Is there a difference between dipole and dipole moment? I’m confused as to whether both terms “dipole” and “dipole moment” are the same or different, does the moment have something to do with the molecular geometry? I know the vectors of the charges cancel out however carbon dioxide has no dipole moment because of 180 degrees, this makes it non polar, but it still has partial charges right? So isn’t it still polar?

The (electric) dipole moment is a mathematically clearly defined quantity, the product of charge difference and distance

A dipole is anything (e.g. a molecule, or a part of a molecule), that has a non-zero dipole moment.

And a polar molecule is one that has at least local dipoles. CO2 is polar if you get close (i.e. in a condensed phase). From a slightly larger distance, it's unpolar, because the centers of the positive and negative charge distributions in it are identical, the two local C=O dipoles cancel each other out.


How was the dissociation constant of water determined? I was studying about pH and pOH of different acids and bases and I read about the dissociation constant of water ($\ce{H2O}$).

It turned out to be $1.00 \cdot 10^{-14}\ \mathrm{mol^2\cdot L^{-2}}$ at $25~\mathrm{^\circ C}$.

Could I get some insights to how was this specific value determined?

I'd dare to say that it was first determined in experiments on electrical conductivity around 1890 by the physical chemist Friedrich Kohlrausch.

In Die Wasserstoffionenkonzentration. Ihre Bedeutung für die Biologie und die Methoden ihrer Messung (Leonor Michaelis, 1922, ISBN: 978-3-642-88800-7) the author writes

  F. Kohlrausch und A. Heydweiller kamen durch wiederholte Destillation des Wassers unter ganz besonderen Kautelen zu einem Grenzwert der Leitfähigkeit des Wassers, der sich durch weitere Reinigungsversuche nicht weiter herabdrücken ließ und daher als die eigene wahre Leitfähigkeit des Wassers gedeutet werden mußte. Sie stellten auf diese Weise die Dissoziation des Wassers zum erstenmal zahlenmaßig fest. Später wurden dann mehrere andere Methoden zur Dissoziation des Wassers gefunden, die wir bald kennenlernen werden, und diese bestätigten die Resultate von Kohlrausch.

This roughly translates to:
Upon repeated destillation of water under extreme precautions, F. Kohlrausch and A. Heydweiller eventually reached a limit for the conductivity […]. This way, they were the first to quantitatively determine the dissociation of water[…]

In a footnote to this paragraph a reference is given:

F. Kohlrausch und A. Heydweiller, Zeitschr. f. physikal. Chemie, 14, 317 (1894).


Home science optical absorption test for approximate caffeine quantity in coffee? I started with How can you measure the caffeine content of a liquid at home? in the cooking stackexchange, but the test strips were only qualitative (yes/no) and don't seem to exist anymore, and the part about rubbing dichloromethane on ones gums frightened me off.

The 2014 NY Daily News article Scientists working on at-home caffeine detection test links to Validation of Caffeine Dehydrogenase from Pseudomonas sp. Strain CBB1 as a Suitable Enzyme for a Rapid Caffeine Detection and Potential Diagnostic Test (Mohanty, S. K., Yu, C. L., Gopishetty, S. and Subramanian, M., J. Agric. Food Chem., 2014, 62 (31), pp 7939–7946) but that's not for home use. I found this nice experiment but that is still serious analytical chemistry.

However the article does mention the use of UV spectrophotometry, and that's something I could imagine playing with at home personally, using UV LEDs or other light sources, a Raspberry Pi camera or some photodiodes, a piece of diffraction grating film, basic physics, duct tape, etc. I'm OK with the optics and electronics, it's the chemistry I need the most help with.

The 1994 paper A Simple Method for Determination of Caffein Content in Tea Samples shows a correlation of caffeine assay from various types of brewed tea with absorption at 440 nm. Strong coffee would contain perhaps 5X more caffein, but being coffee, would probably be 100X or more higher in absorption in blue or UV.

This and this discuss UV absorption of caffeine in a solvent, but I am not sure if there are any solvents that I can find or use at home. Further, while I might be able to find a 275nm LED, it probably limits me to glass optics and expensive detectors, most plastics probably absorb too much there, and most silicon detectors have thin inactive layers that absorb this wavelength of UV; backside-thinned detectors are usually necessary. However, possibly the longer wavelength tail in the direction of 300nm might be useful.

My Chemistry Question: Are there any ways to separate the caffeine in coffee from the organics that make it so dark, using "household chemistry", so that I could try to check  blue or UV optical absorption?

I'm only looking for a way to tell if brewed coffee has a big or a small caffeine "kick" to it, so even something as bad as +/-20% repeatability might do in a pinch. I don't need an absolute measurement - just something that roughly tracks the caffeine content.

Possibly helpful reddit, What wavelength should I measure the absorbance of my coffee to get an idea of how strong it is?, which links to an article I can't read right now, as I am drinking coffee on the wrong side of the paywall: Measurement of caffeine in coffee beans with UV/vis spectrometer.

above: An old Certificate of Analysis from a Caffeine Standards Kit, from Agilent.

above x2: Caffeine absorbance in water, from here (top) and here (bottom).

Home science optical absorption test for approximate caffeine quantity in coffee?

Yes..., except that for "...in coffee?" it's No.
AAI Solutions' Measuring Caffeine Concentration; Applied Analytics Application Note No. AN-019 front artwork certainly looks hopeful...

(click for larger)
After all an Optical Multichannel Analyzer (OMA) in this case, to a home scientist, looks like some light source, a grating, three photodiodes and some tedious calibration, this is really for extracted caffeine in supercritical CO2.
However, from Background in

As different literatures indicated spectrophotometric determination of caffeine is also reported as preferred method of determination such as UV–Vis spectrophotometry because of its relatively low cost, rapidity, high accuracy and reproducibility. But UV–Vis spectrophotometric method cannot be used directly for determination of caffeine in coffee beans extracted with water owing to the matrix effect of UV–Vis absorbing substances in the sample matrix [9]. In aqueous solution of coffee beans it was observed that there is spectral interference from caffeine and chlorogenic acid in the wavelength regions of 200–500 nm. Yet this method requires the extraction of caffeine from the aqueous solution of coffee beans using dichloromethane for the spectroscopic determination. This is necessary since the caffeine spectrum is overlapped with other compounds found in coffee. Hence, the use of dichloromethane limits the wider application of UV–Vis method.
[9] Genetic diversity and correlation of bean caffeine content with cup quality and green bean physical characteristics in coffee (Coffea arabica L.)

In simple language: Coffee is full of all kinds of molecules besides caffein, so there will likely be many broad spectral features in the UV interfering with any simple way to measure one particular molecule's concentration.
You're going to have to resort to some kind of specific extraction. If not supercritical CO2 mentioned above, then something like this:
SPECTROPHOTOMETERS: How can I measure the caffeine content in drinks? with 16 items of apparatus, two reagents including somewhat dangerous to "home scientists" Dichloromethane
or like this: Jenway's 73 SERIES SPECTROPHOTOMETER Application note: A09-010A; The quantitative determination of caffeine in beverages and soft drinks using UV wavelength spectroscopy again with the dichloromethane!
Bottom line
Try not to drink so much coffee that you need to worry about all of this. Listen to "Sir Professor" Martyn Poliakoff's and switch to tea. (note, I'm also the OP)


Why water can't mix with oil or cooking oil? Why water can't mix with oil or cooking oil (both saturated and unsaturated)?

The basic answer is that water molecules attract each other and so clump together forcing almost all of the oil out of the clump.

The attraction exists because water is a polar molecule.  That is, it has a positive end and an negative end.  Thus the molecules tend to clump together just as a bag of small magnets tend to clump together.  Oils, on the other hand, are not polar.

This principle can be expanded.  In general polar materials tend not to dissolve in oils or, if they are liquids, dissolve in them.  The rule in chemistry is "like dissolves like".


Why do we not generally have "enneane"? Most numerical prefixes are based on Greek roots, but for "nine" we use the Latin root instead which seems inconsistent.  Thus, for instance, "nonane" instead of "enneane", or in mathematics "nonagon" instead of "enneagon" (although here the Greek form seems to have more traction) .  Why?  Can we use the Greek root as an alternative?

Perhaps not a complete answer, but this might put you on the right track. It looks as though a lot of the system we have today grew out of recommendations made by August Wilhelm von Hofmann in a paper titled On the Action of Trichloride of Phosphorus on the Salts of the Aromatic Monamines

He actually suggested using the Latin numerals as a prefix for the alkanes with n>3. So pentane was quintane, hexane was sextane, etc. Over time, the Latin prefixes fell out of favor and were replaced with the Greek one's except in the case of nonane (and quadrane which became butane rather than tetrane). A Yale website, which led me to this article and helped me answer a previous nomenclature question about the alkanes (Etymology of saturation degrees (-ane, -ene, -yne) in aliphatic compounds), posits that it might just be that enneane just doesn't sound as good. Plus the trend from the polygons probably reinforced keeping it around.

So not a full answer, but at least it shows that nonane wasn't chosen to break the trend, but remained in spite of new naming convention.

The author of the Yale page does mention that their source was The Name Game by Alex Nickon and Ernest F. Silversmith, so perhaps there is more information there.


Rate equation for an elementary reaction Yesterday, I had a debate with a friend of mine over the determination of the rate of elementary reactions. He said:

  Rate equation of elementary reaction can be directly given by just raising the reactants concentration terms to the respective coefficients of the balanced chemical equation whose rate is to be determined.

and he also gave the example of this reaction

$$\ce{2NO (g) + O2 (g) -> 2NO2 (g)}$$

whose rate is

$$
\ce{[NO]^2[O2]}.
$$

Questions:

Can I write the rate equation of an elementary reaction by just simply looking at the stoichiometric coefficients in the balanced equation and raising them over the concentration of reactants?
Are there any examples of elementary reaction where the exponential powers on the concentration terms in the rate equation comes different than the stoichiometric coefficients of that reaction?

The reaction order with respect to a reactant species cannot be determined from the chemical equation alone. You would have to be given more information. In real life, orders of chemical reactions are found experimentally.


Lewis structures of S2N2 and [S4]2+ This seems so simple, yet I just can't seem to manage. Apparently $\ce{S2N2}$ and $\ce{S4^2+}$ (which are isoelectronic) obey Huckel's rule (at least to some extent) and I was trying to draw the Lewis structures to show that, but I can't manage to draw them without having a diradical, which doesn't seem very likely to exist. And if there are radicals, it doesn't obey Huckel's rule anymore.

Can someone help me draw their Lewis structures that obey Huckel's rule?

Draw each ring with one double bond and the two atoms having no double bond with two electron pairs instead of one.  Verify that the total electron count agrees with two positive charges on $\ce{S4}$ and zero on $\ce{S2N2}$.  The double bond and the extra electron pairs on the two atoms are the six ring electrons in the Huckel rule.


Hartree-Fock vs. DFT vs. Hohenberg-Kohn vs. Kohn-Sham What is the difference between these four approaches?

My current understanding is that Hartree-Fock uses only 1 slater-determine (accounting for antisymmetry) but neglects correlation between electrons. HF orbitals are primarily determined self-consistently.

DFT is the idea that there is a functional (presently unknown) that maps energy in terms of density. Hohenberg and Kohn found a proof that it exists but did not actually find it.

$E[p] = T[p] + E(EE)[P] + E(EN)[P]$

I am rather confused about where kohn-sham theory fits into all of this. Is it's entire purpose simply to define kinetic energy? How does it do this?

The Kohn-Sham approach is one on top of the original DFT idea proposed by Hohenberg-Kohn.
They propose – to find the energy functional and its equations, which, as you said, are unknown – to start with what is known: the functional if there is no interactions. So that is what they do: approximate the kinetic energy and Slater determinant from the known, non-interacting system and add all the terms that are additionally known (i.e. Coulomb, kinetic energies and the electron/nucleus potential) until you are left with the Kohn-Sham equation, which has "the smallest unknown", the exchange & correlation potential.
I recommend to read more in a book or on Wikipedia in case you want a more detailed picture on any/all of the methods.


Small scale gas-to-liquids via Fischer–Tropsch — is it even possible? I'm really interested in the Fischer–Tropsch gas-to-liquids (GTL) process. A lot of people are using wood or other biomass feedstock to make syngas on small scales (farms, shops, etc.).
Practically, what would be required to build a small scale GTL reactor to run a Fischer–Tropsch process on a small scale in a shop? Is that even possible to do safely?

It probably isn't practical to do Fischer-Tropsch on a small scale
Gasification of biomass is easy. And produces syngas, a mixture of carbon dioxide, carbon monoxide, hydrogen and other compounds.
But the Fischer-Tropsch process to convert that mixture into useful liquids (like diesel) is not easy. The process requires carefully designed reactors, controlled heat and specific catalysts which are poisoned by many of the typical minor constituents of biomass originated syngas (especially any containing sulfur). So industrial reactors have to clean the syngas so the catalysts are not poisoned.
It is unlikely that you will be able to build a carefully controlled reactor that guarantees the right conditions (moderate but controlled temperature and pressure, complex geometry, specific catalysts, clean syngas). Even less likely that the result will be remotely economical on a small scale given the required cost and energy consumption. Heck, major chemical and oil firms have been trying for decades and few of their efforts have been  economically viable.
And there is another reason why this is futile on a small scale. Syngas itself is a usable fuel without further processing. It has been used to power engines (when diesel or petrol/gasoline are scarce as they were during WW2). And variants of syngas were once common as a piped domestic fuel for heating and cooking (before natural gas which is mainly methane became far cheaper and less toxic for most customers). So why not just burn the syngas?


What's the chemical nature of a dishwasher's rinsing agent? It seems that a rinsing agent helps the dishes in the dishwasher dry much better than they do without it (which I recently noted on my new dishwasher). What molecules are used in the composition of these rinsing agents? How do they help the plates dry?

According to the MSDS for Jet Dry (a common brand name) it contains:

Sodium 2-phenylpropane-2-sulfonate (Sodium cumene sulfonate)
Propane-1,2-diol
Citric acid

I think it may be safe to assume that the citric acid is in there for a "clean" smell and also to soften the water (thanks to the commenter).  The sulfonate is presumably a surfactant to help break the surface tension of the water, and promote evaporation, and the propylene glycol may help serve as a coolant (and is favored due to low toxicity).


Producing gold nano flakes In terms of producing gold nano flakes, Guo et al. [1] added aniline to the solution of chloroauric acid in ethylene glycol at 95 °C. However, I couldn't figure out the step by step chemical reaction in it. Can anyone please explain me the reaction procedure or guide me to the relevant source?

References

Guo, Z.; Zhang, Y.; DuanMu, Y.; Xu, L.; Xie, S.; Gu, N. Facile Synthesis of Micrometer-Sized Gold Nanoplates through an Aniline-Assisted Route in Ethylene Glycol Solution. Colloids and Surfaces A: Physicochemical and Engineering Aspects 2006, 278 (1), 33–38. https://doi.org/10.1016/j.colsurfa.2005.11.075.

I believe that ethylene glycol in this reaction is used as a solvent and Aniline is used as a reducing agent. Ethylene glycol solution heated with chloroauric acid produces $Au^{3+}$ ions (first 20 min) which is then reduced to $Au^{0}$  (3hr reaction) in the presence of aniline. Because there is no stirring during those 3 hrs (so less agitation of the reaction mixture)there is less formation of gold nanoparticle and more gold nano flakes are formed.

May be this paper will help you, it has a similar mechanism


Which is the better leaving group in the following Cannizzaro reaction? In the given mechanism, is it correct to prefer hydrogen as a better leaving group than the formyl group?

The negative charge on an sp2 hybridized carbon, with the carbonyl group attached to it, seems to be more stabilized than a negative charge on hydrogen. (I have starred the step I am referring to)

If you'd like to read more about this mechanism, you should out this paper:

Swain, C. G.; Powell, A. L.; Sheppard, W. A.; Morgan, C. R. Mechanism of the Cinazzaro Reaction, 1979, Am. Chem. Soc. 3576

They did a kinetic study with deuterium oxide. They gave two equivalent mechanisms:

and

I think that this mechanism, which is essentially what you have shown except with intramolecular hydride shift (it cannot shift through solution) is the way I would represent it. It makes sense that this can happen if you think of the transition state. In my mind, it makes sense that this hydride transfer can occur given the very electrophilic nature of the aldehyde.

This is the transition state:


Why can't O form OF6? At first I thought that the reason is what I was taught, the absence of vacant d-orbitals but then I did some research. Most of the answers stated what I initially thought but one particular answer told that it's a common misconception (d-orbital theory). According to it, there can be two possible reasons-

Not much electronegativity difference between O and F.
The atomic size of Oxygen is not sufficient for it to hold 6 atoms.

So which one is likely to be more appropriate, the d-orbital theory or the electronegativity and atomic size theory?

One possible explanation involves atomic size.  You need a large enough atom in the middle to stabilize all the occupied orbitals in the octahedral structure; sulfur makes it but oxygen does not.
This reference gives the molecular orbitals for sulfur hexafluoride.  A key feature of these molecular orbitals is a doubly degenerate pair, labeled $\mathrm{2e_g}$ in the above source, which effectively contains the "extra" valence electrons in fluorine-based orbitals.
The picture below, taken from a different source due to limited downloading options from the free one, includes one of these combinations (here labeled $\mathrm{3e_g}$, as different nomenclature is used) among others.

If we look carefully we see that the ostensibly nonbonding orbital has antibonding overlap between adjacent lobes and so is technically antibonding.  The same is true of the other component of this degenerate pair.  So to make the orbital effectively nonbonding and keep the octahedral structure from blowing apart, we need a large central atom to act as a "spacer" between orbital lobes.  Apparently sulfur is large enough, oxygen is not.


Relative strengths of hydrogen bonds Order of hydrogen bond strength:
$\ce{O-H...N}$ > $\ce{O-H...O}$ > $\ce{N-H...N}$ > $\ce{N-H...O}$. Why is that so?

According to me $\ce{O-H...O}$ should have the maximum strength while $\ce{N-H...N}$ should have the weakest hydrogen bonds based on the fact that oxygen is more electronegative than nitrogen. It would develop a higher $\delta^-$ charge and the hydrogen attached to it will have a higher $\delta^+$ charge which would make the hydrogen bonding stronger.
I can't reason out the order given in my textbook. After observing it, all I can guess is that oxygen being highly electronegative leads to higher $\delta^+$ charge on hydrogen while nitrogen being less electronegative easily shares its electron cloud with hydrogen, making the hydrogen bond stronger.
I am not sure about this reason as according to this it is more stable for hydrogen to make a hydrogen bond with a less electronegative element and possibly even an electropositive element, (unless and until the hydrogen is attached to a highly electronegative element), which does not occur in normal circumstances.
It would be great if anybody could help me figure out the reason for the above trend.

As a short answer, Hydrogen bonds (for elementary chemistry) are formed with donation of lone pair of a lewis base into LUMO of H attached to a highly electronegative atom. A better donor makes a better bond and more electronegative atom makes a better bond. From my experience, the second factor overrides the first one in most cases.

N is a better donor than O and H in $\ce{O-H}$ is a better acceptor than H in $\ce{N-H}$.


Separating Silica from silicates Does anyone know how one could extract Silica $\ce{SiO2}$ from a one or more of the following compounds in either ground or powder forms?

Plagioclase: $\ce{Ca2Al2Si2O8}$
Pyroxene: $\ce{(Ca,Mg,Fe)2Si2O6}$
Olivine: $\ce{(Mg,Fe)2SiO4}$

Ideally the process would only involve energy and mechanical processes rather than adding many other chemicals.

Mechanically speaking, it is not possible to separate $\ce{SiO_2}$ from these minerals. The only way to do it is carrying out a chemical transformation. Here is an example of such a treatment.
Mix a sample of one of these minerals with $\ce{Na_2CO_3}$ in excess, and heat it in a platinum crucible with a Bunsen burner. When the substance is liquid, wait $10$ minutes. Let it cool down. The mineral is now transformed into sodium silicate $\ce{Na2SiO3}$ plus some oxides and carbonates. If I choose the specially simple olivine $\ce{MgFeSiO_4}$ as an example, the following reaction may happen : $$\ce{MgFeSiO_4  +  Na_2CO_3 -> Na_2SiO_3 + MgCO_3 + FeO}$$ The reaction is somewhat more complicated, because usually $\ce{FeO}$ is oxidized in air into $\ce{Fe_2O_3}$, and $\ce{MgCO_3}$ is partially decomposed into $\ce{MgO}$ and $\ce{CO_2}$. Furthermore, the sodium silicate $\ce{Na2SiO3}$is partly transformed into $\ce{Na4SiO4}$. Fortunately this side reactions do not really modify the rest of the process.
Anyway, put the crucible in a small beaker. Add $20$ mL water. Let it boil for a couple of minutes. When the substance is pulverized, add an excess of diluted $\ce{HCl}$. Heat on a water bath until the substance is dry. Add $2$ mL concentrated $\ce{HCl}$. The following reactions happen : $$\ce{Na_2SiO_3  +  2 HCl ->  H_2SiO_3 + 2 NaCl}$$ $$\ce{MgCO_3 + 2 HCl-> MgCl_2 + CO_2 + H_2O}$$ $$\ce{Fe_2O_3 + 6 HCl-> 2 FeCl_3 + 3 H_2O}$$ After $10$ minutes, add some hot water and filtrate. $\ce{H_2SiO_3}$ remains on the filter. Calcium, magnesium and iron chlorides are soluble into water and are eliminated.  Then $\ce{H_2SiO_3}$ can later be heated to obtain $\ce{SiO2}$ $$\ce{H2SiO3 -> SiO2 + H2O}$$  This is one way for obtaining $\ce{SiO2}$ from the proposed silicates. Other processes are known, using for example fluorhydric acid $\ce{HF}$. But, of course, using this acid is not recommended, because of its toxicity.


What is the correct way to clean spectrophotometry cuvettes? The lab I started working in has a spectrophotometer with cuvettes. The cuvettes need to be inspected and cleaned regularly. The senior people in the lab are in disagreement as to the correct procedure for cleaning them. Some recommend soap and a bristle brush with running water, others just suggest leaving them to soak in an ethanol solution.

The question is twofold;

a) Is there any risk that the brush will scratch the glass?

b) Is there any advantage to the soap method? If not, I am wasting an hour of my boss' time and money.

I always managed to wash out the remaining solution with the corresponding solvent a couple of times, subsequently rinsing with acetone/isopropyl alcohol and leave it dry on air for a few minutes. The method of spectrophotometry requires that you are using homogeneous solution anyway, so there shouldn't be any precipitation/deposition, otherwise you cannot interpret the data with Beer-Lambert law.

Also, don't soak cells in KOH-bath or similar alkaline media. And I'm always strongly against abrasives, especially when it comes to cleaning optics. Even quartz cuvettes can be occasionally damaged with the brush pick. Disposable PS-cells can be easily scratched. Plus, there is a little risk of hydraulic pressure knocking out the cuvette's bottom (like it happens with test tubes when cleaned too vigorously). Leave mechanical cleaning for the professors of mechanics *.

*From a story published in The Oxford Book of Scientific Anecdotes:

  Victor Moritz Goldschmidt, a distinguished geochemist, was reputed to have carried a capsule of potassium cyanide when he was planning his escape from Nazi Germany. When a friend in the engineering department of the university expressed interest, Goldschmidt was supposed to have replied that cyanide was for professors of chemistry; his friend, a professor of mechanics, would have to carry a rope.


White growth on unopened bottle of chlorotrimethylsilane I have a white, moldy-like substance that has been growing near the cap of an unopened glass container filled with TMSCl. It has spread to form a thin, powdery coating on all of the other containers of organic reagents it is stored with. This substance appears to have a corrosive effect on glass, but leaves plastic containers and metal containers unaffected. I'd like to know what it is before I go about cleaning it up.

Chlorotrimethylsilane, like other organosilicon compounds, is easily attacked by nucleophiles and readily hydrolyzes in air yielding hydrogen chloride:
$$\ce{2 Me3SiCl(l) + H2O(l) -> Me3Si-O-SiMe3(l) + 2 HCl(g)}$$
Hydrogen chloride reacting with traces of ammonia $\ce{NH3}$ in the air builds up white solid ammonium chloride $\ce{NH4Cl}$ depositing on the nearby surfaces.
Also, the final product of hydrolysis for the majority of organosilicon compounds is silicon dioxide $\ce{SiO2}$ (sol-gel process similar to olation). Fine powder of silicon dioxide is white as well.


Splitting across a double bond in ¹H NMR Why do the Hs in (E)-1-bromo-2-chloroethene split into doublets instead of staying as a singlet. Wouldn't they be isolated from each other since the double bond does not rotate?

The hydrogens are not 'isolated' - they are still vicinal protons (3-bonds away) and so the J-coupling between them is strong. The strength of a splitting interaction is primarily determined by the number of bonds in between the hydrogens involved.

The angle between them also plays a role but to a lesser extent. The Karplus equation describes the three-bond J-coupling between two protons as a function of their dihedral angle.


Free energy change in collagen hydrolysis In now-deleted comments on this answer, a user with a self-proclaimed chemical engineering background claimed that the reactions that take place when slow-cooking meat are endothermic and require far more energy than actually heating the meat. My intuition says this can't be - the meat would rapidly cool itself down if removed from the heat source, the sous vide circulator would have to work hard the whole time not just during initial heating, the bonds in the collagen are presumably not terribly strong, and, well, heating mostly-water things takes a lot of energy! But I don't want to assume too much.

I believe the primary interesting reaction here is collagen conversion, hydrolizing it into gelatin, and the collagen is something like a few percent of the meat by mass. (When cooking something like this, the rapid reactions have already taken place: the meat looks cooked, and it gets slowly more tender over time as the collagen breaks down.)

So, how much energy does that actually require?

tldr: reaction is exothermic because the heat of reaction is -2.5 Kcal/mol.

(1) Peptides do hydrolize in presence of enzymes. It means that Gibbs energy of hydrolysis is negative. However, Gibbs energy consists of $\Delta$ H and $\Delta$ S*T. $\Delta$ S of hydrolysis is favorable (more disorder), but $\Delta$H can be either lower or higher.  $\Delta$ H is what we are discussing here (the enthalpy of the reaction).

(2) Some endothermic reactions are really very endothermic (like dissolving NH4NO3 in water). Other reactions are mildly endothermic.  Hydrolysis of meat doesn't have a large $\Delta$ H change. "HEATS OF HYDROLYSIS OF AMIDE AND PEPTIDE BONDS" paper has exact value for the reaction: -2.5 Kcal/mol of peptide.  A negative value indicates the reactants have greater enthalpy, or that it is an exothermic reaction. So the reaction is slightly exothermic.


Is blending 15% methanol with petrol a good idea? I just came to know that the Government of India is going to implement a scheme under which 15 percent methanol will be blended with petrol and which will reduce the cost of the fuel by 10 percent.

I want to know that how would this practically affect all the machines where petrol is being used as fuel? Will those machines have to be reengineered to work with the new blended petrol or would those machines work the same way they worked with normal petrol? How would their efficiency be affected?

I think it's important to first understand why methanol is added to petrol to begin with, outside of what I will discuss later on as some of the benefits to atmospheric emissions.

For internal combustion engines, petrol should be of a high "octane rating". This is a measure of the compression ratio of the fuel, and it needs to be high in order for the engine to work efficiently. If the petrol has too low of an octane rating, then something called knocking can occur, which over time can cause very bad engine damage.

In order to improve the octane rating of petrol then, a variety of compounds can be added that boost the overall octane rating (appropriately named octane boosters). In the past this has included tetraethyl lead, which when burned during combustion lead to the accumulation of lead in the atmosphere, which was enormously damaging to the environment and to human health. Methanol, and other alcohols like ethanol, are one such example of an octane booster.

Before I discuss the effects on engine performance, I will say that the use of using methanol blended fuel does have positive environmental impacts over the use of diesel or just straight petrol. Combustion efficiency is improved, thus leading to lower emissions of $\ce{NO}$ and $\ce{NO2}$ (collectively called $\ce{NOx}$ gases), which has positive implications for the atmosphere overall (lower $\ce{NOx}$ generally leads to lower ground level ozone, but this is dependent on VOC emissions of course!), as well as to lower particulate emissions. They also do not contain sulfur, and thus blended mixes have a lower sulfur load - that means less sulfur dioxide emissions.

In terms of redesigning engines on the basic, fundamental level - no change is required. Any internal combustion engine can use blended fuels, it just has implications of the lifetime of the engine. However, methanol can act as a corrosive on certain components of the engine. As blended fuels are increasingly seen as a good way for governments to reduce their $\ce{NOx}$ emissions - particularly after the problem with diesel emissions over the last decade - it seems likely that manufacturers will respond to this by redesigning engines to resist the corrosive effect of methanol within the blend. For now, though, having a small ratio of methanol to petrol - and it seems in this case, India has designed for 15% methanol blend, which should have little impact on engine damage through corrosion - is sufficient.

One last thing to note, however, that may not have come up. Some people might wonder why we don't just a 100% methanol fuel once we design engines that don't corrode due to it. The simple answer is - it's not as fuel efficient as petrol!


Has a carbon compound ever been found having an ionic bond? Though it is highly unlikely, has any carbon compound been found to make an ionic bond and to exhibit ionic properties?

Why unlikely? Ionic compounds of carbon have been known for ages. There are ionic carbides ($\ce{Al4C3}$, $\ce{CaC2}$, etc.), graphite intercalation compounds like $\ce{KC8}$, ionic derivatives of fullerenes and more.

Come to think of it, common $\ce{CaCO3}$ is certainly ionic and at the same time a compound of carbon, but this is most likely not what you want.


Why are Earth's oceans salty? As we all know water is $\ce{H2O}$, it does not contain sodium chloride. The ocean is a solution composed of water and a solute which is salt. I was wondering how the massive amounts of $\ce{H2O}$ in our oceans first mixed with salt? I'm sure the water didn't contain salt when it first got to the oceans so how did it get mixed with salt? Thanks.

Early history of Earth is a question to paleohydrology and paleochemistry. Briefly, water wasn't pure on the early stages of Earth.

Brief history of water on Earth. The planet was hot and seismoactive (lots of vulcanos) in the beginning. Water (as steam) escaped from magma, cooled down and precipitated on the Earth's surface. The surface already had salts on it, so it became saline from the very beginning. A better question is: how come we have fresh water in rivers and saline water in oceans? This separation is due to water cycle on our planet. Rivers bring dissolved salts to the ocean, then evaporates and precipitates over the land. Salts cannot evaporate (at any practical rate), so they stay in the ocean. This is similar to washing dishes. Consider this model: dish gets cleaner - that is a river; drain water gets dirtier - that is an ocean; drain water goes through purification and becomes clean before going to the supply pipe - this is evaporation of water from the surface of an ocean and precipitation over the land in our model.


How to set up equation for buffer reaction? For buffer equations, how can you tell what to react with water?

For example, if I have a 1 M acetic acid solution and 1 M sodium acetate solution (a conjugate acid-base pair) mixed together and am asked to find the pH, how do I know how I should write the equation?

Do I react acetic acid with water to give me the products, or sodium acetate with water to give me a different set of products?

There is a general way to do these sorts of problems. The idea is to consider equilibria of both acid/base and of the water.

When there is solution of a weak acid and it salt, or just the weak acid or just the salt, i.e. pure HA, NaA + HA or pure NaA, then there is no distinction between these types of solution because of the equilibria involved. ( NaA just represents any salt ). The following calculations apply for each or any of these solutions.

Start by defining the equilibrium constants for the reactions. A molecule dissociates as $\mathrm{HA \leftrightharpoons H^+ + A^-}$, but to be general we let the base be [B] instead of $\mathrm{[A^-]}$ and so write $\mathrm{HA \leftrightharpoons H^+ + B}$

$$K_a=\mathrm{[H^+]_e[B]_e/[HA]_e} \tag{1}$$

where the concentrations are those at equilibrium.

There is also the equilibrium $\mathrm{H_2O \leftrightharpoons H^+ + OH^-}$ to consider and

$$K_w=\mathrm{[H^+]_e[OH^-]_e}$$

We know the amounts of acid $c_a$ and base $c_b$ added at the start of the reaction. To obtain the pH at equilibrium the amount $\mathrm{[H^+]}$ has to be calculated. To do this work out the concentration of acid and base in terms of the initial amount and the ionised species.

The total concentration of HA is

$$c_a=\mathrm{[HA]_e+[H^+]_e - [OH^-]_e}\tag{1a}$$

and for the base

$$c_b=\mathrm{[B]_e -[H^+]_e + [OH^-]_e\tag{1b}}$$

The e subscripts are now dropped for clarity. [Note that some authors use mass and charge balance instead to work out the concentrations.]

These values can be substituted into the equilibrium constant equation (1) and using $K_w=\mathrm{[H^+][OH^-]} $. Substitution leads to the general equation

$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - K_w/\mathrm{[H^+]} }{c_a - \mathrm{[H^+]} + K_w/\mathrm{[H^+]}  }\tag{2}$$

which is a cubic equation in $\mathrm{[H^+]}$ that is best solved numerically in the most general case. This equation can be simplified under different conditions as the examples below demonstrate.

Note that when $c_b=0$ equation (2) describes the case of pure HA, when $c_a=0$ it describes the hydrolysis of pure NaA solution.

The figure shows how the pH changes for a weak acid/conjugate base with the p$K_A$ shown and $c_a$ = 0.01 M over a range of $c_b$ concentrations. The curve was calculated by numerically solving eqn 2.  The straight line (coloured grey) is the Henserson-Hasselbalch equation described below in (A). It is clear where this approximate equation fails.

Figure 1. pH vs concentration of base $c_b$. Red line full calculation, straight line (grey) the approximate Henderson-Hasselbalch eqn. The dashed lines show the p$K_A$ and the acid concentration $c_a$ used. The region where the Henderson-Hasselbalch eq. is a good approximation is clear; $c_b$ should be no less than $\approx 0.1c_a$. The plot uses the values in example (B).

Example (A) The Henderson-Hasselbalch eqn is an approximation but limits should be tested.

When the concentrations $c_a$ and $c_b$ are far larger than $(\mathrm{[H^+]} + K_w/\mathrm{[H^+]})$ or $(\mathrm{[H^+]} - K_w/\mathrm{[H^+]})$ these terms  can be ignored without error. This produces

$$\mathrm{[H^+]} = K_Ac_a/c_b,  \qquad \mathrm{pH}=\mathrm{p}K_A - \log_{10}(c_a/c_b)  $$

which is the formula often used. The log form is called the Henderson-Hasselbalch eqn.  It is often the case that $c_a/c_b=1$ then $\mathrm{pH}=\mathrm{p}K_A$.

Example B. What is the pH of a buffer solution consisting of 0.01 M of a weak acid with $K_A$ = 1.5$\cdot$ 10-4, and 0.01 M of its conjugate base?

We could use eqn. 2 but this means solving a cubic which is not necessary. Realising that in acid solution $\mathrm{[H^+] \gg [OH^-]}$ then

$$K_A=\mathrm{[H^+]}\frac{c_b + \mathrm{[H^+]} - K_w/\mathrm{[H^+]} }{c_a - \mathrm{[H^+]} + K_w/\mathrm{[H^+]}  } \to \mathrm{[H^+]}\frac{ (c_b + \mathrm{[H^+]}) }{ (c_a - \mathrm{[H^+]} )}   $$

This equation has to be solved and not approximated to the Henderson-Hasselbalch eqn. since $\mathrm{[H^+]}$ may not be very different to $c_b$.

Letting $x =\mathrm{[H^+]}$ then $x^2+(c_b+K_A)x-K_Ac_a=0$ or

$$x=\frac{-(c_b+K_A)+\sqrt{(c_b+K_a)^2+4c_aK_A}}{2} $$

which gives $\mathrm{[H^+]}=1.45\cdot 10^{-4}$ or a pH = 3.84. This answer is close to that of the HH equation which gives $\mathrm{pH = p}K_A =  3.82$, so as it turns out the HH equation was good enough.


Can we have Aromatic rings in a non-aromatic structure Is it possible to have aromatic rings inside a structure that itself is non-aromatic? Is there an example.

Sounds like a contradiction in terms to me. If a compound has aromatic ring(s), then it is aromatic - that's the definition.
Though aromatic compounds with aliphatic side chains are possible, of course.

Your compound is azulene; look it up on Wikipedia or elsewhere. Indeed, it is aromatic, and so is each of its rings. You may be confused by the fact that cyclopentadiene (which looks just like one of these rings) as an individual compound is not aromatic, but that's another story. See, a 5-membered ring needs an extra electron to be truly aromatic. At the same time, the 7-membered ring has one electron more than it needs. So they exchange that electron and live happily ever after.


Which base is stronger methylamine or propylamine? Which base is stronger methylamine or propylamine? I know that it depends on the strength of inductive effect of alkyl group. Am I right?
But how can I determine the difference between methyl and propyl substituent?

It does not make much of a difference. Both are short chain primary alkyl amines, so chemically much alike ($\ce{pKa} \approx 10.7$ due to the $\ce{-NH2}$ group). In what context do you want to know the answer to this question? That could help me direct my answer better.


Heating of mixture of gases with one gas barely above its boiling point What happens when mixture of two gases is heated in a closed container?

If one gas expands faster than the other and increases pressure of the container enough so that the boiling point of the other gas is decreased below the gas temperature, will second gas start turning into liquid?

This phase change must also give out heat, causing the overall temperature and pressure to further increase.

Is this valid? or mixture of two gases work lot different than two of them separately?

You could certainly have such a situation; that is, heating two compounds in their gaseous state at constant volume such that that the total pressure of the system exceeds the vapor pressure of the lower boiling compound causing it to condense.

As the gas condenses, it will of course give off heat as you said, but it would require continued heat input to sustain the increasing pressure and condensation. I believe that is what you meant in your second question anyway.


dangers of radiation question At a recent flea market, someone was selling a surplus military device that was marked as having 900 millicuries and should be disposed of in a safe way.  I picked it up to read the words.  Whould 900 millicuries be a concerning amount of radiation?

900 mCi is a lot of activity, the problem is that it is impossible to know how bad it will be without knowing which radionuclide is present.

If it is a tritium light with 900 mCi in it, then the external threat is close to zero. Just do not break open the glass vials in a tritium light.

If it is 900 mCi of radium-226 or cobalt-60 then I would suggest to stay a very long way from the object and call the fire brigade or police. They will not have the ability to resolve the problem. What they will do is to seal off the area and then send for expert advice and an emergency team from a nuclear site or some other place where people have experience of working with large amounts of radiaoctivity.

Even a pure beta emitter at 900 mCi is a lot of activity if you are dealing with a high energy beta emitter such as Sr-90 / Y-90.


Why are high pressures used in cracking of long-chain hydrocarbons? If we have a long-chain hydrocarbon, such as decane, and we split it through thermal cracking (say in an industrial plant), we use high temperatures, and high pressures. Cracking produces smaller molecules - alkanes, and alkenes. It might look like this:

$$\ce{CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH3}$$

Which is cracked into products like this:

$$\ce{CH3-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH2-CH3-> 2CH2=CH2 + C3H6 + C8H18}$$

Why do we use high pressure in this case? Le Chatelier's principle states that the equilibrium will shift to resist the change — in this case it would move the position of equilibrium to the left, not the right. In cracking we want maximum amount of products (ie. the right) — therefore surely a low pressure would shift the equilibrium that way.

The reaction is not dependent on molecules colliding — merely enough thermal energy has to be provided to enough molecules to cause them to split — homolytic fission — and become free radicals which react to form smaller molecules. Therefore a pressure requirement to increase the collisions doesn't seem necessary.

Why then do we use high pressure when cracking hydrocarbons?

First of all, cracking is not an equilibrium process, therefore La Chatelier-Brown principle doesn't apply! (irrelevant)
Also, the mechanism is much more complicated than this:

In gas phase, radical reactions generally have extremely complex reaction networks. E.g. burning of methane with oxygen contains thousands if not more elementary reactions. I am pretty sure that gas phase cracking is not a pure first order reaction.
In industry, cracking is generally done on heterogeneous catalyst, therefore not a gas phase reaction. Heterogeneous catalysis is pretty much controlled by the absorbed amount of molecules on the catalyst surface: high pressure, faster reactions.


What is the origin of the transferred oxygendianion in redox reactions? How come the oxygens transferred in redox reactions are always the $\ce{O^{2-}}$anion?

For example, I have this set of rules, and the rules are implicitly referring to the $\ce{O^{2-}}$  anion, a potent base (otherwise the rules wouldn't make sense; for example, it makes sense that a base in acidic solution is protonated to water, and that a strong base in basic solution is leveled to hydroxide ion).

In other words, how come, when electrons are transferred, such as in this unbalanced reaction:

$\ce{ClO_{3}^- + 6I^- \leftrightharpoons 3I_2 + Cl^-}$

The $\ce{O^{2-}}$ anion is formed (if only to be consumed again)? I remember my prof would say that specifically 3 $\ce{O^{2-}}$ anions "disappear".

Why not the $\ce{O^{-}}$ anion or simply $\ce{O}$? Does this have to do with stability?

I think you are misunderstanding the meaning of superoxide and the rules.

Superoxide is $\ce{O_2^-}$

Superoxide is NOT $\ce{O^{2-}}$ as you have written.

The rules are referring to hydroxide which is $\ce{OH^-}$, not superoxide.

The reason hydroxide is used to balance equations in basic aqueous solution is:

$\ce{H_2O \leftrightharpoons H^+ + OH^-}$

There is hydroxide present in aqueous solution, but not superoxide, peroxide, hydroxyl radial, etc.


Drying solvents while taking care of purity I need to dry some solvents like $\text{MeOH, EtOH, and  DMSO} $ and I have to take care of purity of the solvents. These solvents are HPLC grade.

I am interested in freezing point and boiling point of these solvents  with maximum error of $ \pm  0.5  C$, hence, I do not prefer to add silica gel (or other dehydrating agent) in the flask containing solvent.

What are other ways to dry solvents without putting something in them.

There is actually a book on preparing laboratory chemicals you can look up:

  "Purification of Laboratory Chemicals" by WLF Armarego and CLL Chai

However, given that all the chemicals you want to dry are extremely hygroscopic, I'm not sure how long they would stay super dry for calibration purposes (and I'm not sure you need super dry solvents for calibration either). If you need it to last, you may need a large dry box.


Can lead(II) oxidize tin metal in aqueous solution? According to my textbook, element $\ce{X}$ can oxidize element $\ce{Y}$ if $\ce{X}$ was lower in the activity series.
Lead is lower than Tin in the activity series table in the textbook, but the following question contradict that rule and I can't figure out why:

  Will the reaction $\ce{Sn(s) + Pb(NO3)2(aq) -> Sn(NO3)2(aq) + Pb(s)}$ occur?

The book reveals at the end that the answer is no, but doesn't explain why.

Are you certain that your statement from the textbook is correct? The cation of X will oxidize element Y if element Y is more active than X, which is usually labeled (may be towards the top or the bottom of the list). For reference, see the activity series at Wikipedia, where the more active metals are above the less active metals in the list.

  How to use the activity series:

For example, what happens when iron metal is placed into an aqueous solution of copper (II) nitrate?

Let's write the potential equation:

$$\ce{2Fe(s) + 3Cu^{2+} -> 2Fe^{3+} + 3Cu(s)}$$

Since iron is more active than copper, iron will reduce copper cations (and copper cations will oxidize iron metal). The iron will dissolve.

What happens if copper metal is placed into an aqueous solution of iron (III) nitrate?

Since copper is less active than ion, copper metal will not reduce iron cations. No reaction occurs and the copper will not dissolve.

  Your question

Lead is less active than tin in all activity series that I have seen. Lead metal should reduce tin cations. This reaction should work. It is possible that the textbook/solutions manual has a typo. I find them in my books all the time.


Why is hofmann product prefered here? In the following reaction, Hofmann elimination is preferred.

This reaction takes place by E1 mechanism. I thought the Zaitsev product would be major due to 1 extra hyperconjugating structure. In the book, it is given that it is due to sterically compressed alkene, so Hoffmann product is major. I am unable to understand why the sterical hinderance would lead to the instability in alkene as the three methyl groups is one bond away from alkene.

The Hoffmann product is formed as a major product generally in four cases. They are:

When the ${\gamma}$ carbon is 4°
When the leaving group is poor leaving group
When the leaving group is bulky leaving group
When the elimination is internal elimination (${E_i}$ mechanism)

Case 1:

It happens in the example given in the question the hindrance of base in abstraction of proton favors Hoffman product

Case 2:

It happens with poor leaving groups such as flourine (based on the carbanion character on the carbon from which proton is abstracted in the trasition state)

Case 3:

It happens  in  the Hoffmann's exhaustive methylation reaction where the bulky leaving  group  is a 3° amine group

Case 4:
It happens in the treatment of a 3° amine with Caro's acid followed by heating and in heat treatment of xanthate esters

Lastly I want to add that these are general results the yields may vary based on reactant.


Are there any known chemical properties of tritium water that make it unusually different from protium water? I suppose the first question supporting the main question is, has tritium water ever been synthesized in sufficient quantity to test chemical properties?

If so, and apart from the obvious radioactive nature of the molecule, and possibly self-heating nature, what unusual properties does tritium water have over protium water?

For example, deuterium oxide is not (easily) metabolized by living things; it cannot support life in the same manner as protium oxide. Also protium water is blue and deuterium oxide is clear. What color is tritium oxide?

Yes, $\ce{T2O}$ has been prepared and is available in significant quantity.  When relatively pure, the energy released by the radioactive decay process is so intense that $\ce{T2O}$ will boil.  It must be transported in a shielded, cryogenic dewar.

A significant difference between compounds containing an element bonded to protium, deuterium or tritium is the strength of those 3 bonds.  The $\ce{X-H}$ bond will be the weakest and $\ce{X-T}$ bond the stongest.  We can compare these bond strengths by measuring the relative rates at which the $\ce{X-H}$, $\ce{X-D}$ and $\ce{X-T}$ bonds are broken in a given isotopically substituted compound. These rate differences reflect what is known as a primary kinetic isotope effect. In carbon systems (X=C) the maximum primary isotope effects are roughly as follows:
$$\frac{K_H}{K_D} \sim 6-7$$
$$\frac{K_H}{K_T} \sim 13-14$$
In other words, a $\ce{C-H}$ bond may break as much as 13 times faster than a $\ce{C-T}$ bond.

These same effects will also be seen with $\ce{O-H}$, $\ce{O-D}$ and $\ce{O-T}$ bonds.  The primary kinetic isotope effects will be slightly smaller here because the magnitude of the effect is mass dependent and oxygen has a larger mass than carbon.  Still, as you pointed out, the effects can be disastrous in biological systems.  Biological systems cannot survive if the rates for key reactions are slowed down by such large factors.

The blue color of protium water ($\ce{H2O}$) is due to red light absorption around 700 nm. The frequency of an absorption is given by the following equation

$$\nu_e=\frac{1}{2\pi}\sqrt{\frac{k}{\mu}}$$

Here $\ce{\mu}$ is the reduced mass of the system (e.g. the bond involved in the vibration that is producing the light absorption) and is given as

$$μ=\frac{m_1 \cdot m_2}{m_1 + m_2}$$

where $m_1$ and $m_2$ are the atomic masses located at both ends of the bond.  We see that the reduced mass for an $\ce{O-H}$ bond is 16/17, while it is 32/18 for an $\ce{O-D}$ bond. Since, as shown above, the vibrational frequency is inversely related to the reduced mass, we would expect the $\ce{O-H}$ vibration to occur at higher frequency (shorter wavelength) than the $\ce{O-D}$ vibration.  Indeed, while $\ce{H2O}$ absorbs around 700 nm, $\ce{D2O}$ absorbs at higher wavelength (~1000 nm) and is colorless (reference, see p. 82). Given the even larger reduced mass for $\ce{T2O}$ (48/19), it's absorption should be shifted even further out of the visible range, so it too should be colorless.


Sudden pressure rising in a pipe of an old plant transporting 33% ammonia solution We pump 33% ammonia solution along a single pipe line before we introduce it to the mix. We've had two occasions now where the ammonia pressure has rapidly risen in the pipe. It happens when the fluid is at rest, meaning all valve shut and no pumps running. The pressure just escalates. The pipe was rinsed with recycled water, but this is a weak solution of what the ammonia is mixed with anyway.

Is there any material/condition that the ammonia solution could react with to cause this sudden over pressure?

Please note:

The ammonia tanks are outside (+1–2 degrees Celsius).
The internal temperature is (+15–19 degrees Celsius).
There is no pressure release system.
The system is 8yrs old.

Ammonia is more soluble at lower temperatures so the temperature change between the outside of the factory and the inside could raise the pressure inside the pipes.

Gas dissolved in water are metastable systems most of the times. Nucleation points in many cases suffice to promote the release of the gas from the solution (have you ever tried to put some salt in your beer?). It is possible that somewhere in the old pipes there are some areas with higher roughness (e.g. rusty connections) where some nucleation points promote bubble formation and so the ammonia can be released.


Hydrolysis equilibrium of urotropin In a test I was asked why the hydrolysis equilibrium of Urotropine (Hexamethylenetetramine) in neutral solution lies far on the side of the reactants. Well I had no idea and I'd be very glad if somebody could give me an answer on that question.

Small amount of acid is required for hydrolysis.
As a result ammonium salt and formaldehyde are produced.

Update:
It can be explained with enthalpy of reaction Wiki Enthalpy of reaction.
To calculate it, enthalpy of each  reactant required.
From wiki: ΔH(C6H12N4)=-99.2kJ/mol, ΔH(H2O)=-285.8kJ/mol, ΔH(NH3)=-45.94kJ/mol, ΔH(CH2O)=-115.9kJ/mol;
ΔH(reaction)=6*ΔH(CH2O)+4*ΔH(NH3)-ΔH(C6H12N4)-6*ΔH(H2O)=918kJ/mol;

As ΔH(reaction) is large, positive number, a lot of heat required for this reaction, as a result, equilibrium is far on the side of the reactants.


Comparison of chemical shift of methyl group attached to alcohol and carboxylic acid I understand that hydrogens can be deshielded by electron-withdrawing groups, and the more powerful the group, the greater the chemical shift.

However, I ran into a problem when qualitatively compare the chemical shifts of the hydrogens in the methyl group of methanol and ethanoic acid: the carboxylic acid group is stronger than the alcohol group, the oxygens of the carboxylic acid group isn't directly attached to the methyl carbon, unlike the alcohol oxygen.
How would I qualitatively compare the chemical shifts of the methyl groups of methanol and ethanoic acid?

It's a good idea to start with some data. The table below shows relevant $\ce{^1H}$ shifts in $\pu{ppm}$ for three compounds in various deuterated solvents (shift values are from Cambridge Isotope Labs): methanol (MetOH), formic acid (FormAc) and ethanoic or acetic acid (AcAc).

$$
\begin{array}{|c|c|c|c|c|c|}
\hline
\text{Compound} & \ce{CDCl3} & \ce{(CD3)2SO} & \ce{C5D5N}\;\text{or}\;\ce{C5H5N}  & \ce{C6D6}\;\text{or}\;\ce{C6H6} & \ce{D2O} \\ \hline
 \text{Acetic acid} & 2.13 & 1.95 & 2.13  & 1.63 & 2.16\\ \hline
 \text{Formic acid} & 8.02 & 8.18 & 8.54 & 7.24 & 8.22\\ \hline
 \text{Methanol} & 3.48 & 3.20 & 3.57 & 3.09 & 3.35\\ \hline
\end{array}
$$

The protons in methanol are more deshielded than those in ethanoic acid because in methanol an electron withdrawing oxygen atom is directly bonded to the carbon, while in ethanoic acid there is an intervening carbon.

If you remove that intervening carbon things get interesting, as the case of formic acid illustrates. The carbon atom in a carboxylic group is $\mathrm{sp^2}$ hybridized, which makes it more electrophilic than the $\mathrm{sp^3}$ hybridized carbon in methanol. Of course the carbon in a carboxylic group is also bound to two electron withdrawing oxygen atoms, rather than one, which results in greater deshielding. The formic acid proton is also subject to deshielding magnetic anisotropy associated with the delocalized electrons of the carboxylic group. The anisotropy effect is less pronounced in ethanoic acid because the methyl group is further away from the carboxylic group. The ultimate result is that the formic acid proton is very strongly deshielded compared to protons in the other compounds.


Find the displacement of mercury in a vertical tube So we have a 1m tube sealed at both ends. It is initially horizontally placed and the middle 0.1m contains mercury and the two ends contain air. Now the tube is turned vertical, we have to find the amount of mercury displaced.

Let the pressure in the upper end be $P_1$ and the pressure in the bottom end be $P_2$. The area of cross-section is $A$ and the lengths in the upper and bottom end are $(0.45-x)$ and $(0.45+x)$,respectively.Let $x$ be the mercury displacement.

By Boyle's Law
$$P_1V_1=P_2V_2$$
$$P_1A(0.45-x)=P_2A(0.45+x)$$
A single equation is insufficient to solve this, what should I do.

I assume the Hg is a competent barrier between the parcels of air and the parcels both have an initial pressure $P_0$ of 1 bar (760mm of Hg). Each air parcel has an identical initial energy content of $$ P_0V_0 = P_0A(0.45\mathrm{m}) $$

With hydrostatic equilibrium, $\Delta P = P_{\mathrm{bot}} - P_{\mathrm{top}}$ would all be due to the weight of the Hg, IOW, in mm of Hg terms, would be 100mm (0.1m) of Hg. -- This is your second equation. $$\Delta P = P_{\mathrm{bot}} - P_{\mathrm{top}} = 100\, \mathrm{\scriptstyle mmHg} $$From Boyle's Law we know that $$ P_{\mathrm{top}} = P_0 \frac{V_0}{V_{\mathrm{top}}} \quad\quad\text{and}\quad\quad P_{\mathrm{bot}} = P_0 \frac{V_0}{V_{\mathrm{bot}}} $$ Substituting, $$ \Delta P = P_0(\frac{V_0}{V_{\mathrm{bot}}} - \frac{V_0}{V_{\mathrm{top}}}) $$ or $$ (\Delta P) V_{\mathrm{top}} V_{\mathrm{bot}} = P_0V_0(V_{\mathrm{top}} - V_{\mathrm{bot}}) $$ Substituting again, $$ (\Delta P) A^2 ( 0.45\mathrm{m} - x )( 0.45\mathrm{m} + x ) = P_0A^2(0.45\mathrm{m}) ( (0.45\mathrm{m} - x) - (0.45\mathrm{m} + x) ) $$ eliminating $A^2$ and simplifying $$ \Delta P ( (0.45\mathrm{m})^2 - x^2) = -2P_0 x(0.45\mathrm{m}) $$ or in quadratic form $$ x^2 - 2\frac{P_0}{\Delta P}(0.45\mathrm{m})x - (0.45\mathrm{m})^2 = 0 $$ which would be $$ x^2 - (6.84\mathrm{m})x - 0.2025\mathrm{m}^2 = 0 $$ You can take it from there.


How does biodiesel have so much energy density? I find it fascinating that something like cooking oil can be turned into fuel. I know it can't have that energy density off the bat. So what do they do to make fuel for cars? Do they have to add other ingredients, and if so, why?

Where you go wrong is in thinking that cooking oil has less energy density than biodiesel. The Wikipedia page for biofuel energy content list a number of different types of fuel. Along with a long list of other fuels, it lists the following:

Fuel Type   Specific energy (MJ/kg) Energy Density (MJ/L)
Methanol    19.9 – 22.7             15.9
Fat         37.656                  31.68
Sunflwr oil 39.49                   33.18
Olive oil   39.25 - 39.82           33 - 33.48
Biodiesel   37.8                    33.3 – 35.7
Diesel      48.1                    40.3

As you can see, both the specific energy (MJ/kg) and energy density (MJ/l) are about the same as biodiesel and close to those of ordinary diesel fuel. To convert the oil into biodiesel, some methanol is usually added, but that does not increase the energy density.


Why does orthocresol behave as a skin irritant? o-Cresol or ortho-Cresol or 2-methylphenol is a constituent of tobacco smoke, and, like most other phenols, behaves as a skin irritant, causing a burning sensation when it comes in direct contact, or upon inhaling it in.

What is the biochemical reaction that causes our skin to produce such a reaction to orthocresol? Is there any correlation with orthocresol being an acidic compound? Can this be explained by +I effect or Hyperconjugation effect of the $\ce{-CH3}$ group?

Unlike Vinegar or Alcohol, which have different effects (vinegar feels like water to me, and alcohol has a cooling sensation), why do cresols and phenols in general have a stingy sensation on touching?

It is the phenol moiety of ortho-cresol that causes this. Phenol is classified as toxic because of the way it can penetrate the skin and cause damage.

In its neutral state, the phenol molecule is a slighlty polar but generally organic compound. Since our skin is designed to mainly keep water and water-soluble compounds out, it is lipophilic; this in turn helps lipophilic compounds to penetrate it and be absorbed by the body. As soon as phenol has been absorbed, it is again in an aquaeous environment. Here, phenol’s acidic properties ($\mathrm pK_\mathrm a = 9.95$) start playing a role: it can be deprotonated to the phenolate anion. This is hydrophilic and so will be able to dissolve in the water inside your body. If any membrane needs to be crossed, just add a proton to regenerate the lipophilic free phenol.

Another way in which phenol can often react is the single electron transfer mechanism to generate a phenol radical which can then cause havoc in cells. And finally, phenol should also be able to intercalate into DNA further causing damage.

So the stingy sensation is caused by the slight burns that you encounter once phenol has penetrated your skin. since neither ethanol nor vinegar can penetrate your skin as well as phenols can (they are too polar), these react differently and give you a different sensation.


SN1, SN2 or acid-base reaction I am asked to give the type of reaction (SN1, SN2 or acid-base reaction) for the following reactions. I know that both $\ce{MeOH}$ and $\ce{EtOH}$ polar protic solvents, so because the first is an SN2 reaction shouldn't the second reaction be that as well? It says that it is an acid-base reaction as $\ce{MeO-}$ reacts with the $\ce{EtOH}$ to from 1-bromopropane, $\ce{MeOH}$ and $\ce{EtO-}$ and then goes on to from a SN2 reaction.

I would appreciate if someone could explain why the second reaction would be an acid-base reaction at first instead of a SN2 reaction.

In both cases the alkoxide will be in an equilibrium with the alcohol. This will be occurring on a faster timescale than the substitution but in the first case it's irrelevant as  the same species are present on either side of the equilibrium.

Methanol is more acidic than ethanol, however, so the reaction lies in favour of methoxide and ethanol.


Among thioester and ester which has more electrophilic carbonyl carbon? In few reactions I’ve seen that nucleophiles preferably attack thioester’s carbonyl carbon instead of ester’s carbonyl carbon

According to me, the initiation of reaction is decided by the electrophilicity of the carbonyl compound, in esters O being more electronegative makes c=o more electrophilic, thus attack should have been at ester, but many reactions suggest that attack first occurs at thioester.

What makes thioester’s carbonyl carbon more electrophilic?

Reactivity of thioesters with respect to nucleophilic attack

Why is the thioester bond weaker than a regular ester bond?

Let's consider a reaction mechanism like this:

A steady-state analysis will show that the rate of formation of product is

$$r = \frac{k_1k_2[\text{ester}][\text{Nu}]}{k_{-1} + k_2}$$

$k_1$ for the thioester is larger, because the sulfur 3p lone pair overlaps poorly with the C=O π* orbital (which is formed from 2p orbitals). Recall that esters are less electrophilic than ketones because there is an ester oxygen which has a lone pair capable of overlapping with the C=O π*. In this case, since this overlap is diminished, it turns out that thioesters are typically roughly as reactive towards nucleophilic attack as ketones.
$k_2$ for the thioester is also larger because thiolates are better leaving groups than alkoxides.

Depending on the relative rates of $k_2$ and $k_{-1}$, though, the second point may or may not be relevant. For example if $k_2 \gg k_{-1}$, then the rate equation simplifies to $r \approx k_1[\text{ester}][\text{Nu}]$, in which case the comparison of $k_2$ is no longer useful.


does Sodium persulfate evaporate at room temperature I use Sodium persulfate to etch a copper board to manufacture home made Printed Circuit Board (PCB), and I use the same etching tank on the below photo which has no cover on the top. I wonder if it is okay to keep it like that at the room temperature or would it evaporate?

Regardless of evaporation, it is an oxidizer and unhealthy so it is recommended to keep it in a closed container.

With that being said, you most likely have a solution of sodium persulfate in water.  Water will evaporate so again, yes, keep it closed.


Does organic chemistry somehow defy the periodic table? My physics lecturer was introducing the idea of what a Theory of Everything was, and said something like this:

  Other sciences have their own theories of everything too, fundamental axioms from which everything else can be derived from. In mathematics, we have ZFC, and well, the entire framework of abstract mathematics and axiomatic systems. Even in Chemistry, all of modern chemistry can be derived from the periodic table and the laws of atomic structure. Well ... except for two words: Organic Chem.

Is this true? Can organic chemistry somehow not be derived within the framework of the periodic table?

While all chemistry is unified because all chemical compounds are made from things in the periodic table, that isn't like mathematics and much of physics. In these sciences many important complexities are derivable from a simple set of axioms.

Some chemistry is simple to derive or deduce from the periodic table but not all of it. The electronic properties of iron atoms tell you a lot about many iron compounds; the location of alkali metals in the table tells us a lot about their common properties.

But carbon is odd. The properties of simple compounds follow simple periodic patterns. But there are more organic compounds (consisting of carbon skeletons, usually with C-C bonds) than all other chemical put together. The complexity and variety of carbon chemistry is almost unfathomably large. None of this is obvious or predictable from the periodic table. You can't deduce the enormous variety from the simple properties of carbon atoms or anything given in the periodic table.

That is probably the sort of distinction your lecturer was aiming at.


Why doesn't cyclopropyl methyl carbocation stabilises itself by ring expansion? I have been taught that ring expansion stabilises smaller cyclic compounds to a great extent. So why does cyclopropyl methyl carbocation shows such type of resonance

rather than expanding its ring

thereby decreasing it's angle strain, further it will be stabilized by hyperconjugation effect and inductive effect.

Unfortunately, the premise of your question is wrong.

There is no cyclopropylcarbinyl or cyclobutyl cation. They're the same thing. Unlike cations which you might be more familiar with, for example, t-butyl cation, the cation you're looking at here, also called homoallyl, is a non-classical cation.

The open form and the two structures you drew in the second line are not distinct. They're just resonance forms of each other.

You ask does this ion undergo ring expansion? Well, yes and no. Yes, in that there is a contribution from the 4-member ring to the resonance structure. No, in that you can't expand something that doesn't exist.


Reactivity of lithium in water Why lithium is least reactive in water among group one elements though it has the most negative Electrode Potential value?

Potentials only tell you about thermodynamic stabilities. They do not tell you anything about the kinetics of the system, which would correspond to how quickly the system reacts.


Relative reactivities of alkyl halides in an E2 reaction What is the exact order of reactivity of alkyl halides in an E2 reaction?

Organic Chemistry by Paula Yurikanis Bruice says that the order is $3^\circ>2^\circ>1^\circ$. And I think this order is correct because the transition state of the tertiary alkyl halide will be better stabilized by more hyperconjugating structures as compared to the secondary and primary halides.

But my teacher made me write that the order is $3^\circ<2^\circ<1^\circ$ and I even saw the this order in a book.

So I'm kind of confused about the correct order in E2  mechanism.
Any help will be appreciated.

I believe you might be mixing up E1 and E2 elimination reactions. In the first case the reaction first proceeds by formation of a carbocation from the alkyl halide and then a proton extractio. In this case, the order of reactivity is indeed $3^∘>2^∘>1^∘$ as such is the order of carbocation stability - in order the E1 reaction to proceed the carbocation must be formed.

But the situation is completely different in an E2 elimination. In this case the mechanism is the abstraction of the proton that is oriented anti to the halide with the simultaneous formation of the double bond and halide leaving the molecule as anion. That means that in this case the stability of carbocation plays no role and the significant factor are the sterical considerations. If the carbon atoms around the halide make it hard for the base to get to the anti proton then the reaction is less likely to proceed via E2 mechanism.

But bear in mind that usually both E1 and E2 elimination can take place and because of that tertiary halides would usually undergo elimination more readily than primary halides - just not via E2 mechanism.

More details on the E2 mechanism, with pictures: Master Organic Chemistry


Nitric acid plus Hydroiodic acid Why $\ce{I2}$ is formed when $\ce{HI}$ and $\ce{HNO3}$ are reacted?

I know that $\ce{HI}$ is more acidic than $\ce{HNO3}$ so nitric acid will accept protons from $\ce{HI}$, so $\ce{I-}$ (iodide ion ) should be formed and nitric acid on accepting proton would form $\ce{H2NO3+}$ but that is not formed why?

Nitrate is a strong oxidant which oxidizes the iodide to iodine.

$$\ce{NO_3^- + 2I^- + 2H^+\rightarrow NO_2^- + I_2 \uparrow + H2O}$$

Note that the oxidation number of the nitrogen atom in $\ce{NO3-}$ is $+V$ and in $\ce{NO2-}$ it's $+III$, so over all, we have a reduction equation of:

$$\ce{NO_3^- + 2e^- + 2H^+\rightarrow NO_2^- + H2O}$$

On the oxidation side we want to form $\ce{I_2}$ out of $\ce{I^-}$, so the oxidation equation is:

$$\ce{2I^- \rightarrow I_2 + 2e^-}$$


What is an uninegative ligand I can't find a simple definition via google. All I get are research papers, that I don't understand.

I'm pretty sure it just means the charge of the ligand is -1.

https://en.wiktionary.org/wiki/unipositive

Cyanide qualifies, ETDA does not.

Anyway, the word is on its way out. "Monoanionic" seems to be more popular.


Influence of neighboring fluorine and hydroxyl on carbocation Which of the 2-fluorobutan-3-yl and butan-2-ol-3-yl carbocations will be more stable? I know that fluorine has a greater electron-withdrawing inductive effect, but what about the resonance effect of those?

$\hspace{35 mm}$

The answer is actually quite straightforward. Fluorine is more electronegative than oxygen, thus it is less willing to share its lone pair with the carbon bearing the positive formal charge. Thus, the $\ce {-OH}$ group provides greater stabilisation. In fact, the donation of the lone pair by the $\ce {O}$ generates a familiar moiety, an epoxide. The donation by $\ce {F}$ is almost negligible since the positive formal charge on $\ce {F}$ would be highly unfavoured.


Reaction kinetics, relating half life to reaction rate How do I relate the half life to the overall rate of reaction?

I argued that from the data, doubling the partial pressure of either reactant, keeping the other constant, will half the half life.

So try t1/2 =
$\frac{\ce{[1]}}{\ce{[k][PA_0][PB_0]}}$ and since pressure of A is always greater than B, so t1/2 = $\frac{\ce{[1]}}{\ce{[k'][PB_0]}}$ ?

I then wrote:
$$\frac{-\mathrm{d}[PB]}{\mathrm{d}t} = k [P_B]^m $$, where k = k[PA]

Recall the meaning of the half life $t_{1/2}$: At $t_{1/2}$, a concentration (or partial pressure) is decreased to the half of its initial value.

It is crucial to realize that this represents exactly the $\mathbf{-\frac{dp_A}{dt}}$ in your rate equation!

With other words, $-\frac{dp_A}{dt} = k\cdot p_A\cdot p_B^2$ becomes $-\frac{54}{2\cdot100} = k\cdot54\cdot(1.0)^2$ for your first set of data.

Solve for $k$ and do the same for the other data sets. What is the result?


How to determine the product from my attempt of HOCl synthesis This is a follow up to this question. Here is my short procedure:

start with 200mL distilled water
add 7mL of 8.25% sodium hypochlorite (.11M) to the solution. The resulting solution should be 3.7*10$^{-3}$M NaOCl
add bromocresol green indicator.
add acetic acid dropwise, titrate until the solution hits a green color (pH~5).

I think at this point, the solution should contain HOCl and a small amount of dissolved sodium acetate.
$$NaOCl + CH_3COOH \rightarrow HOCl + CH_3COONa$$
I took a pH test with litmus paper and it shows between 5-6. I took a free chlorine test to determine the concentration of HOCl, and it resulted in ~1000ppm, which was what I was aiming for. The chlorine concentration stayed comfortably within its max solubility in water. I used an NaCl meter to determine the concentration in the solution and it returned 0. However, I tested for salts with silver nitrate and it precipitated both white and dark brown/black.
White precipitate indicates presence of chlorine, and dark brown precipitate indicates presence of OH. How did I get AgOH in an acidic solution? Is it simply that the silver rips apart the HOCl to form AgCl and AgOH?
My boss is concerned that I have made dilute hydrochloric acid, not hypochlorous acid. How can I prove that the solution is one product or the other?

Ryan, sorry your approach to generate the desired compound is not the best one. This post also shows the problems with the concept of an "accepted" answer on StackExchange. The answer is also wrong and misleading. Please consult your academic supervisor to help you do a proper literature search. I am not sure why are you doing all this, you did not explain in your two posts. First of all, there is nothing such as pure 8% NaOCl soln. This bleach is a mixture sodium hydroxide and NaOCl. What you are apparently titrating is the free alkali with acetic acid (which you should not be doing). Simultaneously, the hypochlorite ion is decomposing into other products.
The proper place to do inorganic synthesis research is the library, not Q&A site. A good reference is Kirk Othmer Encyclopedia of Chemical Technology to look up bleach and its properties. Any good library must have it.
In your first question you linked a paper "Hypochlorous Acid as a Potential Wound Care Agent" in the Journal of Burns and Wounds. This journal has a zero impact factor which means this is of dubious quality. There was a reason to avoid organic acid such as acetic acid in that paper. Have you read about disinfection by-products in water? Search this topic this too, because disinfection byproducts can be cancer-causing.


How to create 1-100 ppm ethanol gas? I am trying to use TGS 2602 sensor of air contaminants. I would like to test if my circuit is working as intended. For that purpose I need to generate fumes of one of the gases the can be detected by the sensor.

I thought that I would go for ethanol. I tried holding vodka soaked cotton ball near the sensor - to no avail. Later on I tried spilling vodka around the circuit and placing more vodka soaked cotton balls around. I also covered the whole thing with a cardboard box (with gaps at the bottom between table and the box). Again there was no considerable effect. Vodka is essentially ~40% ethanol and water mix (by volume).

Can you say by that short description that I managed to generate between 1 and 100 ppm ethanol gas? If not, can you propose a robust and safe method of producing such a gas?

Please be aware how the sensor looks. It is enclosed in a metallic case with small holes on the top surface.

Waving a vodka soaked cotton ball near the sensor is sufficient to generate the required fumes. It was the circuit that was problematic - I did a subsequent testing with analog reader (voltmeter to be specific).


Finding pH using the -log, why does my calculation differ from the book's answer? An aqueous solution containing 0.0020 mol/L barium hydroxide, what is the pH? I do the following: 14-(-log(0.0020))=11.3pH. The answer is 11.6.

2.0*10^-3 mol/L sodium hydroxide solution, what is the pH? I do the following:
14-(-log(0.002))=11.3pH. The answer is 9.3

When I solve for pH of solutions with acidic solutes, I get the accurate answer, such as in the question: 2.0*10^3 mol/L nitric acid solution, where I do:
-log(0.002)=2.7pH, which matches the answer, 2.7.

My question is, why am I getting the discrepancy between the calculated and the given answer when calculating pH for solutions with base solute? Or am I not doing anything wrong?

Because barium hydroxide is $\ce{Ba(OH)2}$

The book is assuming each mole of barium hydroxide release 2 moles of hydroxide ions, while you are assuming only 1 mole is released.


Can you extract specific compounds from infused oils? I heard a discussion when sitting on the train this morning by a couple of cannabis folk on the way into Denver. This is in regards to coconut oil (for baking).

For example: if someone extracted THC into coconut oil, is there anyway to then separate the THC back out of the coconut oil so you end up having coconut oil in one cup, and THC in the other without affecting degradation? I'm not a smoker by any means, but I thought it was a rather interesting conversation. Feel free to be as technical as possible, as I would like to test this theory among many chemical compounds known in homeopathic medicine.

THC is often extracted from cannabis by mashing up the plant with a solvent such as hexane. The slurry obtained may then be filtered off, and the solvent evaporate to give a crude extract from the cannabis which will contain the THC amongst other organic molecules.  Depending on the extraction solvent, the components of the cannabis plant are extracted in different concentrations, depending upon their relevant affinity for dissolving in the solvent.

Coconut and olive oil have both been used in extractions, as they are safe for human consumption and so can be used directly once isolate without the need to remove potentially toxic (hexane) or flammable (hexane) solvent.

In general, the solvents used are inert, that is, they do not react to degrade the extraction products as you mentioned.

With regard to homeopathy, this too involves the extraction of active components from plant/biological matter. However, in homeopathy, it is generally less of an exacting science (for instance specifically getting THC out of cannabis) and more a case of extracting ALL of the components of a plant, for instance by making a tea. If you are interested in this, you might like to look at the recent Nobel prize winner Youyou Tu (2015, Physiology or Medicine). She realised that boiling the plant in water to create a tea was destroying the functionality of the molecule that chinese medicine was trying to extract, limiting its potency. By exploring alternative extraction methods she was able to get the biologically active compound without degradation leading to a useful drug.


How to decide acidic or basic nature of oxides How can we theoretically decide the acidic or basic nature of an oxide? What are the reasons/ factors which make an oxide acidic, basic, amphoteric or netural?

In general, the electropositive character of the oxide's central atom will determine whether the oxide will be acidic or basic. The more electropositive the central atom, the more basic the oxide. The more electronegative the central atom, the more acidic the oxide. Electropositive character increases from right to left across the periodic table and increases down the column.
The trend of acid-base behaviour is from strongly basic oxides on the left-hand side to strongly acidic ones on the right, via an amphoteric oxide (aluminium oxide) in the middle. An amphoteric oxide is one that shows both acidic and basic properties.
This trend applies only to the oxides of the individual elements in the highest oxidation states for those elements. The pattern is less clear for other oxides.

We define non-metal oxide acidity in terms of the acidic solutions formed in reactions with water. For example, sulfur trioxide reacts with water to forms sulfuric acid.
In sum, acidic oxides are oxides of non-metals, and basic oxides are oxides of metals.

There are three non-metal oxides from the upper right portion of the periodic table, $\ce{CO}$, $\ce{NO}$, and $\ce{N2O}$, which have such low oxidation numbers for the central atom that they give neutral aqueous solutions.

Since the acidity of a cation rises rapidly with charge, d-block elements that exhibit a wide variety of oxidation numbers may have one or more oxides that exhibit only basic properties and one or more oxides that exhibit only acidic properties. The higher the oxidation number, the more acidic the corresponding oxide. Chromium is an example of such an element. $\ce{CrO}$ is basic, $\ce{Cr2O3}$ is amphoteric and  $\ce{CrO3}$ is acidic.


Equilibrium and Free Energy $$\ce{CO2(g) + Ca^2+(aq) + H2O(aq) <=> CaCO3(s) + 2H^+(aq)}$$
Find the equilibrium constant using a free energy table.

Finding the free energy and plugging it into the equation $\mathrm{G = -RT \ln (\mathit{K})}$ is easy, but I am confused about what $K$ it is. Is it $K_c$ or $K_p$, or does it even matter?
This is important because I am supposed to do equilibrium calculations with it later. I guess I have to use $K_c$, because only the concentrations of the aqueous solutions are given, plus $\ce{CO2}$ is in ppm. I think $\mathrm{G = -RT \ln(\mathit{K})}$ gives me $K_p$, so do I have to convert that into $K_c$ afterwards if I want to do equilibrium calculations?

K is properly defined as:

$$K = \dfrac{(a_C)^{c}(a_D)^{d}}{(a_A)^{a}(a_B)^{b}}$$

Activities involve relation of concentrations or parcial presures so K is dimensionless.

Let's find activities.

Ideal gases

$$a_{gas} = \dfrac{P}{P^o} $$  $P^º$ is 1 atm.

Solute in very dilute solutions

$$a_{solute} = \dfrac{C}{C^º}$$ $C^o$ is 1 M

Activities for solvents and pure solids are considered unity. $\ce{H2O}$ and $\ce{CaCO3}$ have unitary activity.

Answer

$$K = \dfrac{\ce{(a_{CaCO{_3}})(a_{H{^+}})^{2}}}{{\ce{(a_{H{_2}O})(a_{CO{_2}{_g}})(a_{Ca{^{2+}}{_{aq}}})}}}$$

$a_{CaCO{_3}}, a_{H{_2}O}$ are unity. It rests to use ideal gas equation and get $C_{CO2}$. $a_{Ca{^{2+}}{_{aq}}}$ and $a_{H{^+}}$ are usual concentrations.

EDIT

The way $K_p$ is converted to $K_c$ is:

$$K_p = K_c(RT)^{\Delta{n}_{gas}}$$

R: gas constant (0.082 L atm mol$^{-1}$ K$^{-1}$). T: temperature in Kelvin (K) at which the equilibrium reaction takes place.

To find $\Delta n_{ gas}$ , simply subtract the sum of the coefficients of the reactants from the sum of the coefficients of the products.
You can deduce that using $P=\frac{nRT}{V}$, $\frac{n}{V}$ is concentration.


Paint changes color when applied to Stucco. Chemical rxn assumed, but what? PROBLEM:
When I apply paint to my home’s exterior stucco (currently white), the paint changes color. The grays change to purple. The whites change to yellow. Greenish-beige becomes orange. I have purchased many samples of paint, and a color change is apparent for all but the darkest colors. The change in color starts happening immediately when I apply the paint, and it becomes more intense over subsequent days. After a few days, there is a radical difference between the paint’s intended color and the new color.

Apparently a chemical reaction is occurring between the paint and a substance in/on the wall, but I don't know what it is, or how to remedy it.

SURFACE FACTORS:

The house is stucco, so perhaps the surface is very alkaline. The house was built in 1971, and several coats of paint have been applied by the previous owners since then, so I would think the alkalinity would no longer be a factor, but perhaps it is. (Would it even be possible for improperly cured stucco to remain highly alkaline for 44 years?)
The previous owner told me that, around fifteen years ago, he painted the house with "pool paint." (Unfortunately, I don’t know what type of “pool paint” was applied, or why it was applied.) My guess is that this was in response to the color-changing problem rather than the cause of the problem, but I don’t know for certain.  (It is interesting that the pool paint has remained white all this time, while I’m unable to get a true white.)

I have painted the wood next to the stucco, and the paint on the wood appears exactly as it should and it remains constant(aside from becoming just slightly lighter as it dries). In other words, the paint only changes color on the stucco. Since the wood was never painted with pool paint, and since the wood is obviously not stucco, I still don’t know which factor (pool paint, stucco, or something else) is causing the color change. However, this does show me the paint color is truly changing on the stucco.

I have some chemistry background, but I'm really stumped by this problem.

Any help in figuring out what’s going on would be very much appreciated.

You haven't told us the color of the stucco you are painting, nor the type of paint.  Chemical reaction is very unlikely, paint that changes color would make for bad paint.  Pigments are chosen to be stable.  Old stucco would be only a little alkaline, but it is porous.  This sounds more like color leaching out of the stucco and into the new paint.  You need paint with better coverage (covers contrasting colors) or better sealing ability.  Seriously, don't ask us, talk to your paint supplier at a paint store.  Maybe get them to refer you to their in-house experts.  You probably will have to spend more for the paint than you are planning on.


How to determine if a compound is likely to become a solid at room temperature? Reasoning: My guess would have to be that it has to do with the strength of the intermolecular forces between each molecule. For example, molecular compounds that have a very low molar mass like $\ce{CO}$ and $\ce{CH4}$. Both of theses molecules have a very small molar mass, a very low London dispersion force, and thus a very low boiling point. This in turns makes it easier to break the bonds that hold each individual atom and more than likely cause them to turn into a gas in normal conditions.

Then when you look at ionic compounds like $\ce{MgCl2}$ and $\ce{NaCl}$, it is more likely for these compounds to become solid because of the strong intermolecular force that they have (ionic bonds).

Is my reasoning correct and is there something else that I should consider?

You are pretty much on the mark with looking at the strength of intermolecular forces. If you wanted to make a general comparison, the tendency to form a solid can be arranged by the strength of the intermolecular forces.$$\text{London Forces}<\text{Dipole-Dipole}<\text{Hydrogen bonding}<\text{Metallic}<\text{Ionic}<\text{Network Covalent}$$
Then, to compare within a group, you need some way of measuring the strength of that type of intermolecular force.

A more mathematical way to look at this if you have empirical values for a compound is to look at the $\Delta G_\text{fusion}$ at room temperature. $\Delta G_\text{fusion}=\Delta H_\text{fusion}-T\Delta S_\text{fusion}$, where $\Delta H_\text{fusion}$ is negative (since it will free up energy in forming the solid) and $\Delta S_\text{fusion}$ is negative (since, whether it started as a gas or a liquid, forming a solid will reduce entropy). Letting $\Delta G_\text{fusion}=0$ (reaction at equilibrium), we find that $T_{\text{fusion}}=\frac{\Delta H_\text{fusion}}{\Delta S_\text{fusion}}$. This allows us to see that when a compound becomes solid is a trade-off between the enthalpy released from forming bonds and the energy needed to reduce entropy at a given temperature.


Why is the boiling point of trichlorofluoromethane lower than that of chloroform? If $\ce{CFCl3}$ has a higher molar mass, how come $\ce{CHCl3}$ has a higher boiling point?
$$
\begin{align}
\mathrm{bp}(\ce{CFCl3}) &= \pu{23.77 °C} \\
\mathrm{bp}(\ce{CHCl3}) &= \pu{61.15 °C}
\end{align}
$$

The main reason acting here is Dipole Moment. The higher the dipole moment of a molecule, the greater will be the intermolecular attractive forces and the higher will its boiling point be.
Dipole moment is the displacement of electron density in a molecule and it is a vector quantity. The net dipole moment of a molecule is the vector summation of all the individual dipole moment of 2 bonded atoms created by the electronegativity difference between them.
So, at first, let's get the structure of the molecules. Both of them are sp3 hybridised and have a tetrahedral structure.

For CHCl3, if you consider the bond between carbon and chlorine, its highly polar. Due to the high electronegativity of chlorine, there will be a net negative pole on the bottom part caused by the three Carbon chlorine bonds. And Hydrogen being less electronegative,however, will not attract the electrons to a great extent. Hence, there will be a pretty significant net dipole moment in CHCl3.

But in the case of CFCl3, yes, there will be a charge separation by the chlorine atoms, but on the upper portion lies the F atom which is very electro-negative. As a result, it will also pull its bond pair of electrons towards itself. Since, the net dipole moment is the vector summation of the individual dipoles, the charge separation of C-F, being in the opposite direction of the charge separation by the 3 chlorine atoms, will reduce the net dipole moment of the molecule.

For this reason, the dipole moment of CHCl3 is more than that of CFCl3 and so, CHCl3 has a higher boiling point.


Need for hybridization I have learned hybridisation as a concept created to explain the same $\ce{C-H}$ bond lengths in $\ce{CH4}$.

I found it easy to find the type of hybridisation in different molecules, but my problem is how to find whether an atom in a molecule under goes hybridisation or not.

For example, $\ce{N}$ goes under hybridisation in $\ce{NH3}$, but there are some examples where $\ce{N}$ won't hybridise (not sure abt $\ce{N2}$, if you may suggest). In $\ce{ClF3}$, $\ce{Cl}$ is hybridised but in $\ce{PCl5}$ $\ce{Cl}$ is not.

I want the reason for undergoing hybridisation.

In general, you should never assume a need for hybridisation unless you have a bonding situation (such as a tetrahedron or related tetravalent structures) which cannot be explained without hybridisation.
As soon as you have a lone pair on any atom, it is more favourable to have this lone pair in an unhybridised s orbital rather than in a hybridised $\mathrm{sp}^n$ orbital. This can be seen for example in phosphane, hydrogen sulphide, etc. in which practically unhybridised central atoms are possible and thus the bond angles show pure p orbitals taking part in the $\ce{A-H}$ bonds.
If you consider very small central atoms (e.g. nitrogen in ammonia) or very large ligands (e.g. chlorine in $\ce{PCl3}$), then having a perfectly unhybridised central atom — meaning a bonding angle of $90^\circ$ — would introduce too much steric strain between the ligands. Thus, the overall system is more favourable if the central atom’s bonding orbitals are partially hybridised, i.e. if the bond angles are enlargened slightly to reduce steric stress. The final bond angle and thus the final central atom’s configuration is a carefully balanced compromise between a hybridisation with a p contribution as high as possible and the ligands separated maximally.

By the way, chlorine is practically never hybridised and it is most certainly unhybridised in $\ce{ClF3}$. Rather, this molecule should be understood as having:

a single $\ce{Cl-F}$ σ bond using a p orbital both of chlorine and of fluorine

a four-electron-three-centre bond using a p orbital of each of the three atoms that form the linear $\ce{F\bond{...}Cl\bond{...}F}$ fragment (i.e. a p orbital of chlorine). This bond can be understood with the following resonance structures:
$$\ce{F-\overset{+}{Cl}\bond{...}\overset{-}{F} <-> \overset{-}{F}\bond{...}\overset{+}{Cl}-F}$$

one chlorine lone pair in a p orbital perpendicular to all $\ce{C-F}$ bonds

one chlorine lone pair in an s-type orbital.


Fake smoke for a fire drill on a boat? Is there any way to generate fake smoke that could be used in a fire drill on boat? Basically the idea is that the captain would set the "fire" and it would start smoking. The crew then has to find it and put it out.

Ideally it needs to smell, if not like real smoke, at least like something.

Also, it needs to billow out and obscure vision the way real smoke does.

I seem to remember a reaction that billows out black smoke but I cannot remember what it was.

Having some level of experience in fire training in various scenarios over a number of years, I highly recommend that you take a commercial approach to generating your smoke, rather than a homemade recipe that compounds your liability. Any smoke you generate from a homemade smoke bomb is likely to be considerably toxic, and should be avoided. Fire fighters would not enter a smoke-filled boat to attack an active fire without the appropriate safety gear which would include correct breathing apparatus.

You can purchase smoke cartridges of various sizes, colours and burn times, and these are designed for this very purpose. Examples can be found using a quick search for fire training smoke cartridges: here, here and here are examples from companies that I in no way endorse. They are pretty cheap for the effects they produce. I use the 20m3 burn capsules for fire fighting training with vehicle fires - these have a burn time of about 5 minutes, and they are very authentic, apart from limited smell. The smell from a fire in a boat is likely to be an acrid toxic affair, quite unlike your typical woodfire smell, so you would be best to avoid this. For better effects, burn a couple together, and you can get burning cups with electronic ignition that allows you to remotely activate it whenever you are ready.

Another option for generating limited visibility that is frequently used is a smoke/fog machine that is used for theatrical smoke and at nightclubs. Search for fog machine or smoke machine. Here is an example - again, I have no affiliation with this company. I have used ones that have come off ebay, and they are very cheap, but do the job nicely. I have used these a lot simulating structural fires and vehicle fires, and they can produce great amounts of 'smoke'. You need to take care that before any water is applied that the smoke machine has been turned off and removed well aware from the area. These also work very well for using with ram fans to visualise wind flow patterns through buildings/structures to understand convection flows and the path of travel for fires.

Be sure to read and understand the SDS of all materials before you use them in an enclosed space such as a boat, and ensure you have a plan to sufficiently ventilate any space between fire drills and afterwards.


Is Cold water heavier than hot water? I have come across this Quora Answer where it was written that cold water is heavier than hot water.

Are there any proofs which support it?

The "counter argument" you link to cites E = mc^2 and the higher energy content of hot water.

First off, note the "c^2" in that formula. That's the speed of light, squared. It is a huge number, meaning that you need huge differences in energy to get tiny changes of mass.

The other way around, you only need little mass to generate lots of energy, something nuclear physicists have been exploiting for quite some time. ;-)

So, while that effect is "true" to some extend, it is tiny. But yes, one molecule of H2O has more mass when hot than the same molecule when cold.

But when we talk about "weight", we usually mean "density", i.e. weight per volume. And here, a different, much more pronounced effect comes into play: Most substances expand when heated, reducing their density. Each molecule might be a little bit heavier, but the molecules are much farther apart, more than offsetting the E = mc^2 effect.

Cooling down increases the density -- the same amount of matter takes up less volume. An effect that home preservers have exploited for generations when canning things: You put hot produce into a container, and close it. The contents cool down and contract, producing underpressure in the container, sealing it. That's the "pop" you hear when opening a glass of marmelade the first time.

So, per molecule hot H2O is (a bit) heavier, but per volume hot H2O is (much) less dense.

At least down to about 4 degrees celsius, where a peculiar quirk to the H2O molecule makes it become less dense again with further cooling. That's why a pond starts freezing at the surface, not the bottom, where the 4 degree water is up to the point where the whole pond is frozen solid. An effect that fish have been exploiting for quite some time now. ;-)