Identifying 5 unknown substances I need help identifying the following 5 substances; G, Y, T, L and J

G is a white solid which is insoluble in water. Consider the following reactions carried out on solid G.
Step 1:
Excess dilute nitric acid is added to a small amount of G in a test-tube. A colourless and odourless gas Y is given off and it is noticed that Y extinguishes a lighted splint.
The solution remaining in the test tube is divided into two different test tubes, sample A and sample B.
Step 2:
A few drops of potassium iodide solution are added to a sample A and a bright yellow precipitate L is observed.
Step 3:
Sodium hydroxide solution is added drop wise to sample B. A dense white precipitate, T, is immediately observed. Precipitate T dissolves when excess sodium hydroxide solution is added to it.
Step 4:
When some G is heated in a hard glass tube, gas Y which is colourless and odourless is given off. The solid J remaining in the hard glass tube changes colour on cooling from red-brown to yellow.

These are my solutions:

G = $\ce{PbCO3}$ because
Y = $\ce{CO2}$ - colourless and odourless gas
L = $\ce{PbI2}$
T = $\ce{Pb(OH)2}$
J = $\ce{PbO2}$

Are they correct or not?

If you don't have any idea:

Start with the hypothesis that your starting material is the salt of some metal.
Some of the test are apparently useful to identify the cation, others tackle the anion.
Is it likely that the flame-extinguishing gas comes from the cation?
If it is not the cation, it must be the anion from which the gas is formed.
Think in the main property of diluted nitric acid. Is it an oxidant in diluted form, or just something else?

I almost completely agree with the solution given in the question, except for the last step.

The three known lead oxides are

$\ce{PbO2}$: dark brown to black, strong oxidant
$\ce{Pb3O4}$: red-orange, used as a rust-proof primer (in German: Bleimenninge)
$\ce{PbO}$: (normally) yellow

When dark $\ce{PbO2}$ is heated, it splits off oxygen to form red $\ce{Pb3O4}$, when it's heated up more, more oxygen is released and yellow $\ce{PbO}$ is formed. Properly done in a quartz tube the colours of the German flag can be produced this way.

In the puzzle, we are not starting with $\ce{PbO2}$ but with $\ce{PbCO3}$.

When $\ce{PbCO3}$(= G) is heated, $\ce{CO2}$ (= Y) is released. The questions mentions a hard glass tube, which suggests that the tube has to withstand high temperatures.

Let's assume that the reaction simply is $\ce{PbCO3 ->[\Delta] PbO + CO2 ^}$

A possible explanation for the colour change would be that at high temperatures, red $\ce{\alpha-PbO}$ is initially formed, which converts to yellow $\ce{\beta-PbO}$ upon cooling.


How does a water molecule bond to an anhydrous salt to form a hydrate? Some sites suggest that water molecules bong to anhydrous salt via a "loose bond". What does that even mean? $\ce{H2O}$ is a polar molecule and salts are made up of ionic bonds. Is it a hydrogen bond between the anions and hydrogen atoms?

"Loose bond" is really a loose term which engenders loose understanding. The bonds that hold water molecules in the structure are usually dative bonds, which is a variety of polar covalent bonds. They are formed between water's O and the metal ion, using spare electron pairs of the former. There is nothing unusual about having covalent bonds in an ionic crystal. Think, for example, of $\ce{S-O}$ bonds in the sulphate anions within $\ce{MgSO4}$ or any other ionic sulphate.

Alternatively, in some relatively rare cases water molecules may be held by hydrogen bonds to the anions, like you suggested. This is the case with that fifth water molecule in $\ce{CuSO4\cdot5H2O}$, as explained by @ringo.


Why can't the electrophile attack the chlorine atom in a halobenzene? Why can't the electrophile attack the chlorine atom in a halobenzene?
I mean, chlorine is a deactivating group, primarily due to its inductive effect.

If we can say that its ortho and para influence on rise of electron density is why electrophiles occupy those positions, why not they bond with chlorine who does take up those electron density at the same time?

Examples of electrophilic substitution could be chlorination of chlorobenzene, nitration of chlorobenzene, sulphonation of chlorobenzene ...

Given the charge distribution and the shape of the HOMO, I would be very surprised if an electrophile does not attack the chlorine. This is from a calculation on the DF-BP86/def2SVP level of theory.

If you account for the hydrogens effectively shielding the negative charges of the carbons, the only place left to go is the chlorine. (Natural Population Analysis, NBO 6.0)

The largest coefficient of the HOMO is at the chlorine (C1 ipso, C4 para):

Alpha occ 29 OE=-0.224 is Cl12-p=0.29 C4-p=0.24 C1-p=0.19 C2-p=0.09 C6-p=0.09

Generally I would expect it to be rather unreactive towards electrophiles.


Will electric field make organic matters (e.g. glucose) ionize in water? After reviewing this post, I have a new question regarding ionization and dissolution.

If applying a strong electric field (DC or AC) in a water solution, will the electric field ionize the organic matters that are not ionizable? For example, an electric field makes $\ce{CH3COOH}$ ionize more? Or makes glucose ionized in water instead of forming hydrogen bonding with the water molecular? Thanks.

The short answer to your question is no, particularly for your example of glucose. Part of the key to your question is the statement: "[...] will the electric field ionize the organic matters that are not ionizable?".  If they are not ionizable, then they are not ionizable :)

You could electrolyze glucose to polyhydric alcohols as in this patent. But in this case you are not ionizing glucose, but creating an entirely different compound (which is still not strongly ionic).

There are compounds that can be electrolyzed to ionic compounds, but that is really a very different process than what you are asking about.  You wouldn't have ionization of the compound, you would be electrochemically converting a non-ionic compound into an ionic compound. See for example, this abstract1 in which the authors electrolyze ethanol to acetic acid, which would then ionize in the aqueous solution to some degree.

But, as in your example of gucose, the ethanol itself would not be ionized. You would simply be producing, via electrochemical reaction, an entirely different, ionic compound.

1) Journal of the Electrochemical Society, "Electrolytic Oxidation of Ethyl Alcohol to Acetic Acid", 1947, volume 92, issue 1, 335-342


Why does the reference electrode in potentiometry have a diaphragm? I am studying about potentiometry and pH-measurement with a glass electrode. I don't understand why the reference electrode has a diaphragm and is immersed in the measured solution. If I got it right, there is no current flowing and the composition of the buffer inside the reference electrode is constant. Why should it then be connected with the solution? Can someone clarify?

This device is basically a galvanic cell (and an amplifier and voltmeter). Galvanic cells are quite similar to electrolytic systems, the difference being wheter one has an external voltage source.  If there is no “salt bridge” (i.e. some way for ions to wander between the two half-cells), the cell immediately stops working. This is certainly true, since one can directly observe it.

This makes sense if one considers that in (about all?) systems only one kind of metal ions is involved at each electrode. If you disconnect the two half-cells, there is a potential difference, but only between the electrode and the electrolyte. (How do you measure such a thing without intruducing another electrode??) The ions would simply stick around near the electrode until the charge makes the reaction unlikely to occur; only if the ions in the electrolyte can mix somewhat can the reaction continue. Keep in mind that the voltmeter does not have infinite resistance but "only" 10MOhms or so. This animation might be quite helpful.
In different terms: One needs a small current (at least for a short time) to measure voltage (keep in mind that the amplifier will have some capacitance) and for this to occur the circuit has to be closed, on one hand via the voltmeter, on the other via the salt bridge.

But, since you do not want to influence the measured system too much, you only provide this almost tight diaphragm — not a large(ish) diameter hole — and almost no current flows. I see no theoretical reason why a very small hole would not work equally well, but I assume this is more annoying for technical reasons (drilling it, clogging, etc.). One could argue that very small currents can flow even without a salt bridge because separate containers are still not infinitely far apart and their fields can influence the other one, but this seems to be negligible; normal glass is already thick enough to limit capacity to a few pF.


Wavelengths of colors emitted by alkali metals during flame test While reading my NCERT chemistry book (Pg no.-300), I came across  a table that mentions colors emitted by alkali metals and corresponding wavelengths.  It mentions that Potassium emits violet color(As far as I know it's more close to lilac) but corresponding wavelength written is 766.5 nm (I mean this is closer to red region) which does not even lie in visible spectrum range...  Does that mean that infrared rays are emitted and book suggests that or is it wrong to say so?

The NCERT textbook is partially wrong for elements after sodium. The flame colors and wavelengths do not match. Keep in mind that for alkali metals, the Bunsen burner flame shows multiple wavelengths for potassium, rubidium and cesium.  It is very easy to excite their outermost electron in such a low temperature flame to higher energy levels. My point is that the table should mention multiple lines for K, Rb and Cs.
What you textbook is showing is showing the wavelength of the strongest intensity for all elements except Cs.
The strongest lines or the so-called resonance lines for the alkali metals are
Li = 670 nm (falls in the red region)
Na = 589 nm (orange-yellow)
K  = 769 nm (deep deep red)
Rb = 780 nm and 794 nm  (still deep red)
Cs = 852 and 894 nm (near infrared, not visible to human eye)
Now you are right, K flame appears violet to us and Cs is indeed blue. As stated above, potassium salts in Bunsen flame also show
K = 404 nm (deep violet)
Now violet and red wavelengths appear somewhat lilac to us.
The same goes for Cs, the strongest emission is near infrared, but the visible spectrum has a faint line of 455 nm - this is bluish. Hence the flame appears blue. The name Cesium also comes from bluish gray. Rubidium comes from red from its red lines in the spectrum.
The source of wavelengths is a very old book by Grotrian.
The Bunsen flame spectrum of Lithium and Sodium is pure Li and Na spectrum. The sodium line is a doublet. I could not find the experimentally measured emission spectrum of K, Rb, and Cs.


How to color bonds by the strain? I need to create a figure of a molecule in which the bonds are colored in a RGB scale according to the values read from a data file. The data file contains, for example, the increase or decrease in the bond length compared to some reference structure. In this way, the intensity in a color range red-white-blue would be given by the difference between the relaxed and unrelaxed bond lengths, similarly to the figure attached.

I have tried to make this figure using jmol, but I didn't have success so far. Could anyone suggest a program that is able to do such figure? Or could someone point out any suggestion of how to do this color maps on chemical bonds?

I will be grateful for any help you can provide.

All the best

StrainViz is the program that might be useful [1].

Reference

Colwell, C. E.; Price, T. W.; Stauch, T.; Jasti, R. Strain Visualization for Strained Macrocycles. Chem. Sci. 2020, 11 (15), 3923–3930. DOI: 10.1039/D0SC00629G.


What has stronger antioxidant potential: BHT or BHA? Butyl Hydroxy Toluene (BHT) has two tert-butyl groups that can stabilize a free radical. Butyl Hydroxy Anisol (BHA) doesn't, though it has the 4-substituted methoxy group (I'm not sure what that does).

What is the stronger antioxidant?

That's a good question. In literature there are some contrasting opinions about the antioxidant power of BHA vs BHT. TBHQ (2-tert-butyl hydroquinone) is also used as antioxidant, although maybe is less common.
There is a paper from 1997 in which the antioxidant activity of BHA, BHT and TBHQ are examined with Miller's test (M. Karamac and R. Amarowicz, Grasas y Aceites, 1997, 48 (2), 83-86), and in this test BHA and BHT gave similar results, although in the initial phase of the experiment BHA seemed to have a higher antioxidant activity than BHT.
In the paper precedent studies are also reported. Quoting from the paper:

Higher antioxidative activity of BHA than of BHT and lower for TBHQ was
demonstrated in thin layer model system of methyl linoleate on cellulose (J. Am. Oil Chem. Soc. 52,174-178). Berner et al. (J. Am. Oil Chem. Soc. 51, 292-296) reported antioxidative index in lard emulsion 30-fold higher for BHA than for TBHQ. Yet, in active oxygen method, antioxidant index for TBHQ was 3-fold higher than for BHA and BHT. Higher antioxidative activity of TBHQ compared with BHA and BHT in model studies with soybean oil and chicken fat in emulsion versus dry oil tests was reported by Cort et al. (J. Am. Oil Chem. Soc. 52,174-178). Relative effectiveness of antioxidants-linoleic acid monolayer on silica gel for TBHQ was almost the same as for BHT, while for BHA almost 3-fold higher (J. Food Sci. 42, 1533-1535).


Why can syn-periplanar E2 elimination happen in molecules like vinyl bromides but seemingly not in other molecules? E2 eliminations occur when the leaving group and the proton are in the anti-periplanar conformation not syn-periplanar. This is beyond a merely a preference as leads to E2 reactions being stereoselective. However, when it comes to vinyl bromides for example (involving a double bond) the reaction can occur in the Z- and E- isomers (i.e both syn-periplanar and anti-periplanar). How can this be the case?

Elimination from vinyl halides occurs faster when the halide and proton have a trans relationship.  When the halide and proton are located on the same side of the double bond (cis), the elimination to form an alkyne is much slower.  The faster elimination from the trans arrangement is consistent with a preferred anti-periplanar arrangement in the E2 reaction.

The fact that elimination still occurs, albeit at a slower rate, when the groups being eliminated are arranged cis to one another suggests that a different mechanism is involved in the "cis" case.  Two possible mechanistic alternatives are

cis elimination from a higher energy (higher energy, therefore less preferred, slower) syn-periplanar conformation, or
an E1CB mechanism where the proton is removed in a first step and a carbanion is generated, and this is followed by halide ejection in a second step.

Experiments in analogous molecules suggest that the E1CB pathway is most often followed in "cis" elimination from vinyl halides.


Hydrogen donor without hydrogen acceptor? Is there a molecule that can donate a hydrogen to a hydrogen bond but not accept one? It would seem that since the donated hydrogen needs to be bonded to an electronegative atom, that electronegative atom will always have lone pairs available to accept a hydrogen bond.

For example, in ethanol, the H in the OH can participate in hydrogen bonding since the the O is electronegative. Then the oxygen with its two lone pairs can also accept hydrogen bonds.

In reverse, things like diethyl ether can accept but not donate hydrogens because the oxygen is surrounded by carbons, not hydrogen.

Are there any exceptions to this apparent rule?

As mentioned in the comments by Philipp, ammonium ions $\ce{R3NH+}$ are a good direct answer to what you are looking for.

The nitrogen in an amide $\ce{RCONHR'}$ is similar - the N lone pair is conjugated into the C=O bond and that makes it a very poor hydrogen bond acceptor, but the N-H lone pair can be a hydrogen bond donor. For more details, see this previous question with an excellent writeup by Jan. Now an amide can of course function as a hydrogen bond acceptor via the lone pairs on oxygen, so it's not strictly what you're looking for, but it was just something related. And it is of course directly relevant to the secondary structure of proteins.


Dependence of mass transfer coefficient on solubility Does solubility of the solute affect the value of the mass transfer coefficient?

The book I am following mentions that the mass transfer coefficient is a characteristic of the constituent and its environment, and solubility is a property of the environment.

No.

In all but highly exotic situations, diffusive and convective fluxes of solutes are linear in the solution-phase concentration of the species:

$$
\begin{align}
J &= -D {\mathrm dC\over \mathrm dx}  & \mathrm{(diffusion)} \\
N_c  &= k_c\left(C_\infty - C_\mathrm{surface} \right)  \quad & \mathrm{(convection)}
\end{align}
$$

Whatever the dissolved concentration $C$ (or concentration gradient ${\mathrm dC\over \mathrm dx}$) of the species is, that is the value you use in both of the above equations.  Whether that concentration (gradient) arises from a species with low, intermediate, or high solubility doesn't matter at all. I suspect your book is using the term environment to refer to things like the geometry, temperature, and fluid-flow characteristics of the system.

The main way I can think of where the solubility might affect the overall mass transport achievable in the system is in its effect on the maximum possible dissolved solute concentration at the surface.  If solubility limitations constrain $C_\mathrm{surface}$ to a low value, then the magnitude of $N_c$ will necessarily be low.  The low solubility is typically assumed not to appreciably change the value of $k_c$, however.


If you accumulate 100 identical spheres how many will end up on the surface? So far I have asked this question in physics, engineering and math sections but have not gotten an answer in three years. If you have 100 identical spheres and group them into A sphere shape, how many will end up on the surface? They can be packed tight or loose in a hexagon pattern or cubicle pattern. They can be symmetrically or evenly spaced. I realize the answer can be off by one or two depending on how you group them, But in general it seems like something that can be answered. Is there a formula that could be applied to any number of spheres? For example what if you have 120 identical spheres, how many will be on the surface then? Imagine an atom with 120 nuclei in the nucleus. Thanks for your help.
EDIT:After seeing some of the responses it reminded me of my thought process a few years ago when I began to think about this. How do you define surface, what type of packing, etc. etc. I too tried the plastic bags and ended up using different colored magnetic balls. Experimenting has worked just fine and I asked the question because I truly thought a simple formula would be available. As for what the surface is I have found another question that actually comes closer to what I'm interested in. No matter how many spheres are accumulated into a spherical shape you can separate them into two categories. Spheres in the (interior) and spheres on the (surface or touching a sphere on the surface). This seems confusing at first but it actually simplifies the surface question by literally filling the gaps.  So really the question is how many spheres are not on the surface or touch a sphere on the Surface? Even still it can be difficult counting spheres when you transfer between hexagonal, cubical or symmetrically spherical shaped. That is why I have hoped for quicker solution.

Experimental approach
Here are 100 marbles in a polyethylene bag:

If a marble touches the bag, it is counted as being on the surface. I did not count, and the bag was not perfectly round. A balloon might be better.
Theoretical approach
This works for large $N,$ and the result is as Ivan predicted (from scaling arguments in 2D and 3D, I would imagine). We neglect the packing defects in the bulk, and treat the surface as having no curvature and no defects. The maximal packing density of spheres is well-known by chemists,
$$P_\mathrm{3D} = \frac{π \cdot \sqrt{2}}{6} = 0.74046\ldots\tag{1}$$
The maximal packing density of circles is less well-known on this site, but you can look it up on Wikipedia:
$$P_\mathrm{2D} = \frac{π \cdot \sqrt{3}}{6} = 0.90687\ldots\tag{2}$$
If you have $N$ marbles with radius $r$, you can estimate the radius $R$ of the enclosing sphere, get its area ($4 \pi R^2$), and calculate how many circles (of area $\pi r^2$) fit in that area. The resulting expression scales with $N$ to the power of two-thirds, as Ivan said:
$$N_\mathrm{surface} = 4 \cdot P_{\mathrm{2D}} \cdot \left(\frac{N}{P_{\mathrm{3D}}}\right)^\frac{2}{3}\tag{3}$$
For example, for a million spheres, about four percent are on the surface. For a single sphere, all are on the surface. The expression for $N = 1$ yields $4.4$, pretty bad. For $N = 100,$ it yields $95.2$ That is a not a bad estimate; from the experimental approach, I see more than $30$ marbles on the surface, and the picture only shows one side.
Approximating a sphere with a cube
If you have 125 spheres that pack in a primitive cubic fashion, you would have a cube with 5 spheres in each direction. The top and bottom, right and left, front and back layer would be on the surface, leaving a 3 by 3 by 3 cube in the center. So 27 spheres would be on the inside and the remaining 98 would be on the surface. When we take those same 125 and pack them tighter and such that they fit in a minimal sphere, the surface area would get smaller, and the packing on the surface would get denser, leading to a similar fraction of spheres on the surface.


Electrolysis of brine without a membrane In the electrolysis of brine sodium hydroxide is normally formed if there is a membrane preventing the transfer of the chlorine gas. So if the membrane is removed,  will the chlorine gas react with NaOH to form sodium hypochlorite? Because I've read that the chlorine will instead just react with the hydroxide ion to produce hypochlorite. Or will it produce both? and where does the sodium ion go if it doesn't?

So if the membrane is removed, will the chlorine gas react with NaOH to form sodium hypochlorite?

Yes.

  Because I've read that the chlorine will instead just react with the hydroxide ion to produce hypochlorite.

Yes. This is just another way to say the same thing.

As for the sodium, it does not go anywhere. It simply continues to stay there in the solution, like it did before the reaction. It does not even "know" about the reaction, or that it now "belongs" to the hypochlorite; it is just a spectator ion. Since the dissolution, anions and cations go their separate ways and do not "care" much for each other. This is always the case with ionic salts.


Are there any general rules for choosing solvents for recrystallization? Re-crystallization is a nice way of purifying a product, but choosing a suitable solvent, if you can't rely on the literature, seems like a lot of trial-and-error.

Are there any general rules on which kind of solvents could be used for re-crystallization?
Which criteria should one use when trying to recrystallize a compound for which no literature on usable conditions are available?

Generally speaking, the best solvent will be dependent on the impurity that you are trying to remove.

The solvent must dissolve both the desired compound and the impurity at a high temperature, but only the desired compound at lower temperatures.  The solubility product of the impurity, as well as the common ion effect should both be taken into consideration.


Order of bond rotation about alkene How to decide which among these has the greatest bond rotation possible? 1 has the greatest and 2 the least, according to me, I think it is due to the presence of bulky groups. What more factors can effect bond rotation?

First, recall that a single bond has the highest free rotation possible (and hence it shows conformational isomerism). A double bond instead does not show any rotation at all (hence, $\ce{abC=Cfg}$ type molecules show geometrical isomerism).

So, it's obvious that, the more double bond character a bond will have, the less likely it is to rotate freely. Here, we're talking about double bond character, because, due to resonance effects, the resonance hybrid in these compounds will only have a partial double bond character.

Thus, you only need to compare the partial double bond character in each compound, and apply the above rule. Bulkiness of the groups isn't really a factor, because the groups actually aren't that bulky, and it isn't a t-butyl group anyway.


How to isolate and determine an unknown molecule in a known heterogeneous mixture using HNMR I am currently working on an assignment as part of my second year lab course. Essentially I have to determine the unknown molecule which is present at 5-15 wt% in 1 gram of Aspartame.

I am planning to do the HNMR of a pure aspartame sample then the HNMR of the heterogenous mixture. Hopefully there will be some distinctive peaks but most likely there will be many overlapping peaks between the Aspartame and the unknown.

My question: Is there any way I can subtract the HNMR of the pure sample from the heterogenous mixture. Therefore I could be left with the HNMR of just the unknown compound.

The software that comes with the instrument (Topspin for a Bruker machine,but Varian has a similar product made by them) will certainly be able to do this, as can third party software such as MestreNova or the free ACD/NMR.  You should check with the NMR service in your department as to what software they have licenses for.

The method you describe of 'subtracting' two NMR spectra, whilst possible, is not however very good in practice. For two spectra (more accurately, for two FIDs) to be subtracted, they must be IDENTICAL. That is, the peaks for aspirin in both the pure sample, and the contaminated one, must be perfectly aligned (same peak shapes, intensity, shift etc), which is practically different due to slight differences during each acquisition.

The easiest thing for you to do is probably to purify the sample, for instance by running some chromatography, or perhaps re-crystallising the aspirin, which would hopefully leave your impurity behind, which you could then analyse.

Failing this, and if you have access/inclination, some 2D NMR experiments would be useful to you. A simple COSY would allow you to see which peaks correspond to which peaks (thus allowing you to see which peaks aren't aspirin). More complex methods (HSQC, HMBC, DOSY) would also be of great help, but are probably beyond your needs for a second year lab course.


What is the colour of chromium(III) chloride solution? I had a question asking about the colour of the solution of chromium(III) chloride in my exam, all my colleagues have different answers, I couldn't find any exact answer after searching!

The color of the solution may vary as long it is not in equilibrium.

According Bjerrum and Bjerrum1 a $\ce{1\mathrm{M}}$ chromium (III) chloride solution in equilibrium at 25° C contains about 13% monochloro and 85% hexaqua ions.

The absorption spectrum published shows minima at $\ce{480 \mathrm{nm}}$ (blue) for the hexaqua and $\ce{510 \mathrm{nm}}$ (green) for the monochloro complex.

When we integrate over the spectrum we get the following colors.

However, it should be noted that the colors vary with spectral optical depth (concentration, thickness of the medium).

1 Bjerrum, M.J. and Bjerrum, J., 1990. Estimation of Small Stability Constants in Aqueous Solutions. The Chromium(III) Chloride System. Acta Chem. Scand. 44: 358-363.


Why is phosphorous pentoxide a dehydrating agent? $$\ce{HNO3 + P2O5 -> 2HPO3 + N2O5}$$
In this reaction, both nitrogen and phosphorus have the same oxidation number before and after the product is formed. Since the $+5$ oxidation of nitrogen is unstable, it has to undergo reduction, but here it remains the same after the product is formed. So what makes the reaction go forward?
Is it because of the dehydrating property of $\ce{P2O5}$?
Why does this compound have this property and why doesn't $\ce{N2O5}$ have this dehydrating property?

First of all, your reaction is wrong. $\ce{HPO3}$ does not exist as a single entity. It is actually a cylic molecule of formula $\ce{(HPO3)_n}$. The simplest such compound is $\ce{(HPO3)3}$ which is called trimetaphosphoric acid. See this answer for more details.

So, what happens is that phosphorus pentoxide is very powerful dehydrating agent and can rip off water molecules from compounds. Phosphorus pentoxide can dehydrate nitric acid to form its anhydride, dinitrogen pentoxide.
$$\ce{2HNO3 ->[P2O5][- H2O] N2O5}$$

Why dinitrogen pentoxide is not used as a dehydrating agent?

My take on this is that hydrolysis of dinitrogen pentoxide is not that exothermic as compared to hydroysis of phosphorus pentoxide. Hydrolysis of phosphorus pentoxide is very exothermic, releasing so much energy to dehydrate compounds to form corresponding anhydrides and phosphoric acid.
$$\ce{P4O10 + 6 H2O → 4 H3PO4   (–177 kJ)}$$
Phosphoric acid is hygroscopic which further sucks remaining water from compound. If we compare all the dehydrating agents, phosphorus pentoxide will be on a different league. Dinitrogen pentoxide is nothing.
Nitric acid also hydrolyses to form nitric acid but the reaction is not that exothermic. Also, nitric acid is not that hygroscopic as compared to phosphoric acid for which it is not that great of a dehydrating agent. But hydrolysis of dinitrogen pentoxide has its importance. A solution of dinitrogen pentoxide and nitric acid is a good nitrating agent and hence $\ce{N2O5}$ is known as a nitrating agent than a dehydrating agent. Also, this hydrolysis reaction is a basis of forming atmospheric aerosols ($\ce{HNO3}$ is a good $\ce{NO_x}$ reservoir).


How can CO2 exist in two different phases? Given a container of $\ce{CO2}$ at a pressure of 100 bar and 295 K, I check the chart below and find:

I'm within the liquid region
I'm before the critical point (304 K)

My question then is, why can I go purchase a tank filled at 100 bar that contains both liquid and gas?  What stops the gas from condensing, since according to this chart, the pressure and temperature map to a liquid phase.

The $\ce{CO2}$ exists in two phases when you buy a partially-filled $\ce{CO2}$ cylinder because there is extra volume (which is needed to allow expansion as the container warms). Therefore, the pressure in the container is not 100 bar, but closer to 80 bar at 293K.
If the container had only enough room for the $\ce{CO2}$(l), then, indeed, there could be no gas phase. But if space is allowed above the liquid, then $\ce{CO2}$ evaporates, filling that space, and the $\ce{CO2}$(g) and $\ce{CO2}$(l) reaches equilibrium pressure.
On the molecular level, consider it an ongoing exchange between the molecules of liquid and gas. For the same reason, there is humidity in the air and oceans full of water, a 1 bar at sea level.


Vaporizing NOx (NO and NO2) with hot metal? So essentially I'm looking at a product that my intuition tells me is kind of similar to a perpetual motion machine (Impossible).

Catalytic Converters use platinum, rhodium, and palladium to convert Hydrocarbons, NOx, and CO.

NOx turns into Nitrogen and oxygen.

Hydrocarbons turn into Water and carbon dioxide

Carbon monoxide into carbon dioxide.

THIS PRODUCT (see link) claims to "vaporize" "toxic fumes". It doesn't seem to explain what they turn into. just that they go away. (in the report it appears that some of it is turning into oxygen as that number goes up?)

Isn't NOx and CO already considered a vapor? how does one further vaporize it?
Simply using hot steel to convert gases seems strange but using hot platinum to convert gasses is also pretty strange when you think about it.
Is this actually a legitimately effective product?

http://www.impulsengine.com/newproducts/how-it-works_ev_muffler.shtml

http://www.impulsengine.com/newproducts/pdf/Impulse_EV16_Muffler_Emission_Results.pdf

Thin wall channels (this would be the "steel" plates perhaps) in the device are coated with a wash coat of aluminum oxide which is porous and increases the surface area. This allows more reactions to take place since the converters also contain precious metals such as platinum, rhodium, and palladium e.g. converting NOX into nitrogen and oxygen, hydrocarbons into water and carbon dioxide, and so on.

These catalytic converters use oxidation and reduction reactions to convert the "toxic" fumes—I'm not certain they are stating to work any differently with this regard than other catalytic converters do though. Oxidation is the loss of electrons and reduction is the gaining of electrons at the molecular level. This in addition to the precious metals help promote the transfer of electrons into the conversion of "toxic" fumes to "non-toxic" fumes.

Furthermore, these can also contain oxygen sensors on them measuring and reporting back to the Engine Control Module the oxygen levels in the exhaust. This will help tell other sensors to adjust air-to-fuel ratio helping to keep the catalytic converter running at optimum efficiency.

I would say if this device has a patent and has held up to the tests thereof, then it's likely legitimate and accurately effective as advertised by the manufacturer.


Why is molar mass of a substance and atomic mass number usually the same? I recognize what atomic mass represent and what molar mass represents by why is it that a substances molar mass and it’s atomic mass are usually the same number

Ex: chlorine has an atomic mass of 35.453 u and a molar mass of 35.453 g/mol

The term mole is a short form of the German word Molekül which was coined by Ostwald in 1890s. At that time, the "atomic mass unit" did not exist. The common term at that time was gram atomic mass or gram-molecule. The original meaning of so-called was defined to be equal one gram molecule i.e, the mass of a molecule expressed in grams. Atomic mass unit came into use by the 1950s.

These values match by definition not by coincidence. Here is how Ostwald coined the concept of a "mole". In those times molecules weights were determined by colligative properties. This is a snapshot of Findlay "Principles of Inorganic Chemistry" (translation of Ostwald's work)


Can entropy increase during bond formation? Recently the question was asked, "Is bond formation “strictly” exothermic?"  As a corollary to that question, what is the relationship between entropy and bond formation?  Does the entropy of the bond formation reaction always decrease when a chemical bond is formed?

I do not think that an entropy decrease is a requirement to form a bond.

Think about a "normal" case: $\ce{HCl}$ (a gas) and $\ce{NH3}$ (another gas) react together to form $\ce{NH4Cl}$, which is a solid. This is an example of bond formation where entropy clearly decreases.

But now think what would happen if two liquid or solid compounds react together to form a gaseous compound. An increase of entropy has taken place. Can't think of a concrete example, but I think this might be the case of, for example, an esterification between a liquid alcohol and a liquid carboxylic acid that form an ester where, due to the lose of $\ce{-OH}$ groups, intermolecular interactions are lower and it can result with the product molecule being more volatile than the original compounds.

It's all about the number of micro states in which the compounds can exist, and gas formation from solids or liquids importantly increases entropy.

(PS: These are just some thoughts, really, I'm no physical-chemist.)


Mechanism of conversion of Cr from FeCr2O4 (chromite) The basic steps involved in the conversion of $\ce{FeCr2O4 -> Cr}$ involves,

$$\ce{FeCr2O4 ->[NaOH / air, \Delta] Na_2CrO4 ->[C, \Delta] CrO3 ->[Al, \Delta] Cr}$$

Basically, I'm having problems in learning so large amount of inorganic reactions, so I thought it would be easy to learn the mechanism behind these reactions, just like we do in organic chemistry.
Therefore, I'm keen to know the mechanism behind first conversion (i.e., from $\ce{FeCr2O4 -> Na_2CrO4}$). I've already tried to search for this, but found nothing related.

My attempt
I know some basic sets of inorganic mechanisms, as follows

Ion-exchange
Decomposition
Water-addition
Redox
Complex reaction

From the above listed, I don't think it would be ion-exchange (bcoz, $\ce{FeCr2O4 -> FeCr+ + CrO4-}$ is very unlikely to happen), water-addition or complex reaction. I guess it could be decomposition or redox, but I'm not sure on the "bond-movement" mechanism part.

The first reaction is a double oxidation. Both Fe and Cr are oxidized by O2 from the air. The reaction is made of two half-reactions. The first half-reaction is not easy to establish, because two elements (Fe, Cr) are oxidized simultaneously, Fe from +II to +III, and Cr from +III to +VI. Sorry to say it : It is one of the most difficult half-equations I ever had to state. It is :    $$\ce{FeCr_2O_4 + 11 OH^- -> Fe(OH)_3 + 2 CrO_4^{2-} + 4 H_2O + 7 e-}$$ The second half-equation is easier to write.
$$\ce{O_2 + 2H_2O + 4 e-  -> 4 OH^-}$$ The overall equation is obtained by multiplying the first equation by $4$ and the second by $7$ and adding the whole. After simplification, it yields :
$$\ce{4 FeCr_2O_4 + 16 OH^- + 7 O_2 -> 4 Fe(OH)_3 + 8 CrO_4^{2-} + 2 H_2O}$$
or, without ions :
$$\ce{4 FeCr_2O_4 + 16 NaOH + 7 O_2 -> 4 Fe(OH)_3 + 8 Na_2CrO_4 + 2 H_2O}$$
This was the first step, defining how to pass from the mineral $\ce{FeCr_2O_4}$ to $\ce{Na_2CrO_4}$, at high temperature, and in the presence of air and $\ce{NaOH}$. The final mixture can be washed with water, which dissolves easily $\ce{Na_2CrO_4}$, as $\ce{Fe(OH)_3}$ is insoluble in water.
Then $\ce{Na_2CrO_4}$ is transformed in $\ce{CrO_3}$ by adding moderately concentrated sulfuric acid : $$\ce{Na_2CrO_4 + H_2SO_4 -> CrO_3 + Na_2SO_4 + H_2O}$$ $\ce{CrO_3}$ is not soluble and can be separated by filtration. Then it is mixed with aluminum powder to get a exothermic reaction when engaged by a match : $$\ce{2 Al + CrO_3 -> Al2O3 + Cr}$$


How to determine methane percentage at home? Suppose I have gas bottle and gas cooker. Suppose I have propane in my bottle and I suspect it is diluted with unknown gas.

How can I measure with simple instruments, how many propane I have (percentage)?

A couple of ideas:

If you think the diluent gas is non-flammable, you should be able to use the adiabatic flame temperature of the gas when you burn it in air to estimate the composition.  You will need to carefully adjust your stove to give the optimal fuel/air blend (that maximizes the flame temperature).  Then you will need to accurately measure the flame temperature.  If you assume a heat capacity for the unknown diluent, you can estimate its concentration in the gas by comparing the measured temperature to the theoretical adiabatic flame temperature for pure propane burning in air.
Use a vacuum pump to empty a container of known volume, and then allow the cooking gas into the container at ~1 atm pressure.  Measure the density of the gas and compare to the density of pure propane.
If you think the diluent gas does not contain carbon: hook up flow meters to the gas bottle (and an air or oxygen stream) to form a torch flame with known gas flow rate.  Collect all of the the exhaust produced by the flame, pass it through a cold trap to freeze out water vapor, and then pass the cold flame exhaust over a bed of anhydrous sodium hydroxide of known initial mass.  It will absorb the carbon dioxide from the exhaust.  Measure the mass increase of the sodium hydroxide.  By comparing the moles of gas that flow into the flame to the moles of CO2 absorbed (as calculated from the mass increase) you can estimate gas purity.
Find a spectrometer and record the IR or Raman spectrum of the gas!  This will probably be more accurate than any of the other methods.


Work done under constant pressure Work done under a constant pressure.

I know $$\Delta U = \Delta V \times \pi_T + C_V\Delta T $$

I thought I can give a try to find work done from above equation under constant pressure for an ideal gas.

$$q_p + w_p = C_V\Delta T$$
$$\implies C_p\Delta T + w_p = C_V \Delta T$$

$$\implies w_p = -\Delta T (C_p - C_V)$$
$$\implies w_p = - R\Delta T $$

Adiabatic work is $w_\mathrm{ad} = C_V \Delta T$

For a adiabatic work under constant pressure $w_\mathrm{ad} = w_{q} \implies C_V = -R \implies C_p = 2R$

Constant $C_p, C_V$ and negative $\gamma$ does not look correct to me.

Which of my steps are incorrect ?

You can't do adiabatic reversible work at constant pressure.  The pressure has to change.  If you want to do adiabatic irreversible work at constant pressure, say by suddenly increasing or decreasing the pressure load to begin with, and then holding it constant during the process, this is sometimes referred to as adiabatic work at constant pressure, but it really isn't.  The final pressure will not be equal to the initial pressure that was present before you changed the applied load.


Why does increasing the temperature of a *solid* increase its entropy? I understand why increasing the temperature of a fluid would increase its entropy, as the particles are free to move, and therefore an increase in kinetic energy would allow the particles to move more, thus increasing the fluid's entropy.

However, this doesn't make sense to me in the case of a solid - increasing the solid's temperature would result in an increase in K.E., but this would only result in the particles vibrating more. As entropy is defined as the 'disorderliness' of a system, and the molecules in a solid always remain in the same 'order'  regardless of temperature, how is temperature proportional to entropy in a solid?

All right, someone bearing the standard of thermodynamics will give you the equations shortly... From a layman to another, here goes my attempt at a simpler explanation.

Entropy may be seen as the "disorderlyness" in some settings, but that is not a very useful way of seeing it. The metaphor is often used, but creates the wrong conclusions when looking closer at it. To me, is best to look at it as a quantity closely related to the number of possible (different) states available to a certain system. It is then obvious that if you increase the temperature, more modes of vibration are available, thus more states, ergo more entropy. The more stuff is in your room,  the less probable is the state where everything is ordered - if all states are equal in probability ;)


Oxygen Preventing the Formation of Large Organic Compounds? My textbook tells me that when carbon bonds with oxygen, the electrons are too close to the oxygen, and therefore they cannot just string together and form large molecules like glucose. It goes on to say that because, during photosynthesis, carbon becomes bonded to less electronegative atoms like hydrogen, it becomes more capable of forming larger compounds. Why is that?

If we take a look at glucose we can see there are oxygen atoms in it, just with extra presence of the hydrogen.
The assumption :

carbon bonds with oxygen, the electrons are too close to the oxygen, and therefore they cannot just string together and form large molecules like glucose.

is wrong.
Take a look at Polymeric carbon oxides

Carbon suboxide spontaneously polymerizes at room temperature into a carbon-oxygen polymer, with 3:2 carbon:oxygen atomic ratio. The polymer is believed to be a linear chain of fused six-membered lactone rings, with a continuous carbon backbone of alternating single and double bonds. Physical measurements indicate that the mean number of units per molecule is about 5–6, depending on the formation temperature.
Carbon monoxide compressed to 5 GPa in a diamond anvil cell yields a somewhat similar reddish polymer with a slightly higher oxygen content, which is metastable at room conditions. It is believed that $CO$ disproportionates in the cell to a mixture of $\ce{CO2}$ and $\ce{C3O2}$ ; the latter forms a polymer similar to the one described above (but with a more irregular structure), that traps some of the $\ce{CO2}$ in its matrix.
Another carbon-oxygen polymer, with C:O ratio 5:1 or higher, is the classical graphite oxide and its single-sheet version graphene oxide.


How would an increase in the concentration of a sucrose solution affect the rate of its acidic hydrolysis? Regarding the acidic hydrolysis of sucrose

$$\ce{\underset{sucrose}{C12H22O11} + H2O + \underset{catalyst}{H+} -> \underset{glucose}{C6H12O6} + \underset{fructose}{C6H12O6} + H+},$$

I think that its mechanism can be described as a 2-step reaction consisting of a step which involves the H+, and then the step which involves the $\ce{H2O}$, and I believe the latter step to be a rate limiting step.

My question in detail is:

  Would an increase in the concentration of a sucrose solution, with a constant $\ce{H+}$ concentration, have a 1-to-1 influence on reaction rate (for example, if doubling the concentration of the sucrose solution would result in a hydrolysis reaction twice as fast), or would the addition of $\ce{H2O}$ in the second step significantly limit the change in reaction rate?

Looking at the reaction mechanism (of which I understand you are aware)
is good to keep in mind that the reaction rate really depends on the amount of intermediate here, provided the second step is rate limiting.

In the first step, $H^+$ is consumed. Since it is not rate limiting, I can assume the intermediate $[sucrose.H^+]$ to be in pseudo-equilibrium with $[H^+]$&$[sucrose]$, on the timescale of the second reaction step.

$$r_2 = k_2[sucrose.H^+]$$

This means adding more sucrose (provided you have enough acid to protonate it) will indead result in the formation of more [sucrose.H^+] intermediate, and thus, higher kinetics.

Btw, I'm simpifying stuff here by not taking into account the reaction backwards for the second step.


Which is the correct name: 2,3-dimethyl cyclopentene or 1,5-dimethyl cyclopentene? According to the lowest locant rule, 2,3-dimethyl cyclopentene should be the correct IUPAC name as the sum of the locants of the substituents(2+3=5) of 2,3-dimethyl cyclopentene is lower than the sum of the locants of the substituents(1+5=6) of 1,5-dimethyl cyclopentene.

But my book says that the correct name of the compound is 1,5-dimethyl cyclopentene. I guess this is due to the fact that methyl gets the 1st locant position. But I am doubtful about this.

What is the correct name?

The correct name is 1,5-dimethylcyclopent-1-ene. The double bond has the highest priority and so these two carbons must be numbered 1 and 2 if possible (yes in this case because the molecule is cyclic). We then have to choose between numbering the methyl substituents 1,5 or 2,3. The lowest locant sum rule that you mentioned doesn't actually exist, even though it is often taught. It is a agglomeration of several rules which works most, but not all of the time. The actual rule we are interested in is the first point of difference rule which states:

  2.2 - The longest chain is numbered from one end to the other by Arabic numerals, the direction being so chosen as to give the lowest numbers possible to the side chains. 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.

Therefore the locant set 1,5 is lower than the locant set 2,3 since at the first point of difference, 1 is lower than 2.


Help interpreting IR spectrum Whilst attempting to create salicylic acid/2-hydroxybenzioc acid, one of our experiments produced a compound with a melting point of over $300^\circ\mathrm{C}$ ( as high as the machine would go) and the IR absorption spectra attached. Can anyone help us identify the compound we have produced?

We used:

$4 \,\mathrm{g}$ of oil of wintergreen
$100 \,\mathrm{mL}$ of $2 \,\mathrm{M}$ aqueous sodium hydroxide solution
Sulfuric acid

The experiment was left for two weeks at which point the crystal pictured had formed.

I do believe something is wrong with this spectrum. I find it highly improbable that a compound would exhibit a single strong absorption peak at low wavenumbers, while showing no other absorbance whatsoever. The fact that there is no overtone of your band also shows me that there was some experimental error when recording this spectrum.


Molecule identification I'd really like to know what compound this formula is and don't know if there's any software for that matter, if anyone could help me :)
(With a sw suggestion or the compound identification)

Many software, both online and downloadable executables, are available for identifying a molecular structure - either by giving the official IUPAC name, or matching up the structure to a large database and returning IUPAC/trade names/properties/etc...

Online: NIST, Pubchem, ChemSpyder, Reaxys

Download: ChemBioOffice Software

A variety of options are available to search for structures such as by molecular formula, structure, identification values, etc... Easiest for most approaches is using any of the software's "Draw" features, to draw the exact structure and then search the database for it. Online options may have difficulty with the exact structure searches on molecules as large as yours above, so as much detail as possible if useful (stereochemistry, for instance, using dashes and wedges).

ChemBioOffice (specifically ChemDraw) is able to give you the IUPAC name, which would then let you look it up in any of the online databases.

If the drawing of the structures is causing you difficulty, you can also search with molecular formulas, but it may take some trial and error to identify the actual structure you want as you look for larger or more complex molecules this way. NIST is a great choice for this approach.

Outside these options, if you have any interest or experience coding, you may be interested in using RDkit (Python module), which has a large array of functions for molecular data interaction. For instance, if you use those above structure "Draw" features online, most will also show a SMILES code which is a way of shorthanding a molecular structure's features; place a SMILES code into a script that accesses RDkit and you can request it to draw the structure, look up information related to it (if it can find the SMILES code in its databases), etc... Warning - it is mostly just development manuals, no tutorials last I checked, so perusing the module can be rather arduous.

As an aside, your molecule in particular gave those online database structure searches a bit of difficulty when I attempted it, so you may need to find a way of giving it more structural information, or perhaps searching for sub structures first (such as just the fused and linked cyclohexanes without the Chlorines or amine group).

However, I found one very similar to your structure, but with aromatics instead for two of the rings: 1,1-dichloroheptadecane


Relating a compound's radii and unit cell dimensions We're asked often to calculate the density of a crystal, and there is a known formula for that related to the unit cell volume.  Now, for simple geometric structures such as the bcc structure (and since it is close packed), I know how to relate the different radii of the constituent compounds with the unit cell's (unique in this case) dimension (the length a), but how do I relate it for such other complex structures?

In general given a certain crystal geometry (cubic , tetragonal , orthorhombic), how does one relate the unit cell dimension with the radii of the constituent atoms?

The ratio of the radius of the cation to the radius of the anion is called radius ratio. The ratio of the radius of cation and the radius of anion i.e, radius ratio ($r_+/r_-$) plays an important role in determining the structures of ionic solids and coordination number of ions. The possible coordination numbers and structural arrangements of anions around  cations for different $r_+/r_-$ values are given below:


Is the salt of a weak acid and a weak base also a weak electrolyte? I'm able to find a lot of information on acids and bases, as well as strong-strong, weak-strong, and strong-weak salts. However all the introductory materials on electrolytes just stop there and say "we won't bother to cover weak-weak salts here". I have also heard blunt statements that all salts are strong electrolytes, but never with satisfactory explanation or even some weak-weak examples.
So are there any salts that are weak electrolytes?

My interest in the subject is mainly directed at finding the van 't Hoff factor, for osmotic effects and tonicity.

When a weak acid such as acetic acid is added to pure water, it does not ionize much, but sufficiently to make the solution acidic. On the other hand, when small amounts of weak acids are added to buffered solution with a pH near neutral, acidic acid dissociates almost quantitatively. So the van't Hoff factor depends on the pH.

The same goes for weak bases used as ammonia - in combination with pure water, you get a basic solution and very little ammonium, but when dissolved in a solution maintaining a neutral pH, ammonia forms the ammonium ion almost quantitatively.

  I have also heard blunt statements that all salts are strong electrolytes, but never with satisfactory explanation or even some weak-weak examples.

If we combine ammonia and acetic acid in aqueous solution and remove the water, we get the salt ammonium acetate. If we dissolve ammonium acetate in pure water, the pH will be roughly neutral, and both ions will remain ions. Therefor, this is a strong electrolyte solution.

  So are there any salts that are weak electrolytes?

In the pure salt, the components are ions by definition. Water is very good at dissolving ions, so the dissolved ions stay ions. To get a salt that is also a weak electrolyte, you would need an uncharged acid and an uncharged base that  would transfer a proton upon forming a salt, but not while in aqueous solution. I am not aware of any such system.


Wavelengths of the visible spectrum Why is data about wavelengths of different colors and the visible spectrum in general so different in different sources? On Wikipedia, the numbers differ by up to $\pm\mathrm{30~nm}$.

The comments on the question discuss this may not be about chemistry.
One could argue that colors are chemistry related because
they are associated with chemical substances: like minerals, chlorophyl, rust, dyes etc.
So colors seem to be properties of substances.
And that's fundamentally wrong.
Color is a function of the visual system of the viewer!
If that sounds surprising, think of color blind people, which see different colors.
Also, there are not just us - there are other animals, with different visual spectrum, partly including what we call UV.
Colors are not well defined
Actually, there are more than enough ways, even standards, to define which part of the rainbow is green, exactly - people working with visual media like print design can tell details.
And there are way more details than anyone could expect.
So, without a common definition of the colors, it's not the wavelengths that vary in different sources - it's the colors themselves.

To illustrate that color is more complex when looking closely, below you see a diagram defining color names. It is part of a formal standard which is in heavy use:

From https://en.wikipedia.org/wiki/CIE_1931_color_space


Dipole moment of water using the definition In many Chemistry books, the dipole moment of molecules is calculated through algebraic formulas. However, another definition of the dipole moment is

$$\vec{\mu} = \sum_i q_i \, \vec{r}_i$$

So, how can we calculate the dipole moment of water, for example using this formula? Is there a difference in interpretation of dipole moment in Physics and Chemistry?

Here is the Definition.

Is there a difference in interpretation of dipole moment in Physics and Chemistry?

Addressing the above part: The definition of dipole moment is the same in physics and chemistry. However, the chemist's dipole moment has opposite directions as compared to the physicist. Which one is right? Of course, the physicists draw it correctly, which is consistent with electrostatics. There was a nice single paged article in the Journal of Chemical Education on this topic. A must read.
Misconceptions in Sign Conventions: Flipping the Electric Dipole Moment
James W. Hovick J. C. Poler
https://pubs.acs.org/doi/10.1021/ed082p889


Why carbonyl groups are strong benzene deactivating group for electrophillic aromatic substitution? In carbonyl group there is a resonance avaibility and also due to inductive effect it can also withdraw electron from benzene ring . As, resonance(e- donating chracter) dominates than Inductive effect , it must be a Benzene activating group.
but accorting to  L.G Wade carbony groups are strong benzene deactivating group for Electrophillic Aromatic substitution
Where am I wrong ? Explain.

A carbonyl group such as an aldehyde is a deactivating group as it withdraws electron density by both the inductive effect (through the sigma bonds) and the resonance effect which involves pi systems.

Here are three resonance forms which show how the aldehyde group takes electron density from the benzene ring in benzaldehyde. There are more resonance forms and some different but related ones account for the reduction in the stability of the cationic intermidate in an aromatic substitution reaction.


IUPAC nomenclature for tricarboxylic acids What is the rule for naming compounds with three carboxyl groups?

I have the two following coumpunds

4-(Carboxymethyl)heptanedioic acid
Pentane-1,3,5-tricarboxylic acid

They both have 7 carbon atoms in their parent chain, yet the second coumpunds is called a pentane. Also both have three carboxyl groups yet the first is a dioc acid. Also in the second one, its name is written as tricarboxylic acid. Why not trioic acid.

All these anomalies are confusing me. Can anybody explain why both are named so?

Systematic names of carboxylic acids are generally formed using the suffix ‘oic acid’ $\ce{{}-(C)OOH}$, or ‘carboxylic acid’ $\ce{{}-COOH}$, or the prefix ‘carboxy’ $\ce{{}-COOH}$. Note that the $\ce{{}-(C)OOH}$ group of ‘oic acid’ formally replaces a terminal $\ce{{}-CH3}$ group whereas the $\ce{{}-COOH}$ group of ‘carboxylic acid’ or ‘carboxy’ is a usual substituent that replaces just a $\ce{{}-H}$.

The relevant rules in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) read as follows.

  P-65.1.2.1 Carboxylic acid groups, $\ce{{}-COOH}$, that conceptually replace a $\ce{{}-CH3}$ group of methane or terminate an unbranched hydrocarbon chain are named by replacing the final ‘e’ of the name of the corresponding hydrocarbon by the suffix ‘oic acid’. No locants are necessary to denote the positions of the carboxylic acid groups in a hydrocarbon chain; locants are used when hydrocarbon chains are modified by skeletal replacement, as shown in P-15.4.3.2.3. Except for formic acid, acetic acid, oxalic acid (see P-65.1.1.1), and oxamic acid (see P-65.1.1.1), systematically formed names are preferred IUPAC names; the names given in P-65 .1.1.2 are retained names for use in general nomenclature.

That’s why the name of the unsubstituted parent structure of the first compound that is given in the question is a heptanedioic acid.

  P-65.1.2.2.1 If an unbranched chain is linked to more than two carboxy groups, all carboxy groups are named from the parent hydride by substitutive use of the suffix ‘carboxylic acid’, preceded by the appropriate numerical prefix ‘tri’, ‘tetra’ etc. and appropriate locants.

That’s why the name of the second compound is pentane-1,3,5-tricarboxylic acid.

  P-65.1.2.2.3 The prefixes ‘carboxy’ and ‘oxalo’
  When another group is present that has priority for citation as suffix, for example, a free valence, or when all carboxylic acid groups cannot be described by a suffix, carboxylic acid groups, $\ce{{}-COOH}$, are indicated by the preferred prefix ‘carboxy’ (also used in general nomenclature). The prefix ‘oxalo’ is recommended as the preferred prefix for $\ce{{}-CO-COOR}$, but cannot be used to lengthen a carbon chain. In general nomenclature, the compound prefix ‘carboxycarbonyl’ may be used, but the compound prefix ‘carboxyformyl’ is not recommended.

That’s why the complete name of the first compound is 4-(carboxymethyl)heptanedioic acid.


Free radical vs radical cation What is the difference between a free radical and a radical cation?  Are they different thing or do they mean same?

A radical cation is simply a free radical with a positive charge. Similarly a radical anion is a free radical with a negative charge.

In other words, radical cations and radical anions are subsets of the free radicals.

That said, the term free radical itself is not entirely clear cut, as illustrated in this question.


Formation of halohydrin vs alkyl dihalide For the electrophilic addition of $\ce{X2}$ to alkene in $\ce{CCl4}$ solvent, a dihaloalkane is formed. This is a halogen addition reaction, with the intermediate being a halonium ion.

However, if instead of $\ce{CCl4}$, $\ce{H2O}$ is used instead, a halohydrin would be formed.

However, in the second case (halohydrin formation reaction), the $\ce{X-}$ ion is still present in the reaction. Furthermore, $\ce{X-}$ is a stronger nucleophile than $\ce{H2O}$. Why is it then that it is $\ce{H2O}$ that reacts with the halonium rather than $\ce{X-}$, the stronger nucleophile?

The halohydrin is usually formed by means of a solvolysis reaction where water is used as the solvent. So, it is present in such high amounts that it overpowers the larger nucleophilic strength of the $\ce{X-}$ ion since the latter is only present in relatively minute quantities. Furthermore, water is a polar protic solvent and thus it has the effect of weakening the nucleophilic strength of $\ce{X-}$ in solution by "encasing" it in a large hydration shell.


Is this Newman projection energy profile for 2,3-dihydroxypropanal correct? Is this Newman projection energy profile for 2,3-dihydroxypropanal correct?

Also, I was to identify R/S configuration for the initial Newman projection. I am pretty sure it is R, but I need help here as well.

There seems to be only one problem. The 240° conformer* will be lower in energy as compared to the 120° conformer as there will be hydrogen bonding between the two hydroxyl groups
.
Rest all looks good.


Can hot food ever emit x-rays or gamma rays? I was just wondering, if heating food up is the result of increasing the energy of bends and stretches in the bonds of the molecules, is it ever possible for tiny amounts of x-rays and gamma rays be emitted?

When we give a molecule enough energy, its electrons can jump to higher orbitals and then return to ground state, releasing EMR with a frequency proportional to the energy gap. So if we heat food, is there a chance that some electron received enough energy to jump up to a really high energy level?

In theory, yes, you can heat objects to a high enough temperature to emit x-rays or gamma rays. You cannot do this to food, and you certainly cannot do this in your kitchen (or probably any kitchen).
Let's take the lowest energy x-ray out there and see what it would take. X-rays range in frequency from $30 \times 10^{ 16}$ to $30\times 10^{10}$ hertz. The energy of one photon of 30 petahertz radiation is:
$$E=h\nu = \left(6.626\times 10^{-34}\mathrm{\ J\cdot s}\right)\left(30\times 10^{16}\ \mathrm{s^{-1}}\right) = 1.988 \times 10^{-16}\ \mathrm{J}$$
This is not a lot of energy! However, a single photon is boring. Let's consider a mole of photons. This will also ease comparison with other phenomena, whose energies are listed per mole of events.
$$1.988 \times 10^{-16}\ \mathrm{J} \times 6.022\times 10^{23}\ \mathrm{mol^{-1}}=1.197\times 10^8\ \mathrm{J\cdot mol^{-1}}$$
In theory, if you could pump that much energy into something, you should get some high energy photons out. In practice, it does not work that way. Other stuff happens first. To simplify our example, let's just consider 1 mole of water (18.0 grams) and heat it up. The fate of basically any other matter will be the same, but the energy required will vary a bit.
First, adding energy heats the water. If we start at room temperature $\left(20\ ^\circ\mathrm{C}\right)$, it takes $80\ ^\circ\mathrm{C}\times 18\ \mathrm{g}\times 4.184\ \mathrm{J\cdot g^{-1}\cdot ^\circ C^{-1}}=6025\ \mathrm{J}$ to heat that water to boiling. It takes 40.66 kJ to convert the water into gas. Neither of these puts a big dent in our energy. It takes further energy to heat the water vapor again, but let's see how far we need to take it.
Once we get enough energy into our sample of water, the molecules start to fall apart.
$$\ce{ H2O(g) -> 2H(g) + O(g)} \ \Delta H^\circ =+920\ \mathrm{kJ\cdot mol^{-1}}\ \Delta S^\circ =0.202\ \mathrm{kJ\cdot mol^{-1}\cdot K^{-1}}$$
By fixing $\Delta G=0$ at equilibrium, we can solve for a temperature at which this reaction becomes spontaneous:
$$T=\dfrac{\Delta H}{\Delta S}=\dfrac{+920\ \mathrm{kJ\cdot mol^{-1}}}{0.202\ \mathrm{kJ\cdot mol^{-1}}\cdot K^{-1}}=4596\mathrm{K}$$
We need to heat our water vapor up an additional 4218 K, which takes $18\ \mathrm{g}\times 1.996\ \mathrm{J\cdot g^{-1}\cdot K^{-1}}\times 4218\ \mathrm{K}=151.5\times 10^3 \mathrm{J}$.
So, we now have pumped nearly 200,000 J into our water sample, atomized it, and heated it to approximately 5000 K. We are now close to the temperature of the outer layers of the sun! Surely we have enough energy at this temperature to produce x-rays. Nope. At 5000 K, we produce minimal x-rays. Most of the radiation is in the visible, UV, and IR (think about what we get from the sun). Below is a plot of black-body radiation as a function of temperature (image by Wikipedia user Darth Kule and released into the public domain):

Okay, so we are far beyond the reality of what can happen in a conventional oven (or almost any reasonable heat source used for food). At this temperature, we can use the Planck Law to calculate the power output ($I$) of x-rays at the temperature. We can also do this at some normal temperatures and for gamma rays. This model is a little goofy, since food is not a black body, but we will at least calculate the max x-ray and gamma ray output.
Rather than grinding through all the maths, I'll just put in a table of some temperatures and watts. 1 watt is not a lot of power. Most lightbulbs produce light in the kilowatts.
$$\begin{array}{|c|c|c|c|}\hline
\mathrm{T\ (K)} & \mathrm{P_{x-ray}\ (W)} & \mathrm{P_{gamma}\ (W)} & \mathrm{notes} \\ \hline
378 & \approx 0 & \approx 0 & \text{boiling point of water} \\ \hline
550 & \approx 0 & \approx 0 & \text{approximate common highest temperature on residential ovens}\\ \hline
700-800 & \approx 0 & \approx 0 & \text{temperature range for wood-fired ovens, tandoors, etc.}\\ \hline
5770 & 4.26\times 10^{-129} & \approx 0 & \text{temperature of the photosphere of the sun}\\ \hline
1.57\times 10^7 & 10.4 & 7.87\times 10^{-54} & \text{estimated temperature of the center of the sun} \\ \hline
\end{array}$$
So, if you could heat your food to the temperature of the sun, it would produce minuscule x-ray radiation. It would also no long resemble food.


Is oxytocin's mirror image still oxytocin? If you were to take the mirror image of oxytocin, does it remain the same or does it change? Say I want a tattoo of it on the left side of my chest, going from the widest part of oxytocin to the smallest part, but it won't fit the way it is depicted below.

Would it still be correct or does something need to be changed? I can just get it on the opposite side but I'd like to know the chemistry of this.

Oxytocin is composed of $\alpha$-amino acids, which with the exception of one (glycine), are all chiral. A chiral molecule's mirror image is called its enatiomer, and has identical chemical properties except in the presence of other chiral molecules, i.e. in the body. This means enantiomer of oxytocin does not have the same biochemical activity as oxytocin, so you were wise to question whether or not this would be the same molecule. In addition to flipping the molecule, you need to change the wedges to dashes and dashes to wedges for it to still be the same molecule:


How to calculate initial and final temperatures during supercritical steam expansion? Ten $\mathrm{kg}$ of steam at $500~\mathrm{bar}$ is expanded at constant pressure until its volume increases to seven times its initial value of $0.01~\mathrm{m^3}$.
  (a) Calculate the initial and final temperature of the steam
  (b) Calculate the heat that must be supplied to carry out the process

An attempt at solving this problem:
$$\Delta H + \Delta E_k + \Delta E_p = Q - W_S\ \ \text{(open system)}$$

Taking out unnecessary terms leaves us with
\begin{aligned}
\Delta H &= Q\\
500~\mathrm{bar} &= 5\cdot10^7~\mathrm{Pa}\\
\Delta H &= H_f - H_i\\
H_i &= (5\cdot10^7~\mathrm{Pa})\cdot(0.01~\mathrm{m^3}) = 5\cdot10^5~\mathrm{J}\\
H_f &= (5\cdot10^7~\mathrm{Pa})\cdot(0.07~\mathrm{m^3}) = 3.5\cdot10^6~\mathrm{J}\\
\end{aligned}

Converting to kJ/kg (10 kg of steam)
\begin{aligned}
\hat{H}_i &= 50~\mathrm{kJ/kg}\\
\hat{H}_f &= 350~\mathrm{kJ/kg}\\
\end{aligned}

And that's where I stopped. I tried looking up the values of $\hat{H}_i$ and $\hat{H}_f$ in steam tables to interpolate the temperatures, but the values are too low. Since the pressure is too high, I also know the ideal gas law $(PV=nRT)$ won't apply to this scenario. Can someone point me in the right direction?

Since supercritical steam is not an ideal gas, you might want to use standard reference data (e.g. provided by NIST).

Initial state
mass m0 = 10 kg
pressure p0 = 500 bar = 50.0 MPa
volume V0 = 0.01 m3
density ρ0 = 1000 kg/m3
temperature T0 = 340.53 K
specific enthalpy h0 = 323.06 kJ/kg

Final state
mass m1 = 10 kg
pressure p1 = 500 bar = 50.0 MPa
volume V1 = 0.07 m3
density ρ1 = 142.86 kg/m3
temperature T1 = 925.86 K
specific enthalpy h1 = 3453.11 kJ/kg

Heat transfer
heat Q = ΔH = (h1 – h0) • m = 31.3 MJ


Seeing solubility in various cases [![enter image description here][1]][1]
Question is in the image 'match' each option having multiple answers. I have invested a considerable time over this but couldn't solve it. Anyone explaining how to see the solubility in those cases will be appreciated. I am having confusion. Or you can give a hint to each given in column 2 or you can also suggest some specific reading.
 I tried this through reaction like agbr+h2o --->agoh+hbr now how do I know its solubility is greater or smaller than expectations!? Similar problems I am facing with other options. [1]: https://i.stack.imgur.com/0Yaot.jpg

A is wrong. $\ce{AgBr}$ is one of the least soluble compounds of $\ce{Ag}$. Why don't you have a look in the table of solubility products to be convinced? Further more, $\ce{AgBr}$ will never react with $\ce{H_2O}$ to produce a strong acid like $\ce{HBr}$ and a base like $\ce{AgOH}$. The reaction goes the other way round. Strong acids react with hydroxydes to produce a salt and water.

C is wrong, because of the Le Chatelier's principle. Try to get the reason your self.

To solve D, you must know that $\ce{Zn(OH)_2}$ is soluble in an excess of $\ce{OH^-}$ ions


Is the kilogram still consistent with the old definition? My old understanding was that a gram is the amount of mass in a cubic centimeter of water.

But then I recently learned that the second was redefined to be some 9 billion cycles of a Caesium atom. Then from this the definition of a meter changed from 1 ten millionth the distance between the equator and north pole, to $1/299 792 458$ the amount of distance that light travels in a second. From this I would have assumed that now a hundredth of that meter would redefine the gram, which would then be multiplied by $1000$ to get the kilogram. But I read that instead the kilogram is some reference mass kept in a safe somewhere.

Are these definitions nevertheless consistent with each other or are we just doing away with the linkage between the centimeter and the gram via the water cube?

The kilogram is likely to be redifined in the near future as a function of the number of atoms of a sphere of ultrapure $\ce{^{28}Si}$, as per the answer to this  Chemistry SE question.
That means that the standard will be the Avogadro number, and the mass will be a derived unit.

Then, any stored mass anywhere will just be an historical curiosity.
This does not mean that it has not been important.
We need exact measurements as Lavoisier taught to all future chemists.
At that time it was the best that was available.
Now we might go to a still better standard.


Understanding how to calculate the pH of a buffer with ice tables Let's say you are asked to calculate the pH of a solution with:

$\mathrm{0.1~M}~\ce{HCl}$
$\mathrm{0.2~M}~\ce{CH3COOH}$.
$K_\mathrm a = 2\times 10^{-5}$

This is how I would do it:

$$\ce{CH3COOH + H2O <=> H3O + CH3COO-}$$

The corresponding ICE table:
\begin{array}{|c|cccc|}\hline ~ & \ce{CH_3COOH}& \ce{H_2O}& \ce{H_3O^+}& \ce{CH_3COO^-}\\ \hline  \mathrm I & 0.2 & & & 0.1\\\Delta  & -x & & x & x\\ \mathrm E& 0.2-x & & x & 0.1 +x\\\hline\end{array}

calculate the pH:

$$\begin{align}
 K_\mathrm a &= \frac{\ce{[H3O+][CH3COO^{-}]}}{[\ce{CH3COOH}]}\\
  &= \frac{x \cdot (0.1+ x)}{(0.2-x)}
\end{align}$$
as $x$ is very small, we leave it out:
$$\begin{align}
 \implies 2\times 10^{-5} &= \frac{0.1x}{0.2}\\
 \implies 2\times 10^{-5} &= \frac{0.1x}{0.2}\\
\implies~~~~~~~~~~~~ x &= 4\times 10^{-5}\\ \implies
 -\log(x) &= 4.40~= \mathrm{pH}\\
\end{align}$$

First of all. Is this correct?
Where do you include $\ce{HCl}$? I didn't write it anywhere. And I see most of the people not writing the second molecule. How/where is it included?
What is the molecule in the first column of the ice table? What does it represent? Do you always write the acid in that column?

I made that exercise up. The original exercise has $0.1$ molar $\ce{HCl}$ and $0.1$ molar $\ce{CH3COOH}$. I changed it in order to clearly see which value is which one at which place in stead of all having the same ones. This is the original exercise:

\begin{array}{|c|ccc|}\hline & \ce{CH_3COOH} & \ce{H_3O^+} & \ce{CH_3COO^-}\\ \hline \mathrm I & 0.1 & & 0.1\\ \Delta& -x & x & x\\ \mathrm E &0.1-x & x & 0.1 +x\\ \hline\end{array}

\begin{align}K_\mathrm a &= \frac{x\cdot (0.1 +x)}{(0.1-x)}\\ &\approx \frac{x\cdot 0.1}{0.1}\\ \implies x&= 2\times 10^{-5}\\\therefore -\log(x) &= 4.70\;.  \end{align}

The answer given is totally wrong. With 0.1 molar HCl, then the pH would be 1.0 from the HCl and virtually none of the acetic acid would be dissociated.

REVISED PROBLEM

  I made that exercice up. The original exercice has $0.1$ molar $\ce{HCl}$ and $0.1$ molar $\ce{CH3COOH}$.

This makes no sense.

The work shown in the second image is for 0.1 molar $\ce{CH3COOH}$ and 0.1 molar $\ce{CH3COO^{-}}$. That work gets the right answer, but the wrong way. At equilibrium: $$[\ce{CH3COOH}] = 0.1 - x$$
but $x$ is so small that $$[\ce{CH3COOH}] \approx 0.1$$ and $$[\ce{CH3COO^{-}}] \approx 0.1$$ which simplifies the mathematics considerably.


Why do the alkali metals form different products upon combustion in air? From Wikipedia's article on sodium:

  When burned in dry air, it forms primarily sodium peroxide with some sodium oxide.

We know that sodium has a strong reducing capacity, so why does it produce a compound in which the oxygen atom is not reduced to the fullest possible extent?

After posting the question, I read up some more and found this, on Chemguide's page describing the oxidation properties of Group 1 elements:

  So why do any of the metals form the more complicated oxides? It is a matter of energetics.

  In the presence of sufficient oxygen, they produce the compound whose formation gives out most energy. That gives the most stable compound.

  The amount of heat evolved per mole of rubidium in forming its various oxides is:

  $$\begin{array}{|c|c|}
\hline
 & \text{enthalpy change (kJ / mol of Rb)} \\ \hline
\ce{Rb2O} & -169.5 \\
\ce{Rb2O2} & -236 \\
\ce{RbO2} & -278.7 \\
\hline
\end{array}$$

This is hardly an explanation, because it just says in a circular way that it's energetically profitable to form such compounds. Why is this energetically profitable?

Can part of the explanation be "because in Group 1, starting from $\ce{Na}$, ion charge densities are low enough to allow peroxides and superoxides to exist"?

I wonder if the following behaviour is related to this: lithium nitrate decomposes to lithium oxide (again the other participant has the oxidation -2), while the rest of the group's elements have their nitrates decompose to a nitrite (again the other participant's oxidation is -1):

  From Chemguide:

  $$\ce{4 LiNO3 (s) -> 2 Li2O (s) + 4 NO2 (g) + O2 (g)}$$

  $$\ce{2 MNO3 (s) -> 2 MNO2 (s) + O2 (g)}$$

Since I will deal with all of the alkali metals in this answer, I think the question should also be broadened. There is no point in covering one single metal (sodium) without touching the others since it is the trend going down the group that we are interested in.

All thermodynamic data is taken from Prof. M. Hayward's lecture notes at Oxford.

So, firstly, some data. There is an increase in ionic radius going down the group, which should not be surprising:

$$\begin{array}{cc}
\hline
\ce{M} & \text{Ionic radius of }\ce{M+}\text{ / pm} \\
\hline
\ce{Li} & 76 \\
\ce{Na} & 102 \\
\ce{K} & 138 \\
\ce{Rb} & 152 \\
\ce{Cs} & 167 \\
\hline
\end{array}$$

This leads to an decrease in the magnitude of the lattice enthalpies of the Group I superoxides, peroxides, and oxides, going down the group:

$$\begin{array}{cccc}
\hline
\ce{M} & \Delta H_\mathrm{L}(\ce{MO2})\mathrm{~/~kJ~mol^{-1}} & \Delta H_\mathrm{L}(\ce{M2O2})\mathrm{~/~kJ~mol^{-1}} & \Delta H_\mathrm{L}(\ce{M2O})\mathrm{~/~kJ~mol^{-1}} \\
\hline
\ce{Li} & -960 & -2748 & -3295 \\
\ce{Na} & -860 & -2475 & -2909 \\
\ce{K} & -752 & -2175 & -2503 \\
\ce{Rb} & -717 & -2077 & -2375 \\
\ce{Cs} & -683 & -1981 & -2250 \\
\hline
\end{array}$$

As for why there is an increase in magnitude going across the table, we have to look at the factors controlling the magnitude of the lattice enthalpy:

$$\Delta H_\mathrm{L} \propto \frac{\nu z_+ z_-}{r_+ + r_-} \tag{1}$$

where $\nu$ is the number of ions in one formula unit, $z_+$ and $z_-$ are the charge numbers on the cation and anion, and $r_+$ and $r_-$ are the ionic radii:

$$\begin{array}{ccccc}
\hline
\text{Formula unit} & \nu & z_+ & z_- & \text{Numerator in eq. (1)} & r_-\text{ / pm} \\
\hline
\ce{MO2} & 2 & 1 & 1 & 2 & 149 \\
\ce{M2O2} & 3 & 1 & 2 & 6 & 159 \\
\ce{M2O} & 3 & 1 & 2 & 6 & 120 \\
\hline
\end{array}$$

The lattice energies of the peroxides and oxides are roughly 3 times those of the corresponding superoxides, because of the larger numerator. The lattice energies of the oxides have a  slightly larger magnitude than those of the corresponding peroxides, because of the smaller anionic radius.

Just looking at the lattice enthalpies, we might think that all metal cations would form the oxides. However, this approach is flawed because it does not take into consideration the energy cost of forming the three different anions from molecular dioxygen. Recall that the lattice enthalpy is defined as $\Delta H$ for the reaction

$$\ce{M+ (g) + X- (g) -> MX (s)}$$

However when we burn a metal in oxygen we are starting from $\ce{M}$ and $\ce{O2}$. So, we have to figure out the energy needed to get from $\ce{M}$ to $\ce{M+}$, and from $\ce{O2}$ to the relevant anion ($\ce{O2-}$, $\ce{O2^2-}$, or $\ce{O^2-}$).

The analysis here is complicated by the fact that the charges on the anions are not the same. Therefore, the reaction

$$\ce{M(s) + O2 (g) -> MO2(s)}$$

cannot be directly compared with

$$\ce{2M(s) + O2(g) -> M2O2(s)}$$

In order to "standardise" the equations, we will consider the reactions per mole of metal. One can loosely think of the combustion reaction as "releasing" some kind of energy within the metal; the most favourable reaction will be that which "releases" the most energy per mole of metal. Therefore, the three reactions we are considering are:

$$\begin{align}
\ce{M (s) + O2 (g) &-> MO2 (s)} & \Delta H_1 \\
\ce{M (s) + 1/2 O2 (g) &-> 1/2 M2O2 (s)} & \Delta H_2  \\
\ce{M (s) + 1/4 O2 (g) &-> 1/2 M2O (s)} & \Delta H_3  \\
\end{align}$$

Now, we construct Hess cycles for all three reactions.

Superoxides
$$\require{AMScd}\begin{CD}
\color{blue}{\ce{M(s) + O2(g)}} @>{\Large\color{blue}{\Delta H_1}}>> \color{blue}{\ce{MO2(s)}} \\
@V{\Large\Delta H_\mathrm{f}(\ce{M+})}VV @AA{\Large\Delta H_\mathrm{L}(\ce{MO2})}A \\
\ce{M+(g) + e- + O2(g)} @>>{\Large\Delta H_\mathrm{f}(\ce{O2^-})}> \ce{M+(g) + O2^-(g)}
\end{CD}$$

Peroxides
$$\require{AMScd}\begin{CD}
\color{blue}{\ce{M(s) + 1/2O2(g)}} @>{\Large\color{blue}{\Delta H_2}}>> \color{blue}{\ce{1/2M2O2(s)}} \\
@V{\Large\Delta H_\mathrm{f}(\ce{M+})}VV @AA{\Large\ce{1/2}\Delta H_\mathrm{L}(\ce{M2O2})}A \\
\ce{M+(g) + e- + 1/2O2(g)} @>>{\Large\ce{1/2}\Delta H_\mathrm{f}(\ce{O2^2-})}> \ce{M+(g) + 1/2O2^2-(g)}
\end{CD}$$

Oxides
$$\require{AMScd}\begin{CD}
\color{blue}{\ce{M(s) + 1/4O2(g)}} @>{\Large\color{blue}{\Delta H_3}}>> \color{blue}{\ce{1/2M2O(s)}} \\
@V{\Large\Delta H_\mathrm{f}(\ce{M+})}VV @AA{\Large\ce{1/2}\Delta H_\mathrm{L}(\ce{M2O})}A \\
\ce{M+(g) + e- + 1/4O2(g)} @>>{\Large\ce{1/2}\Delta H_\mathrm{f}(\ce{O^2-})}> \ce{M+(g) + 1/2O^2-(g)}
\end{CD}$$

Now we need more data. $\Delta H_\mathrm{f}(\ce{M+})$ is simply the sum of the atomisation energy and first ionisation energy:

$$\begin{array}{cc}
\hline
\ce{M} & \Delta H_\mathrm{f}(\ce{M+})\text{ / }\mathrm{kJ~mol^{-1}} \\
\hline
\ce{Li} & 679 \\
\ce{Na} & 603 \\
\ce{K} & 508 \\
\ce{Rb} & 484 \\
\ce{Cs} & 472 \\
\hline
\end{array}$$

and the enthalpies of formation of the anions are

$$\begin{array}{cc}
\hline
\ce{X} & \Delta H_\mathrm{f}(\ce{X})\text{ / }\mathrm{kJ~mol^{-1}} \\
\hline
\ce{O2-} & -105 \\
\ce{O2^2-} & +520 \\
\ce{O^2-} & +1020 \\
\hline
\end{array}$$

Why is the enthalpy of formation of $\ce{O^2-}$ so large? The answer is that, to get from $\ce{1/2 O2}$ to $\ce{O^2-}$, you need to first break the $\ce{O=O}$ bond, then add two electrons to the oxygen. Furthermore, the second electron affinity is often an unfavourable process. For the other two anions, you don't need to break the $\ce{O=O}$ bond.

So now, we can see a trend at work already. Going from the superoxides to peroxides to oxides, the more negative lattice enthalpy favours the formation of the oxide. However, the increasing heat of formation of the anion favours the formation of the superoxide.

When do each of these two factors win out? Well, when lattice enthalpies are comparatively large, we would expect the lattice enthalpy factor to outweigh the heat of formation of the anion. Lattice enthalpies are large precisely when the cation is small, and therefore lithium forms the oxide when heated in oxygen. However, with caesium, lattice enthalpies are smaller, less significant, and the heat of formation of the anion wins out; caesium therefore forms the superoxide.

The trend is of course not black and white. Going down the group from lithium to caesium, we might guess that perhaps there are one or two elements that form the intermediate peroxide. That element is sodium. You could say that the larger lattice energies of sodium salts sufficiently compensate for the formation of the peroxide ion, but aren't enough to compensate for the formation of the oxide ion.

I leave you with the last bunch of numbers, which tabulate the values of $\Delta H_1$ through $\Delta H_3$ for all the elements (all values in $\mathrm{kJ~mol^{-1}}$). You can actually calculate these yourself by plugging the data above into the Hess cycles. It seems that the data is a little different from that given in the Chemguide screenshot you have, but the conclusion is the same, so I'll ignore that:

$$\begin{array}{cccc}
\hline
\ce{M} & \Delta H_1\text{ (superoxide)} & \Delta H_2\text{ (peroxide)} & \Delta H_3\text{ (oxide)} \\
\hline
\ce{Li} & -386 & -435 & \mathbf{-459} \\
\ce{Na} & -362 & \mathbf{-375} & -342 \\
\ce{K}  & \mathbf{-349} & -320 & -234 \\
\ce{Rb} & \mathbf{-338} & -295 & -194 \\
\ce{Cs} & \mathbf{-316} & -259 & -143 \\
\hline
\end{array}$$

As described earlier, the salt with the most negative enthalpy of formation will be preferentially formed. These are bolded in the table.


What is the IUPAC Name of [Cr(NH3)5Cl]Cl2.NH3? What is the IUPAC Name for the complex $\ce{[Cr(NH3)5Cl]Cl2.NH3}$?
I can't even find it through the internet and also by employing a find in "IUPAC's Red Book".

This answer is similar to this answer on $\ce{[Co(NH3)5(H2O)]Cl3}$, Thanks to Loong (Faded Giant) for the references.
According to the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book), the systematic names of coordination entities are generated by using the following principles:

IR-9.2.2.1 Sequences of ligands and central atoms within names
The following general rules are used when naming coordination compounds:
(i) ligand names are listed before the name(s) of the central atom(s),
(ii) no spaces are left between parts of the name that refer to the same coordination entity,
(iii) ligand names are listed in alphabetical order (multiplicative prefixes indicating the number of ligands are not considered in determining that order),
(…)

Therefore the systematic name of the co-ordination entity $\ce{[Co(NH3)5Cl]}$ would be "pentaamminechlorido" since the ammine ligands are named before the chlorido ligands according to alphabetical order.
The ionic composition of the complex salt $\ce{[Co(NH3)5Cl]Cl2}$ may be indicated in several ways:

IR-9.2.2.4 Charge numbers, oxidation numbers and ionic proportions
The following methods can be used to assist in describing the composition of a compound:
(i) The oxidation number of the central atom in a coordination entity may be indicated by a Roman numeral appended in parentheses to the central atom name (including the ending ‘ate’, if applicable), but only if the oxidation state can be defined without ambiguity. When necessary a negative sign is placed before the number. Arabic zero indicates the oxidation number zero.
(ii) Alternatively, the charge on a coordination entity may be indicated. The net charge is written in arabic numbers, with the number preceding the charge sign, and enclosed in parentheses. It follows the name of the central atom (including the ending ‘ate’, if applicable) without the intervention of a space.
(iii) The proportions of ionic entities in a coordination compound may be given by using
multiplicative prefixes. (See Section IR-5.4.2.1.)

According to the above rules, the oxidation state of cobalt in the complex can be written in two ways - as (III) or (3+).
Similarly the chloride ions can be named in two different ways - chloride or dichloride.
Also, see:

IR-5.5 NAMES OF (FORMAL) ADDITION COMPOUNDS
[...]
The names of the individual components of such a generalized addition compound are each constructed by using an appropriate nomenclature system, whether compositional, substitutive or additive. The overall name of the compound is then formed by connecting the names of the components by ‘em’ dashes; the proportions of the components are indicated after the name by a stoichiometric descriptor consisting of arabic numerals separated by a solidus or solidi. The descriptor, in parentheses, is separated from the compound name by a space. The order of names of the individual components is, firstly, according to the increasing number of the components and, secondly, alphabetical. As the only exception, the component name ‘water’ is always cited last. (Note that this represents a change from the rule in Ref. 2 according to which the component names must follow the order given by the formula.) The numerals in the descriptor appear in the same order as the corresponding component names.
[...]

$\ce{CaCl2.8NH3} \equiv \quad$  calcium chloride—ammonia (1/8)

Therefore the systematic names for the compound $\ce{[Co(NH3)5Cl]Cl2.NH3}$ would be:

pentaamminechloridocobalt(III) chloride—ammonia (1/1)
pentaamminechloridocobalt(3+) chloride—ammonia (1/1)
pentaamminechloridocobalt(III) dichloride—ammonia (1/1)
pentaamminechloridocobalt(3+) dichloride—ammonia (1/1)


Preparation of picric acid My textbook says "Nowadays picric acid is prepared by treating phenol first with concentrated sulfuric acid which converts it to phenol-2,4-disulfuric acid and then with concentrated nitric acid to get 2,4,6-trinitrophenol"

My question: Why is this a better method than simply treating phenol with concentrated nitric acid to obtain picric acid?

What I think is that maybe because $\ce{NO2}$ is a stronger deactivating group than $\ce{SO3H}$, it is more difficult to add 3 $\ce{NO2}$ to a benzene ring directly.

But we would also have to substitute $\ce{SO3H}$ for $\ce{NO2}$. Wouldn't that be hard?

The benzene ring in phenol is highly activated toward electrophilic substitution and hence attempts to directly nitrate it result in charring and copious evolution of oxides of nitrogen. The reaction is highly exothermic and difficult to control.
To reduce the reactivity, the phenol is first mono-sulfonated ( some of the product which is substituted may also be used). The products are ortho- and para-isomers. The para-isomer is separated and then nitrated. The nitration is comparatively far smoother (easier to handle).
Ipso substitution of $\ce{SO3}$ groups occur.

Links: Wikipedia, Research paper
Link: (For ipso substitution): PDF, Wikipedia (see subsection: Ipso substitution)


Preparing aqueous solution of ethanol You need to prepare $\pu{250.0 mL}$ of an aqueous solution of ethanol in the concentration will be $\pu{0.500 mol L-1}$. You have $\pu{100 mL}$ of a commercial solution of ethanol in water $(35.0\% \,V/V)$ available. The density of ethanol (pure) under these conditions is $\pu{0.7893 g mL-1}$ and that of water is $\pu{0.9970 g / mL}$.
1.What volume (in $\mathrm{mL}$) will you need to withdraw to prepare this solution?
2.What is the mole fraction of ethanol in the resulting solution?

Answer to (1)
We can prepare $\pu{250 mL}$ of $\pu{0.5 mol L-1}$ solution of ethanol by diluting $\pu{20.85 mL}$ of $35 \%$ $\ce{C2H5OH}$ (density $\pu{0.7893 g/mL}$) with distilled water to a volume of $\pu{250 mL}$.
The dilution is therefore made by adding $\pu{229.15 mL}$ of water.
However, the volume of a liquid obtained by mixing measured volumes of different solutions is not always precisely the sum of the component volumes.
Calculation
Source (B): $35 \%$ solution of $\ce{C2H5OH}$ (density $\pu{0.7893 g/mL}$)
The molar concentration of a solution of a chemical species A is the number of moles of that species contained in one litre of the solution (not in one litre of the solvent).
$$C(A) = \pu{0.5 mol/L}$$
$$\text{Molar concentration} = \frac{\text{Density} \cdot \text{Weight percent} }{ 100\cdot \text{Molecular weight}}$$
$$C(B) = \frac{d(B) \cdot w(B)}{100 \cdot M(B)}$$
$$C(B) = \frac{\pu{789.3 g/L} \times 35 \%}{100 \times \pu{46.0684 g/mol}}$$
$$C(B) = \pu{5.99662675499909 mol/L}$$
The number of moles of solute in the diluted solution must equal the number of moles in the concentrated reagent.
\begin{align}
V(A) \cdot C(A) &= V(B) \cdot C(B) \\
V(B) &= \frac{V(A) \cdot C(A)}{C(B)} \\
V(B) &= \frac{\pu{0.25 L} \times \pu{0.5 mol/L}}{\pu{5.99662675499909 mol/L}} \\
V(B) &= \pu{2.08450525782339 \times 10^{-2} L} \\
V(B) &= \pu{20.85 mL} \\
\end{align}
Answer to (2)
We have
\begin{align}
C(B)  &= \frac{\pu{0.125029667842 mol}}{\pu{20.85 mL}} \\
&=\pu{5.99662675499909 mol/L}
\end{align}
Now we have to add $\pu{229.15 mL}$ of water to $\pu{20.85 mL}$ of commercial solution of ethanol to make it $\pu{250 mL}$.
$\pu{229.15 mL}$ of water has a mass of $\pu{229.15 g}$. This means that there are $\pu{12.7197 mol}$ water because the molar mass of water is $\pu{18.0153 g}$
\begin{align}
\text{Mole fraction of ethanol}&= \frac{\text{Number of moles of ethanol}}{\text{Number of moles of solution}} \\
&=\frac{\pu{0.125029667842 mol}}{(\pu{0.125029667842 mol} +\pu{12.7197 mol}}\\ &
= \pu{0.009734 mol}
\end{align}

Are my answers to both questions correct? If they are incorrect, where am I going wrong?

@ Dhamnekar. Your calculations are correct, provided your assumption (that the volumes are additive) is correct. Nevertheless note that the last value ($x = 0.0097$) has no unit : it is not in moles. And we must also admit that, due to your assumption, x is probably simply $x = 0.010$.
Anyway, next time try to trust your calculations yourself. We are not here to verify correct calculations. We are here to HELP THOSE WHO DO NOT SUCCEED in solving their problems.


How can I dissolve nickel(II) oxide? To insulate $\ce{Pt}$ wire heaters, I plated them with $\ce{Ni}$. After switching on the heater for the first time, green $\ce{NiO}$ is formed which is insulating and protects the $\ce{Pt}$ from shorting. After using the heaters for a week or so, the $\ce{Ni}$ slowly falls away and the $\ce{Pt}$ heater shorts and cuts.

What is the best method of stripping away the balance $\ce{NiO}$ from the used up $\ce{Pt}$ heater?

As mentioned by @Mockingbird, nickel oxide is somewhat soluble in potassium cyanide which will dissolve the oxide to form nickel cyanide, $\ce{Ni(CN)2}$ and potassium tetracyanonickelate, $\ce{K2Ni(CN)4}$  but potassium cyanide is very, very toxic.

Do not use it until and unless you are a professional chemist.


What is the empirical formula for kerosene? Kerosene is a blend of hydrocarbons, so it doesn't have one chemical formula. But I'd like to find an (approximate) empirical formula giving the ratios carbon and hydrogen.

I haven't found anything online, and beggars can't be choosers. My actual interest is in RP-1. There is a NIST document (archive) on its thermophysical properties which gives a spectroscopic breakdown on pages 29–32 showing that even this highly refined kerosene is mind-bogglingly complex chemically.

On p. 28 of the NIST document linked in the question, there is a surrogate mixture formulation of 14 hydrocarbons which closely approximates RP-1. Its empirical formula is $\mathrm {C_{11.8}H_{23.0}}$, and this should be close enough for any sort of kerosene. The properties of this surrogate are given on p. 7.

This means that burning $\pu {1 kg}$ of kerosene in air produces about $\pu {3.15 kg}$ of $\ce {CO2}$ and $\pu {1.26 kg}$ of $\ce {H2O}$ with the extra mass, $\pu {3.41 kg}$, coming from oxygen. Alternately, if you have $\pu {1 kg}$ of a perfect stoichiometric mix of RP-1 and liquid oxygen, burning it would produce $\pu {715 g}$ of $\ce {CO2}$ and $\pu {285 g}$ of $\ce {H2O}$.


Decrease of Bicarbonate to compensate Respiratory alkalosis I am studying this claim

  Compensatory mechanism in respiratory alkalosis is decrease of SB i.e. bicarbonate ions ($\ce{HCO3^-}$).

There are reactions

$\ce{HCO3- \rightarrow H+ + CO3^{2-}}$ (conjugate base)

and

$\ce{HCO3- + H+ \rightarrow H2CO3}$ (conjugate acid)

but since the $pK_a$ for the second reaction is greater

decrease in $\ce{HCO3-}$ yields more acids

so alleviates respiratory alkalosis.

To claim such a thing, I think you must know that the $pK_a$ for the reaction is greater, so change in bicarbonate ions' content results in bigger change in the second reaction.

How does the decrease of bicarbonate ions lead to decreased respiratory alkalosis?

Bicarbonate ion is amphiprotic, as you've recognized. It can react as both a Brønsted base and as a Brønsted acid in water solution:

As an acid:

$K_a = 4.8 * 10^{-11}$

$\ce{HOCO_2^- + H_2O <=> OH_3^+ + CO_3^{2-}}$

As a base:

$K_b = 2.1 * 10^{-4}$

$\ce{HOCO_2^- + H_2O <=> HO^- + H_2CO_3}$ (not really true carbonic acid).

Through K value inspection, we can see that the bicarbonate ion preferentially reacts as a base in water solution. The magnitude of this difference is massive; $K_b$ is more than a million times greater than $K_a$.

So by decreasing the concentration of bicarbonate ion, we can combat respiratory alkalosis.


Calculating dipole moments from atomic partial charges How can I calculate dipole moment of a complex molecule where the partial charges are available from density functional theory? I have the coordinates of each atom from atomistic simulations of an isolated molecule in a box along with the Bader charges. Will a simple charge times distance work in this case? The charges are not equal and opposite in many cases. What should the approach be to calculate the dipole in such cases?

From a theoretical standpoint, that approach is correct:

$$
\mu_i = \sum_{a}^{N_{\text{atoms}}} \sum_{i\in\{x,y,z\}} r_{ia}q_{a},
$$

where the set of atomic partial charges $\{q\}$ could come from partitioning the density in AO space (Mulliken, Lowdin), partitioning the density in real space (Bader, Voronoi, Hirshfeld), partitioning the electrostatic potential (ESP; Merz-Singh-Kollmann or ChElPG), or other methods I'm certainly forgetting. Another interesting method is based on atomic polar tensors [1], called GAPT or IR charges, which are a natural byproduct of harmonic vibrational frequency calculations.

This doesn't mean the results will be of good quality. Those based on partitioning the AO density seem to be the worst overall. Anecdotally, those based on ESP calculations are better, however if your system has a lot of "buried" atoms that aren't solvent surface-accessible, results may be questionable. Real-space partitioning is an attractive method since it is independent of direct basis set effects and surface accessibility. GAPT results also seem to be good, but they require a harmonic frequency calculation, which can be expensive.

Milani, A.; Castiglioni, C. Atomic charges from atomic polar tensors: A comparison of methods. Journal of Molecular Structure: THEOCHEM 2010, 995 (1-3), 158-164. DOI: 10.1016/j.theochem.2010.06.011