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https://chemistry.stackexchange.com/questions/167226/breaking-an-azeotrope-by-using-different-atmospheres
|
Breaking an azeotrope by using different atmospheres?
|
I read the threads about azeotropic mixtures and distillation, and a question arose:
Is it possible to distill an azeotropic mixtures properly (i. e., separate it into its single components) by changing the atmosphere? For example, if you distill under argon atmosphere or gaseous hydrogen chloride?
Best and stay healthy!
Wolf
| 0
|
[] |
https://chemistry.stackexchange.com/questions/167224/problem-on-the-wavelength-of-the-electromagnetic-radiation-needed-to-break-1-mol
|
Problem on the wavelength of the electromagnetic radiation needed to break 1 mol of gas [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed last year.
[Improve this question](/posts/167224/edit)
**Problem.**
To break the bonds in 1 mol of Cl$\_2$ we need E=243.4 kJ of energy. The bonds on this amount of gas can be broken by electromagnetic radiation of a certain length. Determine the wavelength.
**Solution.**
Firstly, there are 34 electrons in the molecule Cl$\_2$. So in 1 mol of Cl$\_2$, there are $n\_e = 34\cdot N\_A = 2.05\times 10^{25}$ electrons, where $N\_A=6.022\times 10^{23}$ mol$^{-1}$ is Avogadro's constant. Now, the amount of energy needed for a single electron is (on average) $E\_e = E/n\_e = 1.19 \times 10^{-20}$ J.
We know that energy is related to the wavelength $\lambda$ by the relation
$$
E\_e = h\frac{c}{\lambda},
$$
where $h=6.63\times 10^{-34}\, m^2 kg/s$ is the Planck constant and $c=3.00\times 10^8 \,m/s$ is the speed of light in vacuum. Thus
$$
\lambda = \frac{hc}{E\_e} = 169\,nm.
$$
[Solution provided: 492 nm]
Am I doing something wrong?
**UPDATE**
I got the mistake thanks to @Domen (+10). I don't have to count all electrons but only those involved in the bonding. There is a single pair of electrons to be broken in a single chlorine molecule. In 1 mole of chlorine there are $N\_A$ bonds, thus $n\_e = N\_A$. So the correct answer is $\lambda = 492$ nm.
| 1
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[] |
https://chemistry.stackexchange.com/questions/167221/how-to-convert-phenylacetic-acid-to-3-phenylpropionitrile
|
How to convert phenylacetic acid to 3-phenylpropionitrile?
|
I understand that phenylacetic acid reacts with $\ce{KSCN}$ can form phenylacetonitrile plus the side product ($\ce{CO2}$ and $\ce{KSH}$). This is Letts nitrile synthesis
However, the problem is to convert phenylacetic acid to 3-phenylpropionitrile.
Is that possible? What reagents do we need?
| 1
|
[
[
"\nYou need to add one carbon, the easiest source of this is as CN-.\n\n\nThree steps.\n\n\n1. Reduce phenylacetic acid to 2-phenylethanol (borane.THF is a good way)\n2. Form the tosylate of the alcohol (TsCl, pyridine)\n3. Treat the tosylate with NaCN in DMSO [reference here](https://www.gaylordchemical.com/synthesis-corner/cyanation/).\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/167218/why-does-the-chlor-alkali-process-work
|
Why does the chlor alkali process work?
|
Chlorine is a stronger oxidiser than oxygen, and sodium is a much stronger reducing agent than hydrogen. Therefore, in the chlor alkali process, when voltage is applied to the electrodes and the solution is one of brine, why don't we end up electrolysing the water instead? We should, in fact, start seeing chlorine gas formation only when the brine is extremely thick, almost like a semi solid.
| 1
|
[] |
https://chemistry.stackexchange.com/questions/167215/how-does-temperature-affect-ionic-ca-in-milk
|
How does temperature affect ionic Ca in milk
|
I’m doing an investigation on the effects of temperature on ionic calcium content in milk through a classic EDTA titration method and have a few questions:
— what is the scientific theory behind the reason why ionic Ca decreases from increased temperature? (with respect to how it is measured through EDTA chelation)
— does this permanently affect Ca content
— how relevant is this with regard to health (e.g. does precipitated Ca have health benefits?)
| 0
|
[] |
https://chemistry.stackexchange.com/questions/167208/does-acrolein-give-positive-iodoform-test
|
Does acrolein give positive iodoform test?
|
>
> Which of the following compound(s) give(s) a positive iodoform test?
>
>
> **A**. Acrolein
>
> **B**. Acetone
>
> **C**. Acetoacetic ester
>
> **D**. 2-Propanol
>
>
>
According to me, the answer should be options **B**, **C**, and **D** as **B** is a methyl ketone, **C** is a β-ketoester and **D** can be converted to methyl ketone by tautomerization.
But the answer in the key is given as **A**, **B**, **C**, and **D**. I cannot figure out why **A** is there. Is the key correct or does it need correction for **A**?
[](https://i.stack.imgur.com/N3qCq.jpg)
I thought of the above mechanism for reaching the answer. Is this reasoning alright? Please approve it, whether it is right or wrong.
| 3
|
[] |
https://chemistry.stackexchange.com/questions/167207/issues-with-chlorine-liquefaction
|
Issues with Chlorine Liquefaction
|
A common problem that occurs when releasing chlorine from cylinders during industrial chlorination processes is that of liquefaction of the gas and ice deposition in and around the cylinders which slow the rate of withdrawal of chlorine from the cylinders. Thermodynamically, its a legit process when chlorine will undergo expansion, with the help of Joule Thompson coefficient, it will cool down.
My issue with the reasoning is that since the expansion is occurring at the valve itself and adiabatically, there shouldn't be cooling inside the cylinder yet I do have pictures of a cylinder with ice deposited all over its surface due to expansion via cooling. How do we explain the fact that the cooling effect is travelling back through the expansion nozzle and more importantly, how do we calculate the cooling rate and total energy removed from the cylinder via expansion through nozzle?
| 3
|
[
[
"\nYou have it backwards. The major part of the cooling occurs within the cylinder itself, with very little cooling (if any) occurring in the valve due to Joule Thomson.\n\n\nThe gas remaining within the cylinder at any time has expanded adiabatically and reversibly to do work in forcing the gas exiting the cylinder ahead of it into the valve. On the other hand, the gas passing through the valve is experiencing both expansion cooling and viscous frictional heating, so that the net effect is much smaller.\n\n\nTo calculate the cooling effect within the cylinder, one merely needs to apply the open system version of the 1st law of thermodynamics to this adiabatic expansion into the valve. For an ideal gas, this would read $$nC\\_vdT=RTdn$$ where n is the number of moles of gas within the cylinder at any time.\n\n\n",
"7"
],
[
"\nGases, that at room temperature condense before reaching sufficient storage pressure, are provided in the liquified form with gaseous headspace, what allows much better storage capacity.\n\n\nLiquid chlorine has the vapor pressure [$\\pu{5830 mm Hg}$ at $\\pu{25^{\\circ}C}$](https://webwiser.nlm.nih.gov/substance?substanceId=338&identifier=Chlorine&identifierType=name&menuItemId=42&catId=48), what is about $\\pu{7.7 atm}$.\nHaving cylinders with just few atmospheres of gaseous chlorine would not make much sense.\n\n\nAdiabatic or Joule-Thompson expansion cooling has minor impact here. Additionally, considering what the output chlorine pressure may be, there may not be much expansion at all, depending on application.\n\n\nThe major effect is evaporation cooling of liquid chlorine, based on its evaporation enthalpy. The similar effect is observed on the outdoor camping LPG burners. When cooking, you can feel how cold the thin wall LPG cartriges become.\n\n\n* The [chlorine boiling point](https://en.wikipedia.org/wiki/Chlorine) is $\\pu{−34.04^{\\circ}C}$.\n* The [propane boiling point](https://en.wikipedia.org/wiki/Propane) is $\\pu{−42.25}$ to $\\pu{−42.04^{\\circ}C}$.\n\n\nTherefore chlorine is even more easily liquified than propane. With an intense output flow, the boiling chlorine may reach deep frost temperature.\n\n\nIf the pressure regulator has equal or lower temperature than the liquid chlorine - e.g. for the minor expansion effect, then the gaseous chlorine will condense there. Similarly as water vapor condenses on glasses, if you enter a vapor chamber in a spa, or if you enter a warm room from cold outside during winter. Mild heating of regulators may help.\n\n\nDew or frost deposition on cylinders indicates another issue, as low cylinder temperature significantly decreases the gas pressure. (LPG cartridges are often kept in warm sleeping bags overnight, as butane boiling point is near $\\pu{0^{\\circ}C}$.) It would help if there was applicable some technical way to keep cylinders at room temperature.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/167205/a-new-green-method-for-purifying-acetylene-with-the-help-of-zeolite-molecular-si
|
A new green method for purifying acetylene with the help of Zeolite molecular sieve
|
I'm investigating for a green method for purification of acetylene in order to reduce the environmental impacts in manufacturing plants. In the current method mercuric chloride is used in the purification material that is indeed harmful to the environment.
The main impurities consists of:
1- Phosphine (Maximum concentration must be 200 ppm)
2- Arsine,
3- Hydrogen sulfide, (Maximum concentration must be 100 ppm)
4- Ammonia
5- water.
Recently, I read that Zeolite molecular sieves are used to purify many gases in industry. Further information can be found in the article below:
<https://www.honeywell-uop.cn/wp-content/uploads/2011/07/UOP-Adsorbents-for-purification-of-olefin-and-polymer-process-streams-brochure.pdf>
If I noticed correctly, this porous material absorbed different gases with respect to its amount of pore diameter. Gases that their molecular sizes are similar or smaller than the pore diameter will be absorbed while other gases will pass from this material. I wonder that if we could use this material for purifying acetylene.
Since the size of the gas molecules in the acetylene gas mixture are different, this seems practical.
What is your suggestion?
Do you think that Zeolite molecular sieves can be a solution for purifying acetylene?
Do you know any papers or articles that is relevant to this subject?
Thank you very much.
| 3
|
[
[
"\nWith respect to \"What is your suggestion?\", I note that apparently hydrogen peroxide does not readily react with acetylene, [see comments here](https://patents.google.com/patent/CN103693622A/en), for example, to quote:\n\n\n\n> \n> A hydrogen peroxide (H2O2) solution reacts with calcium carbide (CaC2), so as to generate the mixed gas of oxygen (O2) and acetylene (C2H2).\n> \n> \n> \n\n\nHowever, the created mix of acetylene and hydrogen peroxide appears per a recent 2018 work to present an enhanced explosion hazard (see [\"Water Vapor and Hydrogen Peroxide as Promoters of Acetylene Explosive Decay\"](https://link.springer.com/article/10.1134/S19907931180402310)), where \"the acetylene explosiveness is largely determined by humidity\" implying the further employment of a drying agent.\n\n\nAlso, on phosphine with hydrogen peroxide, to quote [a source](https://pubag.nal.usda.gov/catalog/666980):\n\n\n\n> \n> Reactive oxygen species for lipid peroxidation may therefore be derived from direct reaction of PH3 with H2O2 as an alternative hypothesis to their respiration-linked formation.\n> \n> \n> \n\n\nAlso, $\\ce{H2S}$ and $\\ce{AsH3}$ both react with $\\ce{H2O2}$ see, for example, [\"Catalyzed oxidation of arsenic(III) by hydrogen peroxide on the surface of ferrihydrite: an in situ ATR FTIR study\"](https://pubmed.ncbi.nlm.nih.gov/12666928/).\n\n\nAs such, I would suggest a possible hydrogen peroxide scrubbing step, per the manner described in the literature, as a feasible green but apparently, not completely safe, route.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/167197/are-the-np-orbitals-of-light-group-2-elements-considered-as-valence-electrons-in
|
Are the np orbitals of light group 2 elements considered as valence electrons in basis sets?
|
[An old mail archive](https://www.mail-archive.com/[email protected]/msg00252.html) says that, in the usually used basis sets, polarisation functions are treated by a single-Gaussian treatment even for valence double-zeta basis sets.
A priori, I've done a few NBO calculations on some simple magnesium compounds, say the cubic magnesium oxide tetramer anchored to the sum of ionic radii, on Gaussian.
The results seem to indicate that the 3p electrons on magnesium are treated as Rydberg, i.e. polarisation, functions on that package. This effect exists on Q-Chem as well, leading to my general conclusion that the np orbitals of lighter group 2 elements beryllium and magnesium are generally treated in the usual basis sets as polarisation functions that are not necessarily accurate and represented by a single Gaussian function, no matter the rest of the basis set's accuracy(i.e. the number of zetas).
However, there do exist circumstances where e.g. a 2p orbital of beryllium act as "legitimate" valence orbitals([such as these cases](https://www.nature.com/articles/nchem.2542)), in which case the treatment of the 2p orbitals of beryllium as Rydberg orbitals should not lead to accurate results. The last paper is behind a paywall that my institution is not affiliated with so I could not check if the authors of the article used "specialised" basis sets(no prizes for guessing out the meaning of the word "specialised" here); even if they did, these basis sets would likely be not directly accessible in the "vanilla" versions of the commonly used quantum chemistry packages and are thus hard to actually use.
My question now follows- do the basis sets included in (some or all) the "vanilla" versions of the commonly used quantum chemistry packages treat the np functions of the two lighter group 2 elements by the same way as they treat the rest of the valence(e.g. by a double-zeta for def2-SVP and by a triple-zeta for 6-311G(d))?; if not, are there any basis sets that do treat them as such available over-the-counter, should that be the right word here, in the literature?
P.S. My question also holds for the np functions of transition metals, which Gaussian treats as valence but Q-Chem treats as polarisation. To quote the quotes on Wikipedia, "Fe(−4), Ru(−4), and Os(−4) have been observed in metal-rich compounds containing octahedral complexes [MIn6−xSnx]; Pt(−3) (as a dimeric anion [Pt–Pt]6−), Cu(−2), Zn(−2), Ag(−2), Cd(−2), Au(−2), and Hg(−2) have been observed (as dimeric and monomeric anions; dimeric ions were initially reported to be [T–T]2− for Zn, Cd, Hg, but later shown to be [T–T]4− for all these elements) in La2Pt2In, La2Cu2In, Ca5Au3, Ca5Ag3, Ca5Hg3, Sr5Cd3, Ca5Zn3(structure (AE2+)5(T–T)4−T2−⋅4e−), Yb3Ag2, Ca5Au4, and Ca3Hg2; Au(–3) has been observed in ScAuSn and in other 18-electron half-Heusler compounds. See Changhoon Lee; Myung-Hwan Whangbo (2008). "Late transition metal anions acting as p-metal elements". Solid State Sciences. 10 (4): 444–449. Bibcode:2008SSSci..10..444K. doi:10.1016/j.solidstatesciences.2007.12.001. and Changhoon Lee; Myung-Hwan Whangbo; Jürgen Köhler (2010). "Analysis of Electronic Structures and Chemical Bonding of Metal-rich Compounds. 2. Presence of Dimer (T–T)4– and Isolated T2– Anions in the Polar Intermetallic Cr5B3-Type Compounds AE5T3 (AE = Ca, Sr; T = Au, Ag, Hg, Cd, Zn)". Zeitschrift für Anorganische und Allgemeine Chemie. 636 (1): 36–40. doi:10.1002/zaac.200900421.", surely some of these entities, such as Ag(-II) and Au(-II), indicate meaningful participation of the np orbitals of the transition metals silver and gold as "genuine" valence orbitals.
| 0
|
[
[
"\nIt is certainly commonplace for orbitals outside the usual valence set to be included in molecular orbital calculations these days. One of the most famous examples actually involves calcium as the alkaline earth metal, with $3d$ as an \"extra\" subshell [[1]](https://doi.org/10.1021/ja808524y). In the \"inverse sandwich\" complex $\\ce{[(thf)3Ca\\{μ-C6H3-1,3,5-Ph3\\}Ca(thf)3]}$, calcium $3d$ orbitals with the correct symmetry are held to overlap with otherwise antibonding orbitals of the organic ligand surrounding the dicalcium core, thus stabilizing these orbitals through forming calcium-carbon bonds. This enables electron transfer into these orbitals and the resultant emergence of calcium(I) in the core.\n\n\n**Reference**\n\n\n1. Sven Krieck, Helmar Görls, Lian Yu, Markus Reiher, and Matthias Westerhausen (2009). \"Stable 'Inverse' Sandwich Complex with Unprecedented Organocalcium(I): Crystal Structures of [(thf)2Mg(Br)-C6H2-2,4,6-Ph3] and [(thf)3Ca{μ-C6H3-1,3,5-Ph3}Ca(thf)3]\".\n*J. Am. Chem. Soc.* **131**, 8, 2977–2985. <https://doi.org/10.1021/ja808524y>\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/167195/additives-to-increase-water-density-and-solubility
|
Additives to increase water density and solubility [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/167195/edit).
Closed last year.
[Improve this question](/posts/167195/edit)
What additives do I need to add to a water-based drink (coffee or tea for example), to increase its density and to make the drink insoluble (like olive oil for example) in salt sea water or inland water, and the drink with additives is still consumable?
A little bit like some bar-cocktails with layered liquids. The idea is to serve and drink tea or coffee under water, whilst scubaing with students or tourists, or to kill boredom during mandatory decos. Not sure why I can't find such an answer elsewhere on the internet, given how cool the demand might be..
| -3
|
[
[
"\nYou can make a drink thicker with [food-safe thickening additives](https://opentextbc.ca/ingredients/chapter/types-of-thickening-agents/), such as pre-gelatinized starches, gelatine, or agar(-agar). Make your beverage *just* thick enough to stay in the glass and resist mixing, and add food coloring, if desired.\n\n\nThere's no safe way to make a water-based drink **insoluble** in water, though. And this doesn't answer the question of **how safe it would be to try to drink while in SCUBA gear under water**. Consider, for example, that at 10 m depth, theoretically, you could exert a force of 2 atmospheres sucking on a straw in the concoction -- likely to pull it up the straw, and also likely to cause damage to the body. After all, if you're in decompression, and form a partial vacuum in your mouth, aren't you inviting air embolism?\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/167193/iupac-names-for-trihalomethanes
|
IUPAC Names for trihalomethanes
|
The drinking water section of the Massachusetts DEP mandates the electronic submission of analytical results. The compound Chlorodibromomethane must be entered when reporting just trihalomethanes and must be entered as Dibromochloromethane when reporting a full list of volatile organic compounds (EPA Method 525).
The IUPAC system says that the substituents should be alphabetized. I would believe the substituents to be alphabetized are bromine and chlorine and that the "Di" prefix is not considered in the alphabetization logic.
Can any cite a specific reference to the proper naming convention?
| 7
|
[
[
"\n[Here's the relevant rule.](https://iupac.qmul.ac.uk/BlueBook/P1.html#140501)\nP-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.\n\n\nYou are right, the di-, tri-, etc prefixes do not matter and only 'chloro' 'bromo' are considered.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/167189/osmotic-pressure-definition-clarification
|
Osmotic Pressure definition clarification
|
currently reading Legninger's Principles of Biochemistry (8th edition if you absolutely needed to know) and under "Osmotic Pressure" it gives the definition to be:
>
> Osmotic pressure, $\Pi$, [is] measured as the force necessary to resist water movement...
>
>
>
I find this way of thinking a little confusing, why should I care about *resisting* the water's force across a membrane? I want to know if it's equivalent for me to interpret this as the force the water EXERTS as it passes through a membrane. Learning from Mr. Isaac Newton, I would take this to be a safe assumption. Please let me know if I am wrong!
| 1
|
[
[
"\nRather as \"...the pressure necessary...\", not force.\n\n\nThe full speed osmosis means zero differential pressure on the membrane. The non zero water net flow is caused by differences of water molecule collision frequencies and therefore of water diffusion rates in both directions.\n\n\nApplying pressure ( externally or hydrostatically ) on the membrane from the side of the (more concentrated) solution compensates lower diffusion speed by counteracting pressure gradient flow.\n\n\nThe particular value of the counteracting pressure, that would cause the zero net water flow, is called the osmotic pressure of the solution.\n\n\nIf the pressure is high enough, it may cause a rupture of the membrane. One of possible demonstrations can be putting a tomato in deionized water.\n\n\nOTOH, if no such pressure is applied and there is full osmotic water(or other solvent) flow, there is no unidirectional mechanical pressure on the membrane.\n\n\n",
"2"
],
[
"\nIn summary, osmosis is the flow of solvent across a semipermeable membrane from a region of lower to higher solute concentration. Thermodynamically the process is driven by the increase in entropy on mixing more solvent with the solute. In solution, the solvent passes both ways through the semipermeable membrane but the solute cannot. Osmosis is not restricted to solution and can occur in the gas phase.\n\n\nUsing the word 'force' is confusing as is 'resistance'. A pressure difference exists across the semi-permeable membrane. This is balanced by applying external pressure, such as from a vertical capillary tube into which the solution moves as its volume increases, and equilibrium thereby restored. The external pressure produced by gravity acting on the vertical column of solution is equal to the osmotic pressure. The thermodynamic description was first given by Gibbs in 1897 but this does not explain anything at a molecular level. In fact many textbooks avoid this aspect completely.\n\n\nOne experimental observation is that the flow due to osmosis is greater than that possible from diffusion and is due to hydrodynamic flow through the membrane. This would be damped out very quickly so there has to be a process that keeps this going. Kramers et al. (Kramers & Myers, Am. J. Phys. v80,p694 2012) suggest the following. The solute collides with the membrane and is repelled, its momentum (now away from the membrane) is then dispersed between the solvent molecules, these being in greatest numbers, and this produces an effective potential barrier to solvent flow which by Boltzmann distribution means that fewer solvent molecules enter the membrane from the solution side, i.e. the membrane partly repels solvent molecules. This does not happen on the pure solvent side so more solvent flows into the solution than flows out increasing the hydrostatic pressure in the solution.\n\n\nAnother explanation based on MD simulations (Lion et Al. J. Chem. Phys. v137, p244911, 2012) is that osmosis is due to the density imbalance of solvent particles across the membrane. They write\n\n\n'the density imbalance of solvent particles across the membrane is maintained by a balance between an outward force-driven flux (due to the higher total density in the solution) and an inward diffusive flux (due to the lower solvent density in the solution). We show that a simple calculation, based on the mechanics of hopping of solvent molecules across the membrane, can be used to derive the same relation between the solvent density gradient and the osmotic pressure as is obtained by standard thermodynamic arguments.'\n\n\nSo it seems that although osmosis is well studied its actual mechanism is still in dispute and much remains unresolved.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/167188/what-are-dinitrophenyl-dnp-derivatives
|
what are dinitrophenyl (DNP) derivatives?
|
what are dinitrophenyl (DNP) derivatives? Reading "The Amino-acid Sequence in the Phenylalanyl Chain of Insulin" paper by Sanger (1951). I am not a Biochem major.. yet
| 1
|
[] |
https://chemistry.stackexchange.com/questions/167079/definition-of-mineralisation-in-context-of-organic-chemistry
|
Definition of mineralisation in context of organic chemistry
|
What does it mean when an organic compound, say [methylene blue](https://en.wikipedia.org/wiki/Methylene_blue), is said to be *mineralised*? An accompanied equation shows the dye being converted into carbon dioxide and oxygen.
The only results I have found on Google are about soil chemistry.
| 3
|
[
[
"\nSimply put: organic → inorganic. Mineralization is often also called *[TOC](https://en.wikipedia.org/wiki/Total_organic_carbon) diminution* (the term “*TOC reduction*” is purposely avoided due to ambiguity in chemical context). From Duffus et al. [[1, p. 1263](https://doi.org/10.1351/pac200779071153)]:\n\n\n\n> \n> **mineralization** \n> \n> Complete conversion of organic substances to inorganic derivatives, often visible as microscopic deposits which may be associated with damage to soft tissue (e.g., in the kidney).\n> \n> \n> \n\n\nRelated entries from [[1](https://doi.org/10.1351/pac200779071153)]:\n\n\n\n> \n> **biomineralization** \n> \n> Complete conversion of organic substances to inorganic derivatives by living organisms, especially microorganisms.\n> \n> \n> \n\n\n\n> \n> **calcification** \n> \n> Form of mineralization in which organic tissue becomes hardened by deposition of calcium salts within its substance.\n> \n> \n> \n\n\n\n> \n> **complete mineralization** \n> \n> Complete breakdown of a complex organic compound to carbon dioxide, water, oxides, and oxidative inorganic products such as nitrate or sulfate.\n> \n> \n> \n\n\n### Reference\n\n\n1. Duffus, J. H.; Nordberg, M.; Templeton, D. M. Glossary of Terms Used in Toxicology, 2nd Edition (IUPAC Recommendations 2007). *Pure Appl. Chem.* **2007**, 79 (7). DOI: [10.1351/pac200779071153](https://doi.org/10.1351/pac200779071153). (Free Access)\n\n\n",
"8"
]
] |
https://chemistry.stackexchange.com/questions/167077/difference-between-change-of-enthalpy-of-neutralization-between-copper-and-manga
|
Difference between change of enthalpy of neutralization between copper and manganese carbonate
|
I was doing a calorimetry experiment to calculate the change in enthalpy during neutralization reactions between metal carbonates and hydrochloric acid. I experimented with 5 metal carbonates: magnesium, zinc, copper(II), manganese(II), and calcium.
I noticed that when I reacted $\ce{MnCO3}$ with $\ce{HCl}$ and $\ce{CuCO3}$ with $\ce{HCl}$, the temperature of the solution increased unusually higher than magnesium, zinc, and calcium carbonates. After calculating the enthalpy change per mole of $\ce{MnO}$ and $\ce{CuCO3}$ when reacted with $\ce{HCl}$, it was unusually higher. For the enthalpy change in neutralization for magnesium, I obtained a value of $\pu{-45.0 kJ/mol}$ while I obtained a value of $\pu{-377 kJ/mol}$ for copper(II) carbonate and $\pu{-240 kJ/mol}$ for manganese(II) carbonate. I have double-checked my stoichiometry calculations and no mistakes were found. Could anyone please explain the large difference in change of enthalpy for copper(II) and manganese(II) carbonate?
| 2
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[] |
https://chemistry.stackexchange.com/questions/167076/why-bromoacetone-is-more-reactive-towards-sn2-than-alkyl-bromides
|
Why are alpha carbonyl halides most reactive towards SN2 reactions? [duplicate]
|
**This question already has answers here**:
[How are SN2 transition states stabilised by adjacent double bonds and carbonyl groups?](/questions/48872/how-are-sn2-transition-states-stabilised-by-adjacent-double-bonds-and-carbonyl-g)
(3 answers)
Closed 5 years ago.
Relative rates of some compounds towards SN2 is given below
[](https://i.stack.imgur.com/WgRon.jpg)
[](https://i.stack.imgur.com/YaqJM.jpg)
The reason for reactivity of alpha carbonyl halides is given as
[](https://i.stack.imgur.com/Zz7su.jpg)
It’s not clear what exactly is happening.
| 12
|
[
[
"\n[I have asked a question previously](https://chemistry.stackexchange.com/q/40084) which concerns the same electrophilic system with the exception that nucleophilic attack happens on the carbonyl carbon and not on the α-carbon.\n\n\n\n> \n> From the organic chemistry textbook by [Clayden, Warren, Wothers and Greeves](https://rads.stackoverflow.com/amzn/click/com/0198503466) on pp. 890-891:\n> \n> \n> \n> > \n> > [](https://i.stack.imgur.com/yDeQN.png) \n> > \n> > The $\\pi^{\\*}(\\ce{C=O})$ and $\\sigma^{\\*}(\\ce{C-X})$ orbitals add together to form a new, lower-energy molecular orbital, more susceptible to nucleophilic attack. But, if $\\ce{X}$ is not a leaving group, attack on this orbital will result not in nucleophilic substitution but in addition to the carbonyl group. Again, this effect will operate only when the $\\ce{C-X}$ and $\\ce{C=O}$ bonds are perpendicular so that the orbitals align correctly.\n> > \n> > \n> > \n> \n> \n> \n\n\nThis interaction when considered in the context of an attack on the carbonyl is the principle of the polar Felkin-Anh model used to describe a certain diastereoselectivity. However it can likewise be used to explain the lowering of the $\\unicode{x3c3}^\\*(\\ce{C-Br})$ orbital which is attacked in a nucleophilic substitution reaction. It may seem as though these statements are contradictory, but the orbital interaction will create a three-centre π-type double-antibonding orbital with only the contribution on oxygen being neglegible due to oxygen’s electronegativity being highest and thus its contribution to high-energy orbitals being lowest.\n\n\nIt will depend on your nucleophile what will actually happen. It can attack both on the carbonyl carbon and on the α-carbon. If the attack on the carbonyl carbon is reversible, only the attack on the α carbon leading to an observed $\\mathrm{S\\_N2}$ process will be observed after the reaction — even if the carbonyl attack is faster. Only if the attack on the carbonyl is not reversible as is the case e.g. with Grignard reagents, a mixture of two products will be observed.\n\n\nNaturally, if $\\ce{X}$ is a very poor leaving group — e.g. $\\ce{OTBS}$ — the attack on the carbonyl will dominate this way or that.\n\n\n\n\n---\n\n\nTo sum up:\n\n\nThe reaction rate is increased because the $\\unicode{x3c3}^\\*(\\ce{C-X})$ orbital and the $\\unicode{x3c0}^\\*(\\ce{C=O})$ orbital linear combine to give a lower-energy orbital. Since this is the LUMO, the LUMO energy is lowered and nucleophilic attacks are facilitated.\n\n\n",
"10"
],
[
"\nMost of the time, a $\\text{S}\\_\\text{N}2$ reaction will involve a negatively charged nucleophile attacking an uncharged substrate (as is depicted in the second picture of the question). This negative charge can be stabilized effectively by the carbonyl group through conjugation. Note that in the example given in the first picture, the stabilization probably also involves the phenyl ring. Of course, the molecular structure must allow for alignment of the orbitals involved.\n\n\nThis mechanism is valid for most examples given in the table of the first picture bar the methoxy-methylchloride (MOM chloride, an old-fashioned protection agent for alcohols). In this case, I would think that the oxygen lowers the $\\sigma^\\ast$ orbital energy such that it may serve better as a point of attack.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/167075/oxidation-of-toluene-ethylbenzene-and-styrene-with-potassium-permanganate-in-ac
|
Oxidation of toluene, ethylbenzene and styrene with potassium permanganate in acidic medium
|
While I was studying my organic chemistry textbook, I came across the following reactions that need more clarifications:
[](https://i.stack.imgur.com/URhgy.png)
Oxidation of toluene (**1**) is the only reaction from the textbook, and I understand it.
I have a doubt about the product of ethylbenzene (**2**) oxidation. I have written the two products — benzoic acid (**A**) and 2‐phenylacetic acid (**B**) — which I'm not sure about. Does the ethyl substituent become a carboxylic or does it become a methyl carboxylic group?
For the oxidation of styrene (**3**) in addition to **A** and **B** I would propose one more product: styrene glycol (**C**) because I was taught that $\ce{KMnO4}$ oxidizes ethylene into diol. So, which of these three compounds will be the main product?
Lastly, are the products of reaction of ethylene with potassium permanganate in basic medium and in acidic medium different or the same? Please correct me if there are any errors.
| 2
|
[
[
"\nAfter asking the question, I did some research about these reactions, and I think it would be valuable if I share it with the community.\n\n\nethene becomes a diol in the presence of basic medium with KMnO4, not when there is an acidic medium. In an acidic medium, it oxidizes completely into a carboxylic acid.\n\n\nStyrene however, always oxidizes completely into benzoic acid. Same happens to any kind of alkyl (alkane, alkene, alkyne) that is combined with benzene.\n\n\nI think this happens because, benzene is so rich with electrons and very unsaturated, KMnO4 is really interested in oxidizing benzene, but as we all know benzene will never oxidize under these normal conditions. Instead KMnO4 tries to oxidize the Carbon atom nearest to the benzene, which is not a part of benzene but has substituted to it, and so, Benzoic acid is made.\n\n\nHope this helps someone who had the same doubts as me...\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/167069/histidine-boc-deprotectection
|
Histidine-Boc deprotectection
|
A question on Fmoc-solid phase peptide synthesis (SPPS). Is Boc-protected Histidine too labile? Merck's [website](https://www.merckmillipore.com/JP/ja/product/Fmoc-HisBoc-OH.CHA,MDA_CHEM-852052?ReferrerURL=https%3A%2F%2Fwww.google.com%2F) shows Boc is labile to treatment with Piperidine. Anyone with such experience?
| 3
|
[] |
https://chemistry.stackexchange.com/questions/167068/are-diffuse-functions-necessary-for-modeling-unbound-conjugated-anions
|
Are diffuse functions necessary for modeling unbound conjugated anions?
|
[According to this paper](https://www.google.com/url?sa=t&source=web&rct=j&url=https://citeseerx.ist.psu.edu/viewdoc/download%3Fdoi%3D10.1.1.494.9684%26rep%3Drep1%26type%3Dpdf&ved=2ahUKEwiQiM7nisD5AhVtmVYBHSszBzkQFnoECCUQAQ&usg=AOvVaw0UqGqgfUfUbiCxUHqODD82), and contrary to popular belief, diffuse functions are far from being necessary for the calculation of the electron affinities of the polycyclic aromatic hydrocarbon(PAH)s, due to delocalisation of the additional charge that overweighs the Coulomb repulsion between the additional electron and the existing electrons. The paper also says that electron affinities of individual atoms do need diffuse functions for their calculation, for the fact that atoms with known positive (i.e. bound) electron affinities could be misinterpeted as having negative (i.e. unbound) electron affinities when diffuse functions are not used. All was fine- or seemed to be.
After reading the paper, I suddenly realised that some highly conjugated neutral molecules are still known to have negative electron affinities- benzene and naphthalene to name a few. Delocalisation of the additional charge would not mandate diffuse functions, while the unbound nature of the radical anion would. So one now has a dilemma- which is generally more important, delocalisation or unboundness, for deciding whether to use diffuse functions or not?
| 1
|
[] |
https://chemistry.stackexchange.com/questions/167067/ozone-sensor-with-range-above-5ppm
|
Ozone sensor with range above 5ppm?
|
I am attempting to find or build a sensor for controlling an ozone generator. To do this, the sensor will need to see the range of at least 0 to 5% (50000ppm). The [commercially available sensors](https://www.digikey.com/en/products/detail/spec-sensors-llc/110-406/7689223) I have found so far are all in the 0 to 5ppm range; primarily used for air quality and safety. It seems like a sensor in this range should exist, but perhaps I am using the wrong vocabulary for searching.
This [website](https://www.aeroqual.com/blog/electrochemical-sensors-hmos) mentions electrochemical ozone sensors as well but looks like they only go up to 20ppm.
Ozone obviously [absorbs UV](https://www.researchgate.net/figure/Absorption-wavelength-of-ozone-135-Advantages-of-optical-absorption-spectroscopy-are_fig1_288835123), but I have not been able to identify this being used outside of a lab environment for sensing concentration. Here is an [expensive lab ozone sensor](https://www.ecosensors.com/product/uv-100-uv-ozone-analyzer-data-sheet/) that goes up to 999ppm. Might be able to build a sensor based on this principle, but maybe I am overlooking some other common atmosphere gas that would give false readings (water vapor, VOCs, etc).
Are these the only three technologies used for measuring ozone? Other creative methods?
| 2
|
[] |
https://chemistry.stackexchange.com/questions/167066/why-did-neutralization-of-weak-acid-calculation-differ-from-empirical-results
|
Why did neutralization of weak acid calculation differ from empirical results?
|
I recently stumbled upon the problem of needing to neutralize a volume of 0.33M acetic acid with 0.1M NaOH. I was curious how many mL of 0.1M NaOH I would need to bring 1mL of 0.33M Acetic Acid solution to a pH of 7.
For my calculations I used the following data:
Inputs
* Ka = 0.000018
* M of Acetic Acid 0.33
* M of NaOH 0.1
* Volume of Acetic Acid (mL): 1
Outputs:
$$
\sqrt{Ka\*0.33} = 0.002437 M of H+
$$
$$
0.002437 /1000mL \* 1mL = 0.000002 mol H+
$$
Volume needed to Neutralize
$$mol H+/M NaOH = 0.000024 L of NaOH$$
$$-> 0.024372 mL of NaOH$$
However, in practice this did not work. My empirical results ended up:
3.1 mL 0.1M NaOH neutralized 1mL 0.33M Acetic Acid
Why did this calculation differ from my observations? How reliable is a disassociation constant here?
| 1
|
[
[
"\nNo need to use dissociation constants.\n$\\pu{1 mL}$ of $\\ce{0.33 M}$ acetic acid contains $\\ce{3.3 10^{-4} mol}$ acid.\nTo neutralize this acid, $\\ce{3.3 10^{-4} mol NaOH}$ is needed, which is included in a volume $\\pu{V = n/c = \\frac{3.3·10^{-4} mol}{0.1 mol/L} = 3.3 10^{-3} L = 3.3 mL}$\n\n\nThis amount will not go exactly to $\\mathrm{p}H$ $7$, as the final $\\mathrm{p}H$ will be around $9$ . But you mentioned that you only want to **approach** neutral $\\mathrm{p}H$. The volume of $\\ce{NaOH}$ necessary to go from $\\mathrm{p}H$ $7$ to $\\mathrm{p}H$ $9$ is negligible in this problem.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/167065/rotten-chicken-smell
|
Rotten chicken smell
|
My boyfriend and I have a camper and he forgot to take a package of chicken out of the refrigerator before parking it. The chicken sat in the fridge which was off in the heat for a few months. We would like to pull the fridge out to dispose of it. However, many other systems run through the fridge. What chemicals will get rid of the awful smell or is it even possible? Thank you!
| 2
|
[] |
https://chemistry.stackexchange.com/questions/167061/is-formation-of-cyclic-carbonate-from-trimethyl-orthoformate-and-vicinal-aromati
|
Is formation of cyclic carbonate from trimethyl orthoformate and vicinal aromatic 1,2-diol possible? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/167061/edit).
Closed last year.
[Improve this question](/posts/167061/edit)
[](https://i.stack.imgur.com/7l2RD.png)
Hello, my question is, "is it possible cyclic carbonate formation between trimethyl orthoformate and vicinal aromatic 1,2-diol"
| 0
|
[
[
"\nTrimethyl orthoformate is at the wrong oxidation level to give a cyclic carbonate. You need phosgene or a dialkyl carbonate. The efficient preparation of catechol carbonate from catechol (benzene-1,2-diol) and dimethyl carbonate is described in this [paper](https://www.researchgate.net/publication/323752125_Process_Systems_for_the_Carbonate_Interchange_Reactions_of_DMC_and_Alcohols_Efficient_synthesis_of_Catechol_Carbonate)\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/167060/can-alkyl-halides-react-with-guanidine-guanidino-groups
|
Can alkyl halides react with guanidine/guanidino groups?
|
I wonder how the π delocalization of guanidino groups affects the reactivity of the terminal nitrogens with respect to alkyl halides compared to, say, primary amines?
| 2
|
[] |
https://chemistry.stackexchange.com/questions/167050/negative-or-fractional-order-of-reaction
|
Negative or fractional order of reaction
|
Let $\ce{A}$ be the reactant and $\ce{P}$ the product at the imaginary elementary reaction $\ce{A->P}$.
Could it possibly have a net total order of $-1?$
$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{A}]^{-1} \tag{1}$$
Or even more, a reaction whose rate depends on the concentration of the product as it follows:
$$-\frac{\mathrm d[\ce{A}]}{\mathrm dt} = k[\ce{P}] \tag{2}$$
Is there any example of an actual reaction that behaves like that?
On the other hand, can we think about non-integer orders of reaction like $1.5$, $0.5$, $-0.5$, $2.368$, $\mathrm e$ or $\pi$ for instance? Any examples?
| 3
|
[
[
"\nWikipedia has you covered for both parts of the question in their article on Rate equation.\n\n\nThey give an example of a [negative order of reaction](https://en.wikipedia.org/wiki/Rate_equation#Negative_order):\n\n\n$$\\ce{2 O3(g) -> 3 O2(g)}$$\n\n\nUnder certain conditions, this reaction is of -1 order in $[\\ce{O2}]$. This covers your question about a reaction being dependent on the product concentration (equation 2 in OP's question) as well as having a negative order (equation 1 in OP's question).\n\n\nThey also give an example of a [fractional order of reaction](https://en.wikipedia.org/wiki/Rate_equation#Fractional_order):\n\n\n$$\\ce{COCl2(g) -> CO(g) + Cl2(g)}$$\n\n\nThis has an order of 0.5 in chlorine.\n\n\n#### Elementary or one-step reaction?\n\n\nIf the reaction is elementary, the reaction orders are equal to the the stoichiometric coefficients of the reactants. I'm not sure what is meant with a one-step reaction. For the examples given in this answer, there are radical intermediates in multi-step mechanisms ([What does a fractional order of reaction mean for the mechanism?](https://chemistry.stackexchange.com/q/49088/72973)) that explain the unusual kinetic features.\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/167049/electronic-configuration-in-d-block
|
Electronic configuration in d-block
|
According to Bohr-Bury rule, 4f and 5d orbital come after 6s orbital. But if I search for the electronic configuration for Os or any atom in d block for that matter, they give the electronic configuration as 4f 5d 6s instead of 6s 4f 5d
Why does this happen ? why is S orbital written in the end even though it has less energy
Can someone please explain ?
| 1
|
[
[
"\nIt doesn't matter the way you write it. 4f 5d 6s is the same as 6s 4f 5d. The first one is ordered by energy and the second by the way the orbitals are filled. Even though, you must note that the 5d${^1}$ fills first in the element lanthanum [Xe] 6s2 4f0 5d1 ^[1] <https://en.wikipedia.org/wiki/Electron_configuration>\n\n\n",
"2"
],
[
"\nWriting 6s orbital at end of electronic configuration indicates that electron will first be removed from 6s orbital then 5d then 4f.\nd blocks are exceptional casses. That's why they behave like that.\n\n\nExample for scandium [Ar] 3d¹4s² indicates that electron will be removed from 4s subshell.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/167044/is-hydrochloric-acid-or-sulfuric-acid-more-corrosive-to-titanium
|
Is hydrochloric acid or sulfuric acid more corrosive to titanium?
|
Which acid is more corrosive to titanium at 20 to 30 °C and similar concentrations: hydrochloric acid $\ce{HCl}$ or sulfuric acid $\ce{H2SO4}$?
I found some info on [AZoMaterials — Titanium - Corrosion by Acids](https://www.azom.com/amp/article.aspx?ArticleID=1240), but it was not exactly apples to apples.
Also, if the acid concentrations are very low (say, less than 0.01%), which one will have worse long term effects on the metal?
| 3
|
[] |
https://chemistry.stackexchange.com/questions/167041/are-all-humectants-sticky
|
Are all humectants sticky?
|
While using a topical lotion based on [ethanol](https://en.wikipedia.org/wiki/Ethanol), water and [propylene glycol](https://en.wikipedia.org/wiki/Propylene_glycol), I noticed that it becomes sticky while drying. Propylene glycol is a [humectant](https://en.wikipedia.org/wiki/Humectant) and is used for this purpose in pharmaceuticals and cosmetics. This made me wonder: Is stickiness a general property of *all* humectants?
| 6
|
[] |
https://chemistry.stackexchange.com/questions/167040/how-to-determine-if-the-ring-compound-has-optical-isomers
|
How to determine if the ring compound has optical isomers?
|
The first column and fourth column of the table below says that the compound has the plane of symmetry and is not optically active. However, I noticed that each side has different arrangement of the H and Cl atoms. Does it still count as being symmetrical? [](https://i.stack.imgur.com/UHeAD.jpg)
| 0
|
[] |
https://chemistry.stackexchange.com/questions/167033/meaning-of-equivalents-in-organic-synthesis
|
Meaning of equivalents in organic synthesis
|
The use of "equivalents" as used in titrations/chemical analysis is an obsolete concept except in some countries. The original meaning of equivalent weights is the weight of a compound that combines with 16 g O or 1 g H and it was developed during the time when electrons did not "exist". With this definition it is easy to determine the equivalent weights, hence equivalents can be calculated directly or indirectly for most of inorganic simple compounds provided a chemical equation is available.
I still occassionally see equivalents in organic synthesis works. For example, one abstract has
*"Deprotonation of pyrazine, pyridazine, pyrimidine, and quinoxaline using an in situ mixture of ZnCl$\_2$·TMEDA (0.5 equiv) and LiTMP (1.5 equiv) was studied.*"
This is another paper title
"*Why Do Catalytic Quantities of Lewis Acid Generally Yield More Product than 1.1 Equiv in the Intramolecular Diels−Alder Reaction with a Furan Diene? "*
It appears that the organic chemist's equivalent is quite different from old-shool's analytical chemist's equivalent. I see this has been asked before, but the answers are not relevant to organic synthesis and it is hard to find a relevant definition in any chemical dictionary or even textbooks with *reference to organic compounds*. Is there a solid reference which defines how equivalents are defined/used in organic chemistry?
| 5
|
[
[
"\n\n> \n> Is there a solid reference which defines how equivalents are defined/used in organic chemistry?\n> \n> \n> \n\n\nI'm not sure about a reference *per se*; but from my (limited) experience as an organic chemist, the usage of 'equivalents' in this context invariably refers to molar equivalents. It's most certainly *not* based on the concept of 'equivalent weights' or normality, which as you noted is obsolete.\n\n\nTypically this is measured with respect to the substrate of the reaction, so in the case you quoted,\n\n\n\n> \n> Deprotonation of pyrazine, pyridazine, pyrimidine, and quinoxaline using an in situ mixture of ZnCl2·TMEDA (0.5 equiv) and LiTMP (1.5 equiv) was studied\n> \n> \n> \n\n\nwould refer to the addition of $\\pu{0.5 mol}$ of $\\ce{ZnCl2.TMEDA}$ and $\\pu{1.5 mol}$ of $\\ce{LiTMP}$ per $\\pu{1 mol}$ of pyrazine (for example).\n\n\nAs for the second quote, the authors are probably comparing the Diels–Alder reaction using catalytic Lewis acid versus (super)stoichiometric Lewis acid. In this case, the Lewis acid is also measured in terms of molar equivalent, relative to the substrate.\n\n\n",
"10"
]
] |
https://chemistry.stackexchange.com/questions/167032/how-to-identify-unknown-insoluble-solid-with-limited-resources
|
How to identify unknown, insoluble solid with limited resources?
|
Looking for suggestions on how to identify an unknown white solid we received as a donation. It was originally labeled as $\text{MgSO}\_4$, but clearly is not. We suspect it might be $\text{CaSO}\_4$. So far, I can say that it is practically insoluble in water. We are a DIYbio lab and so were aren't equipped for complex chemical analyses. We also don't have a fume hood so we can't utilize any reactions that might produce noxious gases. Any ideas would be greatly appreciated if only so I don't have to see the bag sitting on my desk anymore.
[](https://i.stack.imgur.com/QYgIQ.jpg)
| 2
|
[
[
"\nSince you don't know *what* it *might* be contaminated with, it might need to be disposed as *hazardous waste*. It could be contaminated with biological material, or even with radioactive wastes, which would not show on any but the most sensitive *chemical* assay.\n\n\nAs the old riddle goes:\n\n\nWhat do you call 1,000 liters of sewage mixed with 1 ml of *Chateau Mouton-Rothschild 1945*? Sewage.\n\n\nWhat do you call 1,000 liters of *Chateau Lafite 1869* mixed with 1 ml of sewage? It's still sewage.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/167027/solubility-comparison-for-para-nitrophenol-and-phenol
|
Solubility comparison for para-nitrophenol and phenol
|
During lecture with our chemistry teacher, he randomly asked us to compare solubility of phenol and p-nitrophenol, we expected p-nitrophenol to have greater solubility than phenol due to better dipole moment (5.43 D which is 1.70 D for phenol) and better hydrogen bonding but as it turns out phenol is way more soluble (84 g/L) than p-nitrophenol (10 g/L). Teacher was also confused, he didn’t resolve the problem that day and came to class next day saying he did some research and has found the reason but he hasn’t disclosed it yet and has asked us to do research ourself to get the reason.
I think the reason behind this abnormal behaviour is the strong attractions already existing between p-nitrophenol molecules is a lot more stronger than interactions between water molecules and interactions with water molecules are weaker, I am not sure of my answer and would like a better explanation / ideas to think about solubility. What do you guys think should be the reason for this abnormality.
| 2
|
[
[
"\nI asked my teacher and also this is what i suspected , para nitro phenol is less soluble in water than phenol as para nitro phenol is way more polar than water and so the hydrogen bonding and other interactions formed by interaction between para nitro phenol and water are weaker than formed by interaction between molecules of para nitro phenol because of huge polarity para nitro phenol as compared to water but as dipole moments of water and phenol are almost same so they bond better than being alone and hence making phenol more soluble .\n\n\n",
"-1"
]
] |
https://chemistry.stackexchange.com/questions/167024/how-does-adding-oxalic-acid-to-water-change-the-volume-of-the-solution
|
How does adding oxalic acid to water change the volume of the solution?
|
Does $\pu{87.4 mL}$ of $\ce{H2O}$ and $\pu{12.6 g}$ of $\ce{HOOC-COOH.2H2O}$ create $\pu{100 mL}$ of a solution?
So in our practical class our teacher wanted to make $\pu{100 mL}$ $\pu{1 M}$ aqueous solution of $\ce{HOOC-COOH.2H2O}$ [Oxalic Acid Dihydrate]. So first he measured $\pu{12.6g}$ $\pu{(0.1 mol)}$ of $\ce{HOOC-COOH.2H2O}$, then he measured $\pu{(100-12.6) mL} = \pu{87.4 mL}$ of $\ce{H2O}$ on a beaker. Then he mixed the $\pu{12.6g}$ $\pu{(0.1 mol)}$ of solute with the $\pu{87.4 mL}$ of solvent.
I have some confusion about this. Because usually for preparing such a solution I would first add $\pu{12.6 g}$ of solute on a beaker and then keep adding water until the volume reaches $\pu{100ml}$. But I have never seen such a process for creating a solution by measuring a specific volume of solvent and a specific volume of solute and mixing them to get a specific volume of solution. Like: $a$ volume of solvent + $b$ volume of solute = $a+b$ volume of solution.
Can someone explain the theory behind this process?
| -2
|
[
[
"\nThere are 2 **false** assumptions in the proposed procedure:\n\n\n1. Density of the solute is the same as density of water.\n\t* Most solids are significantly denser than water. Density of [oxalic acid dihydrate(OAD)](https://en.wikipedia.org/wiki/Oxalic_acid) is $\\pu{1.653 g cm-3}$\n2. The total volume does not change when the solute is being dissolved.\n\t* The total volume can shrink or expand. This factor is usually less significant than solute density, but e.g. mixing equal volumes of ethanol and water leads to about 5% volume contraction.\n\n\nThese assumptions may be acceptable if the desired molarity is intended to be just orientational and the volume deviation is acceptable.\n\n\nThe initial total volume of water + OAD, before dissolving, would be:\n$$V\\_\\mathrm{tot} = \\pu{87.4 mL} + \\frac{\\pu{12.6 g} }{ \\pu{1.653 g mL-1}} = 9\\pu{5.0 mL}$$\n\n\nGenerally, there is no way to predict the final solution volume (or total volume change) with the given solute mass and solvent volume without knowledge of density of the final solution. If such data is not available, the initial volume can be taken as approximate estimation.\n\n\nFor solution of the given molarity, the standard procedure is proper solute dissolving in such a solvent volume that total volume is somewhat less than final total volume. When dissolved, the volume is then topped to the desired nominal value by the solvent.\n\n\nAlternatives need to know the final solution density:\n\n\n* The procedure calculates the needed solvent volume (or mass) for the target nominal volume.\n* The procedure calculates the solute mass for given solvent volume, leading to not a well rounded final volume.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/167015/does-nitrous-oxide-decolorize-potassium-permanganate
|
Does nitrous oxide decolorize potassium permanganate?
|
My teacher says that nitrous oxide can't form any complex with manganese in aqueous solution, so it can't be oxidised. However, shouldn't it get oxidised as it is present in the intermediate oxidation state of nitrogen?
| 2
|
[
[
"\nYour teacher is partially right. Potassium permanganate solution is used for cleaning/scrubbing nitrous oxide for medical uses. If the nitrous oxide contains nitrogen oxide, NO, as an impurity it will certainly decolorize the solution but if pure nitrous oxide is being bubbled through potassium permanganate solution then the color will stay as it is. This is one of the ways to do chemical analysis of this gas. It is not that nitrous oxide cannot be oxidized or reduced under any conditions because it is an intermediate oxidation state. This is a fallacy and there are no free manganese ions in potassium permanganate.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/167014/how-is-a-barrierless-reaction-confirmed-in-quantum-chemistry
|
How is a Barrierless Reaction Confirmed in Quantum Chemistry?
|
**Background**:
Some quantum chemistry papers explore potential energy surfaces by characterizing critical points with an *ab initio* method. Reactants, products, intermediates, and transition states are found through optimization calculations, and then confirmed using frequency calculations. Intermediates connected by a transition state can be confirmed by intrinsic reaction coordinate (IRC) or some other minimal energy path (MEP) calculation. Here's an example potential energy surface from a paper online:
**Question**:
For any given intermediate and reactant/product pair, how does one confirm (1)
a barrierless connection or (2) a lack of connection? A connection with a barrier is simple: a transition state must exist and connect back to the two species. That's like TS3 connecting H + SO$\_2$ and HSO$\_2$ above. But it's not clear how something like SO--HO and OH + SO can be confirmed, or the lack of connection between HOSO and OH + SO can be confirmed.
**My thoughts**:
(1) Intuitively, an optimization that does very small steps and always goes downhill from the reactant/product to the intermediate of interest may be sufficient to confirm the barrierless connection. At the same time, this feels like a bit of a slippery slope.. If there is a downhill route going from OH+SO to TS', then this could be used to confirm that OH+SO can barrierlessly form HOSO (see crude cyan-drawn path below).
[](https://i.stack.imgur.com/0cfdc.png)
(2) Intuitively, confirming a lack of connection would require some sort of constrained optimization starting from the intermediate which shows that all possible degrees of freedom lead to other intermediates or products.
((Here, I am just assuming we stay on a single, simple, adiabatic potential energy surface. Here is a similar question with smaller scope: [What is a barrier-less reaction in Quantum Chemistry?](https://chemistry.stackexchange.com/questions/24826/what-is-a-barrier-less-reaction-in-quantum-chemistry/104119#104119)))
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[
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"\nThe absence of TS is sufficient to confirm a barrierless path. On the scheme the transition between SO...OH and SO + OH is barrierless because SO+OH is not a minimum on PES. This configuration lies on the slope near the SO...OH minimum or corresponds to a big plate area on the PES\\*, depending on particular orientation and distance between two molecules, so there is no saddle point between them.\n\n\n\\*because there are a lot of possible configurations for SO+OH system with relatively equal energies. Slope corresponds to interacting SO+OH (Van der Waals), plate area corresponds to nearly not interacting molecules\n\n\nIn my opinion it is not correct to consider any existing path as locked on the one side, a system just may have lower probability to follow it starting from a \"bad\" side. This situation may correspond to a high barrier or very narrow path which is not good because it requires concentrating energy only on (for example) one vibrational degree of freedom. If I understood correctly, here transition SO+OH -> HOSO considered as locked, which in my opinion is not correct, I see two possibilities for this transformation:\n\n\n1. stage by stage go backward from SO+OH (slope or plate area) -> SO..HO (minimum) -> TS' (saddle point) -> HOSO (minimum), with barrier because we have TS on this path\n2. directly from SO+OH to HOSO, which in my opinion should correspond to narrow path case because requires quick, relatively hard collision O atom of OH to S atom of SO, because molecules prefer another orientation (partial charges are bigger on H (OH) and O(SO) and corresponding molecular orbitals’ energies should be closer), so this situation may be possible only if two molecules directly collide and rotation of molecules is freezed (which doesn't seems to be highly probable).\n\n\nP.S.\nI may not be right, I will be glad to discuss my point of view\n\n\n",
"2"
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https://chemistry.stackexchange.com/questions/167010/what-do-the-positive-and-negative-signs-on-the-atomic-orbitals-diagram-imply
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What do the positive and negative signs on the atomic orbitals diagram imply? [closed]
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**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed last year.
[Improve this question](/posts/167010/edit)
I was reading up on wave functions and electrons behaving like a wave when they are bound in an atom. In the orbital diagram, the + and - signs don't make any sense to me, and the reasoning that one side of the nucleus taken as positive displacement is probably wrong. Can someone explain it?
| -2
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[
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"\nThe sign on [atomic orbital](https://en.wikipedia.org/wiki/Atomic_orbital) diagram represents the sign of the respective [wave function,](https://en.wikipedia.org/wiki/Wave_function) (a solution of the [Schroedinger equation](https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation)). It is analogical to sign of sin(x) function with positive and negative sections.\n\n\nThere are 2 major implications:\n\n\n* For adjacent atoms, overlapping of orbital parts with the same/opposite sign causes constructive/destructive wave interference, known from classical physics. 2 electrons with the same quantum numbers in the same atom would lead to their wave function mutual cancellation and zero probability.\n* Regions of the given orbital with the opposite signs are separated by 3D \"node surfaces\" with $|\\Psi|^2=0$ and zero probability density of electron occurrence.\n\n\nAn orbital with the main quantum number n has n-1 of such node surfaces. E.g. the orbital 3s contains 2 inner spherical 3D surfaces with zero probability of the electron occurance, having 3 radial probability maxima.\n\n\n",
"3"
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https://chemistry.stackexchange.com/questions/167004/is-anodizing-of-mixed-metals-aluminium-and-cooper-possible
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Is anodizing of mixed metals (aluminium and cooper) possible?
|
I'd like to blacken my [Noctua NH-P1](https://noctua.at/en/nh-p1) passive cooler, made from an aluminium fins and copper heatpipes. Since thermal dissipation is key in my build, I thought about anodizing it, instead of spray paint coating.
The tutorial for aluminium anodization I found on [Wiki-How](https://www.wikihow.com/Anodize-Aluminum) looks quite straight forward, but since my knowledge in chemistry is quite basic, I do not know if and how the heatpipes would interfere the process.
In my case, **not** coloring the cooper at all, would not be a problem, but damaging the heatpipes (which are not solid, but filled with some liquid for ensuring an even better heat transfer) or breaking the soldering between the heatpipes and the cooling fins or even risking to do so, would be a total deal-breaker.
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[
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"\nThere are two separate ways to color copper and aluminum chemically.\n\n\nAluminum and titanium are easily anodized in an acidic water bath. Once anodized, the surface aluminum oxide becomes porous and can be [dyed any color, or it can be colored by interference of light through the anodizing process](https://www.saf.com/faqs/how-do-you-color-aluminum/).\n\n\nPure [copper cannot be anodized](https://sciencing.com/methods-plating-stainless-steel-8741267.html), though some Al-Cu alloys can be. Instead, copper can be [colored by oxidizing (torching) in air to produce interference colors](https://www.youtube.com/watch?v=nQJ_Edt7ORw). However, unlike the *very* hard aluminum oxide ($\\ce{Al2O3}$) coating of anodized aluminum, copper oxides are soft and absorbent, so that touching the unsealed surface leaves fingerprints. The video link also describes sealing the oxidized surface.\n\n\nCopper can also be [colorized through chemical baths and other processes](https://www.getty.edu/publications/virtuallibrary/temp/9780892366385.pdf). Sulfides can produce browns and blacks, for example. Note that some of the chemicals used, such as [\"liver of sulfur\"](https://en.wikipedia.org/wiki/Liver_of_sulfur), are toxic and are not suggested for use at home. Again, a coating to protect the surface is needed.\n\n\nSince this is to be done to a heat transfer device, though, consider that any coating will reduce cooling efficiency. The thicker coating needed on copper might be more problematic. If there is *sufficient extra cooling capacity*, though it might work.\n\n\n",
"2"
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] |
https://chemistry.stackexchange.com/questions/167001/comparing-basic-strength-of-toluidine-isomers
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Is steric inhibition of resonance or steric inhibition of protonation dominant in o-toluidine?
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I wish to know which effect out of steric inhibition of resonance (SIR) or steric inhibition of protonation (SIP) is dominant over the other when comparing basicities of o-toluidine and aniline:
[](https://i.stack.imgur.com/vR6o6.png) [](https://i.stack.imgur.com/Abu0O.png)
**Case 1**, using only SIR effect
We see that due the bulky methyl present on o-toluidine there is inhibition of resonance which increases the availability of nitrogen's lone pair. Hence, we can say o-toluidine is more basic than aniline (as nitrogen's lone pairs are in resonance with benzene).
**Case 2**, using only SIP effect
If we consider that methyl's bulkiness shields the lone pair of nitrogen from protonation in o-toluidine and no such inhibition or shielding happens in aniline, we can conclude that aniline is more basic than o-toluidine.
Both these cases are contradictory. Can anyone please resolve the discrepancy?
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[
[
"\nThe reason for the decreased basicity is essentially the ortho-effect as outlined here: [Ortho-effect in substituted aromatic acids and bases](https://chemistry.stackexchange.com/q/7683/4945).\n\n\nYour question on the other hand asks about a seeming contradiction, but I believe there is a small misconception on your part.\n\n\n\n> \n> We see that due the bulky methyl present on o-toluidine there is inhibition of resonance which increases the availability of nitrogen's lone pair. [...]\n> \n> \n> \n\n\nI don't see how resonance could be inhibited, as that essentially requires that the lone pair orbital of nitrogen would have to rotate significantly towards an in-plane arrangement. The methyl group is simply not big enough for that to happen, and if it were, the lone pair would more likely rotate toward the more shielded side, further reducing acidity.\n\n\nThe highest occupied molecular orbitals (HOMO) of the molecules show this quite nicely, as they are very, very similar.\n\n\n[](https://i.stack.imgur.com/Edv5e.jpg)\n[](https://i.stack.imgur.com/NE6yo.jpg)\n\n\nOn the DF-B97D3/def2-SVP level of theory the RMSD of the core structure (excluding the hydrogen and the methyl group) is negligible with only 0.03 Å.\n\n\nFrom this I would conclude that there is no steric inhibition of resonance.\n\n\nOn the other hand, steric inhibition of protonation is given in 2-methyl aniline, where it is obviously missing in aniline. \n\nAdditionally there is a stabilising non-covalent interaction between the methyl and amine group, which can be analysed with NCIPLOT, and then displayed. The green blob is weakly attractive, red is strongly repulsive, blue would indicate strongly attractive interactions.\n\n\n[](https://i.stack.imgur.com/Q8dZA.jpg)\n\n\n**TL;DR**: The dominant effect is steric inhibition of protonation, as steric inhibition of resonance is negligible (if possible at all).\n\n\n",
"11"
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[
"\nNitrogen’s lone pair can be in a number of different orbital types depending on the geometry of the protons around it. The most important are $\\mathrm{sp^3}$ and pure $\\mathrm p$. The interconversion between the two can lead to the substituents switching sides; therefore, this process is known as nitrogen inversion.\n\n\nNitrogen inversion means that the lone pair is accessible from both sides, so you need to block both sides sterically if you want to inhibit protonation.\n\n\nTo inhibit resonance sterically, introducing an inhibitor on one side is enough since the group must be perfectly in plane for resonance to be significant.\n\n\nThus, aniline is less basic than 2-methylaniline.\n\n\n",
"1"
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https://chemistry.stackexchange.com/questions/166995/if-i-knead-my-dough-long-and-fast-enough-can-it-be-cooked-itself
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If I knead my dough long and fast enough, can it be cooked itself?
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May be Seasoned Advice question, but I thought that it's more of a scientific question. Correct me if I was wrong.
Kneading generates friction between dough chemicals(gluten, starch, etc.), thus generating heat. That's why we constantly chill pastry dough because the heat originated from the dough can ruin butter-dough layers AFAIK.
So... will that heat eventually cook the dough? Will it be heated enough to cook itself?
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https://chemistry.stackexchange.com/questions/166993/why-is-the-standard-electrode-potential-positive-for-half-cells-that-are-easily
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Why is the standard electrode potential positive for half cells that are easily reduced?
|
I was studying electrochemistry from my [school textbook](https://ncert.nic.in/ncerts/l/lech103.pdf).
The cell potential in the book is defined as *the difference between the electrode potentials of the cathode and the anode.*$$E\_\text{cell}=E\_\text{right}-E\_\text{left}$$
We know that electrode potential of an individual half-cell can't have an absolute value and thus we measure it taking the standard hydrogen electrode ($\ce{Pt(s) | H2(g) | H+(aq)}$, SHE) in relation that is assigned a **zero** potential at all temperatures.
As the electrons flow from the anode to the cathode (higher potential to lower potential), the cathode should possess a lower electrostatic potential than the anode.
That means that the electrode potential of a half cell when measured against the SHE should always be **negative** if the reaction is feasible, i.e., electrons flow from SHE to the cathode. Thus, the more negative the value of the electrode potential of the cathode, the more electrons will from the SHE, and the more the tendency of the substance of the cathode to get reduced.
But the values of standard electrode potentials of different half cells in my textbook and [Wikipedia](https://en.wikipedia.org/wiki/Standard_electrode_potential_(data_page)) show that the electrode potentials of substances that get easily reduced is **positive**, e.g., $E^\ominus = 2.87$ V for $\ce{F\_2(g) +2e^-->2F^-}$. Isn't this the complete opposite?
Shouldn't the values of electrode potentials of substances that are easily reduced be **negative** as we just discussed? What am I missing here?
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[
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"\nA large number electrochemicals misconceptions will be solved if we take the electrode potential sign as the *electrostatic* sign of the galvanic cell with respect to the hydrogen electrode. Thus this *sign* will not be affected by how the half cell is written. This was the European understanding of the leading 20th century top-electrochemists. The Germans decided to set the hydrogen electrode at 0 V, the reference point. The Germans also decided to set the hydrogen electrode at 0 V, as the reference point. This is the take of many leading modern electrochemists including A.J. Bard.\n\n\nNow let us take an electrode such as copper. Its half-cell electrode potential is +0.34 V vs. SHE. Now interpret the sign this way: If we make a galvanic cell of copper and SHE, the copper electrode will be \\*positively charged\" = the cathode, and the hydrogen cell will behave as an anode.\n\n\nIn the same way, take the fluorine half-cell (hypothetical), and if we make a galvanic cell of fluorine and SHE, the fluorine electrode will be \\*positively charged\" = the cathode, because its half-cell electrode is + 2.87 V.\n\n\nIn a galvanic cell, the current is from **A**(node) -> **C**(athode) i.e., **A to C**.\n\n\n",
"1"
],
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"\nBlame Ben Franklin or physicists for not knowing what + and - are and having terms like current flow! We chemists know better and think only of electrons altho we do lapse into positive ion flow in Li ion batteries and in acids. No one is perfect.\n\n\nOxidation = Loss of electrons; Reduction = Gain of electrons\n\n\nOxidation is at the ANODE; Electrons LEAVE the anode into the wire. Anions move to the anode.\n\n\nReduction is at the CATHODE; Electrons enter the cathode from the wire. Cations move to the cathode.\n\n\nThose are the simple rules. Unfortunately the physicists still get a say! They defined volts backwards and we are stuck with it [but some references are unstuck so it still can be confusing, so be careful]. A positive voltage means a reaction proceeds as written and its deltaG is negative. Electrode potentials are written [usually] as reduction half reactions. To appease the physicists a positive voltage means the reaction proceeds as written; a negative voltage means it proceeds in reverse. These half reactions are for a reaction under standard conditions with a hydrogen electrode. To get any reaction simply reverse one of the half reactions with its sign and add the equations and the voltages, then correct for differences from standard conditions. If the voltage is positive the reaction goes as written; if negative reverse the equation and change the sign. [or else change the concentrations and the cell voltage. If the cell voltage is close to zero and the electrode reactions reversible the direction of reaction or the cell voltage is easily manipulated.]\n\n\nThe pertinent equation is: DeltaG = -EnF = -E[0]nF +RTlnQ Q is the reaction quotient with the form of the equilibrium constant Keq. At equilibrium deltaG = 0 so \n\nE[0]nF = RTlnKeq = -deltaG[0]. As can be seen all these + and - signs can get confusing. [LOOK THIS UP IN A PCHEM TEXT or two] [The font this site uses substitutes a lower case o for the numeral zero.]\n\n\nTo sum it up a positive cell voltage corresponds to an equilibrium constant greater than 1. and a negative deltaG. The reaction proceeds as written. How far it goes depends on the voltage and the amount of material [the size of the battery]. At equilibrium the cell voltage is zero and the various concentrations satisfy the equilibrium constant\n\n\nTo react lithium and fluorine: galvanic cell: Li = Li+ + e- E= +3v; F- = 1/F2 +e- E= -3. We have two reduction reactions and need one oxidation and one reduction so must reverse one. Lets do Lithium e-+ Li+ = Li E= -3v. Put the two half reactions together Li+ + F- = Li + 1/2F2 E = -6V. Wow it is going nowhere. So we reverse the reaction. Li + 1/2F2 = Li+ + F-; E = 6V! an explosion!! Not so Good so we set up a cell with an anode, Li, a wire to a motor then to a cathode [who knows what, it can't react with F2], surrounded with F2 and a magic electrolyte to carry the Li+ and F- ions formed remotely and desperately trying to get to the cathode and anode respectively. To reverse this a potential greater than 6V is applied in reverse. I do not think that this particular cell has been made not even by accident.\n\n\nWork out some simple cells such as a hydrogen ion concentration cell or electrolytic deposition of copper to get an idea what happens to cell voltages as concentrations change. If you get stuck on conventions or signs remember Li is the strongest reductant and F2 is [almost] the strongest oxidant use these two to determine if a different convention is being used for signs.\n\n\n",
"0"
],
[
"\nEasily reduced redox systems \"suck\" electrons from an electrode, charging it to the more positive equilibrium potential, at which the electron transfer rate is zero.\n\n\nAnd vice versa for easily oxidized redox systems: They are pushing electrons to an electrode, charging it to the more negative equilibrium potential.\n\n\nIn galvanic cells, cathodes have higher potential than anodes, in contrary to electrolytic cells. By other words, when switching the cell mode, electrode names switch places, as the current switches direction.\n\n\nE.g. Imagine the notoriously known [Daniell cell](https://en.wikipedia.org/wiki/Daniell_cell):\n\n\n$\\ce{Zn(s)|ZnSO4(aq)||CuSO4(aq)|Cu(s)}$\n\n\nIf we perform electrolysis on it, with the negative contact attached to $\\ce{Zn}$, the cathode is the left, more negative electrode, as electrons come to $\\ce{Zn}$ and reduce $\\ce{Zn^2+(aq)}$.\n\n\nIf we use it as the galvanic cell, the cathode is the right, more positive electrode, as electrons come to $\\ce{Cu}$ and reduce $\\ce{Cu^2+(aq)}$.\n\n\n\n\n---\n\n\nSHE has the [absolute alectrode potential](https://en.wikipedia.org/wiki/Absolute_electrode_potential) $\\pu{+4.44 \\pm 0.02 V}$.\n\n\nFor Ox/Red redox pair and a formal reaction $\\ce{Ox + n/2 H2 -> n H+ + Red}$, there is direct relation of the standard reaction Gibbs energy and the respective standard electrode potential $\\Delta\\_\\mathrm{r} G^{\\circ}=−nFE^{\\circ}$.\n\n\n",
"0"
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https://chemistry.stackexchange.com/questions/166991/what-will-be-the-product-of-1-eqv-benzene-reacts-with-1eqv-h2
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What will be the product of 1 eqv Benzene reacts with 1eqv H2? [closed]
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**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/166991/edit).
Closed last year.
This post was edited and submitted for review last year and failed to reopen the post:
>
> Original close reason(s) were not resolved
>
>
>
[Improve this question](/posts/166991/edit)
[](https://i.stack.imgur.com/nIbUd.jpg)
how equivalent concept is applied in the above problem, Kindly help me out with the answer and proper explanation.
As, I am getting confused in mole concept and equivalent concept here.
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"\n**The answer makes sense (and doesn't break the law of chemical equivalence) when you consider the reactivity and stability of the intermediates in the reaction**\n\n\nHydrogenation of benzene proceeds in several steps (which is intuitively obvious as a single hydrogen molecule can only add two hydrogens to a benzene ring).\n\n\nBut the intermediate that results from a single hydrogenation step, cyclohexadiene, is *more reactive* than benzene so the further hydrogenation happens faster than further hydrogenation of benzene. The same is true for the product of that reaction, cyclohexene. So, under simple hydrogenations where there is not enough hydrogen to fully hydrogenate the remaining benzene, you tend to get cyclohexane as a product and lots of benzene remains.\n\n\nAs described, the particular reaction shown only has enough hydrogen to turn 1/3 of the benzene into cyclohexane. Hence the mix of products: 2/3 benzene and 1/3 cyclohexane. And this, overall, obeys the law of chemical equivalence.\n\n\nIt is possible to use careful conditions and specific catalysts to change the mix of products. But it isn't easy. Benzene hydrogenation had been known for over 100 years before chemists observed cyclohexene as a product. The first industrial plant exploiting partial hydrogenation didn't open until 1990.\n\n\n",
"6"
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"\nHydrogenation of benzene requires high temperature and high pressure of hydrogen. An equimolar mix of benzene and hydrogen (the term \"*equivalent*\" is unfortunate here) with catalyst and appropriate temperature $T$ and pressure $p$ might form diene but as concentrations increased diene would react much faster than benzene ether by competing for remaining hydrogen or as a source of hydrogen to form the most stable product mix, cyclohexane and benzene.\n\n\nA catalyst will hasten equilibrium and give thermodynamic products unless specifically fitted to a reaction as in enzyme reactions. To form the intermediate alkenes the reaction conditions must be tailored (excess benzene) and the intermediates separated immediately from the reaction, perhaps a flow system allowing only transient contact with the catalyst and with a cold trap or trapping agent downstream, say a reversible Diels–Alder.\n\n\nThe question, as asked, does not arouse the student's curiosity about the chemistry and that is extremely unfortunate.\n\n\n",
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https://chemistry.stackexchange.com/questions/166989/does-water-expand-on-freezing-more-than-any-other-known-substance
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Does water expand on freezing more than any other known substance?
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Several substances expand on freezing. Gallium expands [3.1%](https://www.newworldencyclopedia.org/entry/Gallium) and bismuth [3.3%](https://en.wikipedia.org/wiki/Bismuth). However, both of these are much less than water which expands [9%](http://www.iapws.org/faq1/freeze.html). Is there any substance that expands more than 9% on freezing?
We can generalize this question to any system at thermodynamic equilibrium and constant pressure. Density normally decreases with temperature, but sometimes increases. For example, heating [plutonium](https://en.wikipedia.org/wiki/Plutonium) increases it's density in three places: a solid-solid phase transition, a negative thermal expansion, and melting. These effects combine to make the density increase about 5% between ~320C to 640C, which is still less than cooling 4C water to -0.01C ice.
If we allow substances that dissolve/precipitate and reversible chemical reactions can we beat water? We do not allow super-cooling or super-saturation since that isn't thermodynamic equilibrium.
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"\nSilicon beats it.\n\n\nSilicon has the familiar diamond-type semiconductor only when solid, and like the hydrogen-bonded structure of water ice the directional bonding adds volume to the solid phase. Melted silicon collapses to a structure more typical of a liquid metal, including its electrical conductivity.\n\n\n[Wikipedia](https://en.wikipedia.org/wiki/Silicon) reports the density of solid silicon at room temperature and the liquid at the melting point:\n\n\nSolid at room temperature: 2329 kg/m³\n\n\nLiquid at melting point: 2570 kg/m³\n\n\nThus the solid at room temperature has 10.3% more volume than the liquid at the melting point ... with the thermal contraction from melting point to room temperature included.\n\n\n",
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https://chemistry.stackexchange.com/questions/166988/comparing-the-enthalpy-change-between-the-formation-of-water-in-two-states
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Comparing the enthalpy change between the formation of water in two states
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>
> The enthalpy of the reaction $\ce{H2(g) + 1/2 O2 (g) -> H2O (g)}$ is $\Delta H\_1$ and that of $\ce{H2(g) + 1/2 O2 (g) -> H2O (l)}$ is $\Delta H\_2$. Then,
>
>
> a) $\Delta H\_1 < \Delta H\_2$
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> b) $\Delta H\_1 + \Delta H\_2 = 0$
>
> c) $\Delta H\_1 > \Delta H\_2$
>
> d) $\Delta H\_1 = \Delta H\_2$
>
>
>
According to me, in a reaction forming a gaseous product the pressure would be greater than the one forming a liquid product. Therefore the enthalpy change should be greater for the first reaction. However, the correct answer is a). Where am I going wrong?
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"\nWell in order to convert H2O into liquid we need to lower the energy of system more than H2O of gaseous state (energy of gas is more than liquids and liquids have energy more than solids).\n\n\nThat means more have to be released in order to make it liquid.\n\n\n",
"0"
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"\nSince enthalpy is Q at constant pressure the PV factor can be ignored [this is for the questioner]. The difference is the heat of condensation of the vapor that is released. Therefore the reaction to give liquid water is more negative than that to give gas. I looked it up the heat of formation of water vapor is -57.8 Kcal for liquid water -68.3. Now the problem is to define what is meant by larger? the less negative number or the greater release of heat. As usual test makers do not try to answer their own questions! [I notice that in the comments this dilemma was not resolved. altho Chet tried]. I personally deplore questions that make knowing sign conventions more important than knowing chemistry.\n\n\n",
"-1"
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https://chemistry.stackexchange.com/questions/166982/are-corrosive-and-reactive-synonymous
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Are corrosive and reactive synonymous?
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[This answer](https://chemistry.stackexchange.com/a/9534/402) on a question regarding whether water is corrosive in pure form, the author implies that "reactive" and "corrosive" are the same thing (at least with water.)
"Corrode" seems to mean irreparably damaged which also seems to imply a corrosive reaction is not reasonably reversible whereas "react" can mean several things such as the changing of an oxidation state, or bonding covalently with another element or molecule, to something as seemingly small as a hydrogen bond.
Are "corrode" and "react" truly synonymous or is corrosion just one of many types of reactions?
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"\nCorrosion has an economic and a negative connotation e.g., corrosion scientists are interested in protecting metals/non-metals exposed to the atmosphere to reduce major loses to the city & the government.\n\n\nReaction is a necessary process for corrosion, for instance, eating away or destruction of metals by oxidation. Hence, acids are corrosive to a large number of metals. Even in geology, corrosion implies destruction of rocks mainly by water. Those who live near coastal cities know how cement is corroded (eaten away) by salty water aerosol. There is no oxidation. All senses are negative.\n\n\nIn my humble opinion, if something is reactive, it does not necessarily have a negative tone. Water reacts with sodium, but we would not say, water corrodes sodium because an economic loss/negative tone is not implied. Sulfur reacts with copper and so on. I would not include hydrogen bonding under a chemical reaction but this is really splitting hairs just like chemical change vs. physical change enforced upon undergraduate students.\n\n\nIn short, water can be corrosive or reactive depending on the context by the speaker/author. Acids are corrosive to metals (negative, warning) or aqua regia reacts with most metals (the sense maybe good or bad).\n\n\n",
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https://chemistry.stackexchange.com/questions/166979/how-do-i-derive-metallic-aluminum-without-electricity
|
How do I derive metallic aluminum without electricity?
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I'm laying the foundation for a project called 21st Century steampunk, where I figure out what the world would look like if electricity was never discovered.
I'm wondering if it's possible to derive usable metallic aluminum from naturally occurring substances without using the Hall–Héroult process.
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[
[
"\nOne method, which would also require a non-electrical heating source, involves reduction with carbon. Given a high enough temperature -- meaning over 2000°C -- carbon carries off the oxygen as carbon monoxide and leaves the aluminum behind. See Ref. [[1](https://i.stack.imgur.com/jQNJi.jpg)](<https://doi.org/10.1016/j.energy.2007.06.002>), which includes the equilibrium composition calculation below.\n\n\n[](https://i.stack.imgur.com/jQNJi.jpg)\n\n\n**Reference**\n\n\n1. M.Halmann, A.Frei, A.Steinfeld (2007).\"Carbothermal reduction of alumina: Thermochemical equilibrium calculations and experimental investigation\". *Energy* **32**, Issue 12, December 2007, Pages 2420-2427. <https://doi.org/10.1016/j.energy.2007.06.002>\n\n\n",
"12"
],
[
"\nPrior to the electrolysis (Hall–Héroult) process, elemental aluminium was made by reducing aluminium chloride (AlCl3) with elemental sodium or potassium in a vacuum.\n\n\nSee the wikipedia article on \"History of aluminium\"\n\n\n",
"8"
]
] |
https://chemistry.stackexchange.com/questions/166975/finding-the-density
|
Finding the density
|
35.6g of NaOH is dissolved in 150g of 20% NaOH solution. What is the density od the solution?
I have tried solving this problem, but I 've come across a few obstacles.
Firstly, I found the mass of NaOH in the original solvent which is 30g. The added NaOH weighs 35.6g, meaning there is ultimately 75.6g of NaOH, and that the mass of the solution is 185.6g. The next step would be to find the density of the solution, but this requires the volume of the solution, that is not given in the problem. I also tried to calculate the number of moles to later calculate V using the formula V=n\*Vm, but I am not sure how to calculate the molar mass(M) of the solution.I am not sure if I am to use H2O or NaOH.
Can someone tell me if I am going in the right direction and if my calculations are even in the right direction, and if not, what is the right one?
| 1
|
[
[
"\nYou can simplify your problems by looking at the published tables. As you have been explained in the comments, predicting densities is not an easy job and I don't know if it can even be predicted.\n\n\nThis link provides the densities of NaOH solution as a function of mass percentage.\n<https://wissen.science-and-fun.de/chemistry/chemistry/density-tables/density-of-sodium-hydroxide/>\n\n\nPlot the data in Excel and generate a function. Calculate the final mass percentage of your solution.\n\n\nYour first step is correct, i.e., the initial solution contains 30 g of NaOH, then you add 35.6 g to the solution. The total mass of the solution is 185.6 as you correctly calculated. What is the final mass percentage? 65.6 g NaOH/ 185.6 g solution. This is around 35% wt/wt NaOH. Roughly, interpolating from the given table, the density should be around 1.39!\n\n\nCommas in German data are equivalent to decimals in English, so mass percentage of 5,5 means 5.5.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166972/why-does-the-1h-nmr-spectrum-of-methacryloyl-chloride-show-two-singlets-instead
|
Why does the 1H NMR spectrum of methacryloyl chloride show two singlets instead of two doublets for vinyl protons?
|
Based on what I have understood, for methacryloyl chloride there is a geminal coupling of the vinyl protons since they are not chemically equivalent. Why are two singlets observed for vinyl protons instead of two doublets?
[](https://i.stack.imgur.com/oYGdM.png)
| 5
|
[
[
"\nThe geminal $^2\\!J\\_\\ce{HH}$ coupling in terminal alkene $\\ce{CH2}$ can be pretty small; often on the order of $\\pu{2 Hz}$. The late Hans Reich's [NMR website](https://organicchemistrydata.org/hansreich/resources/nmr/?index=nmr_index%2F1H_coupling#h-coupling06) has a lot of data which you can check out.\n\n\n(And yes, this is in stark contrast to geminal couplings in diastereotopic, $\\mathrm{sp^3}$ $\\ce{CH2}$ groups, which often have large magnitudes!)\n\n\nSo, to observe it in the spectrum, you will need to increase the resolution. The easiest way would be to perform extra zero-filling and see if that helps. The next simplest way is to process the data using a different window function / apodisation function: to increase the resolution, you can use a Gaussian window function, see e.g.\n\n\n* [Why is the Gaussian Window Function used to enhance resolution in 1D NMR spectra?](https://chemistry.stackexchange.com/q/842/16683)\n* and some practical instructions at the bottom of this answer: [Multiplet shape in proton NMR of morpholines](https://chemistry.stackexchange.com/a/57401/16683)\n* and there are numerous resources online which you can find as well.\n\n\nIf none of that works, it's possible that you might need to rerun the spectrum and increase the acquisition time of the spectrum, but I'd consider that to be quite unlikely for a routine $\\ce{^1H}$ spectrum.\n\n\n[That said, this appears to be a spectrum you've taken from someone else's work, so... maybe all you can do in this situation is to take a magnifying glass, and hope that the spectrum was processed and the image created with sufficiently high resolution. The suggestions above obviously only make sense if you are the one who acquired the spectrum.]\n\n\n",
"7"
]
] |
https://chemistry.stackexchange.com/questions/166971/how-exactly-thermal-infrared-e-g-flir-cameras-image-methane-leaks-are-they-m
|
How exactly thermal infrared (e.g. FLIR) cameras image methane leaks? Are they measuring a temperature difference or an emissivity difference or...?
|
The story [Infrared camera shows the impact of vehicle emissions](http://www.vision-systems.com/articles/print/volume-22/issue-5/departments/snapshots/may-2017-snapshots-boston-dynamics-robot-autonomous-vehicles-infrared-imaging.html) says:
>
> In an effort to visually demonstrate the potentially fatal impacts of air pollution, FLIR Systems (Wilsonville, OR, USA; [www.flir.com](http://www.flir.com)) has released a new video containing footage captured by FLIR infrared cameras that shows vehicle emissions and the resulting pollution issues.
>
>
>
>
> [...] To accomplish this, FLIR used its GF320 infrared camera to capture video of the emissions. This camera features a 320 x 240 cooled InSb infrared detector with a spectral response of 3.2 to 3.4 μm. Designed specifically for gas leak detection and electrical inspection, the camera also embeds GPS data into the image, allowing workers to pinpoint the location of the leak or hot spot.
>
>
>
Wikipedia says [InSb is a narrow-gap semiconductor with an energy band gap of 0.17 eV at 300 K and 0.23 eV at 80 K](https://en.wikipedia.org/wiki/Indium_antimonide#Physical_properties) and a focal plane array can image light with a wavelengths in the 1 to 5 micron range. They are often cooled to reduce dark current for better sensitivity. In this case it sounds like they've added an interference filter to selectively pass 3.2 to 3.4 microns.
My question is about what it is they are really imaging here. It is true that gases such as CO2 and methane have strong IR bands, this is why they are important greenhouse gasses. Are they measuring a temperature difference or an emissivity difference or light scattering or absorption or...? How does this technique actually work?
* Youtube [Seeing Air Pollution with FLIR Thermal Imaging Cameras](https://www.youtube.com/watch?v=514Uh6Mq_Cs)
* Youtube [How Dying Gas Wells Are Making One Company Rich](https://youtu.be/6c-WCg2Y7sE?t=487) FLIR imaging of methane leaks from gas wells
[](https://i.stack.imgur.com/O73P5.jpg)
**above:** Image from [this article](http://www.vision-systems.com/articles/print/volume-22/issue-5/departments/snapshots/may-2017-snapshots-boston-dynamics-robot-autonomous-vehicles-infrared-imaging.html).
| 7
|
[
[
"\nShort answer\n============\n\n\nThe cameras you are asking about are called [optical gas imaging (OGI) cameras](https://www.flir.ca/products/gfx320), they are not [thermographic cameras](https://www.flir.ca/browse/professional-tools/thermography-cameras/) per se, but are a modified kind instead.\nHowever, both use the same principles to actually detect the infrared light, the key difference is the incident radiation is filtered to allow only the wavelengths that interact with the methane gas plume (or other analyte gas) to interact with the camera sensor.\nThermographic cameras could be configured to take both kinds of images since the principle of their operation is essentially the same, but they are usually separate devices for practical reasons like cost, increasing image contrast, *etc.*.\nIn the images you provided, you can see the solid objects because they emit, reflect, and absorb infrared light across the spectrum, so they can also be seen by the OGI cameras.\nHowever, other gases are not seen because the IR light they absorb is filtered out.\nFinally, a temperature difference must exist between the plume of gas being imaged and the background scenery because otherwise the amount of infrared radiation emitted will be equal to the amount absorbed by the gas, following [Kirchoff's Law of thermal radiation](https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation).\n\n\nDetailed Answer\n===============\n\n\nMethane detection colour analogy\n--------------------------------\n\n\nCameras for visible light discriminate between colours by placing a filter before the sensor so that only one colour (i.e., red, green, or blue) is absorbed by the sensor at any given pixel.\nThe sum of the RGB signal reproduces the subject's colour.\nAlthough we cannot see them with our eyes, infrared light has different \"colours\" that can be discriminated by a camera sensor with the appropriate absorption range, filters, *etc.*.\nVisible light has different colours depending on the energy (i.e., wavelength or frequency) of the light:\nred light has a longer wavelength (lower energy and frequency) than green, and blue, respectively.\nWe can see an apple is red when the green and blue light in white backlight are absorbed by its skin while the red light is reflected into our eyes (or camera).\nIn the same way, we can see the infrared \"colour\" of different objects by observing the infrared light they absorb.\n[](https://i.stack.imgur.com/jUeSM.png)\n**Figure 1** [FTIR spectrum of carbon dioxide from NIST.](https://webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Type=IR-SPEC&Index=1#IR-SPEC)\n[](https://i.stack.imgur.com/BRVIe.png)\n**Figure 2** [FTIR spectrum of methane from NIST](https://webbook.nist.gov/cgi/cbook.cgi?ID=C74828&Type=IR-SPEC&Index=1#IR-SPEC), notice the lack of overlap between this spectrum and that of carbon dioxide (above).\n\n\nEach arrangement of atoms in a molecule with different nuclear charges (i.e., polar bonds) absorbs a specific wavelength of light because the vibrations are quantum mechanical in nature.\n[These vibrations are characteristic to each molecule, so we can identify them by observing which wavelengths of light they absorb.](http://www2.ess.ucla.edu/%7Eschauble/molecular_vibrations.htm)\nSee the above two Fourier-transformed infrared (FTIR) spectra of carbon dioxide and methane.\nThis is the basis of spectroscopy, which is out of the scope of this answer.\nBut, if we designing a sensor that can discriminate between certain wavelengths of infrared light, a camera can be made that can see how much of a particular wavelength (e.g., 3.5 μm for methane, 10.57 μm for sulfur hexafluoride) is being absorbed relative to the background.\nCouple this principle to a system that can tell where the light is coming from, i.e. add pixels, add some optics to focus the light, and you have a camera.\n\n\nOptical gas imaging (OGI)\n-------------------------\n\n\nThe objects in the field of view (FOV) in OGI are illuminated by the background infrared radiation emitted from the sun and all other warm objects.\nThis is why OGI is also called a \"passive\" thermal imaging technique, in contrast to \"active\" techniques where the scene is illuminated with an artificial light source directed at the subject.\nOGI depicts the relative decrease (or increase) in the intensity $\\Delta I(\\lambda)$ at a certain wavelength $\\lambda$ of this background radiation $I\\_B(\\lambda)$ due to absorption (or emission) by the analyte gas with intensity $I\\_G(\\lambda)$.\nA general OGI equation (Eq. 1) for $\\Delta I$ can be [derived using a radiative transfer model](https://eyecgas.com/wp-content/uploads/2020/05/OGI-101-%E2%80%93-OGI-Equations.pdf):\n\n\n$\\Delta I(\\lambda) = I\\_b(\\lambda) - I\\_g(\\lambda) = [1 - \\tau\\_g(\\lambda)][B(T\\_b,\\lambda) - B(T\\_g,\\lambda)]\\quad(1)$\n\n\nwhere $\\tau\\_g$ is the transmission coefficients of the analyte gas, and $B(T\\_b,\\lambda)$ and $B(T\\_g,\\lambda)$ are the [Planck law functions for blackbody radiation](https://en.wikipedia.org/wiki/Planck%27s_law#The_law) of the background at temperature $T\\_b$ and the analyte gas at temperature $T\\_g$, respectively.\n\n\nWe can obtain the transmission coeffecient of the gas plume in relation to its composition, concentration, and geometry using [Beer's law](https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Spectroscopy/Electronic_Spectroscopy/Electronic_Spectroscopy_Basics/The_Beer-Lambert_Law) (Eq. 2)\n\n\n$\\tau\\_g(\\lambda) = \\exp[-\\alpha\\_g(\\lambda)\\bar{C}\\_g \\ell\\_g]\\quad\\quad\\quad (2)$\n\n\nwhere $\\alpha\\_g$ is the gas's absorption coeffecient, $\\bar{C}\\_g$ is the average concentration of the gas through the line of sight, and $\\ell\\_g$ is the breadth of the gas plume through the line of sight.\n\n\nNow we can see that the gas plume image obtained by OGI systems requires three things to be detectable:\n\n\n1. Absorption (or emission) of infrared radiation in a certain band of wavelengths ($\\lambda$);\n2. Plume that is big enough ($\\ell\\_g$) or concentration that is high enough ($\\bar{C}\\_g$) to absorb (or emit) enough infrared radiation to be detected; and\n3. Temperature difference $\\Delta T = T\\_b - T\\_g$ that is great enough between the background and the gas plume so they absorb (or emit) different amounts of infrared radiation (Eq. 3).\n\n\nEach of these three factors contribute to the overall contrast observed in the image between areas where there is a gas plume and where there is not.\nThe final point may be mysterious to the reader, but it is due to [Kirchhoff's law of thermal radiation](https://en.wikipedia.org/wiki/Kirchhoff%27s_law_of_thermal_radiation).\nPut simply, if the gas were the same (apparent) temperature as the object behind it (in the line of sight), then it would absorb as much infrared radiation as it was emitting, and would appear invisible to the OGI camera.\nIf the gas plume has a lower temperature ($T\\_b > T\\_g$), it will absorb more infrared radiation as it passes through it to the camera relative to the atmosphere around it, thus $\\Delta I < 0$ and the gas plume will appear darker.\nIf the gas plume has a higher temperature ($T\\_b < T\\_g$), then $\\Delta I > 0$ and the gas plume will appear brighter.\n\n\nFor more detail on OGI and its quantitative variant (QOGI), please see this [recent dissertation by Michael Nagorski out of Waterloo](https://uwspace.uwaterloo.ca/handle/10012/18931).\n\n\nCommercial OGI cameras\n----------------------\n\n\nTeledyne FLIR LLC, the company you specifically are referring to in your question and who [appear to make the industry standard cameras](https://dx.doi.org/10.1021/acs.est.0c01285), describe the operation of their OGI and thermographic cameras and sensors in some detail in [US Patent No. 9,276,161 B2](https://patents.google.com/patent/US9276161B2) and [9,007,687 B2](https://patents.google.com/patent/US9007687B2).\nThey have invented arrays of quantum wells of indium antimonide (InSb) that are specially designed to absorb radiation as close to a strong absorption band in the gas plumes that are to be imaged.\nState-of-the-art camera systems include advanced optics and filters to improve the signal-to-noise ratio, thus increasing the contrast of the acquired image. Directly from US Patent No. 9,276,161 B2:\n\n\n\n> \n> Generally, the camera system **100** is used for detecting a gaseous compound in the scene **106** when radiation from the scene **106** is received by the lens **104** and passed to the band pass filter **110** to limit the wavelength range of survey scene energy focused onto the FPA **108**. In one embodiment, the wavelength range is limited to 10.3 to 10.8 μm. Other wavelength ranges are also contemplated. Each sensor element of the FPA **108** generates an analog photo current value according to a photo current responsivity profile and other factors in response to an irradiance generated by the spectrally filtered scene image formed by the lens **104** at the sensor element active surface. The analog photo current values are read out from each photo sensor element and converted to corresponding digital signal values for rendering a video image frame corresponding to the digital signal values.\n> \n> \n> \n\n\nIn the aforementioned patents they give some details on the ways they have improved upon the principles laid out above.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/166960/what-drives-double-displacement-reactions
|
What drives double displacement reactions?
|
I have two questions:
1. I understand that in a displacement reaction the more reactive element displace the less reactive element. But why? In the reaction with Zinc and Copper Sulfate, we form Copper and Zinc Sulfate. But both Copper Sulfate and Zinc Sulfate are soluble in water so how does Zinc Sulfate stay together?
2. What drives double displacement reactions? They are not redox reactions so the reactivity series doesn't apply, so how can we explain why the anions and cations switch places. Is this energetically favorable?
| 0
|
[
[
"\nDissolved salts exist as separated, hydrated ions, moving and reacting independently. (\\*)\n\n\nLike for $\\ce{Zn}$ and $\\ce{CuSO4}$:\n\n\n* The reaction is $\\ce{Zn(s) + Cu^2+(aq) -> Zn^2+(aq) + Cu(s)}$.\n* The ion $\\ce{SO4^2-(aq)}$ is called a spectator/bystanding ion in the reaction context, not participating on the reaction.\n* The reaction can be formally written using full salt formulas, which should not be confused with molecules, as they express just ion stoichiometric ratios: $\\ce{Zn(s) + CuSO4(aq) -> ZnSO4(aq) + Cu(s)}$.\n* The driving force is here different ion affinity to electrons and related redox potential and reaction [Gibbs energy](https://en.wikipedia.org/wiki/Gibbs_free_energy).\n\n\nThe example for \"double displacement\" reaction - precipitation of $\\ce{AgCl}$ from $\\ce{AgNO3}$ and $\\ce{KCl}$:\n\n\n* The reaction is $\\ce{Ag+(aq) + Cl-(aq) -> AgCl(s)}$.\n* The spectator/bystanding ions are $\\ce{NO3-(aq)}$ and $\\ce{K+(aq)}$.\n* The reaction can be again formally written using full salt formulas: $\\ce{AgNO3(aq) + KCl(aq) -> KNO3(aq) + AgCl(s)}$\n* The driving force is here the difference in mutual ion affinity for various cation + anion combinations, leading to precipitation of the least soluble salt - $\\ce{AgCl}$ in this case.\n* It is closely related to energy released by ion hydration versus energy released by ions forming solid ionic lattice, hold by coulombic forces.\n* It again depends on the reaction Gibbs energy.\n* The same case is when all 4 ion combinations form more or less well soluble salts.\n\n\n\n\n---\n\n\n(\\*) At some conditions, hydrated ions may not move independently, but they can have various degree of mutual connection of their hydration envelope, forming [ion pairs](https://en.wikipedia.org/wiki/Ion_association).\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/166954/recovery-of-precipitated-polymer
|
Recovery of precipitated polymer
|
I have recently newly synthesized the polymer. Based on its structure I tried to precipitate its concentrated solution in dichloromethane by dropping it into methanol. It seemed to work. The methanol solution was cloudy, some gooey things that sticked to the wall of the flask were observed. Initially I tried to recover the precipitated polymer by filtering through membrane filter applying reduced pressure (Chmlab, MTF/H Grade, PTFE hydrophobic membrane filter, pore size 0.2 μm, diameter 47 mm). The solution is cloudy orange color, and the filtrate was yellow color. The filtration seemed successful, but shortly after the filtration hardly continued. I am assuming that the precipitated polymer blocked all of the pores of the membrane filter. When I stopped the filtration and checked the filter, there was no powder-like thing on the top surface of the filter, but something seemed to be adsorbed on the surface. I have tried larger pore size, 1~2 μm, but the filtrate has cloudy orange color same to the solution, and shortly after its pores were also blocked. In this situation, do I have to try centrifuging? After centrifuging, can I recover the polymer simply by pouring it onto the membrane filter and washing with methanol?
| 1
|
[] |
https://chemistry.stackexchange.com/questions/166953/solving-two-electron-integrals
|
Solving two-electron integrals
|
I have been trying to teach myself Quantum Chemistry through a book, and I came across this problem and I am not sure how they solve a certain problem.
Essentially, they applied first-order perturbation theory to helium to get a correction term of $\frac{5}{8}Z$ by solving the equation given below in the question which I believe is the Coulomb integral between the two electrons in the $\mathrm{1s}$ orbital of Helium.
[](https://i.stack.imgur.com/LSCfw.png)
The next part of the question is what confuses me. They convert to spherical coordinates and the integral becomes much different.
[](https://i.stack.imgur.com/hcwQw.png)
At this point, I can solve the rest of the problem, but I am stuck on how the conversion worked, and I was wondering if anyone could help me.
Another question I have is that when solving for the electron-electron repulsion energy, would the conversion between the first integral expression to the second be the same if the electrons were in a separate orbital (i.e. For lithium between $\mathrm{1s}$ and $\mathrm{2s}$ electrons)?
| -1
|
[
[
"\nTo introduce a $\\theta$-dependency into the expression for $r\\_{12}$, use that $\\vec{a}\\cdot\\vec{b}=ab \\cos\\theta$, where $\\theta$ is the angle between the two vectors:\n$$r\\_{12}=|\\vec{r}\\_{12}|=|\\vec{r}\\_1-\\vec{r}\\_2|=\\sqrt{(\\vec{r}\\_1-\\vec{r}\\_2)^2}=\\sqrt{\\vec{r}\\_1^2-\\vec{r}\\_2^2-2\\vec{r}\\_1\\cdot \\vec{r}\\_2}=\\sqrt{\\vec{r}\\_1^2-\\vec{r}\\_2^2-2r\\_1 r\\_2\\cos \\theta}$$\nwith $\\theta=\\theta\\_1=\\theta\\_2$.\n\n\nNow switch to a spherical coordinate system. The integral over the full 3d-space changes as follows:\n\n\n$$\\int\\limits\\_\\mathbb{R^3} \\text{d}\\vec{r}\\_2 = \\int\\limits\\_0^\\infty \\mathrm{d}r\\_2\\int\\limits\\_0^{\\pi} \\mathrm{d}\\theta\\_2\\int\\limits\\_0^{2\\pi} \\mathrm{d}\\phi\\_2\\,r\\_2^2 \\sin\\theta\\_2$$\nwhere $r^2 \\sin\\theta$ is the value of the Jacobian determinant of the transformation between spherical coordinates and Cartesian coordinates.\n\n\n",
"2"
],
[
"\nIt's the law of cosines. 1/norm(r1-r2) where both are vectors is the square root of (r1-r2)^2 = r1^2+r2^2-2(r1 dot r2)\nThe last term is the product of the norms times the cosine of the angle between them.\nP.s. sorry I don't know how to format code. First time answering\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/166951/how-can-i-transform-thiamine-sulfate-to-thiamine-hydrochloride
|
How can I transform Thiamine sulfate to Thiamine hydrochloride?
|
I have got thiamine sulfate and want to have thiamine hydro chloride. Can I obtain it by dissolving thiamine sulfate in water and adding calcium chloride water solution as long as calcium sulfate will percipitate out. After it filter the solution and boil off the water? If it's not a good method, can someone tell me a better one?
| 4
|
[] |
https://chemistry.stackexchange.com/questions/166947/how-is-the-entropy-of-mixing-related-to-chemical-reactions-in-solution
|
How is the entropy of mixing related to chemical reactions in solution?
|
When calculating the Gibbs energy of reaction for reaction mixtures at arbitrary concentrations, you use the following expression:
$$\Delta\_\mathrm r G = \Delta\_\mathrm r G^\circ + RT\ln Q$$
The term $RT\ln Q$ has to do with the entropy of mixing, we are [told](https://chemistry.stackexchange.com/a/166895).
How does that work for reactions that are continuously mixed? If I have a reaction like
$$\ce{A(aq) + B(aq) <=> C(aq) + D(aq)}$$
or
$$\ce{A(g) + B(g) <=> C(g) + D(g)},$$
the reactants themselves are already a mixture. In fact, when there is a solvent, even a single reactant is part of a solution, i.e. already in a mixture. If (hypothetically) the reaction goes to completion, you still have a mixture. So why do we have to consider the entropy of mixing if the reaction mixture is already mixed?
| 3
|
[
[
"\nSo, I am gonna use the most basic definition of entropy - the statistic one\n\n\nS = kB *ln* $\\Omega$\n\n\n$\\Omega$ is the total number of microstates possible.\n\n\nIn your case, different microstates can be formed by arranging momenta , energies and also by arranging POSITIONS as any molecule can be found at any possible \"position\" in a solution.\n\n\nI will use basic principles of Permutations and Combinations to find $\\Omega$.\n\n\nSuppose I have 'a' and 'b' moles of A and B initially , while no C and D initially present. After a time 't', x moles of A and B have reacted to form x moles of C and D.\n\n\n\\begin{align}\n\\ce{A(aq) + B(aq) <=> C(aq) + D(aq)}\\\\\n\\end{align}\n\\begin{align} \n\\ce{t=0..... . a. . . . . . b. . . ....0... . . 0 ....... . . . . }\\\\\n\\end{align}\n\\begin{align} \n\\ce{t=t..... (a-x). . . . .(b-x). . ... x... . . x ...... . . . . . . }\\\\\n\\end{align}\n\n\nThe *total* number of molecules remain same in your example , and is (a+b+w) N0 , where N0 is Avogadro's number , and w is the number of moles of water.\n\n\nLet me *simplify* the model by assuming there are (a+b+w) N0 positions available. Any molecule can \"sit\" at any position and can have any of the available value of momentum and energy. Changing either of position or momentum of any *one* molecule will lead to a new microstate.\n\n\nBut there's a catch. Suppose molecule 1 (m1) sits at position r1 and has energy E1 , momentum p1. Another identical molecule (i.e. the same compound) m2 has position r2 , energy E2 , momentum p2. If I exchange *all* - their positions, energies and momenta - I will *not* get a new microstate.\nBut if m2 was not identical, it would surely have resulted in a new microstate.\n\n\nSo, assume I haven't put any molecule at any position. First I assign every position a value of energy and momentum , and write down each possible arrangement in my notepad (my notepad has infinite pages...so don't worry , it won't run out ; ) ). Then I count them. There turn out to be Z possible arrangements of energy and momentum.\n\n\nNow , I start placing the molecules randomly ( there are just A , B and water present for now). How many unique arrangements can I have only due to virtue of positioning? That I can easily calculate. It will be\n\n\n\\begin{equation}\nV\\_{1} = \\frac{[ (a+b+w) N\\_{0} ]!}{( aN\\_{0} )! \\* ( bN\\_{0} )! \\* ( wN\\_{0} )!}\n\\end{equation}\n\n\nSo, total number of microstates ($\\Omega$1) = V1 \\* Z \n Hence, S1 = kB *ln*( V1 Z )\n\n\nNow, after time t , I again do the same. \n The number of arrangements of energies and momenta would still be same , i.e. , Z .\n\n\nBut, the number of arrangements due to positioning will now be ,\n\n\n\\begin{equation}\nV\\_{2} = \\frac{[(a+b+w) N\\_{0}]!} {[(a-x) N\\_{0}]! \\* [(b-x) N\\_{0}]! \\* (xN\\_{0})! \\* (xN\\_{0})! \\* (wN\\_{0})! }\n\\end{equation}\n\n\n$\\Omega$2 = V2 \\* Z \n Hence, S2 = kB *ln*(V2 Z)\n\n\nSo, Change in Entropy $\\Delta$S = S2 - S1 = kB *ln* (V2 / V1 )\n\n\nDividing V2 by V1 , one can easily see it is equal to\n\\begin{equation}\n\\frac{V\\_{2}}{V\\_{1}} = \\binom{aN\\_{0}}{xN\\_{0}} \\* \\binom{bN\\_{0}}{xN\\_{0}}\n\\end{equation}\n\n\n\\begin{equation}\nHence, \\Delta S = k\\_{B} \\* \\ln\\left[ \\binom{aN\\_{0}}{xN\\_{0}} \\* \\binom{bN\\_{0}}{xN\\_{0}} \\right]\n\\end{equation}\n\n\nwhere\\begin{equation}\n\\binom{n}{r}\\end{equation} represents binomial term nCr\n\n\nThis is the increase in entropy due to \"mixing\".\n\n\nI even plotted a graph to verify. Here's the [graph](https://www.desmos.com/calculator/0rrxmw8nvy).\n\n\nI didn't know how to approximate its derivative (as the graph is broken when viewed closer) So I instead compared the graph with another one made by integrating R*ln*(K/Q). Both come out to be same at some value of K. This way, we have also calculated the Equilibrium Constant :)\n\n\n**NOTE**: In the graph, I have also considered change in \"Internal Entropies\" apart from Entropy of Mixing. This \"internal\" entropy arise due to arrangements possible *within* a molecule. It is characteristic of a given compound.\n\n\nThe Black graph is the one made by integrating R*ln*(K/Q) and the Red haze visible behind it is the graph made by the formula we just derived.\n\n\nI took **a** and **b** = 1 \n **SA** , **SB** , **SC** , **SD** (the molar \"internal entropies\" of A , B , C and D respectively) equal to 1 , 1.5 , 2.3 , 2 kJ/mol\\*K (not to scale....just example) \n And **K** came out to be nearly 1.2\n\n\n[](https://i.stack.imgur.com/qTSZ5.png)\n\n\n",
"2"
],
[
"\nI'm a little unsure exactly what you are looking for but I try to show the connection between the Gibbs energy and mixing energy as reaction proceeds. We can start with the chemical potential $\\mu$ and calculate the Gibbs function. Thus,\n\n\n$$\\displaystyle \\mu\\_i=\\mu\\_i^0+RT\\ln(P/P^0)+RT\\ln(x\\_i)$$\n\n\nwhere $P$ is the total pressure, ($P^0$ is unit pressure to make the log dimensionless and is assumed to be included from now on), $x\\_i$ the mole fraction of species $i$ and $\\mu\\_i^0$ is a function of temperature only, i.e independent of composition so is unchanged as the mole fraction changes $x\\_i =0\\to 1$.\n\n\nIn the reaction $A+B=C+D$ the Gibbs function is\n\n\n$$\\displaystyle G=n\\_a\\mu\\_a+n\\_b\\mu\\_b+n\\_c\\mu\\_c+n\\_d\\mu\\_d$$\n\n\nand if we assume a perfect gas mixture, using the first equation produces\n\n\n$$\\displaystyle G=[n\\_a\\mu\\_a^0+n\\_b\\mu\\_b^0+n\\_c\\mu\\_c^0+n\\_d\\mu\\_d^0+RT(n\\_a+n\\_b+n\\_c+n\\_d)\\ln(P)]+RT\\big(n\\_a\\ln(x\\_a)+n\\_b\\ln(x\\_b)+n\\_c\\ln(x\\_c)+n\\_d\\ln(x\\_d)\\big)$$\n\n\nThe first two terms (in square brackets) is the total Gibbs function for the four gases as if each is in a separate container at pressure $P$, the term on the right is therefore the free energy of mixing, $RT\\sum\\_i n\\_i\\ln(x\\_i)$ and if divided by temperature is minus the entropy of mixing.\n\n\nIf the system initially consists of only one mole each of A and B then\n\n\n$$\\displaystyle n\\_b=n\\_a,\\quad n\\_c=n\\_d=1-n\\_a,\\quad x\\_a=x\\_b=n\\_a/2,\\quad x\\_c=x\\_d=(1-n\\_a)/2$$\n\n\nand now $G$ only depends on $n\\_a$ which can vary between $0\\to 1$\n\n\n$$\\displaystyle G= n\\_a(\\mu\\_a^0+\\mu\\_b^0)+(1-n\\_a)(\\mu\\_c^0+\\mu\\_d^0)+2RT\\ln(P)+2RT(n\\_a\\ln(n\\_a/2)+(1-n\\_a)\\ln((1-n\\_a)/2)$$\n\n\nand without loss of generality choose the pressure to be unity, then with some rearranging,\n\n\n$$\\displaystyle G= n\\_a(\\mu\\_a^0+\\mu\\_b^0)+2RTn\\_a\\ln(n\\_a/2)+(1-n\\_a)(\\mu\\_c^0+\\mu\\_d^0)+2RT(1-n\\_a)\\ln((1-n\\_a)/2)$$\n\n\nWhen $n\\_a=1$ so that only reactants are present we find that $G=\\mu\\_a^0+\\mu\\_b^0+2RT\\ln(1/2)$ where $2RT\\ln(1/2)$ has to be the free energy of mixing before any reaction has occurred and similarly when $n\\_a=0$ the same energy of mixing is present for the products there being only two species present in either case. Elsewhere both terms contribute to the mixing free energy.\n\n\nThe shape of the plot of $G$ vs. $n\\_a$ is due primarily to the log terms, if this term were absent the plot would follow the line W-V, see figure. The lower part of $G$ shows that the free energy at equilibrium becomes even smaller than that if there were complete conversion to products, because here there are four species in various proportions not just two. Entropy of mixing depends on the mole fractions is maximal at equilibrium.\n\n\nThe position of the minimum, and so the equilibrium, is determined by $\\mu\\_a^0+\\mu\\_b^0-\\mu\\_c^0-\\mu\\_d^0$ which is the difference in free energies between reactants and product in their pure states. This is the case because the mixing terms are symmetrical in $n\\_a$ but the sloping energy terms biases this. The minimum can be calculated by differentiating $(\\partial G/\\partial n\\_a)\\_{T,P}=0$ or more easily and just as generally via\n\n\n$$\\displaystyle dG=-SdT+VdP+(\\mu\\_a+\\mu\\_b-\\mu\\_c-\\mu\\_d)dn\\_a$$\n\n\nwhere the stoichiometry of the reaction was used ($-dn\\_a = -dn\\_b=dn\\_c=dn\\_d$) and at constant $T$ and $P$ when $\\partial G/\\partial n\\_a=0$, we find that\n\n\n$$\\displaystyle \\mu\\_a+\\mu\\_b=\\mu\\_c+\\mu\\_d$$\n\n\nand substituting for $\\mu\\_i$ and mole fractions leads to the solution for $n\\_a$ at equilibrium just as if $G$ had been differentiated.\n\n\nWhat this shows is that the same relationship exists at equilibrium as in the chemical equation $A+B=C+D$ from which the equilibrium constant is normally obtained and shows the connection to the entropy of mixing as the reaction proceeds. The reaction is therefore seen as a competition between attractive forces causing the reactants to form products on the one hand and increasing entropy forming a mixture on the other.\n\n\n[](https://i.stack.imgur.com/RtvGu.png)\nGibbs function of a reacting system\n\n\n",
"1"
],
[
"\nFor a reaction of ideal gases, such as, say, aA+bB-->cC+dD, the $\\Delta G\\_r$ in your equation is the change in Gibbs free energy in going at constant temperature T from **a** moles of **pure** A and **b** moles of **pure** B at arbitrary pressures $p\\_A$ and $p\\_B$, respectively, to **c** moles of **pure** C and **d** moles of **pure** D at arbitrary pressures $p\\_C$ and $p\\_D$, respectively. If the pressures were not arbitrary, but, instead happened to coincide with those for an equilibrium reaction mixture of A, B, C, and D, then $\\Delta G\\_r$ would be equal to zero, and Q would be equal to the equilibrium constant for the reaction. Under these circumstances, the pure gas pressures would be equal to the partial pressures of the corresponding components in the equilibrium reaction mixture, and could be placed in equilibrium with the components in the mixture through 4 semipermeable membranes which allowed passage of only one of the corresponding components.\n\n\nThe challenge in all this is to figure out how to interpret all this in terms of a reversible process to go from the initial state of pure A and B to the final state of pure C and D so that $\\Delta G\\_r$ can be expressed in terms of the initial and final pressures. This typically is facilitated by the use of a so-called van't Hopf equilibrium box.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166941/why-is-is-that-telluriumvi-fluoride-is-completely-hydrolysed-but-iodineiii-f
|
Why is is that tellurium(VI) fluoride is completely hydrolysed but iodine(III) fluoride isn't, even in hot water?
|
Tellurium(VI) fluoride is completely hydrolysed into tellurium(VI) oxide and hydrofluoric acid, but iodine(III) fluoride isn't, stopping at the iodine(III) fluoride dihydroxide/hydrofluoric acid level, even after extensive warming. Why is it that way, and how do I fix it?
| 0
|
[
[
"\nWhy not? Tellurium hexafluoride also goes through an intermediate:\n\n\n\n> \n> The hydrolysis of tellurium hexafluoride produces the fluoroorthotelluric acids, $\\ce{TeF\\_n(OH)\\_{6–n}}$, n= 1–4, which undergo complete hydrolysis to orthotelluric acid only over a long period of time.\n> \n> \n> \n\n\n**Ref.**: Fraser, G.W., & Meikle, G.D. (**1974**). *Hydrolysis of tellurium hexafluoride*. Journal of The Chemical Society, Chemical Communications, 624-625, DOI: [10.1039/C39740000624](https://pubs.rsc.org/en/content/articlelanding/1974/C3/C39740000624)\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166936/where-does-the-energy-in-coal-bonds-come-from
|
Where does the energy in coal bonds come from?
|
>
> I have been told that when the plants intake carbon dioxide they break carbon and oxygen apart and the carbon is utilised in making the plant's body. This process is driven by the sun's energy so when they decompose and form coal the bonds of coal have the sun's energy stored in it.
>
>
>
The above statement makes absolutely no sense to me.
When the sun's energy was utilised to separate carbon and oxygen how can the coal bonds store it?
If the coal bonds don't store the sun's energy where does the energy released on burning them come from?
| 7
|
[
[
"\nIndeed it's in some way counterintuitive to say that the sun's energy is 'stored' in the coal bonds. Energy never comes from *breaking* bonds, it is released when *forming* bonds.\n\n\nIn the case at hand, the sun's energy is used to break carbon-oxygen bonds through photosynthesis. You get that energy back when carbon-oxygen bonds are formed by $\\ce{C}$ and $\\ce{O2}$ reacting to $\\ce{CO2}$: that's where the energy of burning coal comes from.\n\n\nThe details can be quite delicate, since in most chemical reactions new bonds are formed at the same time as old bonds are broken. It then boils down to the relative energies of those bonds. In the case of burning coal, it costs less energy to break the $\\ce{C-C}$ bonds of coal than what the formation of $\\ce{C=O}$ bonds yield, so the reaction is exothermic.\n\n\n",
"17"
],
[
"\nAs has been pointed out the energy comes in the form of visible light from the sun. The reaction written in summary or overview is\n\n\n$$\\displaystyle \\ce{2H2O \\overset{light}\\to O2 + 4H^+ + 4e^-}$$\n$$\\displaystyle \\ce{CO2 +4H^+ + 4e^-\\to (CH2O) + H2O}$$\n\n\nwhere in the first step, the oxidation of water, the light does not react directly with water. In the second step, $\\ce{CH2O}$ represents carbohydrate and the electrons are carried by proteins.\n\n\nThe overall reactions occur in proteins in the cell membrane of chloroplasts (in green plants) and the first step is absorption of a red photon typically of $700$ nm. This causes an electron to be transferred from a chlorophyll dimer, (the special pair) down an electron transfer chain involving cytochromes and other proteins that eventually reduces the carbon dioxide, the second reaction above.\n\n\nAs the special pair is now short of an electron it recovers this via a metalloprotein grabbing an electron from water. Using X-ray diffraction all these protein structure are now known and ultrafast spectroscopy has measured most if not all the rates of these primary processes.\n\n\nIn energy terms, the couple: CO2-glucose has a redox potential $\\Delta V=-0.43$ eV and for $\\ce{H2O - O2}$ is $+0.82$ eV so the total redox span is $1.25$ eV and as $4$ electrons are involved this is makes $5$ eV in total. A visible photon at $700$ nm has an energy of $1.77$ eV and as one electron is released /photon absorbed then the total energy supplied is $7.1$ eV, which is more than enough for this reaction.\n\n\nHowever, there are losses in the process, and the efficiency rarely exceeds $0.36$ which makes the useable energy available $0.63$ eV/photon which is not enough thus we need $5/0.63\\to 8$ photons in total. This suggest that two photosystems (see Z-scheme) are needed in green plants and this is indeed the case.\n\n\n",
"14"
],
[
"\n**Photosynthesis is a little like the reverse of burning**\n\n\nThe idea that chemicals or their bonds *store* energy needs some clarification. The storage is always *relative* to some other reachable arrangement of bonds that can be achieved by some actual chemical reaction. So, if we say a specific compound \"stores\" energy we need to specify *relative to what*.\n\n\nIf we say coal *stores* energy what we usually mean is that we can burn coal (crudely, impure carbon) in air in a reaction that creates carbon dioxide and a large amount of heat. This is a simple widely observed and widely used reaction that we know releases energy.\n\n\nBut why? The energy is released because the total amount of energy tied up in the bonds of carbon dioxide is less than the energy tied up in the oxygen molecules and the solid carbon of the coal. When we burn coal that energy appears as light and heat we can use to warm our homes. We could, under different circumstances, react coal with other chemicals and get different outcomes (i'm sure coal will also burn in fluorine). But since the air around us has ~20% oxygen, burning coal is a very convenient way for us to produce heat.\n\n\nThe reason we say coal \"stores\" solar energy is because photosynthesis allows plants to drive the reaction in the opposite direction. Crudely (photosynthesis involves some very complex details) plants use the energy from sunlight to allow them to make a range of complicated carbohydrates by taking carbon dioxide from the air and water from the ground. (note photosynthesis also releases oxygen which is the main reason there is any in the atmosphere). The bonds in those more complicated molecules have higher net energy compared to the carbon dioxide and water they are made from and also compared to the products they would create if we burned them in air (wood burns too).\n\n\nCoal is basically old plants that have been transformed by geological processes to impure carbon. So the ultimate reason we can burn it and release energy is because ancient plants captured energy from the sun to make more complicated molecules necessary for their growth. Burning is like the opposite of photosynthesis.\n\n\nCoal \"stores\" energy because the arrangement of bonds in carbon and oxygen have higher energy than the arrangement in carbon dioxide and, in an atmosphere containing oxygen, we can release that difference as heat by burning the coal. But the ultimate reason the coal is the way it is is because plants made other complicated, higher energy, molecules by capturing solar energy to build them. Hence it is not unreasonable to say that coal \"stores\" energy from the sun (though it is oversimplifying a lot).\n\n\n",
"8"
],
[
"\nElectrons around the nuclei of atoms exist in various energy states due to quantum mechanical principles. Due to these principles, they are confined to certain configurations (shells and orbitals). Chemical bonds occur when an electron of one atom can simultaneously take up a vacant position of another atom.\n\n\nFor example, a neutral atom of carbon has six electrons which balance the charge of its six protons; two in a complete inner shell and four in its outer shell. The outer shell of carbon can hold up to eight electrons. A neutral atom of oxygen has eight protons and eight electrons; two in the inner shell and six in the outer shell (which can also hold up to eight electrons). So, an atom of carbon can bond with two atoms of oxygen as follows: two of the carbon atom's outer electrons are shared with one oxygen atom, allowing the oxygen atom to \"complete\" its outer shell, and the other two of the carbon atom's outer electrons are similarly shared with another oxygen atom.\n\n\nWhen a carbon atom and two oxygen atoms are bonded like this, the electrons are in a lower energy state than if each of the atoms existed in an independent, neutral state. Different configurations, such as two oxygen atoms bonded to each other, or a cluster of carbon atoms bonded to each other also each have different energy states (of their outer electrons). Plants have complex arrangements of special molecules which allow them to use energy received from sunlight to break the bonds of carbon dioxide and water molecules to build sugars and release molecular oxygen. Breaking the carbon dioxide and water bonds is all about elevating electrons in those atoms to higher energy states. This is how/where the energy is stored. When carbon and oxygen combine to form carbon dioxide, the electrons drop to lower energy states, which means energy is released.\n\n\nAs plants decay into substances like coal, bonds are reformed and some energy is released - that's what drives the decay process, but the bonds between carbon atoms in coal and in molecular oxygen are still at a higher energy state than in carbon dioxide.\n\n\nThere is one other notion here about activation potential. The resulting oxygen and sugar molecules are stable (they don't immediately react and go back to carbon dioxide and water) because it takes some amount of energy to initiate the transition. It's like a hill with a small dip at the top, you can roll a ball up the hill over a lip and into the dip and it won't roll back down again unless you first lift it back over the lip.\n\n\n",
"1"
],
[
"\n\n> \n> When the sun's energy was utilised to separate carbon and oxygen how can the coal bonds store it?\n> \n> \n> \n\n\nYou use the word 'utilised' as if this means the energy is used; it no longer exists. But energy never disappears, it just changes form. When you take a rubber band and pull the ends apart, you consume (or better: move) energy, which is stored in the rubber band. If you now release one end of the rubber band it will snap together, the energy is released. Imagine there is a rubber band between the oxygen and the carbon atoms\\*, by separating the two, you add energy to the system. By combining the two, you release that energy again.\n\n\n\\*The analogy does not hold for very long, moving the atoms further apart will not increase the energy stored in the system.\n\n\n",
"1"
],
[
"\n[Here's](https://www.sciencedirect.com/topics/earth-and-planetary-sciences/coal-structure) a nicely simplified description of the structure of coal:\n\n\n\n> \n> V.D Macromolecular Nature of Coals\n> Coals are believed to be three-dimensionally cross-linked macromolecular\n> networks containing dissolved organic material that can be removed by extraction.\n> This model offers the most detailed and complete explanation of the chemical and\n> mechanical behavior of coals. It is a relatively recent model and is somewhat\n> controversial at this writing. The insoluble portion of the coal comprises the\n> cross-linked network, one extraordinarily large molecule linked in a three-\n> dimensional array. This network is held together by covalent bonds and hydrogen\n> bonds, the weak interactions that play such a large role in the association of\n> biological molecules. The extractable portion of the coal is simply dissolved in\n> this solid, insoluble framework.\n> \n> \n> \n\n\nThat was taken from a 2003 article in an encyclopedia about coal structure.\n\n\nThe plant matter the coal came from was somewhat similar in nature in that it also contained a lot of cross-linked organic molecules. The above can be read to imply that each lump of coal is a single solid molecule plus dissolved liquids and gasses but that's probably taking it a bit too far.\n\n\nLots of chemical reactions happen under heat and pressure including increased cross-linking of organic molecules. The heat and pressure can be viewed as shifting bonds as well as adding them. In the former case, that's still at least somewhat representative of the original sun-energy donation to the cause.\n\n\nNote that the dissolved organics like methane (natural gas) are probably a mix of decomposed plant material and compounds produced under that same heat and pressure. If you want to get a sense of what those are, read up on making coke from coal for use in steel making.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/166930/for-cannizarro-reaction-is-it-necessary-that-carbonyl-compounds-dont-have-alpha
|
For cannizarro reaction is it necessary that carbonyl compounds don't have alpha hydrogen?
|
Till date, I have encountered carbonyl compounds without alpha hydrogens that can undergo Cannizzaro reaction. But recently I came across the following molecule and I can't understand the reason behind why it shows Cannizzaro reaction.
[](https://i.stack.imgur.com/9AgDV.jpg)
| 0
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[] |
https://chemistry.stackexchange.com/questions/166923/two-solutions-for-balancing-a-chemical-equation
|
Two solutions for balancing a chemical equation? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
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$$\ce{6H2O2 + 2N2H4 → 2N2 + 10H2O + O2}$$
$$\ce{4H2O2 + N2H4 → N2 +6H2O + O2}$$
Which equation from above is correctly balanced? They all seem to be balanced but which one is correct?
| 0
|
[
[
"\nAll your equations are correctly written. But look !\n\n\nThe simplest equation would be $\\ce{N2H4 + 2 H2O2 -> N2 + 4 H2O}$. But plenty of other acceptable equations can be obtained by adding an arbitrary number of times the following decomposition of $\\ce{H2O2}$ : $\\ce{2 H2O2 -> 2 H2O + O2}$.\n\n\nFor example, your first equation has been obtained in doubling my simplest equation and adding once the decomposition of $\\ce{H2O2}$\n\n\nYour second equation is obtained by adding my simplest equation plus once the decomposition of $\\ce{H2O2}$\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/166919/question-about-aluminium-and-complex-ion-formation-in-water-vs-in-acid
|
Question about aluminium and complex ion formation in water vs in acid
|
My textbook states, that aluminium reacts with water after dissolving the outer layer with HgCl2, and forms Al(OH)3, an insoluble compound:
2Al + 6H2O = 2 AI(OH)3 + 3H2
In non-oxidizing acids, it forms a salt and hydrogen gas:
2Al + 6HCI = 2AlCI3 + 3H2
But, aluminium is prone to form complexes, so the next reaction states, that in acidic environment, an aquacomplex is formed:
2Al + 6H3O+ + 6H2O = 2[Al(H2O)6]3+ + 3H2
I have tried to research a chemical equation for this complex formation on the internet, and I only found aluminium and water as reactants. No acid present.
**My question:**
* How does the aluminium aquacomplex form, if there is no acid present and if that's not possible, why does it need an acid for the reaction to happen? What role does the acid have?
Thank you
| -1
|
[
[
"\nMetallic aluminum is not charged. It has to loose three electrons in order to make the ion $\\ce{Al^{3+}}$, which can be complexed later on. An acidic solution is a good way for helping $\\ce{Al}$ to loose its electrons, because it produces $\\ce{H2}$ which is a gas and leaves the solution, allowing the reaction to proceed with efficiency.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166916/pfr-reactor-1-r-vs-conversion-plot
|
PFR reactor: -1/r vs conversion plot
|
A generic reagent A is considered. The behavior equations of a CSTR reactor is the following:
$$
\tau = c\_\mathrm{A,0} \intop\_0^{X\_\mathrm{A,final}}
\dfrac{1}{-r\_\mathrm{A}} dX\_\mathrm{A}
$$
where $\tau$ where it is the filling time, understood as the time required to make a fluid flow rate react whose volume is equal to the reactor volume
$$ \tau = \dfrac{V\_\mathrm{reactor}}{\dot{V\_\mathrm{A}}} $$
$c\_\mathrm{A,0}$ is the initial concentration of A, and $X\_\mathrm{A}$ and it is the conversion of A that we want to obtain
From Octave Levenspiel, Chemical Engineering Reaction, John Wiley & Sons, Third Edition, page 103, you can see how the graph $-\dfrac{1}{r\_\mathrm{A}} = f(X\_\mathrm{A})$ has the shape of a **crescent** curve
[](https://i.stack.imgur.com/QERyn.jpg)
I tried to reproduce this graph, using the formula
$$ -\dfrac{1}{r\_\mathrm{A}} = \dfrac{\tau}{c\_\mathrm{A,0} X\_\mathrm{A}} $$
and what I get is a decreasing curve
[](https://i.stack.imgur.com/IThhe.jpg)
**My question is**: to be able to reproduce the graphic in the text, which formula should I use?
| 0
|
[
[
"\nFrom [http://home.ku.edu.tr/~okeskin/ChBI502/chbi502-Chapter\\_2.pdf](http://home.ku.edu.tr/%7Eokeskin/ChBI502/chbi502-Chapter_2.pdf)\n\n\nAn irreversible reaction is considered.\n\n\n$$\\mathrm{A} \\xrightarrow{k} \\text{Products} $$\n\n\nTo find a relationship between $r$ and $X$, we start from the kinetic law\n\n\n$$-r\\_\\mathrm{A} = k\\ c^n\\_\\mathrm{A}$$\n\n\nwhere $n$ is the order reaction. The conversion is given by the formula\n\n\n$$ X\\_\\mathrm{A} = \\dfrac{c\\_\\mathrm{A,0} - c\\_\\mathrm{A}}{c\\_\\mathrm{A,0}} $$\n\n\nWith a few simple algebraic steps, we get\n\n\n$$ c\\_\\mathrm{A} = c\\_\\mathrm{A,0} (1 - X\\_\\mathrm{A}) $$\n\n\nSubstituting in the kinetic law, we obtain\n\n\n$$-r\\_\\mathrm{A} = k\\ c\\_\\mathrm{A,0}^n (1 - X\\_\\mathrm{A})^n$$\n\n\nMultiplying the first and second members by $-1$, the desired equation is finally obtained\n\n\n$$\\dfrac{1}{-r\\_\\mathrm{A}} = \\dfrac{1}{k\\ c\\_\\mathrm{A,0}^n} \n\\dfrac{1}{(1 - X\\_\\mathrm{A})^n}$$\n\n\n[](https://i.stack.imgur.com/yqQJH.jpg)\n\n\nThis curve has been plotted for $n = 1$, but analogous curves can be obtained for $n \\neq 1$. The only difference is the curve slope, which is obvious since as $n$ changes, the rate dependence on the concentration changes (and therefore from $X\\_\\mathrm{A}$)\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166915/how-to-increase-basis-set-while-running-cluster-calculations
|
How to increase basis set while running cluster calculations
|
I am running (opt+freq)DFT calculations using 6-311G(d) basis set on polymers with different number of monomer units. Upto 2 monomer units optimization is converged but for 4 units max displacement criteria is not converged. Which starting point do I need to consider while increasing the basis set?
| 1
|
[] |
https://chemistry.stackexchange.com/questions/166913/azulene-conjugation
|
Azulene Conjugation?
|
I'm trying to figure out how azulene is an aromatic compound; I understand it is cyclic, flat, and follows Huckel's rule, but I don't understand how it is conjugated.
I have circled what I believe to be an unconjugated area in yellow. Can someone help me figure out how this molecule is conjugated? Thank you.
[](https://i.stack.imgur.com/WSKRL.png)
| 2
|
[
[
"\nIf you insist on neutral atoms then the internal linkage is not conjugated.\n\n\nBUT ...\n\n\nThe notable thing about azulene is that the seven-and five-membered rings can be aromatic when the former is positively charged and the latter negatively charged. That puts six electrons into each ring, and we all know about the $4n+2$ rule (even if technically it should be applied to single-ring systems).\n\n\nIf you allow a positive charge on the seven-membered ring and a negative charge on the five-membered ring, you can render contributing structures with a double bond in the internal linkage.\n\n\n[](https://i.stack.imgur.com/00TeUm.jpg)\n\n\n<http://www.chemspider.com/Chemical-Structure.8876.html>\n\n\nAccording to [Wikipedia](https://en.wikipedia.org/wiki/Azulene), azulene is actually known with a dipole moment having the positive end in the seven-membered ring, and its chemical reactivity (electrophilic in the seven-membered ring, nucleophilic in the five-membered ring) is also consistent with this polar contributing structure.\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/166912/what-is-this-small-peak-that-always-occurs-within-ketone-infra-red-analysis
|
What Is This Small Peak That Always Occurs Within Ketone Infra-Red Analysis?
|
After looking at a bunch of IR spectrums for ketones, I have noticed this tiny peak that I circled on the diagram below. Does anyone know what it actually is and if I can always use it to distinguish a ketone vs aldehyde?
[](https://i.stack.imgur.com/TbCc3.png)
| 3
|
[] |
https://chemistry.stackexchange.com/questions/166905/the-difference-between-the-freezing-point-of-methanol-and-ethanol-at-different-c
|
The difference between the freezing point of methanol and ethanol at different concentrations
|
I would like to ask a chemistry question:
The freezing point of methanol aqueous solution between 0% and 90% is lower than that of ethanol aqueous solution; but why is the freezing point of ethanol aqueous solution between 90% and 100% lower than that of methanol aqueous solution?
Thanks in advance!
| 0
|
[
[
"\nThe OP's claim that \"The freezing point of methanol aqueous solution between 0% and 90% is lower than that of ethanol aqueous solution\" is correct as data given in [Engineering Tool Box (methanol/water)](https://www.engineeringtoolbox.com/methanol-water-d_987.html) and [Engineering Tool Box (ethanol/water)](https://www.engineeringtoolbox.com/ethanol-water-d_987.html):\n$$\n\\begin{array}{c|c|c|c}\n\\text{% MeOH (v/v) or (w/w)} & \\text{The freezing point } (\\pu{^\\circ C}) & \\text{% EtOH (v/v)} & \\text{The freezing point } (\\pu{^\\circ C}) \\\\\n\\hline\n0 & 0 & 0 & 0 \\\\\n13 \\text{ or }10 & -7 & 10 &-4 \\\\\n24 \\text{ or }20 & -18 & 20 & -9\\\\\n35 \\text{ or }30 & -26 & 30 & -15\\\\\n46\\text{ or }40 & -40 & 40 & -23\\\\\n56\\text{ or }50 & -54 & 50 & -32\\\\\n66\\text{ or }60 & -71 & 60 & -37\\\\\n75\\text{ or }70 & -82 & 70 & -48\\\\\n83\\text{ or }80 & -87 & 80 & -59\\\\\n92\\text{ or }90 & -90 & 90 & -73\\\\\n100 & -98 & 100 & -115\\\\\n\\hline\n\\end{array}\n$$\n\n\nHowever the two last data points are simply the comparison of the freeing point of $100\\%$ methanol and that of $100\\%$ ethanol. Evidently, the freeing point of $100\\%$ methanol is greater than the freeing point of $100\\%$ ethanol ([See Table of Melting points of alcohols)](https://www.britannica.com/science/alcohol/Physical-properties-of-alcohols). Thus, there is not an unusual reason for this change in trend.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166898/mathematical-models-of-oscillatory-chemical-reactions
|
Mathematical models of oscillatory chemical reactions
|
I am a mathematician working on real-life models in ordinary differential equations. I want to know if there are any models of oscillatory chemical reactions that consist of three ordinary differential equations where one variable is much slower than the other two. In particular I am interested in bursting behavior, which consists of alternating trains of fast oscillations with periods of rest.
I have been searching and I couldn't find what I am looking for. I found about the Belousov-Zhabotinsky reaction, but in this case we have one fast variable and two slow variables, while I'm looking for two fast variables and one slow variable.
| 7
|
[] |
https://chemistry.stackexchange.com/questions/166897/smarts-distinguish-between-alcohol-and-carboxylic-acid
|
SMARTS: distinguish between alcohol and carboxylic acid
|
I'm new to SMARTS and have to identify substructures with the Python package `rdkit`, so far so good. Let's say I have a SMILES with an carboxylic acid and a alcohol such as
```
OC(C1=CC(CO)=CC=C1)=O
```
and I want to get only the alcohol, then I would try something like this:
```
mol = Chem.MolFromSmiles("OC(C1=CC(CO)=CC=C1)=O")
toSearch = Chem.MolFromSmarts("[#6][OX2H]")
bool = mol.HasSubstructMatch(toSearch)
subStruct = mol.GetSubstructMatches(toSearch)
print(bool)
print(subStruct)
```
which would return:
[](https://i.stack.imgur.com/PUAkK.png)
```
True
((1, 0), (5, 6))
```
However, the substructure search also identifies the OH of the acid which I don't want. How can I exclude that? My idea was to use some kind of recursive smarts for the alcohol to exclude all those "alcohols" where the carbon is bonded to another oxygen like "=O".
| 3
|
[
[
"\nThe simplest change to make for your SMARTS pattern would be specifying the total degree of the attached carbon is 4 with `\"[#6X4][OX2H]\"`. This is enough to discriminate against the carboxylic acid.\n\n\nYou could use the alcohol SMARTS from [here](https://github.com/openbabel/openbabel/blob/master/data/SMARTS_InteLigand.txt#L70), which has the comment\n\n\n\n> \n> nonspecific definition, no acetals, aminals, and the like\n> \n> \n> \n\n\nThe SMARTS pattern is `\"[OX2H][CX4;!$(C([OX2H])[O,S,#7,#15])]\"`, which stipulates that the carbon attached to the oxygen not also be connected to O, S, N, or P.\n\n\nI made a little [interactive widget](https://www.wolframcloud.com/obj/37fe8f9e-7c80-447c-a589-92e94d9197c4) that takes a molecule name or SMILES in the first field and a SMARTS pattern in the second.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/166893/will-my-method-for-catching-lead-dust-work
|
Will my method for catching lead dust work?
|
I am a hobbyist and a crafter working with cast models. These modles are up to 80% lead, with the rest being tin or similar soft metals. They are made in a mould, so they often have mould lines that need to be scraped off with a knife. But doing this can create lead dust which will pollute the immediate working environment. I had the idea of doing any scraping while the model was completely submerged in water. The theory is that small particles will be caught in suspension in the water, and the water can then be filtered or otherwise safely disposed of. (In case this needs illustrating, I have made a video of my method: [link](https://youtu.be/tLe1BQ9tf48)). My question is essentially, "will this method successfully control the spread of lead dust into my working environment, or am I missing something important?"
| 1
|
[
[
"\nAnother option is to buy a small table-top soldering booth. They make acrylic booths with HEPA filtration. You'd put your workpiece in the booth and reach in from the front with your hands. Air is pulled inward and exhausted out the back. For cleanup, I'd suggest something like D-Lead wipes - they contain EDTA or something equivalent that will help to trap and retain lead and other metals in solution. You should also get a colorimetric lead test kit (I've seen them on Amazon) so you can do surface swab sampling in your work area. Color change on the swabs will allow you to visualize the effectiveness of your control measures. Lastly, tell your doctor you are working with lead, and they can arrange a blood lead level test during your annual physical.\n\n\n",
"1"
],
[
"\nIt's risky. The water being filtered off is not good enough to reuse the water. This is because some lead dust might be small enough to fall with the liquid. It is also because if there is any metal higher than lead in the electrochemical series, some lead would fall into solution. Either way, you'd have to dispose of the water safely since it would likely be contaminated.\n\n\nWould it catch the dust? Yeah it would. You'd also need to thoroughly wash the knife. Or maybe it wouldn't. If your lead dust is super small, some amount could be flung and fly. If your dust is more like sand, well it'll probably stay at the bottom of the water.\n\n\nI'm no physicist, but I would think the particles would be larger when cut underwater. So you're probably okay.\n\n\nI would advise placing other precautions such as wearing a mask with a very fine filter. For example a dust mask. It would also be a good idea to wear gloves and long sleeved clothing. But I'm not sure how you would safely clean off your clothing after use.\n\n\nAt that point, even if you get some lead on you, it's not gonna be super bad since there would be very little and since it is metallic. Do note that aqueous and organometallic heavy metals are very, very toxic even compared to the regular metals.\n\n\nAlso, the deeper your lead is, the less likely it is to escape. And whatever tool you're using, you need to make sure it has no lead dust before it leaves the water.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/166891/is-the-entropy-change-positive-or-negative-in-this-reaction
|
Is the entropy change positive or negative in this reaction?
|
Suppose I have a reversible reaction
\begin{align}
\ce{A(g) + B(g) <=> C(g)}\\
\end{align}
with equilibrium constant K.
Its ΔΗ is positive throughout.
Now, suppose I start with 1 mole each of gases A and B, and zero (negligible) moles of C. The temperature is constant.
According to $\Delta G = RT\ln(Q/K)$ , $\Delta G$ will be highly negative at my initial conditions and will increase (become less negative) as reaction proceeds forward.
Also, using the same expression, $$\Delta S = \frac{\Delta H}{T} - R \ln\left(\frac{Q}{K}\right)$$
Hence, $\Delta S$ is highly positive initially and decreases as the reaction proceeds forward. When $\Delta S = \Delta H/T$ (at $Q=K$) , no *net* reaction takes place further and equilibrium is maintained.
I understand it.
But how is $\Delta S$ positive? How can it be explained theoretically?
Theoretically, shouldn't $\Delta S$ be negative as number of gaseous moles decrease on going forward?
Or is *this* type of reaction not possible?
**Source**:<https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.10%3A_Entropy_Changes_in_Gaseous_Reactions>
| 3
|
[
[
"\n\n> \n> But how is dS positive? How can it be explained theoretically?\n> \n> \n> \n\n\nThe reaction entropy rules usually taught first are about the physical state of the pure substances. If you go from a condensed state (solid or liquid) to a gas, the molecules are less constrained. Similarly, if you open up a ring, the molecule itself has more conformational freedom, and this is also reflected in an increase in entropy.\n\n\nThere is a second, concentration-dependent contribution to the entropy of reaction. This is the entropy of mixing. A mixture has more microstates than the separate two pure substances. This is behind the experience that ideal mixtures (when the intra- and intermolecular interactions have the same strength) don't unmix. Entropy increases whem you mix something. This strangely is the case when you go from pure reactant to a mixture of reactant and product as the reaction proceed. Similarly, there is some \"unmixing\" when all the reactant turns into product.\n\n\nThis entropy of mixing is where the $R \\ln(Q)$ (or $R T \\ln(Q)$) comes in. This is what makes the reaction entropy and the reaction Gibbs energy concentration-dependent.\n\n\n\n> \n> Theoretically, shouldn't dS be negative as number of gaseous moles decrease on going forward?\n> \n> \n> \n\n\nYes, going from pure reactants to pure products, the entropy change is negative. However, if reactants and products are in the same phase, you have a very high positive change in entropy as the first couple of reactant molecules turn into product. If you want, you can explain this with kinetics as well, with no back reaction (no product yet) and a robust forward reaction.\n\n\n\n> \n> [OP in comments: What if they are in different phases? Like gases combining to form a liquid?]\n> \n> \n> \n\n\nOne example where all \"reactants\" and \"products\" are in separate, pure phases is ice melting. If you are above the melting point, all ice will melt and the system will not reach a dynamic equilibrium. This is different for water evaporating. In a water bottle that is half-full and capped, after a while you reach equilibrium, and the same amount of water evaporates and condenses, so there is no longer a net change. The entropy of water in the gas phase is dependent on the partial pressure (the \"concentration\"). If you write the equilibrium expression for these processes, you will see the difference (\"leaving out\", or setting to one, pure solids and liquids, but not gases).\n\n\nConversely, if the last few reactant molecules turn into product and there is no more reactant left, the entropy change is negative (this is hypothetical because it is past the equilibrium, i.e. in a direction away from equilibrium). Again, using kinetics, the forward reaction rate has almost \"dried up\", the the reverse reaction rate is robust.\n\n\n\n> \n> Or is this type of reaction not possible?\n> \n> \n> \n\n\nThis type of reaction is possible. The standard Gibbs reaction energy will have a high positive value, though, so the equilibrium constant will be very low. Equilibrium will be reached quickly if you start out with pure reactants.\n\n\n\n> \n> [OP in comments] How do you say that the Standard Gibbs Energy will be highly positive for these type of reactions?\n> \n> \n> \n\n\nThe standard enthalpy of reaction is positive (given) and the standard entropy is negative (less particles on product side). Adding them up according to $$\\Delta G = \\Delta H - T \\Delta S$$ gives a positive value at all temperatures, at least as high as $\\Delta H$\n\n\n### Nomenclature\n\n\nYou have to distinguish $$\\Delta S^\\circ$$ and $$\\Delta\\_r S = \\frac{\\partial \\Delta S}{\\partial \\xi}$$ The former is what you can calculate from tabulated data. The latter is a more formal way of writing your dS, signifying the infinitesimal small change in entropy as the reaction moves forward an infinitesimal amount, i.e. when some reactant turns into some product according to the balanced chemical equation.\n\n\nTextbooks, especially introductory once, are inconsistent in their nomenclature (there is an [entire article](https://www.chem.tamu.edu/class/majors/chem328/CHEM%20328_Chapter%2006.pdf) on this topic for Gibbs energy of reaction). For the nomenclature and symbols for Gibbs energy, see the first slide in this [lecture](https://www.chem.tamu.edu/class/majors/chem328/CHEM%20328_Chapter%2006.pdf).\n\n\n\n> \n> [OP in comments] In your formal notation of dS , what is the physical quantity you took partial differential with?\n> \n> \n> \n\n\nIt is the [extent of reaction](https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Topics_in_Thermodynamics_of_Solutions_and_Liquid_Mixtures/01%3A_Modules/1.14%3A_Excess_and_Extra_Thermodynamics/1.14.5%3A_Extent_of_Reaction) $\\xi$ (greek lowercase xi). It measures the amount that has reacted ($\\xi$ = 1 mol means one mole has reacted if the coefficient is one, two when the coefficient is two and so on, with reactants having negative values and products positive values).\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/166886/what-is-it-about-bleach-that-keeps-pharaoh-ants-away-even-after-drying
|
What is it about bleach that keeps Pharaoh ants away even after drying? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/166886/edit).
Closed last year.
[Improve this question](/posts/166886/edit)
I posted this question [here](https://biology.stackexchange.com/questions/108928/what-is-it-about-bleach-that-keeps-pharaoh-ants-away-even-after-drying) and one of the commentors thought that this Chemistry SE might be able to offer more sure answers.
My apartment high-rise has been afflicted with ants in the past few years. The pest control technician says that it has become a common problem due to warming weather. Various things have been tried, and I am leaving this in their expert hands (or rather, the apartment management is). This question is not asking for further countermeasures.
When I find a place or pathway with many ants, I find that simply wiping with a damp cloth or vinegar doesn't keep them away for long. Wiping is supposed to disrupt their pheromone trails, but I only find somewhat lasting effects if I wipe with undiluted household bleach. I am pleasantly surprised that they stay away even after the bleach has dried. After some internet research, I found that bleach leaves behind a salt residue when dried. **Is it the salt residue that keeps the ants away when wiping otherwise doesn't have lasting effect?**
I've had another corroborative experience that leads me to suspect that salt residue has a repulsive effect on pharaoh ants. I keep a large beer glass of very salty water on hand because of the dental benefits of salt water rinse. I probably use too strong a solution, which likely has drawbacks (something I have yet to research). It is kept on top of the fridge. Around the glass is salt water staining. I never see ants around it, even though ants like water. I admit that this corroboration is not very strong, as I generally do not see ants on top of the fridge anyway.
| 1
|
[] |
https://chemistry.stackexchange.com/questions/166885/priority-of-alphabetic-order-vs-lowest-locant
|
Priority of Alphabetic order vs Lowest locant
|
I am confused about usage priority of lowest locant and alphabetical order rule(s)
For example :
3-ethyl 4,4 - dimethyl hexane
4-ethyl 3,3- dimethyl hexane
Which is considered correct (*or preferred*) **IUPAC** name?
| -3
|
[
[
"\nLowest locant has the priority over alphabetical order while numbering. But when writing name, we write according to alphabetical order. So 4-ethyl-3,3-dimethylhexane is the correct answer. One more thing, we don't consider the prefix 'di' on 'dimethyl' while deciding alphabetical order of the name.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166875/why-will-a-lone-pair-from-h2o-not-attack-a-c-atom-in-elimination-reaction
|
Why will a lone pair from H2O not attack a C+ atom in elimination reaction?
|
In step three of reaction between H2SO4 and alcohol, why will the H2O attack the H atom instead of the C+ atom? they both have positive charge and in nucleophilic substitution reactions the c+ atom is attacked.
(Image taken from [chem guide](https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Alkenes/Synthesis_of_Alkenes/Alkenes_from_Dehydration_of_Alcohols#:%7E:text=The%20dehydration%20reaction%20of%20alcohols,phosphoric%20acid%2C%20at%20high%20temperatures.) )
[](https://i.stack.imgur.com/e8riE.png)
| 0
|
[
[
"\nIf $\\ce{H2O}$ attacks the $\\ce{C+}$ atom, it will produce the original alcohol back. No chemical effect.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166872/would-this-redox-titration-be-possible-as-a-direct-titration-rather-than-a-pseud
|
Would this redox titration be possible as a direct titration rather than a pseudo-back titration?
|
In a general chemistry lab protocol (green chemistry, ISBN-13: 978-1482230208), students determine the content of ascorbic acid in a vitamin C tablet. There is an ascorbic acid standard in a buret, and there is also some ferric ammonium sulfate solution at known concentration, i.e. iron(III). The indicator is salicylic acid, which forms a colored complex with iron(III), but not iron(II).
The suggested protocol is a back titration where you first add a known excess amount of ferric salt to the unknown amount of ascorbic acid and add the indicator, giving a colored solution. Then, you titrate back with the ascorbic acid standard. The endpoint is the disappearance of color.
Is there are reason for the back titration in this case? In other back titrations, there is a step after adding an excess of reactant (for calcium carbonate, you add an excess of HCl and then drive out the carbon dioxide with heat, making the back titration a strong acid/strong base reaction). For the ascorbic acid reacting with ferric salt, I wonder if the endpoint is easier to see, or maybe the kinetics of complex formation motivate this (it says to let the solution sit before starting the back titration).
| 4
|
[
[
"\nThe idea of using green chemistry to titrate vitamin C with environmentally friendly reagents is understandable, but it may not be the most appropriate method. Iodometry has traditionally been the traditional method for ascorbic acid analysis. The main reason for doing back-titration is that ferric ion is a very gentle oxidizing agent but good enough for ascorbic acid. Therefore, we wish to ensure that *all* of the acid in the sample is oxidized by ferric ions.\n\n\nNow note that this is a pseudo-back titration, because the titrant and the analytes are the same! In other words, after adding excess iron (III) to the sample solution, a \"direct\" titration is being conducted between ascorbic acid and iron (III). I assume the titrant concentration is higher than ferric ammonium sulfate to speed up the reaction and ensure completion of the redox process. As an exercise, why don't you ask the students to try direct titration and back-titration and then do a statistical test to see if there is a difference. I mean half the class perform direct titration and half the class do this pseudo-back titration.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/166871/retrosynthesis-of-1s-2r-1-isopropyl-2-4-dimethylcyclohexane
|
Retrosynthesis of (1S,2R)‐1‐isopropyl‐2,4‐dimethylcyclohexane
|
I think that retrosynthesis of (1*S*,2*R*)‐1‐isopropyl‐2,4‐dimethylcyclohexane involves the opening of an epoxide, to control the regiochemistry in order to achieve the *trans*-configuration via an SN2 mechanism — which could be made from a C=C double bond. Would it be plausible to assume that the original cyclohexene is commercially available?
[](https://i.stack.imgur.com/EoCRi.png)
| 0
|
[] |
https://chemistry.stackexchange.com/questions/166863/branch-of-chemistry-that-deals-with-reactions-between-chemicals
|
Branch of chemistry that deals with reactions between chemicals [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/166863/edit).
Closed last year.
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There are branches of chemistry like physical chemistry that hardly involve chemical reactions. Whereas, inorganic and organic chemistries pretty much definitely have a lot of chemical reactions involved.
Is there a collective name for organic and inorganic chemistries? Is there a single word for the branch of chemistry that deals with the possible products given some reactants and reaction conditions?
| -1
|
[
[
"\n\n> \n> Is there a collective name for organic and inorganic chemistries? Is\n> there a single word for the branch of chemistry that deals with the\n> possible products given some reactants and reaction conditions?\n> \n> \n> \n\n\nAs a collective term, *synthetic chemistry* refers to a branch of chemistry that deals with chemical reactions that produce new molecules. Please keep in mind that today these chemical branches are hard to distinguish. A synthetic chemist must know a lot of practical spectroscopy (traditional physical chemistry/analytical chemistry) in order to study his /her reactions. Also physical chemists study a lot of chemical reactions.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166860/how-to-compute-backward-reaction-rate-from-forward-one-and-an-equilibrium-data
|
How to compute backward reaction rate from forward one and an equilibrium data
|
Suppose we have have an ideal gas mixture and a reversible elementary reaction:
$$
\ce{O + H\_2 <=> H + OH}
$$
Its forward reaction rate may be computed as follows:
$$
\frac{d[\ce{H}]}{dt} = \frac{d[\ce{OH}]}{dt} = -\frac{d[\ce{O}]}{dt} = -\frac{d[\ce{H2}]}{dt} = k^{(f)}\left[\ce{O}\right]\left[\ce{H2}\right] = A T^n e^{\left(\frac{E\_a}{RT}\right)}\left[\ce{O}\right]\left[\ce{H2}\right]
$$
where $A, n$ and $E\_a$ are known values. The exponents of concentrations $\left[\ce{O}\right]$ and $\left[\ce{H2}\right]$ are both equal to one due since the reaction is elementary.
Also, there is thermodynamic data for the species - heat of formation, entropy, heat capacity - for all species. For example, for H these values are:
$$
\Delta H^{298K}\_f\left(\ce{H}\right) = 52.1~\text{kcal/mol}, \quad S^{300K}\left(\ce{H}\right) = 27.4~\text{cal/mol K},\quad C\_p(\ce{H}) = 4.97~\text{cal/mol K}.
$$
For simplicity, heat capacity is assumed constant here.
Parameters $A, n$ and $E\_a$ for backward reaction rate $\left(k^{(b)}\right)$ are unknown. According to Warnatz's et al "Combustion" book backward reaction may instead be computed from equilibrium constant, which, in turn, is determined by thermodynamic data:
$$
K\_c = \frac{k^{(f)}}{k^{(b)}}= exp\left(-\Delta\_R\bar{A}^0/RT\right)
$$
**But how to compute $\Delta\_R\bar{A}^0$?**
I assume that it equals to an increment of free energy (Helmholtz function) $A = U - TS$ corresponding to consumption of one mole of each reactant, but not sure how to properly calculate it from thermo data above.
| 1
|
[
[
"\nYou are right that you can determine the backwards rate if you know the forward rate and the equilibrium constant. In fact, you have the right formula\n\n\n$$ K\\_c = \\frac{k^f}{k^b} $$\n\n\nIn general, the equilibrium constant can be calculated from the change of the Gibbs free energy during the reaction\n\n\n$$ \\Delta G^r = - R T \\ln K\\_c$$\n\n\nTherefore in principle you need to calculate the total Gibbs free energy of the reactants and the products and take their difference to get the equilibrium constant. How to do this depends on the nature of the data you have; it might involve quite complicated thermodynamic calculations.\n\n\n**Edit:** I was mistaken previously, the equilibrium coefficient is expressed with the rate coefficients.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166854/does-chromiumiii-oxide-react-with-or-dissolve-in-liquid-sulfur-dioxide
|
Does chromium(III) oxide react with or dissolve in liquid sulfur dioxide?
|
Liquid $\ce{SO2}$ (b.p. −10 °C) is known as a [versatile solvent](https://en.wikipedia.org/wiki/Sulfur_dioxide#As_a_reagent_and_solvent_in_the_laboratory) (yet not particularly pleasant to work with) for a wide range of organic and inorganic compounds having limited solubility in conventional solvent systems. I recall that chromium(III) oxide should chemically interact with sulfur dioxide, but I cannot find a reputable source to back up this fact.
There is a related post on [Reddit — Any tips for a stubborn frit filter?](https://www.reddit.com/r/chemistry/comments/oo8nvm/any_tips_for_a_stubborn_frit_filter_chromium/) describing a problem with $\ce{Cr2O3}$ precipitation being stuck on the glass frit filter. Neither fusion with flux nor dissolution in alkali are suitable because they are guaranteed to permanently damage the frit. Oxide particles seem to be not fine enough to dissolve in strong acids, including oxidizing ones, to a significant degree or at a noticeable rate.
I think it would be interesting to know whether liquid $\ce{SO2}$ could potentially serve as a solvent for similar cases with “uncooperative” $\ce{Cr2O3}$ where relatively mild ambient conditions are expected.
| 5
|
[] |
https://chemistry.stackexchange.com/questions/166852/understanding-grades-of-steel-for-cookware-18-10-vs-18-8
|
Understanding grades of steel for cookware. 18/10 vs 18/8
|
I am looking for a stainless steel cookware and while researching on the best kind of stainless steel cookware, I came across this comment on youtube:
>
> Actually 18/10 steel is better and costlier. You're talking in terms of health hazard where more nickel (10) is considered more harmful than less (8). But in reality more nickel makes a pot sturdier and shinier and less prone to chemical leaching.
>
>
>
So, my question is, is it true that more nickel in case of stainless steel means less leaching? I know 18/10 is better in heat distribution, is sturdier and better than 18/8, but it also has more Nickle. To me, it doesn't make sense. But then again, chemistry was not subject. Or I am worrying too much and just get either 18/8 or 18/10 steel?
| 3
|
[
[
"\nNo stainless steel of the 18/8 family ( about 5 grades) leaches material in normal cooking service. Not to say they are immune to all possible abuse. The element that can make a difference is carbon, higher C can combine with more chrome removing it from solution where Cr provides the primary corrosion resistance. Not to worry, modern steel making generally gets C to 0.02 % max ( L grade). I am trying to say for cookware there is no difference.\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/166850/half-life-of-a-reversible-first-order-reaction-of-the-form-a-b
|
Half-life of a reversible first-order reaction of the form A ⇄ B
|
I am wondering if the half-life of a reversible first-order reaction of the form $\ce{A <--> B}$ is defined as
1. the time in which half of the amount of reactant which should react up to equilibrium, has reacted, or,
2. the time in which half of the total amount of reactant has reacted.
I only found [this](https://www.toppr.com/ask/question/halflife-of-a-reversible-reaction-may-be-defined-as-the-time-in-which-half-of) source.
---
**Example:** Species $\ce{A}$ reacts in first-order kinetics. The initial concentration is $1 \,\pu{mol l^{-1}}$. Its equilibrium concentration is $0.3 \,\pu{mol l^{-1}}$. After $40\,\pu{s}$, its concentration is $0.65 \,\pu{mol l^{-1}}$. After $60 \,\pu{s}$. its concentration is $0.5 \,\pu{mol l^{-1}}$. Is the half-life $40 \, \pu{s}$ (because $0.35 \,\pu{mol l^{-1}}$ of total $0.7 \,\pu{mol l^-1}$ that react, have reacted), or $60 \,\pu{s}$?
| 1
|
[
[
"\nThe [Wikipedia article](https://en.wikipedia.org/wiki/Half-life#In_non-exponential_decay) on half-life states:\n\n\n\n> \n> The term \"half-life\" is almost exclusively used for decay processes that are exponential (such as radioactive decay or the other examples above), or approximately exponential (such as biological half-life discussed below). In a decay process that is not even close to exponential, the half-life will change dramatically while the decay is happening. In this situation it is generally uncommon to talk about half-life in the first place, but sometimes people will describe the decay in terms of its \"first half-life\", \"second half-life\", etc., where the first half-life is defined as the time required for decay from the initial value to 50%, the second half-life is from 50% to 25%, and so on.\n> \n> \n> \n\n\nConcentrations of a reaction approaching equilibrium do not decay exponentially (for one, they don't approach zero as time passes).\n\n\nThe kinetics of a reaction approaching equilibrium can be complicated. If you start without product, kinetics are governed by the rate law of the forward reaction in the beginning. Then, as product is being made, the rate law of the reverse reaction plays into the kinetics as well. Surprisingly, as you get very close to equilibrium, the net forward rate decays exponentially (i.e. apparent first-order reaction) no matter what the rate laws for the forward and reverse reactions are. This is used in [temperature-jump](https://www.nobelprize.org/uploads/2018/06/eigen-lecture.pdf) and other relaxation methods to study fast reactions.\n\n\n\n> \n> [OP] I was wondering if the half-life of a reversible reaction is defined as [...]\n> \n> \n> \n\n\nFor processes that don't follow a pure exponential decay, the half-life is ill-defined. There is no good answer to this multiple-choice question. For example, what do you do when the reaction reaches equilibrium before half of the reactant is used up?\n\n\n",
"3"
],
[
"\nIn the reaction $A\\rightleftharpoons B$ the relaxation rate constant, i.e. the rate of approach to equilibrium is exponential and given by $k=k\\_1+k\\_{-1}$. The forwards and reverse reactions have rate constants $k\\_1,k\\_{-1}$ respectively. The half-life, if you want to define it is, $\\ln(2)/k$.\n\n\nIf the reaction is $A+B\\rightleftharpoons C+D$ then the relaxation time is still exponential but now depends on the equilibrium concentrations.\n\n\nYou can solve the kinetics $A\\rightleftharpoons B$ using $x=[A\\_0]-[A]=[B]$, write down the rate equation $d[A]/dt$, substitute for $x$ and then integrate $dx/dt$ from $0\\to t$. Next represent the equilibrium constant as $K\\_e=k\\_1/k\\_{-1}=x\\_e/([A\\_0]-x\\_e)$ where subscript $e$ is the equilibrium amount. You should find that\n\n\n$$\\displaystyle x=x\\_e(1-e^{-(k\\_1+k\\_{-1})t})$$\n\n\nwhich is a rising exponential as $x$ is the amount of B at time $t$. When $t=0$ this is zero and at long times $x\\to x\\_e$ as expected. Species A falls as B increases (from $1$ to $0.3$ in your example) and with the same rate constant $k$.\n\n\nThe initial amount of A is $1$ and equilibrium amount $0.3$ (so $x\\_e=0.7$) and the half life should be the time to reach half of the range from $1$ to $0.3$ as this is the range that A covers.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166849/is-vapour-pressure-truly-a-colligative-property
|
Is vapour pressure truly a colligative property?
|
Colligative properties are generally defined as
>
> A colligative property depends only on the ratio of the number of particles of solute and solvent in the solution, not the identity of the solute.
>
>
>
[Relative lowering of vapour pressure](https://en.wikipedia.org/wiki/Colligative_properties#:%7E:text=References-,Relative%20lowering%20of%20vapor%20pressure,-%5Bedit%5D) is considered to be a colligative property.
I think, lowering of vapour pressure is not a true colligative property because in certain conditions, it does depend on the nature/identity of the solute.
Consider that we have such a solute mixed with a solvent in a solution that the solution exhibits a negative deviation from Raoult's law, i.e., the attractive forces between solute-solvent are stronger than the attractive forces between solute-solute or solvent-solvent molecules, e.g., mixture of phenol and aniline or a mixture of chloroform and acetone. The solute particles are more attracted to the solvent particles and this inhibits solvent particles from escaping easily from the solution, lowering the vapour pressure of the solvent.
In such a case, lowering of vapour pressure of solvent is clearly dependent on the nature of the solute.
Is vapour pressure truly a colligative property?
| 1
|
[
[
"\nI agree with you that generally, vapor pressure of a solution depends on, not only concentrations of its ingredients, but also on their chemical composition.\n\n\nAddition of ethanol to water will decrease the boiling point of solution relative to water. While, adding acetic acid will increase the boiling point of the solution. One may think a solute that is more volatile than water(as a solvent), will always decrease the solution boiling point. This is what raoult's law predicts. But, this isn't always true and depends on the ideality of the solution. For example, consider Hydrochloric acid. The boiling point of liquid Hydrogen chloride is -85 degrees of centigrade and is more volatile than water. But after it dissolves in water, the boiling point of solution can even increases up to 108 degrees(in some concentrations) This is because the interactions between particles in Hydrochloric acid is stronger than pure water. This stronger bonding arises from dissociation of Hydrogen chloride molecules into ions. When the intramolecular forces and sizes of the components in a solution are near to each other the solution behaves more ideally. One example is a solution of n-pentane and n-hexane.\nIn addition, when a solution is diluted the solution becomes more ideal. \n\n\nBack to your question, that wondering about vapor pressure of a solution to be a colligative property.\nI think the vapor pressure in solutions that, the solute is **non-volatile** may only depend on solute concentration. If so, in this case the vapor pressure of the solution is a colligative property. But if the solution behaves ideally. Let me explain the reason for it.\nAt the surface of the solution, a number of solvent particles are exchanged by solute particles. In an ideal solution, the solute particles bond to solvent molecules, just as solvent molecules bond to each other (same intramolecular forces and molecular size). So regardless of the identity of desired solutes, the chance of solvent molecules to escape from the surface will decrease with respect to concentrations of solute. As the solute is non-volatile, the total vapor pressure is equal to solvent vapor pressure. And, that can be predicted by only the concentration of the solute(a colligative property). Conceptually, the ideality of the solution is an important factor in the answer of your question. Ideality is not an absolute property of a solution. Any solution can be considered to be non-ideal with sufficient precision in measurement. Thanks. \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166846/does-aqueous-koh-favor-sn2-or-sn1-with-secondary-halides
|
Does aqueous KOH favor SN2 or SN1 with secondary halides?
|
I have read that aq. KOH is *equally reactive* towards SN1 and SN2 with secondary halides. I'm aware that aq. KOH will participate in SN2 as well because of the lack of steric hindrance in secondary halides, but since aq. KOH is a polar protic solvent, shouldn't it *favor* SN1 comparatively?
| 1
|
[] |
https://chemistry.stackexchange.com/questions/166841/in-bulletproof-vests-why-do-we-use-ceramic-plates-instead-of-metal
|
In bulletproof vests, why do we use ceramic plates instead of metal?
|
In many articles, it has been mentioned that bulletproof vests also contain ceramic plates for extra protection. Wouldn't it be better to use metal instead? Since ceramic breaks easily.
| 0
|
[
[
"\nI think [Wikipedia](https://en.wikipedia.org/wiki/Ceramic_armor) provides a very good summary:\n\n\n\n> \n> Unlike metals, ceramics are never used alone, as standalone materials, in armor systems; they are always combined with a ductile backing or support layer of metal or fiber reinforced plastic composite materials, and this ceramic-faced assembly is called ceramic armor. Ceramic materials, like glass, have high hardness and compressive strengths but low tensile strengths. Bonding a ceramic tile to a metallic or composite backing material, with high strength and good ductility, delays or mitigates tensile failure upon impact, and forces the ceramic to fail in compression.\n> \n> \n> \n\n\nSo ceramic would never used on its own in a bulletproof vest. It will always have metal or something with similar strength.\n\n\n\n> \n> Ceramic armor systems defeat small arms projectiles and kinetic energy penetrators by two main mechanisms: Shattering and erosion. When a hard steel or tungsten carbide projectile hits the ceramic layer of a ceramic armor system, it is momentarily arrested, in a phenomenon known as dwell. Depending on the thickness and hardness of the ceramic layer, the projectile core is then either shattered, fractured, or blunted. The projectile's remnants continue to penetrate the comminuted ceramic tile at a reduced velocity, which erodes those remnants and reduces their energy, length, and mass. The metal or fiber reinforced plastic composite layer behind the ceramic layer then arrests the projectile's fragments or its eroded remnant, and absorbs residual kinetic energy, typically via plastic deformation.\n> \n> \n> \n\n\nEven though ceramic breaks easily as you said, it has high compressive strength, i.e. it can resist compression to a large extent. When combined with metal backbone, this helps to reduce the kinetic energy of the bullet and thereby reduce the damage.\n\n\nAs an aside, I should add that this idea that bullets can only be stopped with a hard material like metal is not always true. While metals can stop bullets, they are also heavy and reduce mobility. Stopping bullets requires that its kinetic energy be absorbed and dispersed accross a larger area so that the impact is no longer felt. This can be done with materials other than metal. For example, during the Vietnam war, [flak jackets](https://en.wikipedia.org/wiki/Flak_jacket) (armor) without any metal plating was used. These would be made from multiple layers of ballistic nylon, along with fibreglass plates, and had much less weight than metal plated armors. The multiple layers of strong fabric and the fibreglass would spread the impact of incoming projectile across a larger area. Note, however, that these jackets were only effective against grenade fragment, shotgun pellets, and possibly small pistols. They would not work against rifles or anything stronger.\n\n\nIn the modern day, polymeric materials like Kevlar are used to protect against bullets. There are also many researchers working in the material science side of chemistry to find new materials that can be light but withstand impact.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166838/what-makes-a-compound-high-or-low-energy-and-why-does-that-relate-to-reactivity
|
What makes a compound high or low energy and why does that relate to reactivity? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Add details and clarify the problem by [editing this post](/posts/166838/edit).
Closed last year.
[Improve this question](/posts/166838/edit)
After reading about thermodynamic stability, it now makes sense to me how the enthalpy of a reaction relates to the thermodynamic stability of a compound.
However, when looking for a clear way to define thermodynamic stability, a few sources mentioned how high-energy compounds are more reactive and therefore less thermodynamically stable. What actually makes a compound high, or low, energy and how does this relate to reactivity?
| -1
|
[
[
"\nStronger bonds make more stable compounds both kinetically and thermodynamically. Kinetically because bonds must be broken for reactions to happen. Thermodynamically because bonds must be formed. A second factor is the reaction must be entropically favored. A third factor is the available energy thermal, photolytic or electrical must be sufficient to either not break bonds, stability, or to break bonds, reactivity.\n\n\nBonds are formed when electrons are shared between atoms, electrons are attracted to two or more nuclei. This can be a weak attraction such as intermolecular attractions; weak molecular bonds such as found in molecules such as $\\ce{F2},$ $\\ce{I2},$ $\\ce{Li2};$ stronger bonds such as $\\ce{C-H},$ $\\ce{C-C},$ $\\ce{C-O},$ $\\ce{C-F},$ $\\ce{C-Cl},$ $\\ce{H-H},$ most ionic bonds and metallic bonds; and very strong bonds such as $\\ce{C=O},$ $\\ce{C=C},$ and the triple bonds in $\\ce{N#N},$ $\\ce{C#O},$ $\\ce{HC#N},$ $\\ce{HC#CH}.$ Double and triple bonds have unique properties because of their orbital structures.\n\n\nAll bonds are stable under low enough energy conditions or possibly a molecule in isolation. If energy is added: Heat, Photons, electricity, etc. Bonds will break. The energy to break sufficient bonds to cause a reaction is called the activation energy. The bonds will either reform or form new bonds. If the reaction is entropically favored by removal of energy stronger bonds will form and the reaction will proceed. The final products are determined by the strength of the bonds in the various compounds and the distribution of the energy involved.\n\n\nMolecules with weak bonds are considered to be high energy; they tend to be reactive. Molecules with strong bonds are considered to be low energy; they are more stable. Reactions happen in the direction of forming stronger bonds and increased entropy. The latter is mostly manifested in the disposition of the energy evolved in the formation of the stronger bonds.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166837/does-the-nature-of-the-h-a-bond-tell-us-about-the-deltag-of-an-acids-dissoci
|
Does the nature of the H-A bond tell us about the $\Delta$G of an acid's dissociation?
|
When I first learned about acid strength trends in general chemistry, I remember it was explained in terms of the character of the H-A bond. The more polar the H-A bond, the easier it was to heterolytically split it into H$^{+}$ and A$^{-}$. This explained why acid strength increased from left to right across the periodic table: as you went from left to right, electronegativity increases, so the H-A bond becomes more polar, hence more acidic. The strength of the H-A bond also contributed to the acidity; the weaker the bond, the easier to break it into A$^{-}$ and H$^{+}$, so the higher the acidity. This explained why acidity increased as you went down a group; the H-A bond became weaker, leading to higher acidity.
However, I have also seen acidity strength explained solely through the stability of the conjugate base, without mention of the nature of the H-A bond. When viewing acidity through the base stability point of view, we can explain the increase in acidity from left to right as a result of charge stabilization: in the conjugate base, a more electronegative atom is better able to stabilize the excess negative charge, therefore the conjugate base becomes more stable as the electronegativity of the atom attached to the acidic proton increases. The trend down a group can also be explained in terms of charge stabilization: as the size of an atom increases, any excess charge on the atom is spread out due to the diffuse orbitals, leading to charge stabilization. I think because I so often saw acid strength just explained in terms of base stability, I started to think that the H-A bond explanation was just a simplified version of the base stability argument.
However, I recently saw this question, which made me reassess my thinking: [](https://i.stack.imgur.com/inJYg.png)
Here, you can not use base stability to explain the acidity difference, because both acids have the same conjugate base. So in this case, any difference in acidity must come from the stability of the acid, not the base. In other words, the differences in the $\Delta$G's for the two dissociations must be due to a difference in the Gibbs Free Energy of the acids. This makes me think that the discussions in general chemistry about the H-A bond were really explaining the $\Delta$G of acid dissociation, which was independent of the stability of the base. Now, I am starting to think that weak and polar H-A bonds lead to more acidity because they tend to create a more exergonic reaction (if this is the case, could someone explain why?).
This leads to my overall question: **is explaining acidity through the nature of the H-A bond just another way of explaining it through conjugate base stability, or does the H-A bond tell us something about the $\Delta$G of the acid dissociation that we would not know from looking at conjugate base stability? Is it just coincidental that the nature of the H-A bond and the stability of the conjugate base contribute to the same trends in acid strength?**
| 2
|
[
[
"\nThis is a good question, and Ill try my best!\n\n\nThe acidity of an acid is solely dependent on the nature of the HA bond. When learning about acidity and basicity, the concept of conjugate base stability is invariably taught to help explain the strength of various acids. But, molecules don't have brains. They arent thinking, \"When I lose my proton, I'll be super stable, so I wanna be really acidic!\"\n\n\nThe *electronics* of the starting acid are what determines its acidity, and as a result, things that are more acidic have relatively stable conjugate bases. The stability of the CB only *helps* explain its acidity (and 99% of the time, it works).\n\n\nOH vs CH? Oxygen is more electronegative, pulling electron density away from H, so the electron pair done (the base) can use its electrons to grab that hydrogen more easily.\n\n\nCarboxylic acid vs alcohol? Oxygen is donating electron density to the carbonyl, resulting in the oxygen of the OH bond being more electron deficient. But electronegative oxygen wants electrons, so it pulls even MORE electron density from H, making the proton even more grabbable than in an alcohol.\n\n\nThiol vs alcohol? Yeah sure, oxygen is more electronegative, but the orbital overlap between sulfur and hydrogen is garbage, that is, the energy gap between H's s orbital and sulfur sp3 orbital is much larger than that of an alcohol. Thus the S-H bond is less stable kinetically, and so bases come and snatch that hydrogen.\n\n\nTo answer your question completely, it is **NOT** coincidental that the acidity of HA correlates with the stability of the conjugate base. However, rather than thinking that HA is more acidic *because* the CB is more stable, think about how the two concepts go hand in hand and influence *each other*.\n\n\nHere's another example: The iodide in tert-butyl iodide doesn't leave *because* it forms a more stable tertiary carbocation; the molecule doesn't *know* that it can form a stable tertiary carbocation. Rather, it leaves because the adjacent sigma C-H bonds donate electrons density to the C-I antibonding orbital, significantly labilizing the C-I bond. Thus, with enough energy, iodine leaves, forming the tertiary carbocation.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166833/corrosiveness-of-deionized-vs-distilled-water-for-use-in-laboratory-water-bath
|
Corrosiveness of Deionized vs Distilled water for use in Laboratory Water Bath [duplicate]
|
**This question already has answers here**:
[Is pure water very corrosive?](/questions/7467/is-pure-water-very-corrosive)
(7 answers)
Closed last year.
I've read conflicting descriptions of the corrosiveness (or lack thereof) of very pure water. For a precision water bath, it's stated that one should fill the bath with distilled water rather than deionized of a certain purity [as it would be corrosive](https://tools.thermofisher.com/content/sfs/manuals/D01203%7E.pdf):
>
> 5.2 Filling Bath
>
> NOTE: DISTILLED WATER IS RECOMMENDED; IT WILL
> NOT CORRODE THE BATH CHAMBER
> AND IT REDUCES THE NEED FOR FREQUENT
> BATH CLEANING.
>
>
> DO NOT USE 18 MEG DEIONIZED WATER.
>
> NOTE: If this is the only
> source of treated water available; mix 50% with regular tap water.
>
>
>
Why is this? I'm struggling to find a specific mechanism by which purer water would corrode the bath.
| 1
|
[
[
"\nYou can stop struggling. Your chemical instincts are correct; the purer water is, the less corrosive.\n\n\nThe main factor in the corrosivity of water solutions is the ion content, the ability of the water to carry a current in galvanic reactions. Lower ions, less current, less corrosion. The second factor is the presence of dissolved ions etc. to participate in various and sundry reactions. The less of these the fewer possible reactions. It is best to use water of the highest quality available in water baths, humidifiers, drip coffee makers, CPAP devices, plating baths, even laboratory dishwashers, and definitely as a chemical reactant or solvent. Printed circuit board manufacturing uses only the highest quality DI water in any aqueous process [and they probably wish they could get rid of the pesky 10^-7M H3O+ and OH-]. [Which they do by inventing these super nonionic detergents and no residue fluxes and, once upon a time, freon rinses.]\n\n\nThe increased corrosiveness of pure[r] water is a myth that I think originated with the water supply companies that found that certain plumbing structures literally fell apart when high calcium water was replaced by softened water. The pipes were partially held together by buildup of CaCO3 deposits and the common ion effect prevented dissolution of the deposits. Remove Ca++ from the water the deposits dissolved and everything leaked or even fell apart. This was probably part of the problem in Flint, MI. A change in water disturbed decades of protective coating in the pipes; the problem was deeper than removal of a coating.\n\n\n",
"1"
],
[
"\nNASA published a 23-paged reported in the 60s on the corrosion of metals in deionized water at 38 $^o$C. The report is titled \"Corrosion of Metals in Deionized Water at 38° C (100° F)\" By Barbara Alice Johnson, 1969.\n\n\nAs one can see, we cannot make sweeping statements about corrosion in deionized water. Some metals will corrode and some will not in deionized water. Long time ago, someone told a story of detecting measurable amounts molybdenum in human blood. They were using very sensitive analytical technique and it puzzled everyone as to why Mo is being detected. It turned out that the metallic syringe needle leached a few atoms of Mo in a very short period of drawing blood. Yes, deionized or distilled water will leach some metallic ions but corrosion means *measurable metal loss to the extent that it causes economic damage*.\n\n\nThus, corrosiveness distilled vs. deionized water is like a popular urban myth. What the manufacturer probably want is some level of slow scaling in the bath to \"protect\" metallic heaters.\n\n\n[](https://i.stack.imgur.com/fXhj9.png)\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166832/what-compounds-smell-like-burning-rubber-or-tar
|
What compounds smell like burning rubber or tar?
|
**Background:**
A couple of friends of mine live next door to each other in a 10-year old block of flats in London, UK. They say that in the early hours of the morning, a toxic smell appears in their flats, which smells a little bit like burning rubber or tar. It also makes their eyes burn and affects their breathing. Their attempts to inform the housing authority ended in dead ends, as the housing authority didn't seem to believe them about the smell.
In order to try to get to the bottom of this, I had some of their air sampled inside one of the flats (once when the smell was present, and once when it was not present), and sent it for Gas chromatography–mass spectrometry at a lab.
The lab only tests for about 70 specific volatile organic compounds, and none of those compounds showed much concentration change between the two samples.
Presumably, whatever is causing the smell is not one of those compounds.
**Questions**:
1. What compounds smell like burning rubber or tar?
2. If the lab sends me the raw data from the GC-MS test, how easy is it to identify the presence those compounds from the data?
| 0
|
[
[
"\nCheck local history:\n\n\n1. Was the land used as a dump or filled in with [waste wood](https://bjrbe-journals.rtu.lv/article/view/bjrbe.2018-13.424)? Creosote is derived from tar, and it is used as a wood preservative. If creosoted wood, such as railroad-ties, was used in landfill, that could lead to a strong odor.\n2. [Coal ash or clinker](https://www.nrdc.org/onearth/coal-ash-hazardous-coal-ash-waste-according-epa-coal-ash-not-hazardous-waste) can have a strong smell. Was a coal-burning power plant situated where those flats are now located? Was waste from coal-burning buried there?\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/166831/in-a-20wt-ammonia-solution-what-are-the-species-in-solution
|
In a 20wt% ammonia solution, what are the species in solution?
|
I have bought a 20wt% ammonia solution. It is described as having a specific gravity of 0.92.
I worked out its Molarity earlier today, and I think it was approximately 10M
My question can I assume that the solute remains effectively as ammonia molecules because the dissociation constant is so small?
In my ammoniated solution I have:
NH3 + H2O and I want to know what concentration of NH4$^+$ and OH$^-$ I have.
So my equation is NH$\_3$ + H$\_2$O <<<>>> NH$\_4^+$ + OH$^-$
I have a dissociation constant of ~ 10$^{-5}$.
So if I assume that from 1 mole of ammonia, x moles of ions form I get
10$^{-5}$ = $\frac{x^{2}}{1-x}$, so x = 3.16 x 10$^{-3}$ i.e.there are 0.003 moles of ammonium ions in solution.
Could someone please check? If this is the case, I am going to ignore their presence
Thank you
| 0
|
[
[
"\nThanks for your question. Your equation and calculations are correct.\nFirst, I have to say that the variable in equilibrium equations is **concentration**. By multiplying concentration into volume of your solution, you can compute the number of moles of the ion or extra. So in your example of ammonia solution, by considering the concentration of ammonia be 1M and equilibrium constant be equals to 1.8 \\* 10^-5, the concentration of ammonium ion would be calculated 0.00423M and the concentration of remaining ammonia after dissociation would be (1-0.00423= 0.99577M). We can express this amount of dissociation by *dissociation degree*. I think this is what you are looking for. It is the ratio of dissociated solute to the entire solute that had been dissolved. In the 1M Ammonia solution, the dissociation degree is 0.42% (just a note, in ammonia solution, ammonia doesn't dissociate. Actually, water will dissociate and acts like an acid by giving a proton to Ammonia.)\nIf I understand correctly, your claim was this that, Whenever the dissociation constant of an acid or base be law, the dissociation degree will be law too(most of the solute remain undissociated). That seems to be correct, but it's not!\nActually, the dissociation degree not only depends on equilibrium constant, but also to solute concentration. The equation bellow demonstrate the relation between dissociation degree, initial solute concentration and equilibrium constant that is called the law of dilution (<https://en.wikipedia.org/wiki/Law_of_dilution>).\n\n\n[](https://i.stack.imgur.com/2Iacx.png)\n\n\nWhere the square brackets denote concentration, and c0 is the total concentration of electrolyte and the degree of dissociation of a weak electrolyte is α.\nLet's see how α percent change by concentration of ammonia solution. I have solved that equation for several concentrations as you can observe below.\n\n\n11M (~20% weight percent) : 0.128%\n\n\n10M : 0.135%\n\n\n9M : 0.142%\n\n\n8M : 0.15%\n\n\n7M : 0.161%\n\n\n6M : 0.17%\n\n\n1M : 0.42%\n\n\n0.1M : 1.34%\n\n\n0.01M : 12.6%\n\n\nAs you can observe the degree of dissociation is increased by dilution of ammonia solution. So, for your propose you have to consider concentration too.\n\n\nOne last thing is, these calculations are only assumptions of what concentration is going to be measured. Divisions may observe when another salt for example is in solution. For more precise computing of the degree of dissociation and extra, you need to consider other possible equilibrium too.\n\n\nTake care about reactions with ammonia, that is capable to form explosive and or toxic compounds in some circumstances.\n\n\nA site about degree of dissociation:\n<https://www.nagwa.com/en/explainers/950137956236/>\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/166829/why-is-active-sec-pentyl-group-called-so
|
Why is active sec-pentyl group called so?
|
My teacher as well as [Wikipedia — Pentyl group](https://en.wikipedia.org/wiki/Pentyl_group) use the following names for the “pentyl” groups:
| | Group name | Structure with attachment point |
| --- | --- | --- |
| **1** | active *sec*-pentyl | [active sec-pentyl](https://i.stack.imgur.com/S1fT0.png) |
| **2** | *sec*-pentyl | [sec-pentyl](https://i.stack.imgur.com/ngrqH.png) |
| **3** | active pentyl | [active pentyl](https://i.stack.imgur.com/an6Ir.png) |
If a group is attached to the second carbon of pentyl, then it is called “active *sec*-pentyl” (**1**). If it is attached to the third carbon, then it is called “*sec*-pentyl” (**2**). In case **3** it is called “active pentyl”.
What is the difference between them, and why are they called so?
| 1
|
[
[
"\nI, as orthocresol♦ questioned in the comments section, don't know where these terms came from. Thus, any attempt to rationalize these terms would be a speculation. For clarity, it could be argued that each of the structures of these alkyl radicals are drawn in such a way that each of them is attached to a functional group such as hydroxyl or halide or etc. For sake of an argument, let's assume the functional group is hydroxyl, thus the compound from the first radical would be 2-pentanol (or pent-2-ol). Thus, let's try to give a meaningful rationalization to the terminology in the question:\n\n\n1. Why it is called 'active sec pentyl' (case 1): If you carefully look at the alkyl radical in case 1, you would realize it is a secondary radical so that it is a *sec*-pentyl group. Thus, the compound is **optically active** alcohol (in our case), pent-2-ol, which can exists in both (*R*)- and (*S*)-forms. Thus, I assume 'active' part comes to indicate optically active compounds.\n2. Why it is called 'sec pentyl' (case 2): If you carefully look at the alkyl radical in case 2, you would realize it is also a secondary radical so that it is also a *sec*-pentyl group. Thus, the relevant compound in our case is pent-3-ol, which can exists in only one form, because it is **not optically active**. Thus, I assume 'active' part is dropped from 'active sec pentyl' to indicate the radical forms only the non-optically active compounds, yet it is a 'sec pentyl.'\n\n\nThe compound related to \"case 3' is 3-methylbutan-1-ol, which is **optically active** so that it has been given the terminology, 'active pentyl.'\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/166827/retrosynthesis-of-2e-1-2-4-dimethylphenyl-3-thiophen-2-ylprop-2-en-1-one
|
Retrosynthesis of (2E)-1-(2,4-dimethylphenyl)-3-(thiophen-2-yl)prop-2-en-1-one
|
I am struggling to find a way to disconnect this molecule and I am unsure of the order:
[](https://i.stack.imgur.com/EYBMo.png)
First, I was thinking of disconnecting the C=C of the α,β-unsaturated carbonyl. Then for the thiophene group disconnecting the formyl substituent, then disconnecting the thiophene either side of the sulfur atom to give a 1,4-diketone.
Then for the benzene substituent: disconnecting the carbonyl first (via Friedel–Crafts acylation), as the methyl groups will be *ortho*/*para*-directing.
Would this be the correct order?
| 7
|
[
[
"\nThere are several approaches to this molecule, but I think the key disconnection is the *E*-double bond. This looks like it can be formed by a [Horner-Emmons-Wadsworth](https://en.wikipedia.org/wiki/Horner%E2%80%93Wadsworth%E2%80%93Emmons_reaction) reaction between the widely commercially thiophene-2-carboxaldehyde and a ketophosphonate derived from 2,4-dimethylacetophenone. HEW reactions give predominantly *E*-alkenes which is the reason for the choice.\n\n\nThe ketophosphonate can be prepared in good yield from the commercially available 2,4-dimethylbenzoyl chloride by reaction with diethyl phosphonoacetic acid/MgCl2/Et3N, method described [here](https://www.tandfonline.com/doi/abs/10.1080/00397910008087362?journalCode=lsyc20).\n\n\nIt is possible to prepare both starting materials, but it is an important principle of synthetic chemistry that your time as a researcher is more valuable than the cost of advanced materials. If a starting material is available, buy it.\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/166490/relationship-between-molar-flux-and-density-of-current
|
Relationship between molar flux and density of current
|
I am studying the mass transport for a charged species in a stirred solution. So, I have obtained the Nernst-Planck equation:
\begin{equation}
\mathbf{J} = -D\nabla C - \frac{zFDC}{RT}\nabla \phi + C\mathbf{v}
\end{equation}
where $\mathbf{J}$ is the molar flux, $C$ is the molar concentration of the charged species, , $z$ is the ion valence, $R$ is the gas constant, $F$ is the Faraday constant, $D$ is the diffusion coefficient, and $\mathbf{v}$ is the stirring rate of the solution.
Now, I want to relate $\mathbf{J}$ with the current $\mathbf{I}$. By definition the current sign is given by the movement of the positive carriers. I have found this relation in many electrochemistry books:
\begin{equation}
\mathbf{J} = -\frac{\mathbf{i}}{zFA}
\end{equation}
where $A$ is the area. I have nearly understood this relation, but I can't understand the sign minus. Why? If $z>0$, the carriers are cathions, positive charges. Therefore their movements should be in the same direction of the current. Where is my mistake?
| 1
|
[] |
https://chemistry.stackexchange.com/questions/166483/difference-between-number-of-hydrogen-bonds-formed-and-number-of-hydrogen-bonds
|
Difference between number of hydrogen bonds formed and number of hydrogen bonds in a molecule [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
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Closed last year.
[Improve this question](/posts/166483/edit)
I would like to know how to count the number of hydrogen bonds formed in a molecule and the number of hydrogen bonds that a molecule has. For example, water forms two hydrogen bonds with its two hydrogen atoms, while accepting two hydrogen bonds from other molecules' hydrogen atoms, which makes a water molecule have four hydrogen bonds.But some molecules such as ammmonia have only one lone pair at the electronegative atom, while the number of hydrogen atoms in the molecule is higher than the number of receptors in the molecule, while in some cases, the number of lone pairs in the electronegative atom is higher than the number of hydrogen atoms? How do I find the number of hydrogen bonds formed? How do I find the total number of hydrogen bonds in a molecule?
| -3
|
[
[
"\nThe number of hydrogen bonds formed by an atom depend on several factos:\n\n\n1. Presence of small, highly electronegative elements ($\\ce{O}$,$\\ce{F}$,$\\ce{N}$, and sometimes $\\ce{Cl}$)\n2. Number of hydrogens covalently bonded to such elements\n3. Spatial considerations\n4. State-dependent factors (Temperature, Pressure, etc.)\n\n\nYou are confusing covalent bonds present in water with hydrogen bonds. In the stucture ice (phase?), each water molecule does form four hydrogen bonds in a tetrahedral geometry ([wiki](https://en.wikipedia.org/wiki/Hydrogen_bond)).\n\n\nYou may also find this [article](https://techiescientist.com/does-nh3-have-hydrogen-bonding/#:%7E:text=Hydrogen%20bonding%20is%20the%20intermolecular%20forces%20acting%20between,partial%20positive%20charge%20develops%20on%20the%20hydrogen%20atom.) useful.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166481/stable-conformation-of-2-3-dimethylsuccinic-acid
|
stable conformation of 2,3-dimethylsuccinic acid
|
I was researching for answering this [question](https://chemistry.stackexchange.com/questions/166474/kolbes-electrolysis-reaction-for-alkenes-production) on Kolbe's electrolysis of 2,3-butanedicarboxylic acid (succinic acid). Turns out, this compound is not very well researched in particular, and shows few results on search engines. I did find some results for 2,3-dimethylbutanedioic acid (2,3-dimethylsuccinic acid). [PubChem](https://pubchem.ncbi.nlm.nih.gov/compound/11848) shows the anti-conformer as the standard structure.
[](https://i.stack.imgur.com/amm9e.png)
Now, the compound has two stereogenic carbons and 3 isomers:
1. R-R
2. R-S
3. S-S
Do any of these isomers under any conditions show higher stability for the syn-conformer? I ask because of the possibility of $\ce{H}$-bonding:
[](https://i.stack.imgur.com/JkoWH.png)
Would the stability due to $\ce{H}$-bonding be enough to overcome the steric repulsion?
---
As told in comments, the above schematic has a 7-membered ring, which are generally less stable than 6-membered rings. This prompted me to draw another diagram which has two 7-membered rings instead:
[](https://i.stack.imgur.com/grxTj.png)
I see that the rings are somewhat strained, but that doesn't mean that, at least, one of the two hydrogen bonds does not form.
| 2
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[] |
https://chemistry.stackexchange.com/questions/166480/why-dont-we-take-activities-in-the-rate-law
|
Why don't we take activities in the rate law?
|
I am slightly confused by the fact that concentrations and not activities are taken in the rate law which is different to equilibrium and seems to lead to a discrepancy. For example, for the reaction $aA(g)+bB(aq)+cC(g)⇌eE(g)+fF(aq)+gG(aq)$, the value of $K\_{eq}$ is:- $$K\_{eq} = \frac{(P\_{\ce{E}})^e[\ce{F}]^f[\ce{G}]^g}{(P\_{\ce{A}})^a[\ce{B}]^b(P\_{\ce{C}})^c}$$
where all quantities are dimensionless.The above expression follows from the definitions of activities of gases and solutes.
Now, the rate law for the forward reaction will be $v\_f= K\_{fwd} [A]^a[B]^b [C]^c$ and for the backward reaction will be $v\_b= K\_{bwd} [E]^e [F]^f[G]^g$ where [X]=$ \frac{P\_x}{RT}$. Since, at equilibrium, $v\_f=v\_b$, so $K\_{fwd} [A]^a[B]^b [C]^c=K\_{bwd} [E]^e [F]^f[G]^g$. Since $K\_{eq}$=$\frac{K\_{fwd}}{K\_{bwd}}$ so we have $$K\_{eq} = \frac{[{\ce{E}}]^e[\ce{F}]^f[\ce{G}]^g}{[\ce{A}]^a[\ce{B}]^b[{\ce{C}}]^c}$$
which is incorrect. So, do we take activities in the rate law(and concentration is a common misquotation) or is there something else I'm missing?
| 0
|
[] |
https://chemistry.stackexchange.com/questions/166478/alkynes-product-with-bromine-water
|
Alkynes product with bromine water [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
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[Improve this question](/posts/166478/edit)
Has anyone analyzed the products of the reaction of any simple alkyne (eg.: acetylene) with bromine water? Most online references state that a dihalo-addition or tetrahalo-addition happens, but the only experimentally-based reports I found are more than 90 years old and state that dihalooxy-addition happens (although with hypobromous acid), leading to α,α-dibromoaldehydes and α,α-dibromocarboxylic acids (I expect α,α-dibromoketones possible with internal alkynes). Any new data on that?
| 1
|
[
[
"\nThere are 4 kinds of reactions which occur for sure:\n\n\n1. addition of dihalogen to alkene and alkyne in carbon tetrachloride gives vicinal dihalide(alkane and alkene respectively).\n2. addition of hydrohalic acid to alkene and alkyne gives monohalogenated alkane and alkene respectively.\n3. addition of dilute hydrobromic acid to alkene gives alpha-bromo alkanol\n4. addition of hydrohalous acid to alkene also gives the same product\n\n\nIf such kind of reactions are possible then I think that the reaction about you are talking should also be possible via tautomerism of alkenol, but I can't see how can carboxylic acids form.\n\n\n[Addition of HOBr across C=C](https://chemistry.stackexchange.com/q/42599/124759)\nI think this could be helpful\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166475/does-zinc-oxide-lose-its-uv-blocking-properties-when-mixed-with-oil
|
Does zinc oxide lose its UV blocking properties when mixed with oil?
|
Zinc oxide provides strong protection against UVA and UVB radiation. Does $\ce{ZnO}$ retain this protection when it is dissolved in oil?
When $\ce{NaCl}$ is dissolved in water, the $\ce{Na+}$ and $\ce{Cl-}$ ions are pulled apart by the water molecules. I would assume that a similar thing happens when $\ce{ZnO}$ is dissolved in oil, where the $\ce{Zn^2+}$ is separated from the $\ce{O^2-}$ ions.
Also, according to [Wikipedia — Zinc oxide](https://en.wikipedia.org/wiki/Zinc_oxide),
>
> ZnO reacts slowly with fatty acids in oils to produce the corresponding carboxylates, such as oleate or stearate.
>
>
>
But that still doesn't tell me whether that would lead to a loss of the UV protective properties.
| 3
|
[
[
"\n$\\ce{ZnO}$ is not soluble in oils. But it can form dispersion in oils. $\\ce{ZnO}$ may react with traces of fatty acids in oils, forming respective zinc salts. But for UV screening context, it is negligible negative effect. $\\ce{ZnO}$ does not lose in such a case its properties, there is just little less of it there.\n\n\nThe major factor in UV screening context is difference in refraction indexes of $\\ce{ZnO}$ ( or $\\ce{TiO2}$ ) particles and the fluid (liquid or gas). The bigger difference means stronger light scattering and longer average UV path through the given layer.\n\n\nWith a longer optical path comes also better UV absorption by the used liquid, more exactly by UV absorption additives.\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/166474/kolbes-electrolysis-reaction-for-alkenes-production
|
Kolbe's electrolysis reaction for alkenes' production
|
I came across a question in which the major product was asked for Kolbe's electrolysis of potassium salt of 2,3-dimethylbutane-1,4-dioic acid. The answer was given as trans-(but-2-ene). I couldn't understand how does the trans isomer form as the specific major product from this reaction, following free radical mechanism. Would someone please help me with this. Thanks!
| 2
|
[
[
"\nIt has little to do with mechanism and more to do with the fact that stable products are, in general, the major products in most of the chemical reactions.\n\n\nIn this case, trans-(but-2-ene) is more stable than cis-(but-2-ene), due to steric reasons.\n\n\n",
"2"
],
[
"\nGenerally during electrolysis of dicarboxylic acid we will get a few different products but the main product is unsaturated hydrocarbon. Maybe at this point, based on ChemElectroChem, 2020, 7, 4874 - down of page, the mechanism reaction is like in drawing:\n\n\n[](https://i.stack.imgur.com/bWi0f.png)\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/166468/iapws-if97-empirical-equations-for-water-and-steam
|
IAPWS-IF97 empirical equations for water and steam
|
IAPWS-IF97 provides empirical equation the dynamic viscosity and the thermal conductivity for water, are there any formulas by IAPWS-IF97 for the density and the specific heat capacity of water?
I found "some" empirical formulas for both, the density and the specific heat capacity of water by going through some papers, but I wonder about recommendations by the IAPWS-IF97 if there is any.
Do you have any recommendation for similar equation for steam?
| 1
|
[] |
https://chemistry.stackexchange.com/questions/166463/calculation-of-the-seperation-factor-of-ethanol-water-rectification-based-on-t
|
Calculation of the Seperation factor of Ethanol / Water rectification based on two saturated vapor temperatures
|
I want to calculate the efficacy of a Water / Ethanol rectification based on two temperatures with help of a vapor-liquid equilibration diagram.
The first temperature is measured right above the liquid.
The second temperature is measured right before condensation.
The rectification factor doesn't has to be extremely exact.
The calculation is carried out by a computer. The curves of equilibrium diagram is calculated by regression based on a few data points.
[](https://i.stack.imgur.com/S1nES.jpg)
[](https://i.stack.imgur.com/PP9O3.jpg)
[](https://i.stack.imgur.com/vatyS.jpg)
| 0
|
[] |
https://chemistry.stackexchange.com/questions/166459/how-do-i-measure-the-percentage-of-co2-in-the-air
|
How do I measure the percentage of CO2 in the air? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed last year.
[Improve this question](/posts/166459/edit)
What tools can, in a very simple home environment, be used to easily measure CO2 levels in the air, for use in a home school science experiment?
| -1
|
[
[
"\nYou may determine the amount of $\\ce{CO2}$ in air by using a saturated solution of calcium hydroxide $\\ce{Ca(OH)2}$ ($\\pu{0.0234 M}$ at $\\pu{20°C}$). If you take $\\pu{25 mL}$ of this solution and titrate it with $\\ce{HCl 0.05 M}$ and thymolphtalein as indicator, you will need $\\ce{23.38 mL HCl}$.\n\n\nNow if you put $\\pu{50 mL}$ of the same calcium hydroxide in a $5$ liters flask, close it and stir for half an hour, part of the calcium hydroxide will react accordingly to $$\\ce{Ca(OH)2 + CO2 -> CaCO3(s) + H2O}$$ So the solution gets turbid. Titration of $\\pu{25 mL}$ of the obtained solution with the same $\\ce{HCl}$ solution will give you less than $\\ce{23.38 mL}$, as $\\ce{CaCO3}$ will not be destroyed by $\\ce{HCl}$ if the indicator is thymolphtalein ($p\\pu{Ka = 10}$). So the difference between the two volumes of $\\ce{HCl}$ gives you the amount of $\\ce{CO2}$ in the air of the big bottle.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/166457/cant-figure-out-whether-lithium-or-strontium-is-in-my-substance
|
Cant figure out whether Lithium or Strontium is in my substance? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
---
**Homework questions** must demonstrate some effort to understand the underlying concepts. For help asking a good homework question, see: [How do I ask homework questions on Chemistry Stack Exchange?](https://chemistry.meta.stackexchange.com/questions/141/how-do-i-ask-homework-questions-on-chemistry-stack-exchange)
Closed last year.
[Improve this question](/posts/166457/edit)
So for uni I got a substance and I have to basically figure out which anions and cations are in it. I get two chances to guess otherwise I get a new substance and have to start from scratch.
Anyway yesterday I made my first guess:
Cations: Kalium, Ammonium and Lithium
Anions: Chloride
My anion was correct however they said either CO3, NO3, or SO3 is in there as well. I tested negative for all 3 of them, so not sure how to continue here.
My main issue though is the cations:
-Ammonium, this one Im sure about. I did the test with those two circle glasses where you put ph-paper on one side and NaOH with my substance on the other and then put them together. Did that multiple times, the paper turned blue in seconds every time.
-Kalium, semi sure on this one. I mixed it with HOAc and Na3(Co(NO2)6 and it turned yellow-orange. I put in the centrifuge after and its precipitation was yellow as well. Any opinions on the thrustworthiness of this test?
-Lithium, now this one I feel is the wrong one, because I did the flame colour test and as you all know its dark red looks the same as Sr. Since I have absolutely no idea how to differentiate them I just guessed Lithium. Please if you have any idea how test for either of them, help me out because I dont know how continue. I might just guess Sr for my second try.
| -3
|
[
[
"\nLithium produces a red purple flame with only one big line in the spectrum. Strontium produces an orange flame with plenty of lines in the red, orange, yellow and green region of the spectrum. Lithium has no lines in the green. The color of the flames are not the same. Sr is orange red. Lithium is more red purple.\n\n\nYou may also test the solutions of your substances. Solutions of lithium salts do not react when adding a sulfate solution (like sodium sulfate $\\ce{Na2SO4}$). Solutions containing strontium ions produced a white precipitate of insoluble strontium sulfate $\\ce{SrSO4}$.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/166453/do-the-electronegativity-of-atoms-affect-the-molecular-geometry-of-molecules
|
Do the electronegativity of atoms affect the molecular geometry of molecules?
|
I would like to know whether the electronegativity of atoms will affect the molecular geometry of a molecule.
For example, the molecule $\ce{CHFCl2}$. Originally I thought that as fluorine was the most electronegative, it would have a stronger partial negative charge, so the shape of the molecule would not be tetrahedral. I also thought that as carbon has a higher electronegativity, there would be a partial positive charge at the carbon atom. Can anyone explain whether the electronegativity of atoms in a molecule can affect the shape of the molecule? In $\ce{CHFCl2}$ why can $\ce{CHFCl2}$ even have a tetrahedral shape although I think the charges cannot cancel out?
| 0
|
[
[
"\nUnder special circumstances, yes. For example, the NIST Web Book entry for tris(trifluoromethyl)amine (not the preferred IUPAC name, written here in an acceptable name for the sake of intuitivity) clearly gives a trigonal planar coordination geometry at nitrogen, while e.g. the valence isoelectronic trimethylamine has the VSEPR-expected (3BD1LP) triangular pyramidal geometry at nitrogen- this is due to negative hyperconjugation from the nitrogen lone pair to the C-F antibonding orbital, which is promoted by the so high electronegativity of fluorine. [This effect is quite general, and is called the anomeric effect(Warning: PDF auto-download)](https://www.google.com/url?sa=t&source=web&rct=j&url=https://eprints.whiterose.ac.uk/126810/1/AAM.pdf&ved=2ahUKEwiZ2u61q_34AhWPhVYBHazPAo0QFnoECBQQAQ&usg=AOvVaw0l3Y0hEC0TCYJ25HK_OBot)\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/166452/is-hydroxylamine-monodentate-or-ambidentate-ligand
|
Is hydroxylamine monodentate or ambidentate ligand?
|
>
> What is the total number of geometric isomers of the complex $\ce{[Pt(Cl)(py)(NH3)(NH2OH)]+}$?
>
>
>
If $\ce{NH2OH}$ acts as an ambidentate ligand, then there should be six geometric isomers. Or is it strictly a monodentate ligand?
| 2
|
[] |
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