversimplified) answer is that dissociation of an ionic compound is facilitated by ion-dipole attractions between the ions of the compound and the polar water molecules.
The water molecules (dipoles) are attracted to the ions  and cause the crystal lattice to destabilize and ions to dissociate. The charged ions in solution are surrounded and stabilized by the water molecules (dipoles).
Note: Some ions migrate as ion pairs, but for a strong electrolyte most or all the ions will be dissociated and surrounded by water molecules.

As you seem dissatisfied with the answers received so far, I assume you want to fully understand the mechanism involved in dissociation. Unfortunately, this is rather complex.
There is an excellent  paper by Ballard & Dellago [1] that explains their work on the subject, but you will probably need a bit more physical chemistry / thermodynamics knowledge to understand it fully.
Reference

Ballard, AJ & Dellago, C "Toward the Mechanism of Ionic Dissociation in Water," *J. Phys. Chem. *B 2012, 116, 45, 13490–13497
Publication Date:October 19, 2012
https://doi.org/10.1021/jp309300b