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https://chemistry.stackexchange.com/questions/161646/lennard-jones-potential-repulsion-by-nucleus-nucleus-or-pauli-repulsion
|
Lennard-Jones Potential repulsion by nucleus nucleus or Pauli repulsion?
|
I read different things, some textbooks say that the repulsive part of the lennard jones potential is due to the repulsion of the nuclei when they are too close, others say it is due to the pauli repulsion between the electrons.
Or can I imagine that the pauli repulsion creates an antibonding sigma orbital, which means that the electron density between the nuclei is no longer so high and therefore the nuclei repel each other?
| 2 |
[
[
"\nThe short answer is: the Lennard-Jones potential is a completely heuristic model, and the short-range part of the potential is not physically meaningful except in the sense that it becomes very repulsive very quickly. Pauli repulsion is known, also heuristically, to decay as an exponential, so using a very steep polynomial works reasonably well. As to whether it is modelling the nuclear repulsion or the Pauli repulsion, the answer is both. Also, these two effects are closely related when discussing intermolecular interactions because exchange has the effect of de-shielding electrons and hence enhancing nuclear repulsion.\n\n\n\n\n---\n\n\nTo be more explicit, the Lennard-Jones potential,\n$$\nV(r)=4\\epsilon\\left[\\left(\\frac{\\sigma}{r}\\right)^{12}-\\left(\\frac{\\sigma}{r}\\right)^{6}\\right]\n$$\naimed to provide a simple functional form that reproduced the interactions between neutral gas molecules. In short, any potential that is repulsive at short-range, has a minimum at some intermediate distance, and goes to a constant as $r\\rightarrow\\infty$ is a heuristic model for the medium-range to long-range interactions between nonpolar molecules. What I mean by medium-range is something like twice the diameter of the molecule in question.\n\n\nThe special thing that the Lennard-Jones potential does is get the correct asymptotic form of the attractive interaction. Namely, it is known from the [London Dispersion formula](https://en.wikipedia.org/wiki/London_dispersion_force) (and other ways) that the dispersion interaction decays as $r^{-6}$. Not only that, but the induced dipole interaction between two nonpolar molecules also decays as $r^{-6}$.\n\n\nSo, this brings us to the short-range part of the potential. As you mention, there are two effects which result in short-range repulsion: nuclear-nuclear repulsion and so-called Pauli repulsion due to the electron exchange. Note that classically, the reason exchange results in repulsion is that it de-shields electrons in the internuclear region and hence results in increased nuclear repulsion.[1]\n\n\nIn fact, many theories predict that exchange repulsion decreases exponentially with distance, so using a polynomial distance dependence is somewhat arbitrary. Due to this, there are other potentials that choose to use an even steeper polynomial which goes as $r^{-14}$, which tends to give somewhat better results in simulations.\n\n\nNow the question is why does the Lennard-Jones potential choose the specific polynomial of $r^{-12}$? The answer is rather funny. Back in the day when computers were very slow and floating point operations were very expensive, one needed to be very mindful of how expensive calculations were. Now, calculating exponents can be rather expensive. To calculate $(\\sigma/r)^6$ requires a division and six multiplies. It would be a shame to have to do even more multiplication operations for the short-range part of the potential, which is the least physically meaningful part anyways. Hence, the exponent 12 was chosen because it is simply the product $(\\sigma/r)^6\\cdot (\\sigma/r)^6$. Hence, one can save many floating point operations by doing this.\n\n\nNote that this story about choosing the exponent 12 for computational speed is mentioned on the wikipedia page and I have been told this story by multiple scientists I trust, but I haven't looked closely enough at the original paper to know if this is really what Lennard-Jones had in mind.\n\n\n\n\n---\n\n\n**References:**\n\n\n[1]: Rackers, J. A., & Ponder, J. W. (2019). Classical Pauli repulsion: An anisotropic, atomic multipole model. The Journal of chemical physics, 150(8), 084104.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/161639/oxidation-of-sulfuric-acid-to-marshalls-acid-and-its-half-cell-reaction
|
Oxidation of sulfuric acid to Marshall's acid and its half cell reaction
|
>
> $\ce{H2O2}$ can be prepared by successive reactions:
>
>
> $\ce{2NH4HSO4 -> H2 + (NH4)2S2O8}$
>
> $\ce{(NH4)2S2O8 + 2H2O -> 2NH4HSO4 + H2O2}$
>
>
> The first reaction is an electrolytic reaction and second is steam
> distillation. What amount of current would have to be used in first
> reaction to produce enough intermediate to yield $\pu{102g}$ pure
> $\ce{H2O2}$ per hour. Assume efficiency 50%.
>
>
>
I was solving this question on Faraday's Laws Of Electrolysis when I stumbled across a conceptual flaw of mine. I realized that the sulfate anion in the first reaction is at a +6 Oxidation State and so is Marshall's acid, as far as what I know of Redox Reactions and their balancing we look at the number of electrons exchanged and thus formulate the half-cell reaction.
However, this logic of mine failed in the above question as the oxidation state of the central atom is unchanged, which makes me wonder how to calculate the **valency-factor/N-factor** and correspondingly the equivalent weight. I know this is a conceptual shortcoming of mine and that the aforementioned logic is very 'methodical' per se, which is why it fails.
If someone could please point out where I am going wrong it'd be highly appreciated.
| 2 |
[
[
"\nForget about the equivalent weights. Work with moles, and only with moles. Look how it goes.\n\n\n$102$ g $\\ce{H2O2}$ is $\\ce{\\frac{102 g}{34 g/mol} = 3.00 mol H2O2}$. The production of $1$ mol $\\ce{H2O2}$ requires $1$ mol $\\ce{(NH4)2S2O8}$. Now we will show that the production of $1$ mol $\\ce{(NH4)2S2O8}$ requires $2$ moles electrons. Before doing this, we will first show that the ion $\\ce{HSO4-}$ from $\\ce{NH4HSO4}$ is at least partly decomposed into the following two ions according to the following equation : $$\\ce{HSO4^- <=> H+ + SO4^{2-}}$$ One of these ions ($\\ce{H+}$) is reduced in $\\ce{H2}$ at the cathode, and the other one ($\\ce{SO4^{2-}}$) is oxidized at the anode according to : $$\\ce{2H+ + 2 e- -> H2}$$ $$\\ce{2 SO4^{2-} -> S2O8^{2-} + 2 e-}$$ This shows that $2$ electrons are needed to produce $1$ mole $\\ce{(NH4)2S2O8}$, and later on $1$ mole $\\ce{H2O2}$.\n\n\nThen Faraday law gives you the intensity I needed to produce $3.00$ mol $\\ce{(NH4)2S2O8}$ in $1$ hour = $3600$ s.\nIt is :\n\n\n$$\\pu{I = \\frac{3.00 ~mol~·~2~·~96500~ As/mol}{3600~ s} = 160 A}$$\nThis value is obtained if the yield is $100$%. As the yield is $50$%, the intensity must be twice the previous value. This is $320$ A.\n\n\n",
"5"
],
[
"\nIn the event that this question is actually more than just a theoretical exercise, the general electrolysis half-reactions, as cited by Maurice in my opinion, may not actually be in accord with this particular's complex reaction system per a review of the possible underlying mechanics.\n\n\nMore precisely, here is a suggested overview of reaction mechanics that support my comment:\n\n\n\n> \n> $\\ce{H2O = H+ + OH-}$\n> \n> \n> \n\n\n\n> \n> $\\ce{Electrolysis of HSO4- => .H + .SO4-}$\n> \n> \n> \n\n\n\n> \n> $\\ce{NH4+ = H+ + NH3}$\n> \n> \n> \n\n\n\n> \n> $\\ce{.H + .H = H2}$\n> \n> \n> \n\n\n\n> \n> $\\ce{.H + .SO4- = HSO4-}$\n> \n> \n> \n\n\n\n> \n> $\\ce{.SO4- + .SO4- = S2O8^{2-} }$\n> \n> \n> \n\n\n\n> \n> $\\ce{ 2 NH4+ + S2O8^{2-} = (NH4)2S2O8 }$\n> \n> \n> \n\n\nAlso, a slow reaction, introducing a powerful radical:\n\n\n\n> \n> $\\ce{.SO4- + H2O = .OH + H+ + SO4^{2-} }$\n> \n> \n> \n\n\n\n> \n> $\\ce{HSO4- +.OH = H2O + .SO4- }$\n> \n> \n> \n\n\nSo, while one may claim seemingly that only that 2 electrons are required, my analysis as outlined above which is subject to kinetics, suggests some possible reversed reactions resulting in reduced efficiency.\n\n\nAs a result, not surprisingly, more than 2 electrons may actually be required to produce the single mole of $\\ce{(NH4)2S2O8}$ in an experiment, and as such, I would recommend qualifying Maurice analysis with the words \"at least\", if one is writing up this experiment to account for observed results. More interesting is the provided statement \"Assume efficiency 50%\", which appears supportive of my take on the reaction system.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/161636/why-do-ccl4-and-ch4-have-the-same-bond-angle
|
Why do CCl4 and CH4 have the same bond angle?
|
I was reading about the molecular shape of compounds. I learned that the electronegativity of the central atom and the terminal atom in a molecule both play a role in determining bond angle. In $\ce{NH3}$ and $\ce{NF3}$, $\ce{F}$ having higher electronegativity than $\ce{H}$, $\ce{NF3}$ has a smaller bond angle compared to $\ce{NH3}$.
Applying the same logic, it was expected that $\ce{CCl4}$ would have a smaller bond angle than that of $\ce{CH4}$. Surprisingly, I found that both of them have the same bond angle. Why is this the case?
| 1 |
[
[
"\n\n> \n> In $\\ce{NH3}$ and $\\ce{NF3}$, $\\ce{F}$ having higher electronegativity than $\\ce{H}$, $\\ce{NF3}$ has a smaller bond angle compared to $\\ce{NH3}$.\n> \n> \n> \n\n\nBoth of these compounds have a lone pair on the central atom. So the bound electrons and the lone pair (if you are using the simple \"electrons pair up\" model) compete for space.\n\n\n\n> \n> Applying the same logic, it was expected that CCl4 would have a smaller bond angle than that of CH4.\n> \n> \n> \n\n\nAll electrons around carbon are involved in bonding, so all four pairs are the same. To apply the electronegativity argument, you should compare the [distinct bond angles in $\\ce{CH2F2}$](http://www.uwosh.edu/faculty_staff/gutow/P-Chem_Web_Posters/AndyNickWebsite/CH2F2/CH2F2.html) or in $\\ce{CH2Cl2}$.\n\n\n",
"4"
],
[
"\nWelcome to Stack exchange chemistry.\n\n\nConsider for a moment what is known as the isolobal concept, there are a series of atoms and groups which all present the same types of orbitals (or at least close to identical orbitals) and the number of electrons.\n\n\nConsider for a moment a methane molecule, if we were to break a C-H bond then the carbon atom would only have seven valance electrons. The carbon in an alkane such as methane has rehybridized its orbitals to give us four sp^3 orbitals. These are arranged in a tetrahedron around the carbon.\n\n\nA covalent bond is formed by sharing the one electron in the sp3 orbital of the carbon with an atomic orbital from another atom that has the right geometry to overlap with the sp3 orbital. The sp3 orbital has two pear-shaped lobes, one is large and one is small. These have opposite signs of the wavefunction.\n\n\nA hydrogen atom in the ground state has a single electron in an s orbital, this is a sphere-shaped orbital that can interact with the sp3 orbital to form both a bonding and an antibonding orbital. We will only concentrate in this answer on the bonding orbitals.\n\n\nThe sphere-shaped s orbital has the right geometry to interact with the sp3 orbital and it can result in the formation of an occupied (2 electrons in it) bonding orbital between the carbon and the hydrogen. This will be a sigma bond (single bond)\n\n\nIf we change to chlorine, then the outermost orbital (for the valence electrons) of the atom has also rehybridized to give us four sp3 orbitals. Three of these are occupied with two electrons while one in an isolated chlorine atom only has one. The orbital with only one electron can interact with the sp3 orbital on the carbon (bearing only one electron) to form two new molecular orbitals. One is antibonding and one is bonding.\n\n\nIf the bonding orbital between the carbon and the chlorine is occupied with two electrons then we have a bond. The C-Cl and C-H bonds will be different in length. But the angle between them will be dictated by the arrangement of the sp3 orbitals around the carbon atom.\n\n\nIf you still do not understand it then I would suggest that you fall back to VSEPR theory. As it is the festive season go and grab an orange and four cocktail sticks. Stab them into the orange in such a way that they are the greatest angle apart. You should find that the tips of them form a triangle-based pyramid (tetrahedron). It will not matter if you put grapes on the points of the cocktail sticks to represent hydrogen atoms in the methane. Or apples to represent the chlorine atoms in carbon tetrachloride. You will still have the same arrangement of the atoms in your model.\n\n\nYou can then hang it on the tree as a decoration or pull it apart and eat the fruits. When I can not lay my hands on my molecule modeling kit made of plastic balls and straws I tend to grab oranges and then draw atoms with a marker pen on the skin.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/161628/what-material-properties-are-unpredictable-not-reproducible-but-can-be-measure
|
What material properties are unpredictable (not reproducible) but can be measured consistently and at a low cost?
|
I'm looking for a process to create a material which has some easy to measure properties.
These properties should be consistent over a long period.
It should be very hard (expensive) to predict/reproduce a material which results in the same measurable properties. Every material product should generate an unique measurement result and impossible (or very expensive) to product a second product with the same measurement result.
What process, material and/or measurement could be used?
Some context:
The goal would be block-chain backed physical cash currency.
The process should result in a 'coin' (material product).
The 'coin' can be 'read' (measurement of some over time consistent material properties) resulting in a 'coin-id' (measurement).
The producer of the 'coin' reads the coin-id and adds a block to the chain which contains the 'coin-id' and his signature of the 'coin-id' and spends some (crypto) currency value. The physical coin represents the spend (crypto) currency value. For ease of use the signature could be attached to the coin (bar or qr-code).
The blockchain also contains the certificates of the coin producers.
The coins can be exchanged in the physical world without changes on the blockchain.
A receiver of a coin can read (measure) the unique coin-id and scan the signature.
A receiver knows the certificates of the coin producers. A receiver can check the veracity by checking if the coin-id is signed by a known coin producer certificate.
*Addendum*
Let the cost to produce a coin be **c**.
Let the probability of a duplicate measurement be **p**.
Let the value represented by the coin be **v**.
Because the represented value should be a lot bigger than the cost of a coin. The value should be a factor **f** bigger than the cost.
$v=cf$
The minimal value to make forgery too expensive:
$v=\frac{c}{p}$
The maximal **p** should be:
$p=\frac{c}{v}$
So given a $c=0.1$\$ and $v=100$\$
then the maximum $p=\frac{0.1}{100}=0.001$
Or put otherway around: given a more realistic forgery probability of $p=10^{-10}$
Maximum $f=\frac{1}{p}$, $f=10^{10}$
So a coin given $c=0.1$\$ could have a maximum value of $v=0.1\*10^{10}=1.000.000.000$\$
| 6 |
[
[
"\nImpossible for a homogeneous material, however ...\n\n\nbasically *every* nonhomogeneous material fits your description. Say the pattern of microphase separation in a copolymer. Or the arrangement of filler particles in a composite.\n\n\nYou just have to think of something that is not only easy to make, but also easy to measure, i.e. take a digital photo of (microscopes and MRT or µCT scanners are out, I guess):\n\n\nPour resin of say four colours into a round bin, so you have four differently-coloured pie pieces. Now take a fork and run it through the bin a few times, perhaps in a \"random\" fashion (rotation, speed, direction, duration).\n\n\nLet resin set, and you get a disk (\"coin\") with a pattern that is impossible to regenerate, even if someone stole the random number, which you delete immediately after use, that initialised the fork movement. They'd get something that looks somewhat similar, but clearly distinguishable. If they tried ten thousand times, they might get one or two that look similar enough to fool your algorithm, but that forgery is uneconomic, so you're safe.\n\n\nOf course someone could forge those coins by printing a photographed pattern onto a coin, just like you can photocopy a dollar note.\n\n\nThe bigger problem I see is to make sure the \"measurement\" never creates a false negative (negative==forged) outcome, *and* that it doesn't require you to save a multi-megabyte dataset for every single coin.\n\n\n",
"10"
],
[
"\nThe OP asks: What process, material and/or measurement could be used?\n\n\nAnswer: Snowflake maker, dihydrogen monoxide, photograph (with metric ruler included). \"Snowflake maker\" is a vapor condenser from steam onto a cooler surface or simply from the vapor.\n\n\n[](https://i.stack.imgur.com/GihzC.jpg)\n\n\nIt is well-known that no two snowflakes are alike. The unscientific media (Ref 1) propose the opposite, but the scientific consensus is that the probability of finding two identical snowflakes is zero (Ref 2).\n\n\nAn advantage of Snowflake Bitcoin is that it requires a physical repository (a bank, but not a snowbank) with a temperature low enough to prohibit melting and sublimation (which would allow the snowflakes to grow or diminish). Antarctica might not be cold enough; the actual snowflakes might have to be stored in outer space, but that cost could be prohibitive. An alternate bank could be physical storage of the actual photographs, or digital images with a very high resolution.\n\n\nRef 1. <https://www.nbcnews.com/id/wbna16759121>\n\n\nRef 2. <https://www.loc.gov/everyday-mysteries/meteorology-climatology/item/is-it-true-that-no-two-snow-crystals-are-alike/>\n\n\n",
"5"
],
[
"\nAs to what process, material and/or measurement could be used, the same that exists for minting metal alloy coins.\n\n\nThe material could be composed of said four (or more) alloys with the coin displaying an ID. The latter can be a mathematical function of associated chemical and electrical properties of said coin.\n\n\nFor example, an alloy of Ni, Sn, Fe and Zn may work based on the there actual (or predicted) respective variations in an anodic index ([see this Table](https://www.zygology.com/cms/upload_area/pdf/Zyg-Anodic-Index.pdf)).\n\n\nSo, a simple chemical (like density) and electrochemical measurements (electrical resistance) may provide in conjunction with engraved ID an authenticity check.\n\n\nThis concept differs from government issued coins in the respect they are crafted for corrosion resistance (to increase circulation life span) while these coins are minted for identification purposes primarily,\n\n\nCoins are known historically to be robust instrument for currency, and said coins can also be made completely virtual by mathematical fitting algorithms (applied to a provided data string) based on actual coins with specified varying compositions, from which the ID is generated.\n\n\nNote: this answer complies with the need to be measured consistently over time at a low cost.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/161627/why-is-1-3-dimethyl-cyclohexane-a-meso-compound
|
Why is 1,3-dimethyl-cyclohexane a meso compound?
|
[](https://i.stack.imgur.com/LVwUL.jpg)
So i was trying to find the plane of symmetry in 1,3-dimethyl-cyclohexane in chair conformation, but then i realised the plane is present only at a particular orientation [see figure]. So why do we consider this to be a meso compound?
| 3 |
[
[
"\n1,3-dimethylcyclohexane is not exactly a compound, but **three** different compounds, depending on the configuration of the chiral centers. Two of them (RR and SS) are enantiomers of each other, and the third one (RS) is meso.\n\n\nWhy meso?\n\n\nWell, because it has no enantiomeric forms which couldn't be transformed to each other by single-bond rotations. That's what is important, and the symmetry plane is just one possible *evidence* of this situation. It is sufficient, but not necessary.\n\n\nWe can put it differently: you seem to know full well that all conformations are in fact one compound, since you can't separate them. Also, at least one of the conformations is surely a meso compound, since it has a mirror plane. Then what about any other conformation, possibly having no mirror plane? Is it a different compound? No.\n\n\n\n\n---\n\n\nOn a side note, mirror plane is not necessary at all, and [Wikipedia](https://en.wikipedia.org/wiki/Meso_compound) gives us an example:\n\n\n\n> \n> A meso isomer need not have a mirror plane. It may have an inversion or a rotoreflexion symmetry such as S4. For example, there are two meso isomers of 1,4-difluoro-2,5-dichlorocyclohexane but neither has a mirror plane, and there are two meso isomers of 1,2,3,4-tetrafluorospiropentane (see figure). In fact, a meso compound may have no symmetry in some conformations...\n> \n> \n> \n\n\n[](https://i.stack.imgur.com/iNzkf.png)\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/161622/inhibitors-of-hydration
|
Inhibitors of hydration
|
Superglue (Cyanoacrylate) polymerizes via hydration, and using an activator such as a borax solution speeds up that process. I need to mix these 2 without them reacting. (Or at least not noticeably.)
So, I was wondering if there are any inhibitors that can stop this reaction at least temporarily. I'm not well versed in chemistry, so if I got a concept or anything wrong with my question, please tell me. Thanks.
| 0 |
[
[
"\nThere are [inhibitors and \"stabilizers\" such as acids](https://patents.google.com/patent/US5290825A/en). You might experiment with acetic acid (of course, \"glacial\", not as a water solution), or $\\ce{FeCl3}$, which would make the polymer dark or opaque. Another inhibitor is [pyrogallol and boric acid](https://patents.google.com/patent/US4182823A/en), which likely would be compatible borates.\n\n\nThe reference above also cautions that an *excess* of inhibitor can prevent polymerization, so test various inhibitors and various concentrations to see what works for your intended use.\n\n\nIf you do find a good inhibitor, to help others, please add an answer here or at least a comment on what worked in that application.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/161619/reaction-kinetics-of-exothermic-reaction
|
Reaction kinetics of exothermic reaction
|
A colleague said we can’t dissolve a salt (whose solvation enthalpy is exothermic) faster if we increase the temperature (the solubility equilibrium product is not reached) because Le Chatelier‘s principle would favor the reactants.
For example, imagine dissolving NaOH(s) in distilled water, can’t this be accelerated by elevating the temperature?
I read a similar question on here, see: [Factors that influence the kinetics of an irreversible exothermic reaction](https://chemistry.stackexchange.com/questions/91147/factors-that-influence-the-kinetics-of-an-irreversible-exothermic-reaction)
How I see it is the following:
As Le Chatelier‘s principle only applies to systems in equilibrium, and we only dissolve a salt (whose enthalpy happens to be exothermic), no equilibrium is established. The same goes for exothermic reactions in general. As long as no equilibrium is yet established, the process should be accelerated by an increase in temperature, regardless whether it’s exothermic or endothermic, as more molecules can overcome the activation energy barrier in general (for example in the Arrhenius equation).
It is important to note that this is about kinetics, that is, how quickly the NaOH dissolves, in a non-saturated solution, that is about 1M, not about how much NaOH a saturated solution can contain at a given temperature.
Is my reasoning correct?
| 2 |
[
[
"\nTypically, rates of uncatalyzed simple reactions increase with temperature. There are well-known examples where this is not the case (such as enzyme-catalyzed reactions where the enzyme denatures at high temperatures, or reactions with an intermediate that is at rapid equilibrium with the reactant in an exothermic step).\n\n\nIn this case, if the $\\ce{NaOH}$ is soluble at both temperatures, it is likely that the increasing rate in the forward direction will make a bigger difference that the also increasing rate in the reverse direction.\n\n\n\n> \n> As long as no equilibrium is yet established, the process should be accelerated by an increase in temperature, regardless if it’s exothermic or endothermic, as more molecules can overcome the activation energy barrier in general (for example in Arrhenius equation).\n> \n> \n> \n\n\nIn this case, the reaction goes to completion rather than attaining equilibrium (if you stay below the solubility limit). Even for reactions that attain equilibrium, however, the rate with which they approach the equilibrium constant is proportional to the [sum of the forward and reverse reaction](https://en.wikipedia.org/wiki/Temperature_jump) (for a one-step reaction closes to equilibrium, it is possible to show this in a quick derivation).\n\n\n",
"4"
],
[
"\nOur current understanding of chemical kinetics is that all reactions with an activation energy are increased in rate by an increase in temperature. The question that is pertinent is \"Which reaction is accelerated faster?\". A temperature change at equilibrium influences the equilibrium and the overall composite rate. Whether the overall direction of a given reaction is accelerated by an increase in T is determined by the actual value of delta G; if negative it is accelerated as written.\n\n\nThe choice of NaOH as an illustration is a bit unfortunate. As anyone who has tried it knows adding water to solid NaOH results in a severe exothermic reaction even to a boiling solution. Yet NaOH is more soluble at higher T. This is contradictory to the idea that an exothermic reaction is lessened at a higher T. What happens is that the dissolution reaction becomes endothermic as the concentration increases; don't ask me how or why.\nThe general answer: Displaced from equilibrium any reaction is accelerated by an increase in T provided there is an activation energy. At equilibrium the increase of both rates moves the equilibrium in the direction of the reaction with the higher activation energy. If the activation energy is vanishingly small the problem becomes removal of energy and the reaction doesn't exist.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/161618/how-to-validate-choice-of-dft-functionals-and-basis-sets-for-atomic-charge-calcu
|
How to Validate Choice of DFT Functionals and Basis Sets for atomic charge calculation of Fluorocarbons
|
I need to calculate the charges (electron densities) on each atom of a set of partially fluorinated hydrocarbon molecules by NPA. The question arises:
What is the currently accepted / best practice method for validating these calculations as a function of Functional and Basis Set for DFT based calculations? Since no "experimental" values for electron densities are available, I have tried evaluating the "correctness" of the chosen functional and basis set by correlating the predicted bond lengths and bond angles to experimental values. However, this seems to be far too simplistic to provide a meaningful estimation of the correctness of the choice.
| 5 |
[] |
https://chemistry.stackexchange.com/questions/161617/producing-hydrogen-sulfide-difficult-to-do-much
|
Producing hydrogen sulfide—difficult to do much?
|
When I was young (nearly sixty years ago), my "chemistry set" taught me how to heat sulfur and candle wax to stink up the house. It had **no** warning that hydrogen sulfide is toxic and explosive in large quantities. **Is that because people weren't as litigious back then, or because producing large quantities is hard to do?**
The reason I ask is that near here, we had a major issue with hydrogen sulfide coming out of an SUV and affecting three first responders who tried to rescue the driver (deceased). It was announced as a suicide, but I thought if a fellow wanted to end it all, that doesn’t seem like an easy way to do it.
And there is at least one known case of [deaths in an SUV from hydrogen sulfide](https://www.atlantictraining.com/blog/florida-turnpike-deaths-h2s/). I’ve read a few other reports of problems from people leaving the vent hose off of a battery in the passenger area of a car. Maybe [this week's incident](https://www.klcc.org/disasters-accidents/2021-12-20/hazmat-incident-closes-part-of-main-st-in-springfield) was not a suicide. (Maybe putting a lead-acid battery inside a passenger compartment is a really stupid thing to do.)
| 3 |
[
[
"\nThe problem with H2S is that it is both highly toxic (albeit not quite at the level of HCN but surprisingly to me, too close) and with uncommon properties such as at a threshold of exposure, a suppression of the ability to detect any smell.\n\n\nThe latter is particularly pernicious as one can, apparently, achieve a toxic dose while not even realizing that one is continued to be exposed to the extent of having actually received a fatal dose (reputedly, a 10 minute exposure at 50-100 ppm destroys the sense of smell and thus results in the loss of ability to sense its presence through smell).\nAlso, pathways of fatal ingestion also relate to a limited extent to skin contact, albeit sewer workers wearing a gas mask with extended exposure could, for example, actually fall victim to this deadly poisonous gas.\n\n\nThe victims with fatal doses have actually been referred to as \"the walking dead\" as they will, progressively sicken over a course of hours, suffering irreversible organ damage leading to death.\n\n\nI agree that general knowledge and labeling on H2S is deficient, and that there is an issue, as you noted: \"It had no warning that hydrogen sulfide is toxic and explosive in large quantities\".\n\n\nI also view this a failing relating to the quality and availability of knowledge in the field of inorganic chemistry. In the case of poisoning in a car with a battery, this knowledge base apparently extends to electrochemistry as well.\n\n\nAs such poisoning from H2S is, in certain locales (Japan being an exception), in my opinion, can be more likely be the result of absence of knowledge and proper labeling than intentional suicide.\n\n\nSome confirming [references include](https://www.nature.com/articles/srep20831) to quote:\n\n\n\n> \n> Hydrogen sulfide is a highly toxic gas—second only to carbon monoxide as a cause of inhalational deaths. Its mechanism of toxicity is only partially known and no specific therapy exists for sulfide poisoning....No antidote is currently available for sulfide poisoning and treatment is largely supportive.\n> \n> \n> \n\n\n[Per the CDC](https://wwwn.cdc.gov/TSP/MMG/MMGDetails.aspx?mmgid=385&toxid=67), to quote:\n\n\n\n> \n> Hydrogen sulfide is well absorbed through the lungs; cutaneous absorption is minimal. Exposure by any route can cause systemic effects...However, although its strong odor is readily identified, olfactory fatigue occurs at high concentrations and at continuous low concentrations. For this reason, odor is not a reliable indicator of hydrogen sulfide's presence and may not provide adequate warning of hazardous concentrations...Inhalation of high concentrations of hydrogen sulfide can produce extremely rapid unconsciousness and death.\n> \n> \n> \n\n\n",
"3"
],
[
"\nI doubt the article is accurate. Hydrogen sulfide is noticeable at less than 1 ppm and up to 100 ppm causes only sinus irritation in a half hour exposure. It is very dangerous as 150 ppm can cause loss of smell sense in minutes. H2S is common in oil/ gas production so oil companies and governments have extensive rules concerning it. In particular it causes stress corrosion cracking of high strength steels so NACE MR 01-75, now ISO 15656 has detailed rules for H2S. Internal combustion engines do not produce it in quantity. I would like to see some technical analysis of how a lead acid battery can produce H2S in quantity. Also prior to 1940 , it was not unusual for an auto battery to be located under the floor of a car. I have seen a later model Cadillac with the battery under the back seat so apparently that is not a unique location. I have never seen another report blaming battery location for mysterious deaths. I got some information from SPE Sour Gas Design Monograph , 1993 ; I doubt it is readily available. Pardon my ramble but I like H2S it provided me a large portion of my career. PS ; From what I find on the net , in the US sulfur is limited to 15 ppm in gasoline, there are exceptions for off-road and other categories.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/161616/average-distance-of-1s-electron-from-nucleus-in-he%e2%81%ba-ion
|
Average distance of 1s electron from nucleus in He⁺ ion
|
>
> What would be the average value of $r$, i.e. $\langle r\rangle$, in the $\mathrm{1s}$ orbital of $\ce{He+}?$
> $$
> \text{a}.~\frac{3}{2}a\_0 \qquad
> \text{b}.~\frac{3}{4}a\_0 \qquad
> \text{c}.~3a\_0 \qquad
> \text{d}.~\frac{1}{2}a\_0
> $$
>
>
>
I have written the normalized wavefunction of $\mathrm{1s}$ orbital of $\ce{He+}:$
$$R\_{(1,0)} = \frac{2 \sqrt 2}{\sqrt{\pi a\_0^3}} \times a^{-2r/a\_0},$$
but I could not proceed further.
| 2 |
[
[
"\nThe average value of $r$ can be found out, in case of $\\text s$ orbitals by multiplying the volume of each thin spherical shell by the probability density at that $r$ and adding them up.\n\n\nThis can simply be accomplished by using integration as a limit of sum. We know that the probability density $\\mathrm dP/\\mathrm dV = R^2$. So,\n\n\n$$\n\\begin{align}\n⟨r⟩ &= \\int\\_0^\\infty r\\,\\mathrm dP \\\\\n⟨r⟩ &= \\int\\_0^\\infty rR^2\\,\\mathrm {d}V\\\\\n⟨r⟩ &= \\int\\_0^\\infty r\\left(\\frac{2 \\sqrt 2}{\\sqrt{\\pi a\\_0^3}} \\times e^{-2r/a\\_0}\\right)^2\\left(4\\pi r^2\\,\\mathrm {d}r\\right)\\\\\n⟨r⟩ &= \\frac{3a\\_0}{4}\\\\\n\\end{align}\n$$\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/161612/can-oxidation-and-reduction-be-described-in-terms-of-transfer-of-oxygen-in-the-f
|
Can oxidation and reduction be described in terms of transfer of oxygen in the following way?
|
After reading and thinking about oxidation and reduction a bit, I tried to come up with a way to explain it to myself… Could someone verify if this is technically correct?
Consider 2 elements A and B where A is more electronegative than B.
(Also O represents Oxygen)
If in a reaction
$$\ce{AO + B -> BO + A}$$
So assuming that O is more electronegative than both A and B
In AO,
* Oxidation Number of O = -2
* Oxidation Number of A = +2
So we could say that O is “negatively charged” (or partially negative if it’s a covalent bond) while A is “positively charged” ( or partially positive if it’s covalent) and hence this is why they are attracted to each other (This part I’m not sure). Now given that A is more electronegative than B, when B comes near AO, A attracts electrons from B and oxidises B.
* Oxidation Number of A gets reduced to 0 (A is reduced)
* Oxidation Number of B increases to +2 (B is oxidised)
Now we could say that A is now “neutral” and B is “positively charged” while O remains “negatively charged”. Hence O is attracted to B forming BO.
Thus
$$\ce{AO + B -> BO + A}$$
From the above, we can thus say that a loss in oxygen causes reduction and a gain in oxygen causes oxidation.
Is this how redox reactions work or am I exaggerating everything?
| 0 |
[
[
"\nYour explanation reminds me of the historical definition of the words \"oxidation\" and \"reduction\". In the early 19th century, oxidation was the reaction an element with oxygen. The only thing that the chemist could measure with precision was the weight, the mass, before and after the reaction. When iron or copper gets oxidized, it produces oxides which are heavier than the element. Now when one of these oxides reacts with hydrogen, it produces the original metal, but its weight is reduced. So the chemist said that this oxide has been reduced in metal by hydrogen. The oxide has lost some mass to be transformed into the metal. Its mass is reduced. So the element is also \"reduced\".\n\n\n",
"1"
],
[
"\nAn easier approach would be to think about redox-reactions as an exchange of electrons, rather than of oxygen. This is more general and equally allows oxygen to be oxidized (e.g., fluorine is even more electronegative than oxygen, as in [oxygen fluorides](https://en.wikipedia.org/wiki/Oxygen_fluoride)).\n\n\nRetain that an oxidation is the removal of electrons, and reduction is the addition of electrons. The «who gets the electrons» depends on the partners reacting with each other. In an allegory, think about an ox whose horns pierce and remove material from a tree (*oxydation*). And think of reduction as bringing electrons back (based on the Latin root *reducere*) to heal the wound.\n\n\n",
"0"
],
[
"\nYou can find examples where your thoughts fit what is actually happening. For example, a reaction of magnesium oxide with barium (I don't know if this reaction would actually happen):\n\n\n$$\\ce{MgO(s) + Ba(s) -> Mg(s) + BaO(s)}$$\n\n\nHere, it is correct to speak of the negative and positive charges because we are looking at ionic compounds. You could show the ions explicitly:\n\n\n$$\\ce{Mg^2+ + O^2- + Ba -> Mg + O^2- + Ba^2+}$$\n\n\nOn the other hand, you could have a redox reaction involving molecular compounds:\n\n\n$$\\ce{H2O + C -> H2 + CO}$$\n\n\nHere it does not make sense to talk about charges of atoms in the compounds (partial charges maybe, but they are also not necessarily correlated with oxidation states). The easiest way to figure out oxidation states in this case is to define the oxidation state of oxygen as -2 (except when oxygen is bound to itself or to more electronegative atoms), and use that to assign +1 to hydrogen in water and +2 to carbon in carbon monoxide.\n\n\n\n> \n> From the above, we can thus say that a loss in oxygen causes reduction and a gain in oxygen causes oxidation?\n> \n> \n> \n\n\nNo, unless this is the only change and the binding partner of the oxygen is less electronegative, as in your examples. Here is a counterexample:\n\n\n$$\\ce{F2 + O2 -> O2F2}$$\n\n\nWhen you add water across a carbon-carbon double bond, this is usually not called oxidation either, even though one carbon gains a bond with oxygen and is oxidized. The reason it is not called oxidation is that at the same time, the other carbon gains a bond with hydrogen and is reduced.\n\n\n\n> \n> Is this how redox reactions work or am I exaggerating everything?\n> \n> \n> \n\n\nThis is how a subset of redox reactions work. In many redox reactions, there is no oxygen at all, e.g.:\n\n\n$$\\ce{Fe^2+ + Cu^2+ -> Fe^3+ + Cu+}$$\n\n\nFor a general modern definition of a redox reaction, we look at the transfer of electrons (as opposed to transfer of protons in acid/base reactions).\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/161607/what-molecules-other-than-n-2-have-wrong-orbital-ordering-when-calculated-usin
|
What molecules other than N$_2$ have wrong orbital ordering when calculated using Hartree-Fock?
|
N$\_2$ is a canonical example of this aspect of the failure of HF method, for instance, see table 4.9 in "Modern Quantum Chemistry" by Szabo and Ostlund. Here it is shown that in HF, the HOMO of N$\_2$ is of $\pi$ symmetry, whereas correlated method and experiments show that its HOMO should have $\sigma$ symmetry, which becomes HOMO-1 in HF.
| 3 |
[] |
https://chemistry.stackexchange.com/questions/161606/are-double-spike-isotopic-standards-suitable-for-quantification-by-isotope-dilut
|
Are double spike isotopic standards suitable for quantification by isotope dilution mass spectrometry?
|
I would like to apply isotope dilution mass spectrometry (IDMS)\* to quantify lead at trace levels. A good spike solution for the process would be the [NIST SRM 991](https://www-s.nist.gov/srmors/view_detail.cfm?srm=991) which is an enriched material of almost only lead-206, but this material is now discontinued and the replacement is a radioactive material [NIST SRM 983](https://www-s.nist.gov/srmors/view_detail.cfm?srm=983), which I think, will be a little difficult to import.
The NRC of Canada has produced the lead-204 and lead-207 double spike isotopic standard [BLED-1](https://nrc.canada.ca/en/certifications-evaluations-standards/certified-reference-materials/list/145/html) which contains 49% and 33% the two isotopes, respectively, and I was wondering if it could be suitable for IDMS considering that the natural relative abundance of those isotopes is low. I have only performed IDMS using the (single spike) [NIST SRM 991](https://www-s.nist.gov/srmors/view_detail.cfm?srm=991), and although I think the same IDMS equations will apply to each pair of isotopes, would like to know please if there is something I am not taking into account.
\*) An overview about this technique is provided by Vogel, J.; Prizkow, W. Isotope dilution mass spectrometry — A primary method of measurement and its role for RM certification. *MAPAN* **25**, *2010*, 135–164, [doi 10.1007/s12647-010-0017-7](https://doi.org/10.1007/s12647-010-0017-7). The author's copy on [ResearchGate](https://www.researchgate.net/publication/225942041_Isotope_Dilution_Mass_Spectrometry_-_A_Primary_Method_of_Measurement_and_Its_Role_for_RM_Certification).
| 8 |
[
[
"\nYes, I think it would be suitable, although I am not an expert in IDMS either. It seems like the new standard contains two stable isotopes. This should give you a built-in check on the data you collect: if you treat Pb-204 as the spike isotope, and calculate the concentration of Pb in your sample from it, you should get the same (or very similar) answer as if you treat Pb-207 as the spike isotope.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90887/carbocation-rearrangement-involving-three-membered-rings
|
Carbocation rearrangement involving three membered rings
|
**Question:**
>
> Taking into account of various carbocations and, as well as the rules governing mechanisms of carbocation rearrangements, which reaction is most likely to occur during the given reaction?
>
>
> 
> 
>
>
>
---
My take on the question was this mechanism:
I'm torn between this mechanism and option (b). How do I judge between anchymeric assistance and stabilization of the intermediate through 5 alpha hydrogens. How will the carbocation rearrange?
**Source**: MS Chouhan, Chapter: Alkenes, Q.17
This inadequate solution was given:
| 8 |
[
[
"\nUnfortunately you have opened the can of worms that is cyclopropylmethyl carbocations.\n\n\nIt is known that the parent unsubstituted cyclopropylmethyl carbocation is non-classical, and leads to mixtures of cyclopropylmethyl, cyclobutyl, and homoallyl derivatives. However, the presence of substitution can alter the electronic structure quite significantly. There is too much on the topic to discuss here; the interested reader may find reviews for themselves in George Olah's books *Onium Ions* and *Carbonium Ions*.\n\n\nIn this case, I would say that it is virtually impossible to predict which major product will be formed. Perhaps that's not the answer you wanted, but I don't want to propagate fake knowledge and claim that we can predict something that we can't.\n\n\nLooking up this reaction on Reaxys, I find that it has not been done before. However, the reaction of the corresponding *alcohol* with HBr has been reported before. In different people's hands and under different conditions, it gives different major products; both the cyclopropylmethyl bromide and the homoallyl bromide have been isolated. [Apparently not the cyclobutyl bromide, though; why? Again I don't have a good answer.]\n\n\n",
"15"
],
[
"\n**Disclaimer:** This is a *very simplistic* view of the cyclopropyl methyl cation and its reaction given above with $\\ce{Br-}$, and is **completely wrong**. In real conditions, there are a lot of complexities and one can also expect cyclobutane derivatives to form. The only reason why I let this answer remain is that it matches with the answer and explanation given by the OP in their book.\n\n\n\n\n---\n\n\n\n> \n> How will the carbocation rearrange?\n> \n> \n> \n\n\nThe following carbocation:\n\n\n[](https://i.stack.imgur.com/sa8Vq.png)\n\n\nwill **not** rearrange. It is a cyclopropyl methyl cation. This is *undoubtedly* the **best** method of stabilization especially since here there is no case of tribenzylic cation or tropylinium cation.\n\n\nThe ring will not expand to form cyclobutane. No alkyl shift occurs. The $\\ce{Br-}$ will attach at the very position shown above. The correct answer should thus be (d).\n\n\n**PS:** The reason why the rearrangement of the cyclopropyl cation to a cyclobutyl cation is unfavorable is explained in detail [here](https://chemistry.stackexchange.com/questions/28939/carbocation-rearrangement-due-to-ring-strain-and-back-bonding/28944#28944) (remark by @ron: \"The stabilization of a cyclopropylcarbinyl carbocation is very dependent upon the relative orientation of the cyclopropane \"banana bonds\" and the adjacent cationic center\"). \n\n\nThese two links [(1)](https://chemistry.stackexchange.com/questions/19090/what-is-the-reason-for-the-exceptional-stability-of-the-cyclopropylmethyl-carboc) and [(2)](https://www.quora.com/Why-is-cyclopropyl-methyl-carbocation-exceptionally-stable) are for further reading regarding exceptional stability of cyclopropyl methyl cation.\n\n\n\n\n---\n\n\n",
"9"
]
] |
https://chemistry.stackexchange.com/questions/90886/expansion-under-isobaric-adiabatic-condition
|
Expansion under isobaric adiabatic condition
|
>
> A 3 mole sample of a triatomic gas at $\pu{300 K}$ is allowed to expand under isobaric adiabatic condition from $\pu{5L}$ to $\pu{40L}$. The value of change in enthalpy is:
>
>
> 1. $\pu{12.46 KJ}$
> 2. $\pu{-14.965 KJ}$
> 3. $\pu{-24.62 KJ}$
> 4. $\pu{-10.24 KJ}$
>
>
>
In case of adiabatic process, there is no transfer of heat so $Q=0$. Again for an isobaric process, pressure is constant that is $\Delta P=0$.
So, change in enthalpy:
$$\Delta H = \Delta U + \Delta(PV) = Q + W + P\Delta V + V\Delta P = W + P\Delta V = -P\Delta V + P\Delta V = 0$$
Thus, we see that change in enthalpy is zero. But the answer matches with none of the options. What is incorrect in my solution?
| 0 |
[
[
"\nIf, at time = 0, we suddenly drop the external force per unit area to $P\\_{\\text{ext}}$ and hold it at this value until our adiabatic system equilibriates, the first law of thermodynamics for this situation reads:\n\n\n$$nC\\_\\mathrm v(T\\_2-T\\_1)=-P\\_{\\text{ext}}(V\\_2-V\\_1)=-\\frac{nRT\\_2}{V\\_2}(V\\_2-V\\_1)=-nRT\\_2\\left[1-\\frac{1}{8}\\right]$$\n\n\nSolving this equation for $T\\_2$ yields: $$T\\_2=\\frac{T\\_1}{\\left[1+\\frac{7}{8}\\frac{R}{C\\_\\mathrm v}\\right]}$$\n\n\nBut, for $C\\_v=3R$, this gives:$$T\\_2=0.774T\\_1=\\pu{232 K}$$\n\n\nSo, $$\\Delta H=3(4R)(232-300)=\\pu{-6.78 kJ}$$\n\n\nSo the final temperature is higher than in the adiabatic reversible case, and the decrease in enthalpy is less.\n\n\n",
"2"
],
[
"\n**I figured out the problem!**\n\n\nThere is a misprint. The word *isobaric* should be deleted. Without it, the question makes complete sense. Here's how:\n\n\nSince it is now a *reversible* process, apply $TV^{\\gamma-1}=\\text{constant}$, with $\\gamma=1+\\frac2f$ and $f=6$ to get $T\\_2=\\pu{150K}$. Now, $\\Delta H=n\\cdot C\\_p\\cdot\\Delta T=3\\cdot(1+\\frac62)\\cdot R\\cdot(-150) \\pu{J}=-\\pu{14.965 KJ}$ as given a correct answer in your book.\n\n\nThank me later! ;)\n\n\n\n\n---\n\n\n*Note: This answer has been completely rewritten. The previous version of this answer can be found in the revision history.*\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90885/how-to-decide-what-pathway-a-reaction-will-follow-elimination-addition-to-a-ca
|
How to decide what pathway a reaction will follow, elimination, addition to a carbonyl or nucleophilic substitution?
|
There is a substrate such as this,
[](https://i.stack.imgur.com/h6QUr.png)
and it is treated with alcoholic KOH.
I know that a hydrogen in $\alpha$-position to the carbonyl group will be abstracted, leaving a carbanion. My query lies in deciding what pathway will the substrate follow from hereon. I see three options: An E1cB elimination of the bromide, aldol condensation by nucleophilic addition to the carbonyl group and SN2 with the 1 degree bromide.
Which pathway will be favored and why?
I'd be glad if someone could specify the conditions for one pathway being predominantly favored over another (even if I haven't provided information regarding that, like temperature, solvent, etc) or some empirical data on one pathway being faster or slower. Any degree of help will be appreciated, thank you.
Edit: The answer in the book where this question is from says that nucleophilic addition should predominate (providing neither justification nor data), which contradicts my guess that elimination would dominate because of it being an internal mechanism.
| 7 |
[
[
"\n1. The nucleophilic substitution with another equivalent of the primary bromide will never be favourable as the nucleophilicity of the enolate generated is pretty low (the charge is delocalized!)\n2. Even if **E1cb** was a possible route, you would end up with an extremely reactive vinyl aldehyde, which would likely participate in a Michael addition with an enolate in the reaction medium. So, isolating the product from this route would be very unlikely, even though I cannot back my claim up immediately.\n\n\nIf you ask me, both the Michael addition **and** the Aldol reaction should compete, given the nature of the base the problem has used. Had it been something like LDA (or any other directed Aldol conditions), the greater stability of the enolate would probably have precluded **E1cb** as a possible route.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90881/how-to-determine-crown-ether-cavity-size
|
How to determine crown ether cavity size? [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 5 years ago.
[Improve this question](/posts/90881/edit)
I want to determine the cavity radius/ dia of crown ether. Which technique should I follow?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/90877/what-is-the-order-of-volatile-nature-of-group-15-hydrides
|
What is the order of volatile nature of Group 15 hydrides? [duplicate]
|
**This question already has answers here**:
[Why is the boiling point of stibane higher than that of ammonia?](/questions/64191/why-is-the-boiling-point-of-stibane-higher-than-that-of-ammonia)
(2 answers)
Closed 5 years ago.
Among NH₃,PH₃, AsH₃, SbH₃ the increasing order of volatility given in my textbook is as follows.
>
> NH₃< SbH₃< AsH₃< PH₃
>
>
>
I know that among all these hydrides NH₃ alone is capable of forming Hydrogen bonds thus it is least volatile, but I’m not sure how to compare others.
| -1 |
[
[
"\nRecall that volatility is inversely proportional to the boiling point. The boiling point itself is directly proportional to the intermolecular forces. In the above case, we have two major intermolecular forces: H-bonds and dipole-dipole attractions.\n\n\nNow, it is factual info that ammonia's intermolecular forces surpass even stibine's intermolecular attractions, so its volatility is the least.\n\n\nThus, we are only left with phosphine, arsine, and stibine. Here, recall that intermolecular forces are van Der Waal's forces, and thus **their magnitude is directly to their size.** ([read the \"why\" of it here](https://chemistry.stackexchange.com/questions/10383/why-is-mass-proportional-to-the-strength-of-a-dipole-dipole-attraction-meaning)). Hence, stibine is the second-least volatile (as it has maximum van der waal's force due to its bigger size), followed by asrine, and finally phosphine.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90868/how-to-display-diffusion-to-a-young-audience
|
how to display diffusion to a young audience?
|
I need to display some science topic pertaining to the atomic bombs, or atomic history in some way to children of approximately middle school level. So I've decided that something interesting to cover would be gas diffusion (as used to isolate U-235 for the atomic bombs). I tried to see if there were any non-toxic, easy to obtain colored gases, but I couldn't find any. So the next thing I thought of is liquids.
Now I'm wondering if I can do something similar with liquids but I'm not sure how liquids would behave. Could I set up an effusion device, fill it with water and red food dye on one side, and water with blue food dye on the other and open a hole between the two and have them slowly diffuse? Will the diffusion of these two side be related to the molar mass or concentration of the dyes? Or would it follow some other law?
| 2 |
[
[
"\nAs you mention most colored gases (Cl2, Br2, NO2, I2) are toxic and being a gas difficult to handle. Therefore not suitable for such demonstration. Maybe, you could try two small colored smoke bombs in a transparent container although the colour is not really a gas (more like fine particulate) in this case.\n\n\nI would use liquids with the most common diffusion demonstration being a drop of blue ink in a glass of water. However, we need something fancier than that so how about this: Prepare a solution of red dye in water and a solution of blue dye and sugar or salt in water as well. In a 120 mL volumetric cylinder (needs to be large and for people to be able to see) add the sugar or salt blue solution and then on top of that layer with the red water solution by adding it slowly and carefully. Due to the larger density of the blue solution the two solutions will not mix instantly. You should end up with a bottom blue solution and a top red solution. Diffusion will start where the two liquids come in contact giving you a purple band in the middle which will expand over time with the speed being depending on the concentration of the bottom solution in sugar or salt (you ll need to tune this according to your presentation). You could even show how diffusion becomes faster with temperature by heating with a hairdryer but the effect will be masked a bit by convection.\n\n\nThis is how we grow crystals of compounds in the lab as well: Dissolve your compound in a heavy solvent such as DCM and layer it with a light solvent the compound doesnt dissolve in such as hexane. As the hexane slowly diffuses in the DCM solution the solubility slowly drops resulting in slow formation of high quality crystals suitable for X-Ray diffraction studies.\n\n\n",
"3"
],
[
"\nSpray perfume in the air. Ask the people around you to raise their hands when they can smell it.\n\n\nThe people closest to you will raise their hands first. People farther away will raise then hands after that.\n\n\nA perfect example of diffusion.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/90855/can-someone-explain-how-do-the-mechanism-for-this-reaction
|
Can someone explain how do the mechanism for this reaction? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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**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 5 years ago.
[Improve this question](/posts/90855/edit)
[](https://i.stack.imgur.com/Zgjxp.png)
Do you protonate the oxygen first?
| -3 |
[
[
"\nI think the mechanism will be something like this, \n[](https://i.stack.imgur.com/V8TV0.jpg)\n\n\nIn first step the hydrogen attached to $C\\_2$ as the carbocation on $C\\_1$ is more stable.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/90852/can-the-change-in-the-entropy-of-the-surroundings-always-be-obtained-by-dividing
|
Can the change in the entropy of the surroundings always be obtained by dividing heat transferred by the temperature at which the transfer occurs?
|
>
> Consider $\pu{1 mol}$ of an ideal monoatomic gas going through reversible isochoric heating from $\pu{100 K}$ to $\pu{1000 K}$.
>
>
> Calculate $\Delta S\_\pu{sys}, \Delta S\_\pu{surr}.$
>
>
>
$$\Delta S\_\pu{sys} = nC\_v \int\_{\pu{100 K}}^{\pu{1000 K}} \frac{dT}{T} = \frac{3}{2}R\ln10$$
Now, as I have
$q\_\pu{sys} = 1350R$.
I can put $\displaystyle \Delta S\_\pu{surr} = \frac{-q\_\pu{sys}}{T\_\pu{surr}}$ to get
$$\Delta S\_\pu{surr} = -\frac{27}{20}R$$ but in a reversible process $\Delta S\_\pu{total} = 0 $ but that gives me $$\Delta S\_\pu{surr} = -\frac{3}{2}R\ln10$$
Where am I wrong?
[](https://i.stack.imgur.com/5Ym9A.png)
Image source: [*Physical Chemistry*, 10th ed. by Atkins and de Paula, p. 116.](https://books.google.co.in/books?id=sWTYAwAAQBAJ&printsec=frontcover#v=onepage&q=hot%20spot&f=false)
Here, $\displaystyle q\_\pu{sur} =$ heat supplied to surroundings.
| 3 |
[
[
"\nI disagree with the explanation given in the text. The entropy change of the surroundings is given by the equation presented in your title only for an *ideal constant temperature reservoir*. Such an ideal reservoir is characterized by (a) an infinite capacity to absorb heat without its temperature changing and (b) infinite thermal conductivity, so that temperature is uniform throughout the reservoir, including at the interface with the system (where all the heat transfer occurs). So all the heat transfer between the system and the reservoir occurs at the reservoir temperature $T\\_{Res}$. In this way, all the irreversibility of the process occurs within the system, irrespective of whether the process is reversible, and the entropy change of the system always satisfies $ΔS≥Q/T\\_{Res}$. \n\n\nNone of the entropy generation for the process occurs within the ideal reservoir, and the entropy change of the ideal reservoir is thus always $-Q\\_{sys}/T\\_{Res}$, irrespective of whether the process is reversible. I should also mention that, although often not stated, the Clausius inequality is supposed to be expressed in terms of the temperature at the interface with the surroundings $T\\_{Res}$ where all the heat transfer is occurring. \n\n\nTypically, all the surroundings reservoirs you will encounter in most thermodynamics texts will be ideal constant temperature reservoirs. \n\n\nThe description in your post is inappropriate because, if the reservoir has a constant volume (i.e., constant mass), its temperature can change between the beginning and end of the process. Its entropy change will not be given then by the equation presented. In addition, the description omits the requirement that the reservoir most have infinite thermal conductivity (so that the temperature in the reservoir is spatially uniform).\n\n\n",
"4"
],
[
"\nThe *surroundings* that are assumed in the question are **not** the same surroundings as in the image. The surroundings in the image are essentially at a constant temperature while it is assumed in the question that *the surroundings are heating the gas reversibly*. The gas is heated by some heat source which is in thermal equilibrium with the gas.\n\n\nLet us assume that the surroundings which are not at a constant temperature are heating the gas. We assume that our universe consists of only gas plus the surroundings. The surroundings have some internal source of energy distributed uniformly so that no local hot spots occur. Now,\n\n\n$$\\Delta S\\_\\pu{gas} = n C\\_V \\int\\_{\\pu{100 K}}^{\\pu{1000 K}} \\frac{\\operatorname{d}T}{T} = \\frac{3}{2} R\\ln10 \\tag1\\label{(1)}$$ \n\n\nAnd for a reversible change, $\\Delta S\\_\\pu{total} = 0$ and thus $\\Delta S\\_\\pu{surr} = - \\Delta S\\_\\pu{gas}$.\n\n\nWe can also verify this in the following way:\n\n\n$\\pu{d}q\\_\\pu{gas} = n C\\_{V} \\cdot \\operatorname{d}T$ and from the First law of thermodynamics, $$\\pu{d}q\\_\\pu{surr} = - n C\\_{V} \\cdot \\operatorname{d}T = - \\pu{d}q\\_\\pu{gas} \\tag{2}\\label{(2)}$$\nThus, $$\\Delta S\\_\\pu{surr} = \\int^{\\pu{1000 K}}\\_{\\pu{100 K}} \\frac{\\pu{d} q\\_\\pu{surr}}{T} = - n C\\_V \\int\\_{\\pu{100 K}}^{\\pu{1000 K}} \\frac{\\operatorname{d}T}{T} = - \\frac{3}{2} R\\ln10 \\tag{3}\\label{(3)}$$\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90845/hofmann-rule-and-deviation-from-e2-mechanism
|
Hofmann rule and deviation from E2 mechanism
|
I've learnt that
>
> The best reaction conditions for synthesis of alkene by dehydrohalogenation are those that promote an E2 mechanism.
>
>
>
(From "Organic Chemistry" by TW Graham Solomons)
But, in the same book it is given that when 2-bromo-2-methylbutane follows Hofmann elimination with *tert*-BuOK as follows:
[](https://i.stack.imgur.com/HLOq8.png)
Clearly, it isn't following E2 mechanism.
When given such reactions, should I go with E2 mechanism or Hofmann rule, and why?
| 5 |
[
[
"\nThe difference between the different eliminations like E1, E2 or Hoffmann elimination is the proper choice of the base and proper solvent medium.\n\n\nLet's take your example i.e. 2-bromo-2-methylbutane, and choose different bases and reaction media for observing differences between E1, E2 and Hoffmann elimination.\n\n\n\n\n---\n\n\n**Why not E1?**\n\n\nFirst let's consider the reaction as:\n\n\n$$\\ce{2-bromo-2-methylbutane + EtO^- (in EtOH) -> }$$\n\n\nClearly, this reaction is conducted in a highly polar solvent where the carbocation - formed when $\\ce{Br^-}$ leaves - gets stabilised. The base then abstracts the hydrogen of $\\ce{C3}$ to form 2-methylbut-2-ene by an E1 mechanism.\n\n\nIf you use the same substrate but this time in a highly non-polar solvent, the possibility of E1 decreases as carbocation formation won't get extra stability. But as the base is not sterically hindered, it can still abstract hydrogen to form the Saytzeff product, but there is a case where it can also form the Hoffmann product.\n\n\n\n\n---\n\n\n**Why not E2?**\n\n\nThe E2 mechanism follows an \"anti-periplanar orientation of hydrogen and leaving group\". Suppose the reactant's stereochemistry is such that there is no anti-$\\ce{H}$ on the more substituted adjacent carbon for forming the Saytzeff product, but there is an anti-$\\ce{H}$ on a less substituted adjacent carbon, which will lead to the Hoffmann product. In that case, the major product can be the less substituted alkene due to stereospecificity of E2 reaction.\n\n\n\n\n---\n\n\n**Why E1cb?**\n\n\nNow consider your above given case. Here, the base $\\ce{t-BuO^-K+}$ is sterically hindered and is a very strong base. So, prior to any carbocation formation, this base will abstract hydrogen from the less hindered position, and form a carbanion. Also, a less substituted carbanion is more stable than a more substituted one (due to a smaller +I effect). Both these effects lead to formation of a less substituted alkene via an E1cb mechanism.\n\n\n\n\n---\n\n\nSo, depending upon which product is needed as a major one and the proper stereochemistry of reactant, you should choose whether to go with an E1, E2, or E1cb mechanism by choosing the proper base (bulky or less hindered) and proper solvent (polar or non-polar).\n\n\n",
"5"
],
[
"\nThe anti-elimination is not controlled by the bulk of the base. What is controlled by the bulk of the base is the accessibility of the bases to either of the two methylene hydrogens or the six methyl hydrogens and consequently the distribution of the Hofmann and Saytzeff alkenes. From the data below you can see that an increase in the bulk of the base favors the Saytzeff alkene.\n\n\nThe selectivity values correct for the number of hydrogens available for each mode of elimination. For potassium ethoxide, assume that the ratio is 70/30 to make the math simple. The percent 70 is divided by 2 (70/2 = 35) and 30 is divided by 6 (30/6 =5). The ratio 35/5 simplifies to 7/1 selectivity. This means that the rate constant for Hofmann elimination is seven times greater than for Saytzeff elimination, both of which are irreversible. In the presence of potassium t-butoxide the rates of the two eliminations are about equal. The base derived from 3-ethylpentan-3-ol favors Saytzeff over Hofmann elimination by a factor of 3.\n\n\n[](https://i.stack.imgur.com/2O3Jj.jpg)\n\n\n",
"2"
],
[
"\nHint: When the reactant is bulky , or it involves fluorine , then hoffmann rule is preffered over the normal saytzeff rule .\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/90842/can-polypropylene-be-separated-from-polyethylene
|
Can polypropylene be separated from polyethylene ? [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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Closed 5 years ago.
[Improve this question](/posts/90842/edit)
What techniques are available for sorting on a large scale?
| -2 |
[
[
"\nPractically only by mechanical means. Recycling numbers:\n\n\n2 – HDPE (High-Density Polyethylene) \n\n4 – LDPE (Low-Density Polyethylene) \n\n5 – PP (Polypropylene) \n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/90841/reaction-of-electromagnetic-radiations-with-the-air-pollutants
|
Reaction of electromagnetic radiations with the air pollutants [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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My friends, who are working on an IoT based technology project, explained their idea to me which is roughly like **reducing air pollution** by converting pollutants to oxygen ($\ce{O2}$) in any reaction, using waves and in particular microwaves. I have two questions in mind:
1. How do waves such as infrared, ultraviolet, microwave and others react with pollutants or in general with a gaseous mixture?
2. Which electromagnetic waves would be better in terms of factors such as time, temperature, rate of reaction and cost?
The pollutant mixture may contain all the pollutants described [here](https://en.wikipedia.org/wiki/Air_pollution#Pollutants), but the project is much more concerned with those that are more harmful than others to humans and life on earth.
| 1 |
[] |
https://chemistry.stackexchange.com/questions/90836/why-oxides-with-the-element-in-higher-oxidation-state-are-more-acidic
|
Why oxides with the element in higher oxidation state are more acidic
|
We see this trend of acidic-basic property of oxides in most cases:
$\ce{Mn2O7}$ is acidic, $\ce{MnO}$ is basic, the intermediate oxides are progressively less acidic and more basic, as the oxidation state of $\ce{Mn}$ decreases.
What is the reason of this general trend in almost all cases?
I have seen many explanations of this, most of them involving the proton releasing capacity of $\ce{E-O-H}$ bond, in the hydrolysed oxide. But most oxides of transition metals do not hydrolyze in water. So, I am seeking an alternative explanation without having to involve the hydrolysis of oxide. Also, most explanations on the internet lack any reference, so I prefer a reference.
| 4 |
[
[
"\nIt does not appear to be always true. According to [Wikipedia](https://en.m.wikipedia.org/wiki/File:Copper_in_water_pourbiax_diagram.png), copper(I) becomes acidic, forming an anionic hydroxide complex, at about 1.5 pH unit *lower* than copper (II). The higher oxidation state of copper qualifies as less acidic.\n\n\n",
"1"
],
[
"\nA simple answer for this would be to say that $\\ce{Mn2O7}$ has +7 oxidation state. Thus it is electron deficient and behaves as a lewis acid.\n$\\ce{MnO}$ has +2 oxidation state and it is comparatively electron rich. Thus it behaves as a lewis base. So compounds in the middle are amphoteric.\n\n\n",
"1"
],
[
"\nIn the bonds $M^{z+}-O-H$, where $M$ can be $Mn$ or $Cr$, the atom $M$ is positively charged. When this charge $z$ is large, $M$ repells the $H$ atom stronger than if z is small. So the molecule containing these bonds is a stronger acid. For example, $Z$ = $6$ in $H\\_2CrO\\_4$ and $Z = 7$ in $HMnO\\_4$. They both contain at least one $M-O-H$ bond. As $Z$ is high in these molecules, they are strong acids. Molecules containing $H, O$ and the same atoms $Mn$ and $Cr$ at a lower oxidation state exist. But they are not acidic in water. Examples : $Mn^{2+}$ makes $Mn(OH)\\_2$ which is insoluble in water and not acidic. $Cr^{3+}$ and $Cr^{2+}$ both produces $Cr(OH)\\_3$ and $Cr(OH)\\_2$ which are also insoluble and not acidic. \n\n\n",
"-2"
]
] |
https://chemistry.stackexchange.com/questions/90830/addition-of-acid-to-3-methyl-2-3-dihydropyran
|
Addition of acid to 3-methyl-2,3-dihydropyran
|
[](https://i.stack.imgur.com/ynt52.png)
When a general acid, H-X reacts with the alkene in an addition reaction, which regiochemistry is expected?
In addition reactions X attaches to the most substituted carbon whilst H bonds to the carbon with fewest alkyl groups. In this case I would believe the **third** carbon to be the most substituted carbon due to it being bonded to three other carbons. And the **second** carbon to be where the hydrogen will bond. However, apparently it should be the other way around. And this would be the compound formed:
[](https://i.stack.imgur.com/MWIzL.png)
Why is the second carbon the most substituted in this case? Does it have to do with oxygen being more electronegative and thus it is seen as a higher prioritized substituent?
| 2 |
[
[
"\nApparently, you're using the Markovnikoff's rule here. Recall that it is only a *basic* rule intended for simple unsaturated hydrocarbons, and **fails** at more advanced examples involving heteroatoms or rings.\n\n\nIn this case, the first step is the addition of $\\ce{H+}$ ion from the acid to the pi bond, which leads to the formation of a carbocation. The carbocation's positive charge is most stable next to the oxygen atom. Can you tell why?\n\n\nIn the second step, the $\\ce{X-}$ ion will simply attach at the position where the positive charge was the most stable, leading to a product as you've shown in the second image.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/90828/rate-of-formation-of-hydrate-in-a-carbonyl-compound
|
Rate of formation of hydrate in a carbonyl compound [closed]
|
**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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When water in presence of any acid is added to a carbonyl compound, it leads to the formation of its hydrate. But how do we determine the rate of reaction? Is it done by checking the amount of partial positive charge on carbonyl compound (by electron withdrawing group attached to carbonyl carbon)?
| -2 |
[
[
"\nThe formation of the hydrate of any carbonyl compound has the rate determining step as the nucleophilic addition of a water molecule to the electophilic carbon atom of the carbonyl group. \n\n\nHence, any electron withdrawing group, like fluoro or nitro, attached in the chain next to the carbonyl group will favor the *product* side of the equilibrium. \n\n\nOn the other hand, extra branching at the alpha position due to methyl/t-butyl groups will cause steric hindrance, while electron releasing groups will reduce the carbon's electrophilicity, both favoring the *reactant* side of the equilibrium instead.\n\n\nIt is important to note that while the hydration of alkenes goes to completion, the hydration of carbonyls does not. Hence, ignoring a few exceptions like formaldehyde, ninhydrin, chloral, etc., an *equilibrium* is setup during hydration of carbonyls. Therefore, it is better to talk about the *position of equilibrium* in this process rather than the *rate* of the reaction, as correctly pointed out by @Mithoron.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90826/difference-between-ld-and-sr-in-naming
|
Difference between L&D and S&R in naming [duplicate]
|
**This question already has answers here**:
[What is the difference between D and L configuration, and + and −?](/questions/44260/what-is-the-difference-between-d-and-l-configuration-and-and-%e2%88%92)
(2 answers)
Closed 5 years ago.
What is the difference between L&D and S&R? Can we say S-alanine instead of L-alanine?
| 0 |
[
[
"\nThe main difference between L, D configuration and *S*, *R* configuration is that the first one is relative configuration while the second one is absolute configuration. \n\n\nWhen you are distinguishing L-alanine from D-alanine, you only know that the $\\ce{-NH2}$ group on the chiral carbon of alanine is on the left hand side, while in D it is on the right hand side relative in a Fischer projection. There you can't be specific about other functional groups (like $\\ce{-COOH}$ or $\\ce{-H}$) position. \n\n\nBut when you distinguish *S* form of Alanine from the *R* form, it is done by proper arrangement of other two functional groups and a hydrogen atom. If any one of those is switched to some other position, the absolute configuration may change from *S* to *R* or from *R* to *S*. So, the configuration is more rigid in case of *R*-*S* configuration. \n\nAs the L-D configuration doesn't specify other group's position, you can't always say *S*-alanine instead of L-Alanine unless you know total spatial arrangement of all functional groups of alanine.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/90821/does-titaniumii-oxide-conduct-electricity
|
Does titanium(II) oxide conduct electricity? [duplicate]
|
**This question already has an answer here**:
[How is solid titanium(II) oxide an electrical conductor?](/questions/81724/how-is-solid-titaniumii-oxide-an-electrical-conductor)
(1 answer)
Closed 5 years ago.
In my book, it is given that $\ce{TiO(s)}$ is an electrical conductor. But I'd think that since $\ce{TiO(s)}$ is ionic and in solid state, it shouldn't conduct electricity. So, why does it conduct electricity?
| -1 |
[
[
"\nI'm not really convinced that $\\ce{TiO\\_{$1\\pm x$}}$ is best described as a mixture of Ti metal with oxygen.\n\n\nThe way I was taught it is that in $\\ce{TiO}$ the oxidation state of $\\ce{Ti}$ is $+2$; hence the $\\ce{3d}$ band is partially filled and the electrons in that band can act as charge carriers. $\\ce{TiO}$ has a notoriously wide range of stoichiometries, but the presence of extra or fewer vacancies doesn't affect the argument (as long as the average oxidation state of $\\ce{Ti}$ is less than $+4$, the $\\ce{3d}$ band will be partially filled).\n\n\nFor the $\\ce{Ti 3d}$ orbitals to form a band, they need to have sufficiently good overlap; this is ensured by the fact that (1) $\\ce{Ti}$ is early in the $\\ce{3d}$ block, so $Z\\_\\mathrm{eff}$ is low and the orbitals are relatively radially extended; (2) the ordered vacancies in $\\ce{TiO}$ allow for a more compact structure and closer $\\ce{Ti-Ti}$ contact. See e.g. Smart/Moore *Solid State Chemistry*, 4th ed., p 262.\n\n\nI'm not a solid state chemist and I do not know if this is still a simplification, but I think it is an improvement over treating $\\ce{TiO}$ as $\\ce{Ti}$ metal with oxygen in it.\n\n\n",
"3"
],
[
"\nFrom Wikipedia,\n\n\n\n> \n> Titanium(II) oxide ($\\ce{TiO}$) is an inorganic chemical compound of titanium and oxygen. It can be prepared from titanium dioxide and titanium metal at 1500 °C. It is non-stoichiometric in a range $\\ce{TiO\\_{0.7}}$ to $\\ce{TiO\\_{1.3}}$ and this is caused by vacancies of either $\\ce{Ti}$ or $\\ce{O}$ in the defect rock salt structure. (citation given below)\n> \n> \n> \n\n\nBasically, for your purposes, this is basically titanium metal with oxygen dispersed throughout its crystal structure. Hence, the titanium metal constituent conducts electricity.\n\n\n(Citation: \nHolleman, Arnold Frederik; Wiberg, Egon (2001), Wiberg, Nils, ed., Inorganic Chemistry, translated by Eagleson, Mary; Brewer, William, San Diego/Berlin: Academic Press/De Gruyter, ISBN 0-12-352651-5)\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/90819/calculate-the-magnitude-of-nonadiabatic-coupling
|
Calculate the Magnitude of Nonadiabatic Coupling
|
I'm in the great position to be able to calculate nonadiabatic couplings with TDDFT.
The result of such a calculation is a displacement for every atom in x, y and z.
```
Atom X Y Z
---------------------------------------------------
1 -0.000000 -0.000000 -0.160841
... ... ... ...
N 0.000000 -0.000000 0.009654
---------------------------------------------------
```
How do I calculate the magnitude of that coupling?
(It would be great, if you'd be as tutorial-like as possible.)
| 3 |
[] |
https://chemistry.stackexchange.com/questions/90672/most-activated-position-on-para-terphenyl-for-eas
|
Most activated position on para-terphenyl for EAS
|
**Para-terphenyl**: *it doesn't look pretty with all those math-y numbers, but those are going to come helpful in answering my question!*
[](https://i.stack.imgur.com/7Z2wG.png)
A question asked me to tell the expected product when this reacts with $\ce{Br2/FeBr3}$. Now, I have done such questions with all types of fancy organic molecules ranging from benzene to picric acid to parantiroanisole etc. I know that I am supposed to find the *most activated position*1 for EAS. But I have never done more than one phenyl ring at once.
So, in this question, I have drawn the eight resonating structures (without charge separation) of paraterphenyl manually. Here they are:
[](https://i.stack.imgur.com/3iq2v.png)
But, as you can see, every position seems to be activated identically. Each position has their own share of electrons through a pi bond, but it ends up being completely symmetric.
So, how do I find the most activated position in this compound?
---
1:Clarification: By "most activated position" I mean the position with maximum electron density. For example, in aniline, those are the ortho and para positions.
| 14 |
[
[
"\n\n> \n> So, how do I find the most activated position in this compound?\n> \n> \n> \n\n\nStart by drawing resonance structures of the various possible intermediates (sigma complexes) formed when $\\ce{Br^{+}}$ attacks the different ring positions in *p*-terphenyl. Whichever intermediate has the most resonance structures is likely to be the most stable (lowest energy) intermediate. The lowest energy intermediate will have the smallest activation energy and consequently be the kinetically favored product (barring steric effects).\n\n\nI've drawn just a few of these resonance structures in the following diagram.\n\n\n[](https://i.stack.imgur.com/pzkoI.png)\n\n\nIn the top row I've drawn a few of the possible resonance structures for electrophilic attack at the *para* position on a terminal phenyl ring. You can draw many more resonance structures delocalizing the charge around the various rings, **but notice that it is possible to delocalize the positive charge over all 3 aromatic rings**. In the second row I've drawn a few resonance structures showing charge delocalization when electrophilic attack occurs in the center ring. **Importantly, in this case charge can only be delocalized over 2 of the aromatic rings**.\n\n\nThis suggests that **based on resonance effects, electrophilic attack on a terminal benzene ring would be preferred over attack at the central benzene ring.**\n\n\nFurther, we would expect *ortho* and *para* attack on the terminal rings to be preferred since the phenyl substituent acts as an activating *o-p* director in electrophilic aromatic substitution. The rate of attack at the *ortho* position might also be decreased somewhat due to the steric effect of the adjacent phenyl (actually biphenyl) substituent.\n\n\nIn a paper by Shafig and Taylor (*J. Chem. Soc., Perkin Trans. 2*, **1978,** *0*, 1263-1267. DOI: [10.1039/P29780001263](https://doi.org/10.1039/P29780001263)), they explore your question by studying the electrophlic protonation of tritiated *p*-terphenyl. After correcting the rates for statistical effects (*e.g*. there are twice as many *ortho* positions on the terminal rings as there are *para* positions) they find that:\n\n\n* the *para* position (your positions $10, 16$) reacts at a relative rate of $273$\n* the *meta* position (your positions $9, 11, 15, 17$) reacts at a relative rate of $1.54$\n* the *ortho* position (your positions $8, 12, 14, 18$) reacts at a relative rate of $176$, and finally,\n* the $4$ identical positions on the middle ring (your positions $2, 3, 5, 6$) react with a relative rate of $59.1$\n\n\n**These results support our predictions** that\n\n\n1. attack on the terminal rings will be preferred over attack on the central ring and\n2. that attack at the *ortho* and *para* positions is preferred over attack at the *meta* position on the terminal rings.\n\n\n",
"9"
]
] |
https://chemistry.stackexchange.com/questions/90668/biomolecules-enzymes
|
Biomolecules : Enzymes [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
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My class $12$ board exams are approaching and i'm having a doubt in this question:
This is from my class $12$ NCERT exemplar book
>
> Activation energy for the acid catalysed hydrolysis of sucrose is $\pu{6.22 kJmol^-1}$, while the activation is only $\pu{2.15kJmol^-1}$ when hydrolysis is catalysed by the enzyme sucrase. Explain
>
>
>
I've found nothing so good from the web. All I know is this:
Enzymes, the biocatalysts, work even under mild conditions, they do not require elevated temperatures or other physical conditions. While with ordinary chemical catalyst, this might be the case. Hence, activation energy for chemical catalyst reaction is little more than enzyme catalyst.
Is this reason valid? Or is there any other reason?
| -1 |
[
[
"\nI'll just quote the official answer given in the NCERT Exemplar book. It reads:\n\n\n\n> \n> Enzymes, the biocatalysts, reduce the magnitude of activation energy **by\n> providing alternative path** (my emphasis). In the hydrolysis of sucrose the enzyme sucrase\n> reduces the activation energy from $\\pu{6.22 kJ mol^–1}$ to $\\pu{2.15 kJ mol^–1}$.\n> \n> \n> \n\n\nThe key thing to note here is an *alternate path* is provided to the reaction. The values given in the question ($6.22$ and $2.15$) are only to distract you from the systematic answer. You are not supposed to rote learn them. Focus on the lowering of activation energy due to an alternate pathway. Here is a graph depicting such a situation:\n\n\n[](https://i.stack.imgur.com/cVaxC.gif) *([source](https://www.chemguide.co.uk/physical/basicrates/catalyst.html))*\n\n\nThe mechanism how the catalyst *exactly* achieved this lowering of activation energy, that too specifically for sucrose, would be difficult to describe unless this is a research paper, and is surely not required from a 12th grader. All that you are required to answer is what I put in the yellow blockquote above.\n\n\nI hope it helps!\n\n\n",
"5"
],
[
"\nWell, in general, no matter if biocatalyst or homogeneous or heterogeneous catalyst in organic chemistry or the industry the role of the catalyst itself is to lower the activation energy of the reaction. Often people have two mistakes when they think about that. One is that they think, that the catalyst is not changed in the reaction and the other one is that the catalyst can change any thermodynamic energy level or so. \n\n\nAbout the first point, the catalyst takes part in the reaction. It is true that in general, it leaves the reaction as it entered (there is decomposition but we ignore that for a second) but inbetween the catalyst will take part in the reaction itself. And therefore we are speaking of a completely different reaction at all, which leads to the second point. The energy level of starting material, transition state (so activation energy or reaction barrier) and product is set. The catalyst does usually not influence this. Therefore you have to cross another reaction path.\n\n\nFor example, let's take the hydrogenation of ethylene, you add $H\\_2$ to ethylene to create ethane. If you do this with a metal catalyst on the metal's surface you will end up in an activation of the hydrogen itself fist. So the hydrogen molecule adsorbs on the surface of the metal and the bond is then cleaved. And those two separated hydrogen atoms can now hydrogenate the ethylene, which also adsorbs to the surface. So metal can help here and lower the overall energy of this reaction. The new path consists of several steps like coordination, cleavage, reaction and desorption from the catalyst. And if you choose this correctly then those energies are lower than for the uncatalyzed one.\nAnd in biocatalysis, it's basically the same. You can check for example the catalytic triad to see how many amino acids and the different sites in the catalytic center of the enzyme help to perform the reaction. \n\n\nAbout your specific reaction if have no idea, I am no biochemist, this is the only thing I could find about that. \n[](https://i.stack.imgur.com/mxfr5.png)\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/90667/are-certain-logp-values-eluted-when-100-methanol-is-used-as-the-solvent-in-a-c1
|
Are certain logP values eluted when 100% methanol is used as the solvent in a C18 column?
|
I have used a C18 sep-pak to concentrate metabolites in a chemically complex solution. The column was first washed with pure water, 75% methanol:water mix, and finally the compounds eluted with 100% methanol.
What I am wondering is what chemical properties that I can infer from this information? I know I would have washed the very polar compounds away, and in pure methanol, I would expect compounds of a semi-polar nature to be eluted. Can I assume that these compounds would have a specific LogP range? I cannot find an answer online anywhere, so I guess the answer must be no, and if so, why not?
There is one (or more) compound of a biologically active nature in this complex mix. I am wondering if I can use the elution conditions to understand some of the chemical properties of the compounds eluted. Further, I did FT-ICR-MS on this mix, and I would like to look at a list of compound hits from the mass spec databases and understand how feasible a particular hit is based on its polarity.
| 1 |
[
[
"\nHPLC is often used to determine lipophilicity in industry, as the original methods to determine logP by partitioning between octanol and water was not particularly amenable to high-throughput assay. \n\n\nLipophilicity values measured by HPLC are often reported as cLogP / cLogD (where c is relating to the fact that the value has been determined by some form of chromatography). \n\n\nExperimentally, a C18 column is used with an acetonitrile/water solvent system, buffered appropriately. Unlike with the octanol/water procedure, different companies run these HPLC assays differently, and as such they aren't as comparable (i.e. if you ran 20 compounds through a GSK assay, and the same 20 compounds through a Pfizer assay, the precise values for cLogD obtained in each case would vary, even if the general trend remained the same).\n\n\nIn terms of *you* doing this yourself, you'd first need to find a training set of compounds with known and reliable values. If you know paracetamol comes of at x mins and has a literature logP of y, and ibuprofen comes off at z mins and has a literature logP of w, you can interpolate. When you then run an unknown sample, its retention time using your method can be used to predict the cLogP/D (keep in mind that you'd need a far greater training set than 2 compounds). \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90664/how-to-explain-the-products-of-the-nucleophilic-aromatic-substitution-of-1-chlor
|
How to explain the products of the nucleophilic aromatic substitution of 1-chloro-3-methylbenzene with NaOH?
|
I guess this reaction is a nucleophilic aromatic substitution where $\ce{ONa-}$ is the nucleophile $\ce{Cl-}$ the leaving group, is that right?
[](https://i.stack.imgur.com/sBmBT.png)
Methyl group is a ortho-, para- director. So why should I have a meta product here? Don't we need an electron withdrawing group ($\ce{NO2}$ , carbonyl) in a NAS-reaction which is at ortho or para positions to the leaving group?
I am a bit confused about the shown picture.
| 1 |
[
[
"\nWhat is going on here is a benzyne process. Under the conditions of intense heat the hydroxide removes one of the protons either side of the Cl. Cl- then leaves to create a triple bond, giving a benzyne. As you might imagine this is a highly reactive species and reacts with the OH- with no regioselectivity; either end can get hydroxylated and you have two possible benzyne intermediates. Thus you get the mixture of products. \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90663/reason-behind-conversion-of-acid-to-acid-bromide-in-hell-volhard-zelinsky-reacti
|
Reason behind conversion of acid to acid bromide in Hell-Volhard-Zelinsky reaction [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 5 years ago.
[Improve this question](/posts/90663/edit)
Why in Hell-Volhard-Zelinsky (HVZ) reaction the acid is first converted to acid bromide by using $\ce{PBr3}$ and then alpha halogenation is carried out?
| 4 |
[
[
"\nThe $\\alpha$ halogenation occurs through the enol form. \n\n\nThe formation of the acyl halide is key because carboxylic acids do not form enols readily since carboxylic acid proton is removed before the $\\alpha$ hydrogen. Acyl halides lack the carboxylic acid hydrogen. \n\n\nThe phosphorus halides form the acyl halides readily because phosphorus has a huge affinity for oxygen. \nSometimes, phosphorus tribromide is not enough to form the acyl halide but, since there is also bromine in the media, phosphorus pentabromide is formed which can form acyl halides even of solid acids. \n\n\nThe halogenation is performed by the $\\ce{X2}$, either bromine or chlorine. \nWhen $\\ce{PBr3}$ and $\\ce{Cl2}$ is used, the *alpha*-chloro acid is obtained and not the bromo acid. \n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/90661/basicity-of-nitrogen-heterocyclic-compounds
|
Basicity of nitrogen heterocyclic compounds
|
>
> **Question** :
>
>
> The order of basicity for the following compounds is :
>
>
>
[](https://i.stack.imgur.com/fIFHc.png)
The logic I've applied is that since **4** has a conjugated system, the lone pair of N will be delocalised and hence would be the least basic. Conjugate systems/Resonance is not seen in the other three compounds
I'm not entirely sure about how to differentiate between the other three. For morpholine **2**, If we see the stability of the conjugate acid formed, the O would exert (very less) -I effect, which should destabilize the positive charge on N (on formation of conjugate acid). Now between **1** and **3**, **1** has a five other C atoms attached to the N, while **3** has four. Now CH3 is a +I group, hence more CH3 in **1** should stabilize the positive charge on N more than **3**.
I don't think Inductive Effect should matter much, considering Inductive Effect becomes very ineffective with increase in chain length. But this is the only reasoning that strikes me. So ultimately, order of basicity becomes **4** < **2** < **3** < **1**. Is my reasoning and the final answer correct?
| 3 |
[
[
"\nNitrogen inversion is not relevant to this discussion. A faster rate of inversion does not correspond to a greater delocalisation in the electronic structure, as lone pair inversion is not equivalent to resonance. The lone pair does not exist as a delocalised smear; it exists in two different localised forms.\n\n\n[](https://i.stack.imgur.com/2O3lu.png)\n\n\nThe PDF on nitrogen inversion in the comments does not even bring up the topic of basicity, and quite rightly so.\n\n\nInductive effects explain why morpholine **2** is less basic, and resonance effects explain why pyrrolidone **4** is even less basic. However, since the p*K*aH's of piperidine **1** and pyrrolidine **3** are so similar (11.22 and 11.27 respectively), I think it is pointless to try and qualitatively rationalise a difference in their basicities. It is like trying to explain why the temperature in Spain one day is 11.22 °C and the temperature the next day is 11.27 °C. The relative basicity could well be different if you switch the solvent, which would invalidate any reasoning based on - for example - s-orbital character of the nitrogen lone pair, which I think is the only plausible qualitative argument here.\n\n\nSimply looking at the data, though, the order you want is **4** < **2** < **1** < **3**.\n\n\n",
"7"
],
[
"\nTo compare between **1** and **3**, let's take into account the interaction between the non-bonding electron pair on nitrogen and the anti-bonding MO of the C-H bond (σ\\*C-H).\n\n\nIn order for the interaction to be effective, the lone pair on nitrogen and the C-H bond must be planar. In the six-membered ring, there is always a C-H bond next to the nitrogen lays in the parallel position against nitrogen's lone pair, which is the axial one. In the five-membered ring, the C-H bond does not totally lie on the same plane with the lone pair of nitrogen, so the interaction is less effective.\n\n\nThis interaction decreases the energy of the non-bonding MO, which consequently decreases the basicity of the nitrogen. As this interaction is more stressed in the six-membered ring, its basicity is less then that of the five-membered ring.\n\n\nBasicity: **4**<**2**<**1**<**3**\n\n\n",
"0"
],
[
"\nI would like to answer it,\n\n\n**KEY POINT:** More electron density on N, more availability of the lone pair, more good base.\n\n\n$(1)$ The nitrogen is directly connected inside the ring, surrounded by lots of carbon atoms. N being more electronegative than C, snatches the electrons towards itself as the nearest carbon atoms shows +I effect, it drops down as we move away from N. So, electron density on N increases, which means the lone pair is more available for donation. Hence, it's a good base!\n\n\n$(2)$ The ring is the same but this time, a highly electronegative O came now. As it is more electronegative than N, it will snatches the electrons towards itself from the downside ring. Thus, decreasing the electron density on N. Hence, it's now no more willing to donate its lone pair. It is more tightly bound to the N as the repulsions from the electrons around it are decreased. Hence, not so good base.\n\n\n$(3)$ Everything still the same, but the number of carbon atoms reduces to $4$ around it. Hence, +I effect reduces compared to $(1)$. So, electron denisty reduces comapred to $(1)$. Basic character reduces compared to $(1)$.\n\n\n$(4)$ Less basic than $(3)$ as N got a $\\ce{sp^2}$ hybridized carbon around it which is more electronegative than the previous. Hence, that +I effect reduces. So, a weak base than $(3)$ . But, the O here isn't connected directly to the ring. So, it will not effect any other C atom to perform their +I effect. Hence, the density on N reduces a bit, whereas in $(2)$, it reduces a lot as the O here will effect every C to not to give their best at +I effect, it will also snatches electron density around N a bit. Hence, it's the least Basic. So, correct order:\n\n\n$\\ce{(1)>(3)>(4)>(2)}$\n\n\n*Actually I do not at all like chemistry, but i don't know why i posted this as my ever first unexpected answer on this site. This is probably because this topic is my favourite by mistake from my class $12$ chapter: Amines.*\n\n\n",
"-3"
]
] |
https://chemistry.stackexchange.com/questions/90660/chemical-tests-given-by-glucose-biomolecule
|
Chemical tests given by glucose (biomolecule)
|
We know that glucose gives Tollen's test because it's a reducing sugar, and having a hemiacetal group, on hydrolysis can be converted into an aldehyde group, giving the positive test.
But, when we come to tests like the Schiff's test or with sodium bisulphite, it does not show any reaction which are shown by aldehydes. Why? Is the answer that hemiacetal group *cannot* be converted into aldehyde or anything else? Probably there shouldn't be any water used in these reactions, so no hydrolysis, no conversion from hemiacetal to aldehyde. Is there any water present in Tollen's reagent?
| 2 |
[
[
"\nThis are two different types of reactions. \n\n\nTollens is a redox reaction in which silver is reduced and glucose is oxidized. \nThis reaction depends on the redox potential of glucose only. \n\n\nSchiff and bisulphite tests are addition reactions. The reagents condense with the aldehyde group. \nThis occurs also in the case of glucose but it is not sufficient to give a positive result. \nIn the case of bisulphite, the adduct must precipitate, but it is too soluble. \nIn the case of the Schiff test, the adduct must dehydrate, but it forms the glucosamine derivative instead. \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90658/intensity-of-surface-plasmon-resonance-absorption-peaks-for-a-nano-rod
|
Intensity of “surface plasmon resonance “ absorption peaks for a nano-rod
|
The SPR absorption spectrum of a nano-rod is seen to have two peaks with the longer wavelength peak corresponding to the absorption by the longitudinal plasmons and the other to the transverse plasmons. When the intensities are compared.. longitudinal plasmon peaks are almost always more intense than transverse ones. Why?
| 2 |
[
[
"\nThe intensity of a plasmon resonance is related to the magnitude of the transition dipole moment. Longitudinal resonances have a larger transition dipole moment compared to transverse resonances, and so have more intense absorption peaks. You can rationalize this by imagining the change in dipole moment from a longitudinal versus transverse electron density oscillation.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/90653/does-95-8-silver-tarnish-as-fast-92-5-silver
|
Does 95.8% silver tarnish as fast 92.5% silver?
|
I've been trying to find out whether 95.8% (Britannia silver) would tarnish as fast as 92.5% (Sterling silver)? I haven't been able to find much on tarnish for Britannia silver.
Given that both silvers are kept in the exact same environment, would the Britannia silver tarnish at the same rate as Sterling silver? If not which of the two tarnishes quicker?
| 9 |
[
[
"\nIf you look here: <https://blog.centimegift.com/does-sterling-silver-tarnish/>, it says that pure silver is very tarnish resistant. Sterling silver on the other hand being mixed with some Cu to enchance the metal's mechanical properties tarnishes as a result of the Cu in it. Therefore, considering that the amount of Cu in Britannia silver is smaller, it should tarnish more slowly than sterling silver. In practice i dont know if there will be a significant difference as the increse in silver purity is not that much.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90652/voltaic-cell-and-nernst-equation
|
Voltaic cell and Nernst equation
|
>
> It is given voltaic cell
> $\ce{Cu}|\ce{CuSO4}(\pu{10^-3 M})||\ce{CH3COONa}(10^{-2})|\ce{Pt}(\ce{H2})(\pu{1 atm})$. Find the potential at $\pu{50^\circ C}$.
>
> Given $K\_\mathrm{a}$ for $H= 1.8 \times 10^{-3}$.
> $E\_{\ce{Cu^2+/Cu}} = \pu{0.336 V}$
>
>
>
When voltaic cell starts working: $\ce{H2}$ becomes $\ce{2H+}$. In the other hand $\ce{CH3COO-}$ hydrolizes forming $\ce{CH3COOH}$ and $\ce{OH-}$. We know $K\_\mathrm{a}$ and we can find $K\_\mathrm{b}$. And from that we can find the concentration of $\ce{OH-}$ and after that using the formula $\mathrm{pH}=14-\mathrm{pOH}$ we can find the concentration of $\ce{H+}$.
Now how we can find the concentration of $\ce{Cu^2+}$ in order to replace it into Nernst equation?
\begin{align}
E &=E^\circ - \frac{RT}{nF} \cdot \ln\{\text{Qc}\}
\text{, where}&
\text{Qc} &= \frac{[\ce{H+}]}{[\ce{Cu^2+}]}.
\end{align}
| 1 |
[
[
"\nA correct standard notation for the galvanic cell would also include the physical state of each species involved. As such, the correct notation for the given cell is:\n\n\n$$\\ce{Cu(s)}~|~\\ce{CuSO4(aq)}~(10^{-3}~\\mathrm{M})~||~\\ce{CH3COONa~(aq)}~(10^{-2}~\\mathrm{M})~|~\\ce{Pt(s)}~(\\ce{H2(g)})~(1~ \\mathrm{atm})$$\n\n\nRecall that aqueous solutions of salts are generally accompanied by their molarity in solution. Also, metals are usually solid, while gaseous species are accompanied by a reference electrode (platinum in this case) and their pressure value. \n\n\nIn your case, notice that the ionic salt $\\ce{CuSO4}$ is assumed to be a $100\\%$ dissociated ionic salt, hence, its aqueous solution will produce $[\\ce{Cu^{2+}}]=[\\ce{SO4^{2-}}]=10^{-3}\\mathrm{M}$. That should answer your final query.\n\n\nApart from that, your expression for $\\mathrm{Q\\_c}$ is wrong. It should be $\\displaystyle\\mathrm{Q}=\\frac{p\\_{\\ce{H2}}}{[\\ce{Cu^{+2}}]}$. Since $\\ce{H2}$ is present in gaseous form, use its *pressure* value in the expression for $\\mathrm{Q}$ (and not molarity which doesn't exist)\n\n\n**PS:** The concentration ($\\pu{10^−3M}$) that's given to you is the concentration of copper ions after equilibrium has been achieved for the interconversion of $\\ce{Cu+2(aq)}$ and $\\ce{Cu(s)}$.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90644/naming-tetrahalogenated-benzene
|
Naming tetrahalogenated benzene
|
[](https://i.stack.imgur.com/34x4J.png)
ChemSketch generates the IUPAC name of the compound above as "2-bromo-1-chloro-5-fluoro-3-iodobenzene" but I have a different answer.
I know that I have asked a [related question](https://chemistry.stackexchange.com/questions/68624/alphabetization-rule-in-case-of-consecutively-placed-substituents) before as well. The key takeaway from orthocresol's answer was that:
>
> first, order the prefixes by alphabetical order, and then choose the locant set in order to minimise the locants
>
>
>
I have applied the same here: the prefixes ordered in alphabetical order are "bromo, chloro, fluoro, iodo". Now, I have to choose a locant set that minimizes the locants. A possible locant set is "1-bromo-2-chloro-4-fluoro-6-iodo" which correctly locates the substituents. Also, the locant set "1246" is lower than "2153", as noted in [this answer](https://chemistry.stackexchange.com/questions/27095/iupac-nomenclature-smallest-sum-of-locants/28009#28009). So, according to me, the preferred IUPAC name should be "1-bromo-2-chloro-4-fluoro-6-iodobenzene".
What exactly am I doing wrong here?
| 5 |
[
[
"\nThe locant set ‘1,2,4,6’ is not lower than ‘2,1,5,3’.\n\n\nAccording to the current version of *[Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)](http://dx.doi.org/10.1039/9781849733069),* the lowest locant set is defined as follows:\n\n\n\n> \n> **P-14.3.5** Lowest set of locants\n> \n> \n> The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference; for example, the locant set ‘2,3,5,8’ is lower than ‘3,4,6,8’ and ‘2,4,5,7’. \n> \n> (…)\n> \n> \n> \n\n\nEspecially note that the locants are cited **in order of increasing value** when they are compared.\n\n\nThus, the locant set ‘2,1,5,3’ is regarded as ‘1,2,3,5’, which is lower than ‘1,2,4,6’.\n\n\nTherefore, the correct name is **2-bromo-1-chloro-5-fluoro-3-iodobenzene.**\n\n\n\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/90643/how-to-determine-the-ph-range-at-which-maximum-hydrate-is-present
|
How to determine the pH range at which maximum hydrate is present?
|
**Question:**
>
> Find the $\mathrm{pH}$ range at which maximum hydrate is present in a solution of oxaloacetic acid (given $\mathrm{pK\_a}$ of $\ce{-COOH}$ in the left is $=2.2$ and on the right is $=3.98$):
>
>
> [](https://i.stack.imgur.com/SKBVP.png)
>
>
>
I understand that the hydrate will be formed at the carbonyl group position, resulting in a product like this:
[](https://i.stack.imgur.com/DhkEw.png)
I know that the gem-diols are usually stablized by H-bonding, as is the case here as well (between the lower hydroxy and left acid group). I can also see that loss of the $\ce{H+}$ ion from the carboxylic acid group on the left will create an $\ce{O-}$ ion which will have an even stronger H-bonding. However, I am not sure how I can use this fact to relate the given $\mathrm{pK\_a}$ values and the expected $\mathrm{pH}$ range.
I need help completing my train of thought. Thank you!
| 1 |
[] |
https://chemistry.stackexchange.com/questions/90641/why-does-cobalt-have-no-negative-charge
|
Why does cobalt have no negative charge?
|
I would like to know why cobalt cannot have a negative charge (or at least why a negative charge for cobalt isn't typical). I am not sure where I have gone wrong in my reasoning.
The electronic configuration for cobalt is:
$\ce{Co}$: $[\ce{Ar}] 4s^2 3d^7$
Clearly, you could give cobalt a $+2$ charge in order to get a more stable ion as follows
$\ce{Co^{2+}}$: $[\ce{Ar}] 3d^7$
And from there, you could go further and remove $2$ more electrons from the $d$-orbital to become even more stable like this:
$\ce{Co^{4+}}$: $[\ce{Ar}] 3d^5$
If this logic is correct, then could I apply this same thinking but for adding electrons? Such that
$\ce{Co}$: $[\ce{Ar}] 4s^2 3d^7\to$ $\ce{Co^{3-}}$: $[\ce{Ar}] 4s^2 3d^{10}$
Since the $d$-orbital has been filled, the atom should have become more stable. Is it true?
Thus, am I able to argue that Cobalt has $3$ main ion charges: $+2, +4, -3$?
| 0 |
[
[
"\nCobalt can be found in various oxidation states including negative ones such as Co(-I) in NaCo(CO)4. In this instance however the excess electron density on the cobalt center is relieved by backbonding with the CO ligands and therefore such a low formal oxidation state of the Co atom is stabilized. On the other hand the free Co$^-$$^1$ ion cannot hold the extra electron first due to the low electronegativity of Co as TAR86 hinted out in his comment and second due to the fact that the extra electron is placed in a d orbital which is not very penetrating and thus shielded by all other electrons. Hence, it is unlikely to encounter compounds such as [Na]$^+$[Co]$^-$ where there are no ligands on the cobalt to stabilize the excess electron density. The opossite applies when you remove electrons from Co (and any other transition metal): the atom becomes positively charged and the nucleus attracts the remaining electrons even more effectively (Higher Zeff). Therefore, positive oxidation states are more common.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/90634/is-it-possible-to-form-an-alloy-of-gold-and-zinc
|
Is it possible to form an alloy of gold and zinc?
|
I know you can dissolve gold in mercury, but i do not have access to any. Would it be possible to dissolve it in molten zinc instead mercury.
| 0 |
[
[
"\nYes. [There are alloys and intermetallic compounds up to $\\ce{44\\%Au:56\\%Zn}$](https://www.sciencedirect.com/science/article/pii/0022508879902820). This phase diagram is from [here](https://www.researchgate.net/publication/253702430_Size_effect_on_the_mechanical_properties_in_zinc_oxide).\n\n\n\n\n\nThat said, crystal structure changes with composition, so appearance on solidification might be hard to predict.\n\n\n[Are you trying to emulate Hevesy?](https://www.npr.org/sections/krulwich/2011/10/03/140815154/dissolve-my-nobel-prize-fast-a-true-story)\n\n\n",
"5"
],
[
"\nGold looks pretty soluble in molten zinc. The melting temperature increases up to about 625 C with 50% gold.Charcoal on the surface of the melt will partly protect the zinc ; I have seen charcoal used on lead but not zinc.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/90628/why-we-need-halogens-of-different-periods-in-wurtz-fittig-reaction-which-is-t
|
Why we need halogens of different periods in wurtz-fittig reaction? + which is the correct mechanism?
|
For the mechanism of wurtz-fittig reaction I got two different ones on the net.
1. From another [question](https://chemistry.stackexchange.com/questions/81983/does-wurtz-fittig-reaction-involves-sn1-or-sn2) on stackexchange: [](https://i.stack.imgur.com/1eIwx.png)
2. From tutorvista:[](https://i.stack.imgur.com/Jfs88.png)
They are not really the same. In the first one the alkyl group is acting as the nucleophile and as Sn reaction do not occur at sp2 hybridized orbitals, it goes through an SnAr reaction. But in the second one the aryl group is acting as a nucleophile. So the same reasoning does not apply here, because here Sn2 is happening at a primary sp3 hybridized carbon whereas the aryl group is just the nucleophile. So is the second mechanism wrong or I am missing some point? Neither of the mechanisms really show why we used two different halogens in the alkyl and aryl group (or maybe I can't understand), in fact the first mechanism is using Cl on the alkyl group whereas the second mechanism is using iodine (which is a far better leaving group). But the aryl group has bromine on both. Really confusing if I am to search for explanations myself. Please clear me up.
| 2 |
[] |
https://chemistry.stackexchange.com/questions/90625/why-does-the-alkyl-group-anti-to-the-hydroxyl-migrate-in-beckmann-rearrangement
|
Why does the alkyl group anti to the hydroxyl migrate in Beckmann rearrangement?
|
**Background:** Many organic reactions involve the migration of an alkyl group from one position to the adjacent one. The migration of the alkyl group is decided by its *migratory aptitute* *i.e.* electron-richness. It generally follows the priority order of hydride > phenyl > higher alkyl > methyl.
**Main question:** The Beckmann rearrangement also involves an alkyl migration. However, this migration is *not* governed by migratory aptitude. In fact, the $\ce{-R}$ which is in *anti*-position to the hydroxyl group in the oxime migrates, irrespective of its migratory aptitude! My question is why is this so? And are there other organic reactions which have such a rule for alkyl migration?
[MasterOrganicChemistry](https://www.masterorganicchemistry.com/reaction-guide/beckmann-rearrangement/) and [Wikipedia](https://en.wikipedia.org/wiki/Beckmann_rearrangement) don't even mention the word "trans" or "anti" anywhere. My textbook obviously doesn't mention anything either, hence I am driven to ask this question.
| 5 |
[
[
"\nOximes can undergo the Beckmann rearrangement. Oximes also exist as stable *syn* and *anti* isomers. In the figure below, $\\ce{R\\_1}$ is *anti* to the hydroxyl group; another isomer exists where it is *syn* the the hydroxyl group.\n\n\nIn the first step of the Beckmann rearrangement the protonated hydroxyl group makes for a good leaving group. As the hydroxyl group begins to depart, the $\\ce{R}$ group which is *anti* to the leaving group begins to migrate and form a bond with nitrogen.\n\n\n[](https://i.stack.imgur.com/mgLW1.png)\n\n\n([Wikipedia](https://en.wikipedia.org/wiki/Beckmann_rearrangement))\n\n\nThis step is said to be concerted (rather than stepwise) with the $\\ce{N-O^{+}H2}$ bond being broken and the new $\\ce{R-N}$ bond being formed at (more or less) the same time. Whether you think of the $\\ce{R}$ group as participating in something similar to an SN2 reaction or whether you think of the $\\ce{R}$ group as starting to bond to the available $\\sigma^{\\*}$ orbital of the elongating $\\ce{N-O}$ bond (equivalent descriptions), you see that having the $\\ce{R}$ group approach the nitrogen from a 180° angle from the breaking $\\ce{N-O}$ bond would be preferred. **Hence, the $\\ce{R}$ group *anti* to the $\\ce{N-O}$ bond migrates preferentially**.\n\n\nThere are many reactions showing this geometric preference, the [Baeyer-Villiger oxidation](https://en.wikipedia.org/wiki/Baeyer%E2%80%93Villiger_oxidation) being an example.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/90618/does-langmuir-model-predict-evaporation-rates
|
Does Langmuir model predict evaporation rates?
|
[Slide #16 of this presentation on amalgams](https://www.fda.gov/ohrms/dockets/ac/06/slides/2006-4218s1-03.pdf) says that "Langmuir's equation" provides a specific "theoretical maximum evaporation rate" for liquid Hg as a function of temperature and surface area.
The only common equation bearing Langmuir's name is [his adsorption model](https://en.wikipedia.org/wiki/Langmuir_adsorption_model). I can vaguely imagine that model being applied in reverse to characterize evaporation rates, but I can't see or find how in general, much less in the specific case asserted in the paper. Can someone provide a reference or example of how that (or any other model) provides a "theoretical maximum evaporation rate" of a particular liquid?
| 2 |
[
[
"\nThat most likely refers to the [Hertz-Knudsen equation](https://en.wikipedia.org/wiki/Hertz%E2%80%93Knudsen_equation), also known as **Knudsen-Langmuir equation**. It is the most simplified model of evaporation in which no discussion is given into the diffusivity and the chemical potential gradient (let alone other driving forces like gravity and pressure gradient): one just assumes the leaving mass flux is dictated by the system's thermal energy. This model then suggests a direct relation between evaporation rate and vapor pressure, temperature and molar mass.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/90615/how-to-filter-out-sulfur-dioxide
|
How to filter out sulfur dioxide [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Update the question so it focuses on one problem only by [editing this post](/posts/90615/edit).
Closed 5 years ago.
[Improve this question](/posts/90615/edit)
Lets say I have a container that contains a mixture of air and sulfur dioxide. I want to be able to filter out the sulfur dioxide from the air, how can I do this either through the form of a filter or a chemical reaction?
| 0 |
[
[
"\nJust add your gas mix in a container with NaOH solution and shake well or bubble it continuously through the solution. SO2 dissolves in water to form H2SO3 (and H2SO4 in the atmosphere when it rains resulting in what we call acid rain) which will be neutalized by NaOH to give dissolved Na2SO3. Your air should be freed of most SO2. This is how it is done in the industry. There are many other ways as well but most rely in neuralizing the SO2. Google Flue-gas desulfurization for more details.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/90613/can-bronze-be-electroplated-with-zinc
|
Can bronze be electroplated with zinc?
|
I had to have a motorcycle part cast in bronze that was originally steel. I want to electroplate it with zinc. Will this work? Thanks
| -2 |
[
[
"\nBronze is mainly copper (with a bit of tin in it) which can be electroplated with Zn and then heated to form Brass (Copper-zinc alloy): <https://www.stem.org.uk/resources/elibrary/resource/33571/zinc-plating-copper-and-formation-brass-%E2%80%93turning-copper-%E2%80%98silver%E2%80%99>. \n\n\nAlso this question has been discussed extensively here: <https://www.finishing.com/126/37.shtml>.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90598/is-internal-energy-equal-to-3-2nrt-for-an-ideal-monoatomic-gas
|
Is internal energy equal to (3/2)nRT for an ideal monoatomic gas?
|
I came across a Khan Academy video providing the following relation for the internal energy of the system $U$:
$$U=\frac32nRT,\label{eqn:1}\tag{1}$$
where $n$ is the amount of gas, $R$ is the universal gas constant and $T$ is the temperature of the system.
I know internal energy is a state variable and only its *change* can be measured, not the value at a particular state. Also, isn't \eqref{eqn:1} given for the average kinetic energy
$$\bar{E\_\mathrm{k}} = \frac 32nRT,\tag{2}$$
not for the internal energy $U$? I'm confused.
| 5 |
[
[
"\nMolecules have internal energy due to intermolecular interactions, as well as translational kinetic energy, rotational energy, vibrational energy, electronic energy (and if you care to include them, nuclear energy and the mass-energy of the protons, neutrons and electrons themselves). \n\n\nBy saying \"ideal\", energy due to intermolecular interactions is eliminated. \n\n\nBy saying \"monoatomic\", energy due to rotation and vibration is eliminated. \n\n\nSo only translational kinetic energy and electronic energy remains. \n\n\n(3/2)nRT is the translational kinetic energy, and since almost all atoms are in the ground electronic state at low temperature, it is a good expression for internal energy as long as the temperature is low enough that essentially all atoms are in the electronic ground state. \n\n\n",
"10"
],
[
"\nFrom a statistical standpoint, the mean energy of a system is given by\n\n\n$$\\langle E \\rangle = E \\cdot P(x) = \\frac{\\int\\limits\\_{-\\infty}^{+\\infty} E \\mathrm e^{- \\beta E}}{\\int\\limits\\_{-\\infty}^{+\\infty}\\mathrm e^{-\\beta E}},\\tag{1}$$\n\n\nwhere $\\beta = 1/(k\\_\\mathrm{B}T)$ and $P(x)$ is the probability of the system being at a particular energy $E$.\nNow, if your energy dependence is quadric in some variable, this is if $E=ax^2$, where $a$ is just some constant, the mean energy becomes (thanks to some pretty cool math and Gaussian integrals)\n\n\n$$\\langle E \\rangle = \\frac{\\int\\limits\\_{-\\infty}^{+\\infty} {ax^2\\mathrm e^{- \\beta ax^2}}}{\\int\\limits\\_{-\\infty}^{+\\infty}\\mathrm e^{- \\beta ax^2}} = \\frac{1}{2 \\beta}= \\frac{1}{2}k\\_\\mathrm{B}T.\\tag{2}$$\n\n\nYou should note that this result is independent of $x$ and $a$. Actually, it is easy to show that if instead of one variable, the energy depends on $n$ quadratic variables, often called the modes of the system, each mode contributes the same amount of energy $k\\_\\mathrm{B}T/2$ to the system (all you have to do is repeat the calculations for $\\left.E = \\sum\\_{i=0}^n {a\\_i}x\\_i^2\\right)$ — this is known as the **equipartition theorem**. Hence, the mean energy of a system with $n$ quadratic modes becomes\n\n\n$$\\langle E \\rangle = \\frac{1}{2}nk\\_\\mathrm{B}T.\\tag{3}$$\n\n\nIn an ideal monoatomic gas the internal energy of the system $U$ is just its kinetic energy $E\\_\\mathrm{k}$ (there is no energy due to vibration, rotation and intermolecular interactions), and therefore\n\n\n$$E=E\\_\\mathrm{k}=\\frac{1}{2}mv^2 = \\frac{1}{2}mv\\_x^2 + \\frac{1}{2}mv\\_y^2 + \\frac{1}{2}mv\\_z^2.\\tag{4}$$\n\n\nThe energy is the sum of three quadratic modes, so the internal energy of the system is simply (substituting $n$ by $3$ in our expression for $U$)\n\n\n$$U = \\frac{3}{2}k\\_\\mathrm{B}T = \\frac{3}{2}nRT.\\tag{5}$$\n\n\nP.S. This might have been a bit more calculus then you'd asked for, but it is the reason this expression exists. I always find it helpful to know where things come from, so I hope this helps you as well.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/90595/why-is-hydroxide-ion-the-strongest-base-only-in-water
|
Why is hydroxide ion the strongest base only in water?
|
My professor mentioned that hydroxide ion is the strongest base when we only consider aqueous solutions. Then why don't the strongest bases outside of aqueous solution, like superbase, remain the strongest in water?
| 0 |
[
[
"\nIf you add a superbase to water in small enough quantity that it still makes sense to think of water as the solvent, all (or essentially all) of your superbase will be converted to its conjugate acid by the reaction\n\n\n$$\\ce{B-}+\\ce{H2O} \\to \\ce{HB} + \\ce{OH-}$$\n\n\nThis will have a very large equilibrium constant (by the assumption that $\\ce{B-}$ is a superbase) and will also be driven heavily to the right by the high concentration of water. If you add enough of your superbase, eventually you can have appreciable amounts of $\\ce{B-}$ sticking around in the solution, but at this point it no longer really makes sense to call the solution \"aqueous\".\n\n\n",
"7"
],
[
"\nA pH electrometer measures changes of [HO-] or [H3O+] ions (from that of neutral water) due to reaction of bases or acids with water. The reactions are expected due to solvent levelling (i.e. H2O). On the basic side, water levels to HO-, as a result of its own auto-ionisation process.\n\n\nIn another solvent, a given base, may exhibit a different ability for deprotonating the solvent molecules, which will to some extent be governed by that solvent's auto-ionisation process (which will differ from that of water). Relative to its behaviour in water, this may be expressed as an increase (stronger basicity) or decrease (weaker bacisity). Availability of the proton changes with solvent, and the resulting BH+ or (solvent)- may be stabilised by the system, or not.\n\n\npK's are solvent dependent. For example, in the solvent DMF, picric acid is stronger than HBr, but the reverse is true in water.\n\n\nWhile a \"super-acid\" is anchored to a species having a Hammett acidity of -12 or lower, a \"super-base\" is not anchored to base strength, but is more of an acknowledgement that acidity changes with solvent. As defined, the notion of a superbase is misleading, since it is only a statement concerning structure (combine known bases to form a new base) or that such a species is a stronger base than in water (which it may, or, may not be, in varieties of solvents).\n\n\nIn short, there is nothing \"super\" about a given superbase, and one should not expect it to behave as a stronger base in general, although in some solvents it MAY be stronger, while in others, weaker.\n\n\n",
"1"
],
[
"\nIn this example we use the concentration of [H+] Ions solvated by H2O to form the Hydronium ion. Since [H+] also happens to be how we are defining the term acid in this example... The conjugate hydroxide Base [OH]- , the [H]+ acid, and the solvent [H2O] are in equilibrium. Because the [H+] in pure water should be 0, it has a pH of 7. The hydroxide ion is the most perfect species to attack and disassociate H+.\n\n\nWhen you change the solvent to a non aqueous solution, H+ may not be present...if you’re dealing with a bronsted laury acid. In such cases it’s better to redefine acid and base in terms off a compounds ability to transfer a proton to another compound (acid). Hydrogen ions or hydroxide ions don’t have to be present to have strong non aqueous acids or bases. Like n-butyl-Li or antimony pentafloride.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/90587/enrichment-of-d2o
|
Enrichment of D2O
|
The Wikipedia page for deuterium oxide (heavy water) says that its boiling point is 374.5 K and the one for ordinary water is 373.13 K.
Is there a way of distillation which we can use to enrich deuterium oxide?
| 8 |
[
[
"\nFractional distillation can work, but the separation per round of distillation is very low so you would need a very large multi stage process to achieve significant separation.\n\n\nNobody uses direct distillation to produce heavy water as there are better chemical processes that achieve higher enrichment per stage, for example the [Girdler sulphide process.](https://en.wikipedia.org/wiki/Girdler_sulfide_process)\n\n\n",
"17"
],
[
"\nFor a comparison of different methods, including distillation and the Girdler processes, have a look at\n[Heavy Water: A Manufacturers’ Guide for the Hydrogen Century](https://cns-snc.ca/media/Bulletin/A_Miller_Heavy_Water.pdf)\n\n\nIn this 14 pages document a survey of the different available processes is given. \nThe distillation of water, as suggested by the OP is included. \nThat process is the easiest but requieres huge energy input to perform. \nThe distillation must be done under vacuum because the separation factor is higher than at normal pressure. \n\n\nThe processes that are used are based on exchange reactions. \nAmong these the most used is the Girdler process based on $\\ce{H2S}$ but other possibilities exist. \nHowever, this is the only one that does not require a catalyst. \n\n\nOther processes are also considered briefly. \nAn interesting point is that the estimated cost of production is given as 300 \\$/kg $\\ce{D2O}$. \n\n\nA table with the characteristics of the different processes is included. \n\n\n[](https://i.stack.imgur.com/I78Xb.jpg)\n\n\n",
"7"
]
] |
https://chemistry.stackexchange.com/questions/90586/why-is-the-repulsion-between-lone-pairs-and-bonded-pairs-stronger-than-the-repul
|
Why is the repulsion between lone pairs and bonded pairs stronger than the repulsion between the bonding pairs and the bonding pairs? [duplicate]
|
**This question already has answers here**:
[why are lp-lp repulsions greater than bp-lp and bp-bp?](/questions/32211/why-are-lp-lp-repulsions-greater-than-bp-lp-and-bp-bp)
(2 answers)
Closed 5 years ago.
>
> My book says "The reason for this difference is that the electron pair in a bond is further from the nucleus of the central atom than the electron pair in a lone pair."
>
>
>
I don't get it , how does the distance make any difference ?
I'm , taking AS level chemistry
| -1 |
[
[
"\nFrom Coulomb's law [Wikipedia](https://en.m.wikipedia.org/wiki/Coulomb%27s_law)\n$$\\pu{ F \\propto \\frac{1}{r^2} }$$\nIf your electrons are closer to the center then the distance between them is small. \n\nA small distance would lead to a greater repulsive force as the dependence of force is inverse square to distance between charges. \n\n\n",
"0"
],
[
"\nI dont understand exactly the quote either, maybe because it is isolated from the rest of the text but one way to see this is by observing that an electron pair in a bond is \"trapped\" between the two nuclei. Hence, it wont be able to expand too much in space. In contrast a lone pair is constrained only by one nucleus. Hence, it is more likely to be found further away in space compared to a bonded pair. Hence a lone pair which is nearby a bonded pair will push that bonded pair further away compared to another bonded pair. This is the reason why the H-N-H angles in NH3 are smaller than the H-C-H ones in CH4.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/90581/rearrangement-of-an-oxyallyl-cation
|
Rearrangement of an oxyallyl cation
|
I am supposed to predict the product of rearrangement of this cation, having been told that it's an oxyallyl cation:
[](https://i.stack.imgur.com/GEUCn.png)
I did have a look at the [Wikipedia entry](https://en.wikipedia.org/wiki/4%2B3_cycloaddition) (since it mentions "oxyallyl cation") and not even once does it mention such a compound as I have shown above. This question is from my high school text book chapter on "Aldehydes and Ketones", and I'm not sure I'm even supposed to know what to do here in the first place.
Anyways, I did guess that the negatively charged oxygen atom will donate its lone pair to the positively charged carbon atom1, forming a three member hetero-ring. But, my book gives the correct answer as:
[](https://i.stack.imgur.com/zJSTq.png)
I'm not sure where they dumped the second oxygen atom (and from where they brought an extra carbon atom) to form this all-carbon ring. Also, is that $\ce{-OMe}$ at the ortho position only for ornamental purposes, or does it affect the mechanism in some way?
---
**Source:** MS Chouhan; Advanced Problems In Organic Chemistry; 11th ed; page 218 in solution manual, Q91 chapter on Aldehydes and Ketones
---
1: Yeah I know that all these atoms are in conjugation and the charges are only partially negative/positive.
| 4 |
[
[
"\nCount carbon and oxygen atoms in the book's answer versus the original molecule, and infer that the book got it wrong. One of the carbon atoms in the book's answer should be oxygen. Question for you: Which one? \n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/90580/effect-of-rain-on-water-and-soil-samples-contaminated-by-petroleum
|
Effect of rain on water and soil samples contaminated by petroleum [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
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I am about to perform research on the impact of petroleum fire disaster on the soil and water sources within the region of disaster, using heavy metals and total petroleum hydrocarbons as impact markers.
My supervisor suggested I collect samples during raining season and dry season and compare.
My question is, would there really be any significant difference between the two samples in the rainy and dry season?
| 0 |
[
[
"\nIt has much sense since the rain will modify the concentrations of both the heavy metals and the petroleum. \n\nThe initial quantities will be redistributed. \nOne part will penetrate deeper in the ground, another will be moved due to water runoff and what is left will be measured in your samples. \n\n\nIf possible, you should take the first samples as soon as possible and the the next ones towards the end of next season. \nIt is also important to get the value of the total precipitation during the rainy season. \nThis will serve to estimate how long it will take to eliminate the effects of the disaster. \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90573/why-does-nitrogen-have-a-maximum-covalency-of-4
|
Why does nitrogen have a maximum covalency of 4?
|
As nitrogen has $1$ lone pair and $3$ electrons, either it should have maximum covalency of $5$ or $3$. But why does it have a maximum covalency of $4$ instead?
Why did it leave $1$ electron? Why did it have to break a lone pair?
| 11 |
[
[
"\nRecall that covalency is the number of shared electron pairs formed by an atom of that element. Nitrogen's *maximum* covalency is indeed $4$. And no, it does *not* break up its lone pair.\n\n\nI'll give you a simple example. Have a look at the Lewis structure of the ammonium ion:\n\n\n[](https://i.stack.imgur.com/Q3jJp.png)\n(*[source](http://www.chemhume.co.uk/ASCHEM/Unit%201/Ch3Chemstr/chapter_3__chemical_bonding_andc.htm)*)\n\n\nNotice that nitrogen's octet is complete as soon it bonds with three $\\ce{H}$ atoms (aka forms ammonia). The fourth covalent bond is actually a *coordinate* covalent bond, formed when that nitrogen atom's lone pair gets donated to a proton. \n\n\nThis is also the maximum covalency for the nitrogen atom, since it has no more unpaired electrons that could be paired up with other atoms, to form more covalent bonds. \n\n\n**Homework exercise:** Can you now deduce the maximum covalency of nitrogen's elder brother, oxygen? \n\n\n\n\n---\n\n\n**Caution: extra stuff ahead** *this will help those with knowledge beyond high school level. If you're at or below high school level, go back home and play with your cat.*\n\n\n**After-thought:** Valency is a useless term that doesn't help you do any chemistry. Different sources define it differently (two definitions on [Wikipedia](https://en.wikipedia.org/wiki/Valence_(chemistry)#Modern_definitions)). It just helps fill high school textbooks with more pages, but it becomes irrelevant with the introduction of better terms like coordination number, which actually tells you something about the structure of the molecule. While \"valency\" maybe useful in basic beginners course at getting a grip, there are limitations to this point of view.\n\n\nHere's an example of such a contradiction. You may expect elements of period $3$ and above to display higher covalencies. One such example is phosphorus. Though it belongs to the same group as nitrogen, it can form compounds like $\\ce{PCl5}$, (apparently) increasing its maximum covalency to $5$ instead. The reasons for these are usually attributed to hypervalency/octet-expansion, but they are *wrong and obsolete concepts*, having been superseded by [newer concepts](https://chemistry.stackexchange.com/q/18427/5026). In fact, $\\ce{P}$ still has a *covalency of **four*** in $\\ce{PCl5}$, since there are only four shared pair of electrons (the non-bonding electrons don't count). This reinforces the idea that coordination numbers are a better and more useful term than valency. ($\\ce{P}$ now has a coordination number of $5$ in $\\ce{PCl5}$)\n\n\n",
"27"
],
[
"\nThe nitrogen has an electronic configuration $\\ce{1s^2 2s^2 2p^3}$, therefore it could be expected to make 3 covalent bonds accommodating 3 electrons in orbitals $\\ce{p}$. However, there is less repulsion of the electron domains when the $\\ce{2s}$ orbital hybridizes with the $\\ce{p}$ orbitals to form 4 $\\ce{sp^3}$ orbitals. In one of the orbitals there will be a pair of paired electrons that can make a coordinate bond. The other 3 orbitals will make 3 covalent bonds accommodating 3 electrons.\n\n\nNote:\nIn the ammonia molecule the electron pair alone is in an orbital $\\ce{sp^3}$, different from the isolated atom.\nThere is no distinction between the bonds (covalent and coordinate) in the ammonium ion.\n\n\n",
"1"
],
[
"\nSimple answer would be to say that nitrogen bonds with any element with only 4 orbitals,1s and orbital and 3p orbital.So its covalency is limited to 4.\n\n\n",
"1"
],
[
"\nNitrogen has maximum covalency 4, because we know that maximum electrons a shell can accommodate is $2n^2$, and hence it can have only 8 electrons. So covalency cannot exceed more than 4.\n`\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/90571/why-does-hydration-of-ions-depend-on-their-charge-to-area-ratio
|
Why does hydration of ions depend on their charge to area ratio?
|
The following is written in my book *Concise Inorganic Chemistry* by J. D. Lee.
>
> Hydration energy, hydrated radius and hydration number of a particular ion depends upon the charge per unit area. Hence their order is the same for the following cations in the given examples:
>
>
> $\ce{Li+(aq) > Na+(aq) > K+(aq) > Rb+(aq) > Cs+(aq)}$
>
>
>
And other similar examples are also given. My question is, why does the charge to surface area ratio matter? The force felt by the water molecules due to any of these cations should be the same due to the shell theorem. Also, a cation of bigger size can accommodate more water molecules in its vicinity due to its size. Then still, why does an increase in charge to surface area ratio correspond to an increase in hydration tendency?
| 12 |
[
[
"\nMany people have done experimental, theoretical, statistical, molecular dynamics, etc. studies of solvation free energies of ions. I encourage you to just search solvation free energies of ions on google scholar and you'll find a bunch of very interesting and quite readable papers.\n\n\nOne of these papers ref. [[1](https://i.stack.imgur.com/UaY1v.png)] ([here for pdf](http://www.elch.chem.msu.ru/rus/wp/wp-content/uploads/2015/08/s411_2014_1.pdf)) provides a lot of data and a very simple model which can basically be used to answer your question. In this paper, they derive a simple model which has only one fitting parameter. From this, they can determine the effective cavity size of an ion in solution. This cavity size is given by $r+\\Delta r$, where $r$ is the radius of the ion and $\\Delta r$ is the thickness of the shell.\n\n\nEssentially, the value of $\\Delta r$ is determined by the fitting parameter, and also determines the number of water molecules within the cavity of the ion. A portion of the data table is shown below with actual experimental values on the far right and parameters determined from the model in the other columns.\n\n\n[](https://i.stack.imgur.com/UaY1v.png)\n\n\nJust focus on the alkali metals which you list in your question. Notice as we go further down, the value of $\\Delta r$ approaches zero and the actual radius of the ion is also the size of the cavity. This, however, is very much not true for the smaller ions. The Hydrogen ion being an extreme example where a solvation number of 12.0 is obviously unphysical. This, however, does provide a clue for a reasonable interpretation of $\\Delta r$. $\\Delta r$ is a measure of how far away a molecule must be for it to not \"notice\" the effect of an ion in solution. Thus, somewhat counter-intuitively, smaller ions interact *effectively* with a larger number of water molecules. The reason for this is that as the ions get larger, solvating them requires a greater reorganization of the hydrogen-bonding network in liquid water. This means that one must lose hydrogen bonds in order to solvate the ion.\n\n\nThus, the reason the charge to area ratio dictates the solvation energy and the solvation number is because a smaller ion will bring the same charge into solution, but due to its smaller size, will only require a small reorganization of the hydrogen bonding between waters, which minimizes losses in already stable interactions. This lack of reorganization is what leads to large values of $\\Delta r$ and hence large values of $n$ in the model described above.\n\n\nLet me be clear, $n$ is not simply the number of water molecules pointing towards an ion. This would not be measurable experimentally, or well-defined theoretically. Rather, $n$ will always be some kind of average value which tells you the effective number of water monomers which energetically benefit from the presence of the ion in solution. So, now the picture is clear. The smaller cations do not cause too large of a first hydration shell, so that the second hydration shell is closer to the ion than for a larger ion. Hence, the free energy and the solvation number are larger for smaller ions of the same charge.\n\n\nAs a bonus, ref. [2] is an interesting paper which approaches this problem from molecular dynamics. They give plots showing the relative effects which van der Waal's forces and electrostatics have. Note that the smaller ions are the only ones which are stabilized effectively by van der Waal's interactions as these scale closer to $r^{-5}$ rather than $r^{-1}$\n\n\n\n\n---\n\n\nReferences:\n\n\n[[1](https://i.stack.imgur.com/UaY1v.png)] Marcus, Y. (1991). Thermodynamics of solvation of ions. Part 5.—Gibbs free energy of hydration at 298.15 K. Journal of the Chemical Society, Faraday Transactions, 87(18), 2995-2999.\n\n\n[2] : Grossfield, A., Ren, P., & Ponder, J. W. (2003). Ion solvation thermodynamics from simulation with a polarizable force field. Journal of the American Chemical Society, 125(50), 15671-15682.\n\n\n",
"7"
]
] |
https://chemistry.stackexchange.com/questions/90560/why-does-cono264-ion-have-3-unpaired-electron-as-opposed-to-1
|
Why does [Co(NO2)6]4- ion have 3 unpaired electron as opposed to 1?
|
Why does $\ce{[Co(NO2)6]^4-}$ ion have 3 unpaired electron as opposed to 1?
The book says that it has 3 unpaired electrons. I thought $\ce{NO2-}$ was a strong field ligand, making $\ce{Co^2+}$ a $\ce{d^7}$ low spin ion. Six electrons go in the $\mathrm{t\_{2g}}$ orbital and 1 in the $\mathrm{e\_g}$ orbital.
Please explain the bonding in this compound (preferably with a suitable MO diagram to go with it).
| 6 |
[
[
"\nAccording to Elliot, Hathaway & Slade (1966) cited by [this paper](http://scripts.iucr.org/cgi-bin/paper?S0567740880005535), Co(II) with nitrite ligands is low spin.\n\n\n[](https://i.stack.imgur.com/ir2ram.jpg)\n\n\nSo maybe it is just a typo. For d7 low spin, you would expect one unpaired electron.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/90558/electron-energy-levels
|
Electron Energy Levels? [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
---
**Want to improve this question?** Update the question so it focuses on one problem only by [editing this post](/posts/90558/edit).
Closed 5 years ago.
[Improve this question](/posts/90558/edit)
New to chemistry; In my book it talks about electrons in atoms moving from one energy level or shell to another and denotes this by n. How does this exactly happen, do electrons move to different energy levels when they bond with different atoms? It also talks about something along the lines of photons. Also please explain ground level. It says Hydrogen is in ground level when n=1 but what about other atoms where n>1? For instance, carbons has electrons where n>1 does this mean all the electrons other than the first shell are NOT in ground level? Please explain in layman terms as Im not that fluent in this discipline.
| -2 |
[
[
"\n\n> \n> Please explain in layman terms as Im not that fluent in this discipline.\n> \n> \n> \n\n\n**Introduction:** According to Bohr's model, an atom is surrounded by shells. They are depicted by $K,L,M,...$ and are denoted by $n=1,2,3,...$ respectively.\n\n\nWhen an electron occupies ground state level, it means that the electron occupies the lowest energy state possible. Since the lowest energy occurs closest to the nucleus, each atom tries to occupy the lowest $n$ possible (while also satisfying the rule that in each shell the maximum number of electrons can only be $=2n^2$)\n\n\n\n> \n> For instance, carbons has electrons where n>1 does this mean all the electrons other than the first shell are NOT in ground level? \n> \n> \n> \n\n\nNo, note that different electrons have different ground levels. An electron may have ground level $n=2$. If it gained extra energy and moved to $n=3$, we would consider it to be in excited level now. But, an electron that was *already* in $n=3$ is *not* considered to be in excited level.\n\n\n\n> \n> It also talks about something along the lines of photons.\n> \n> \n> \n\n\nNote that a photon with wavelength $\\lambda$ has energy $E=\\frac{hc}{\\lambda}$. A photon beam (having several such photons) when incident on an atom can excite its electron to higher energy levels. You'll study this in detail in the chapter on \"Atomic structure\" (topic \"Atomic spectra\") in higher classes. Do not worry about this for now.\n\n\n\n> \n> Do electrons move to different energy levels when they bond with different atoms?\n> \n> \n> \n\n\nThis is a very complicated topic. It will unleash hybridiation, MOT, and too many other things. It's like opening a dam of Niagara falls. Please don't confuse yourself with bonds and excitation for now.\n\n\n\n\n---\n\n\nNote: this was a pretty dumbed-down explanation of what all you asked. None of it will actually hold true in higher sciences, but, it will work for the class you're in currently.\n\n\n",
"1"
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] |
https://chemistry.stackexchange.com/questions/90557/how-to-calculate-the-electrochemical-potential-of-a-cell-using-known-half-cell-p
|
How to calculate the electrochemical potential of a cell using known half-cell potentials?
|
>
> What is $E^\circ$ for the half-reaction representing the oxidation of $\ce{HOCl}$ to $\ce{ClO3-}$ given the equations below?
> \begin{align}
> \ce{ClO3- +6 e− + 6 H+ &= Cl− +3 H2O},& E^\circ &= \pu{1.446 V}\\
> \ce{HOCl + 2 e− + H+ &= Cl− + H2O},& E^\circ &= \pu{1.481 V}\\
> \end{align}
>
>
>
It seems too simple, but would the answer be to take
$$E^\circ = \pu{1.481 V} - (\pu{-1.446 V}) = \pu{2.927 V}?$$
| -1 |
[
[
"\nFor adding up equations, you can't directly add $E^0$,unless you know that the no. of electronic transfres are same in both the reactions given.So, it is safe to calculate by $\\Delta$$G^0$ calculation, \n\nFor $HOCl$ reaction , you have, \n\n$HOCl + 2e^- + H^+ = Cl^- + H\\_2O$ -- $\\Delta$$G^0$= -$nFE^0$ = $-2\\*F\\*1.481$ \n\nBy inverting the First reaction, sign of $E^0$ will be changed and thus, \n\n$Cl^- + 3H\\_2O = ClO\\_3^- + 6e^- + 6H^+$ --- $\\Delta$$G^0$=-$nFE^0$ = $-6\\*F\\*(-1.446)$ \n\nAdding the two reaction , \n\n$HOCl + 2H\\_2O = ClO\\_3^- + 4e^- + 5H^+$ -- $\\Delta$$G^0$= - $(4\\*F\\*E^o)$ \n $\\Delta$$G^0$ will be added and for the whole reaction, \n\n$E^0(total) $ = \n $\\Delta$$G^0$(total)/$nF$= $-F\\*(2\\*1.481-6\\*1.446)/(-4\\*F)$)=$-1.4285$ $V$. \nSo, $E^0$ for the oxidation from $HOCl$ to $ClO\\_3^-$ is $-1.4285$ $ V$. \nThus, the answer is different from simply adding the $E^0$ for reactions which is not correct for all the cases(is correct when electron transfer is same for both reactions).\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/90553/why-dichlorobenzene-doesnt-show-geometricalcis-trans-isomerism
|
Why dichlorobenzene doesn't show geometrical(cis/trans) isomerism? [duplicate]
|
**This question already has answers here**:
[Why does 1,3-dimethylcyclopent-1-ene have geometric isomerism while 1-bromo-2-chlorobenzene doesn't?](/questions/84133/why-does-1-3-dimethylcyclopent-1-ene-have-geometric-isomerism-while-1-bromo-2-ch)
(3 answers)
Closed 5 years ago.
As it has restricted rotation due to the presence of a double bond, it should show geometrical isomerism. Then, why *doesn't* it show cis/trans isomerism?
| -2 |
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"\nThere are dichlorobenzene isomers but not cis-trans. You can have ortho (where the Cl atoms are close together), para (where the Cl atoms are opposite) and meta where Cl atoms are one C atom apart. Lets examine the ortho-dichlorobenzene which part of it \"looks\" the most like cis-1,2 dichloroethene (which also has a trans isomer). Is there any way you could make the Cl atoms in ortho-dichlorobenzene to be as far apart as in trans-1,2 dichloroethene without breaking any bonds? The answer should be no and this is why cis-trans isomerism in common substituted aromatic rings makes no sense.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90551/are-all-degenerate-d-orbitals-identical
|
Are all degenerate d-orbitals identical?
|
This discussion spun off from the comments to another question. The basic idea of that question was that electrons don't have a *preferred order* of filling in $p$ orbitals, i.e. the first electron will fill in the $p\_x$ orbital just as well as it will fill in the $p\_y$ or $p\_z$ oribtal. This is because in the absence of electromagnetic fields, *all p-orbitals are degenerate*, and the coordinate axes are things we arbitrarily assign, so, the degenerate $p$ orbitals are all identical.
I wonder if the same can be said about $d$-orbitals. Four of them ($d\_{xy}$, $d\_{xz}$, $d\_{yz}$, $d\_{x^2-y^2}$) feel identical to me, as they all have four perpendicular and coplanar lobes (but I could be wrong). The fifth one - $d\_{z^2}$ - doesn't look identical to me. Its positioning of the lobes is *wildly* different from the other four.
This makes me wonder,
>
> Are all degenerate $d$-orbitals identical?
>
>
>
By identical, I mean that the electron filling order has no preference. (but this might *not* be the best definition of identical orbitals please mention if you have a better one)
---
**Remarks:**
1. I didn't mention $f$-orbitals because they have even ridiculous [orbital structures](https://upload.wikimedia.org/wikipedia/commons/b/bb/F-orbitals.png) (4 types have six lobes, 2 have 8 lobes, and 1 seems to be a advanced version of $d\_{z^2}$ with two rings. Crazy!), but in case their answer is remotely similar, please consider mentioning them in your answer for completeness.
2. I have talked about *atomic* orbitals and not *molecular* orbitals (just in case that made the answer different).
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[
"\nYes, they are identical.\n\n\nOne thing that we don't really teach well with orbitals is thinking about the symmetry of the orbital with respect to the name of the orbital.\n\n\n$p\\_{x}$ has the same symmetry as the function $f(x,y,z)=x$.\n\n\nLikewise, $d\\_{x^{2}-y^{2}}$ has the same symmetry as $f(x,y,z)=x^{2}-y^{2}$.\n\n\nWhat about $d\\_{z^{2}}$? You should note that $d\\_{z^{2}}$ is really $d\\_{2z^{2}-x^{2}-y^{2}}$.\n\n\nImagine taking $d\\_{z^{2}-x^{2}}$ and $d\\_{z^{2}-y^{2}}$, summing them up, and renormalizing. So the symmetry is like: $z^{2}-x^{2} + z^{2} - y^{2}=2z^{2}-x^{2}-y^{2}$.\n\n\nWhat you get is a big lobe of the same phase along the $z$-axis. And around it in the $x$-$y$-plane (but smaller), you see a smear of the opposite phase that's shaped kind of like a torus. That's why it looks different. It's really two of the other looking orbitals put together.\n\n\nWhy don't we just use $d\\_{z^{2}-x^{2}}$ and $d\\_{z^{2}-y^{2}}$ instead? Because they're not linearly independent with the 4 other \"normal\"-looking orbitals. $d\\_{x^{2}-y^{2}}$ summed with $d\\_{z^{2}-x^{2}}$ is $d\\_{z^{2}-y^{2}}$. The 5 conventional orbitals are just the nice linear combinations of the 5 spherical harmonic solutions to the angular part of the Schrodinger equation. See [here](https://en.wikipedia.org/wiki/Cubic_harmonic#The_d-orbitals).\n\n\nYou don't seem to have a problem with $d\\_{xy}$, $d\\_{zx}$, $d\\_{yz}$, and $d\\_{x^{2}-y^{2}}$ all being degenerate. So hopefully, you can see that $d\\_{z^{2}}$ is not really different and therefore also of the same energy as the other 4.\n\n\nEDIT:\n\n\nBased on comments, it's also important to point out that the pictures of orbitals with smooth surfaces are not real pictures of orbitals. These are boundary surface diagrams where we've drawn the surface that represents a constant probability of finding the electron for that orbital. This basically means that electron densities for $d\\_{z^{2}-x^{2}}$ and $d\\_{z^{2}-y^{2}}$ will smear together in the $x,y$-plane (the orbitals are the same phase) to create something that is symmetrical and shaped like a torus.\n\n\nThis analysis extends fully to $f$ orbitals. Just identify the full functional name for each orbital and apply the same analysis here.\n\n\n",
"14"
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[
"\nBy identical i guess you mean to be able to be transformed to one another by a symmetry operation. So:\n\n\nI think it is the same reason why the orbitals 2s, 2px, 2py, and 2pz are all degenerate in the H atom. You can clearly see the symmetry between the 3 p orbitals but how can the spherical 2s be symmetric with the rest? The answer is that the Coulomb potential has a hidden symmetry that shows up in spaces of dimension higher than 3. Hence, in the same way that a 2 dimentional being would not be able to see the symmetry between the 3 p orbitals (the px and py will look like two disks that touch when projected to a plane while the pz will look like a simple disk) we cant see the symmetry between the s and p orbitals in the H atom (or dz2 and the rest ds) since we are only 3 dimentional beings. Ref: Molecular quantum mechanics by Atkins, p.95. For this reason you need to look at the math behind this. Unfortunaltely, being no physical chemist i wouldnt be able to show the math behind this (or it would take me a while figuring it out). \n\n\n",
"1"
]
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https://chemistry.stackexchange.com/questions/90546/how-are-these-molecules-chiral-and-achiral
|
How are these molecules chiral and achiral?
|
[](https://i.stack.imgur.com/JUGv4.png)
I always thought that a molecule couldn't be chiral if it had a plane of symmetry. Doesn't the chiral molecule here have symmetry diagonally across the OHs?
The achiral one doesn't look symmetrical to me, so I also don't know how it is achiral.
| 1 |
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"\nRegarding the **first molecule:** I understand that the simplistic Fischer projection - with all its bond depicted by solid lines - can be very confusing at first glance in determining optical activity. You may believe it to be optically inactive, hoping for a center of symmetry to exist, but it is not so.\n\n\nYou must keep in mind how the **molecule actually exists in 3d**! Recall the rules of Fischer projection, the horizontal bonds project outward, while the vertical bonds extend inward. Hence, the molecule would actually look like this:\n\n\n[](https://i.stack.imgur.com/NTNKj.png)\n\n\nNow, I hope you can tell, there is neither any center of symmetry nor any plane of symmetry here. Hence, the molecule is **chiral.**\n\n\nPS: Your second molecule is chiral by similar logic, as also mentioned in the comments, and as you yourself noted.\n\n\n",
"7"
]
] |
https://chemistry.stackexchange.com/questions/90542/why-did-permanent-marker-come-off-of-glass-after-rubbing-it-with-my-fingers
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Why did permanent marker come off of glass after rubbing it with my fingers?
|
No, my fingers don't look stained but I accidentally made a stroke of permanent marker on glass and I was looking for nail polish remover to use, couldn't find it. So I started rubbing the stroke of permanent marker with my fingers and it started coming off. I was able to get it off with 3 fingers, with the thumb getting most of it off.
I wonder, is it because permanent marker dissolves in skin oil that I was able to get it off? Or was it the small amount of sweat? Or both? Or was it just that I was rubbing my fingers on the permanent marker and the shear force got it off?
I mean yes, this was just a few minutes after the accidental stroke that I got it off with my fingers but that would have been plenty of time for it to dry, I think.
| 1 |
[
[
"\nGlass is a very smooth surface and therefore things adhere to it less effectively such as the ink in your permanent marker. Maybe your marker is not of a very good quality or maybe you didnt leave it long enough for the ink to stabilize on the surface.\n\n\nWhy dont you conduct an experiment to try your various hypotheses by varying one factor while keeping the rest constant? For exaple:\n\n\n1) Try different brands of markers on the same glass surface leaving them to dry for the same time and then rubbing them in the same way.\n2) Use the same marker but leave it to dry for different amount of time and then rub it in the same way each time.\n3) Use the same marker on different surfaces and rub in the same way. \n4) Try with different amounts of dilute saline solution to simulate sweat or cooking oil to simulate skin oil etc.\netc.\n\n\nOr even better: Try at glass surfaces that have been etched with different sandpaper grits.\n\n\nYour experiment is reproducible though, i tried as well and small amounts come off easily.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90534/how-to-determine-if-a-reaction-proceeds-via-sn1-sn2-e1-or-e2-mechanism
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How to determine if a reaction proceeds via SN1, SN2, E1 or E2 mechanism?
|
>
> [](https://i.stack.imgur.com/JDB1u.jpg)
> Give the main product and reaction type: SN1, SN2, E1 , E2.
>
>
>
As the 1-bromohexane is primary and the nucleophile a strong unhindered base, the reaction should be a SN2 reaction. However, the solvent $\ce{EtOH}$ is a polar protic solvent which favours SN1/E1 reactions. In addition, the temperature of the reaction is quite high and high temperatures favour the formation of elimination reactions, thus the reaction should be E1.
But an E1 doesn't require a strong base, or else an E2 reaction would take place at the beginning. So does that make this reaction an E2 reaction? Am I thinking correctly on how to consider all the different parts of the reaction?
| 1 |
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[
"\n$\\ce{EtO-}$ is a strong base, as well as a strong nucleophile, and the temperature is not too high. Therefore a mixture of products is most likely to be formed, where substitution will lead to the major and elimination will lead to the minor product. \n\nAlso, the reactant is a primary halide. So, in substitution, SN2 will be favoured, whereas in elimination E2 will be favoured. \n\nSo, final product will be,\n\n\n* Ethoxy hexane (via SN2) (major)\n* Hex-1-ene (via E2) (minor). (Maybe, a very little amount of hex-2-ene may also form by very very small fraction of E1).\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90531/lowering-the-boiling-point-of-wax
|
Lowering the boiling point of wax
|
Wax comes in all kinds (paraffin, bee wax, palm oil wax, gel wax, ...) but since it's usually a mixture of different carbonated chains, its physical properties vary a lot from one manufacturer to another.
A good candle is usually one made of a wax with a high melting point, such that it takes longer to burn.
I am trying to make a *bad* candle, more specifically I am trying to make a candles that produces as much wax vapor as possible. To make it happen, I am looking for a wax that has the lowest possible boiling point.
**How can I lower the boiling point of a given wax** (paraffin wax for instance since it's the cheapest and easiest to buy) **by adding other substances inside ?** One can easily find on the internet how to *increase* the BP (by adding stearin for instance), but not the reverse.
Ideally the additional substance should not be added in too great proportion, since that would lower the wax composition and thus the amount of wax vapor emitted.
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"\nHow about mixing your paraffin wax with some paraffin oil which in practice is liquid wax (just shorter parafinic chains) in the appropriate ratio to get whatever melting point you like? \n\n\nThe boiling point and the amount of wax vapours is a different story since as Karl said wax is a mixture so even if it could be boiled (most likely under vacuum, otherwise it will autoignite) it will fractionally distill so it wont have one boiling point really. Also the amount of wax vapour will be small in all cases since the wax will burn to CO2, CO and H2O (however you could have some lower boiling point hydrocarbons flying around if you mixed your wax with them but that wouldnt be wax vapours then). \n\n\nIf you really wanted to lower the boiling point of your liquid you would need to mix it with something that will form a low boiling azeotrope mixture. No idea what that would be for wax.\n\n\nAnd lastly, taking James' thought below a step further, if you would like a really really \"bad candle\" just use one of those old oil lamps. The way they work is exactly the same as a candle but the waxy bit is a liquid so it needs to be in a container. Then you have a wick just like candles do. Only difference is you dont need to melt the wax first in order for it to be sucked in the wick since it is already a liquid.\n\n\n",
"3"
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[
"\nWhenever you just mix two substances (dissolve in each other), you lower the melting point and raise the boiling point. That's a fundamental principle. If it was the other way round, the substances *would not mix*, or at least not mix well. \n\n\n(Azeotropes of compounds with very different polarity (water/ethanol) are the exeption confirming that rule, but the acchievable raise in Bp. won't be large. And I doubt you find one you can make a candle from.)\n\n\nThat sounds bad, but: Paraffin wax *already is* a mixture, so the desired effect might be reached by fractionating it, instead of adding sth. else. \n\n\nYou might try to slowly crystallise a molten sample of paraffin wax, and pour off the residual liquid when the sample is about half solid. Now the residual solid should have a higher melting point than the original wax.\n\n\n(Note: This is purely my educated guessing, no idea if the effect is large enough or if some secondary kinetic effect might completely preclude it.)\n\n\n",
"2"
],
[
"\nTwo points: melting point and boiling point. Hmmm. If you lower the boiling point, you will likely also lower the melting point, so your candle will be really bad - it will drip away.\n\n\nWell, let's take a low melting \"wax\", like petrolatum (Vaseline). Mix in a stiffening agent like fumed silica or cellulose fiber (it will burn off when the flame hits it). The wick should be thick to hold the melted petrolatum.\n\n\nThis might resemble a torch, with a huge flame. To moderate it somewhat, you could blend paraffin with the petrolatum.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/90530/sigma-and-pi-coordination-in-complexes-symmetry
|
Sigma and pi coordination in complexes (symmetry)
|
According to the Miessler Tarr and many molecular orbital diagrams for octahedral complexes on the internet will σ-donors/acceptors only influence the $e\_g$-orbital, while π-donors/acceptors influence the $t\_2g$-orbitals. This can also be shown using the corresponding character table, but it's been a while since I used these, so I will just believe its correct. Still, I was wondering, is there no case of it ever mixing, for example, that a π-donor/acceptor does not coordinate with both lobes but in a σ-fashion? I thought about this case, and there is also an example mentioned in the book, where they show how a σ-donor and π-acceptor-ligand (I suppose they mean one with both abilities) would look like. Here they draw the lines from the π-orbitals or rather π\*-orbitals in this case to the $t\_2g$-orbital and the σ-orbitals are connected to the $e\_g$-orbital.
One example where I know this behavior, just vice versa would be chloride as a ligand, and I have also seen schemes, where it was drawn to coordinate in a π-fashion to the metal's d-orbitals. But in this case, you have both properties. What about a hydride for example, as pure σ-ligand? Would it only influence the $e\_g$-orbital or is there a way it could also somehow weakly influence the $t\_2g$-orbital?
| 0 |
[] |
https://chemistry.stackexchange.com/questions/90529/what-is-the-greenish-substance-formed-after-the-electrolysis-of-two-graphite-ele
|
What is the greenish substance formed after the electrolysis of two graphite electrodes in a sodium sulphate solution?
|
Today at school we tried the electrolysis with two graphite electrodes in a sodium sulphate solution. We put the electrodes in two test tubes and filled them with some more sodium sulphate so that the solution could react with the electrodes and after that, we quickly turned them over and put them in a beaker filled with sodium sulphate. After that, we added some bromothymol blue in it to see where the hydrogen and oxygen would have gone. Afterwards, we turned the power on and saw that it was producing gas and everything was normal at first (1st image).
Some time after that we didn't see the volumes between the two test tubes changing so we waited some more. Later, we saw that the solution started assuming a green color and something started forming on the surface (2nd image).
It became more and more green so we decided to turn the power off. Then we noticed that on the bottom of the beaker was something yellowish and brownish (3rd image). We checked if the wires were consumed by the solution or if they had reacted with it but nothing was wrong with them. I have no idea what it is and my teacher told me and my lab group to find an explanation to that. What is the problem here?
List of the materials used:
* 1 beaker
* 2 test tubes
* two electrodes connected to wires (graphite)
* $\ce{Na2SO4}$ (sodium sulphate)
* bromothymol blue
| 4 |
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[
"\nI think that what possibly happened here is that the copper wire that connects the cable with the graphite electrodes is exposed in the solition (if you look carefully in Fig.1 there seem to be a lot of bubbles forming near the red cable insulation at the bottom of the tube which escape out of the tube). So copper is oxidising to give copper cations which give the coloured material in the solution. \n\n\nThis is just a theory though. To confirm, repeat the experiment first with copper electrodes then with your graphite electrodes but ensure that only the graphite is in the solution (so get rid of the tubes). If using the copper elecrodes gives the same result as above (blue green stuff) while using the graphite electrodes just gives O2 and H2 it means that the above theory is likely. Finaly you could isolate the greenish solids by filtration, dry them and do a flame test. If the flame becomes green it suggests presence of copper further confirming the above.\n\n\nThe only thing that doesnt fit with the above is that Cu2+ ions are blue and soluble in solution not green, so maybe the green stuff also (or only) result from some reaction with the indicator. Hence, it would be good to repeat the experiment with and without the indicator to see what happens (that makes 4 experiments in total).\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/90523/why-enthalpy-change-delta-h-is-equal-to-heat-transfer-at-constant-pressure
|
Why 'Enthalpy change' (Delta H) is equal to 'Heat transfer at constant pressure' (Qp)? [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/90523/edit).
Closed 5 years ago.
[Improve this question](/posts/90523/edit)
Why 'Enthalpy change' (∆H) is equal to 'Heat transfer at constant pressure' (Qp)?
∆H = ∆U + ∆pV, here only expansion work done by the sustem is added. If non expansion work is done on the system then it should be,
∆H ≠ Qp.
What's wrong? Explain.
| 4 |
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[
"\nIn,Thermodynamics,Work is defined as \n\n$W$= $-\\int\\_{V\\_1}^{V\\_2}pdV$. So, if $dV$=0, there is no work done on the system,as $V\\_1=V\\_2$.\n \n\nComing to the enthalpy,you are right. \n\n$\\Delta$H=$\\Delta$U + $\\Delta$($PV$).But if you apply the product rule then, \n\n$\\Delta$H=$\\Delta$U + $P$$\\Delta$V + $V$$\\Delta$P. \n\nAt, constanst presurre,$\\Delta$P=$0$. so at constant pressure,$\\Delta$H=$\\Delta$U + $P$$\\Delta$V. \nBut,according,to first law of Thermodynamics, \n$Q$= $\\Delta$U - $W$.At,constant pressure, the $p$ in the integral comes out.So, \n$W$= $-p\\int\\_{V\\_1}^{V\\_2}dV$= -$p$($V\\_2-V\\_1$) = -$p$$\\Delta$V.So, first law in constant presuure becomes, \n$Q\\_p$= $\\Delta$U - (-$p$$\\Delta$V)= $\\Delta$U+$p$$\\Delta$V= $\\Delta$H. Thus, it is justified why Enthalpy change equals Heat change at constant pressure.\n\n\n",
"2"
],
[
"\nEnthalpy is a intensive property of the material that has nothing to do with the specific process that the material is subjected to.\nIt just so happens in a constant pressure condition, involving only P-V work, that the change in enthalpy is equal to the heat added.\n\n\nIf non expansion work is done on the system, then ∆H≠Qp. It should be mentioned while writing '∆H = ∆U + p∆V =Qp', that 'it is applicable when only p-V work is done'.\n\n\nThanks to @ChesterMiller\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/90519/is-pyridine-a-reducing-agent-if-so-is-it-a-strong-or-weak-reducing-agent
|
Is pyridine a reducing agent? If so, is it a strong or weak reducing agent?
|
My question relates to Darzen's procedure to prepare monohalides by reacting an alcohol with thionyl chloride in the presence of pyridine.
Pyridine is the reason why this reaction follows an SN2 mechanism. That means pyridine is a reducing agent since it abstracts the hydrogen from electropositive oxygen (while the reaction is in process) before the electronegative chlorine (from $\ce{SOCl2}$) can. So my question is,
>
> Is pyridine a reducing agent? If so, is it a strong reducing agent or a weak one?
>
>
>
| -2 |
[
[
"\nPyridine has two roles in this reaction but neither of them are as reducing agent.\n\n\nFirstly it mops up $\\ce{H^+}$ (it does not deprotonate OH, it is not a strong enough base for that).\n\n\nSecondly it acts as a nucleophilic catalyst by reacting with thionyl chloride to give $\\ce{[pyridinium - SO2 - Cl]^+}$ which is more reactive than thionyl chloride towards ROH.\n\n\n<http://www.chemtube3d.com/Pyridine%20-%20Nucleophilic%20Catalyst.html>\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/90518/reactions-of-tertiary-alkylhalide-with-aqueous-potassium-hydroxide
|
reactions of tertiary alkylhalide with aqueous potassium hydroxide? [duplicate]
|
**This question already has answers here**:
[Why does alcoholic KOH prefer elimination whereas aqueous KOH prefers substitution?](/questions/15728/why-does-alcoholic-koh-prefer-elimination-whereas-aqueous-koh-prefers-substituti)
(5 answers)
Closed 5 years ago.
When a tertiary alkylhalide reacts with aqueous koh then why #E2#reaction occurs .As tertiary alkylhalide undergo #E1#reaction What makes the difference?
| -1 |
[
[
"\nThere is a significant difference between two mediums. \n\nFor Substitution reactions we need a $Good Nucleophile$, not a $Good Base$.Because we want the attacking species to preferably attack the electrophilic atom(In most of the cases it's Carbon).If it's a good base it will take the proton attached to the Carbon which will not help the Substitution reaction. The medium is aqueous KOH(amount of water is large compared to OH-), in which $H\\_2O$ is the dominant Nucleophile,as it is not so strong base,so it attacks the Carbon and in the deprotonation step, OH- acts as a base to take H+(as, water is a weak acid, so, conjugate OH- is a strong base).\n \n\nNow, if you consider alcoholic KOH medium(generally $EtOH$ medium),There is predominant OH- and little amount of $EtO-$ which are both Strong bases. For,the eliminantion reaction we need specifically a proton acceptor. Thus, highly basic medium supports elimination.\n \nNow, if we consider Substitution and Elimination in case of Tertiary alkylhallide, there is always competition between $S\\_N1$ and $E\\_1$ .So, the reaction medium and maintaining $Temperature$ is very important. If your solvent medium acts as a good Nucleophile(bad base) then $S\\_N1$ will occur.But if your medium is highly basic,$E\\_1$ occurs. \n\nSometimes, there are solvent mediums such as $MeO-$ or $EtO-$( in some cases) which can act as both Nucleophile and Base.Then, \n\nlowering the temperature follows $S\\_N1$and at higher temperature $E\\_1$ occurs for tertiary alkyl hallides.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90515/is-it-right-to-compare-the-molarity-of-a-substance-with-the-probability-of-its-m
|
Is it right to compare the molarity of a substance with the probability of its molecules reacting?
|
I was watching [this](https://www.khanacademy.org/science/chemistry/chemical-equilibrium/equilibrium-constant/v/keq-intuition-mathy-and-not-necessary-to-progress) video on Khan Academy (about the equilibrium constant), and the analogy that they made really got me thinking. I wanted to verify whether it was completely correct, or just an analogy to explain reaction rates to high school students. I'm well aware that this is an equilibrium constant, and the probability of molecules reacting relates to kinetics. However, this is how it was explained in the video, and I'm trying to make sense of it.
Would anyone please guide me as to whether this analogy, of the molarity and the probability of a molecule reacting, is really true *and mathematically correct* or whether its just a simplification for students? (I'd recommend that you watch the video first before trying to answer.)
Also, please note that I'm not saying that what Khan Academy is teaching is wrong, I'm just trying to clarify whether what they're teaching is a simplification, or a scientifically accurate explanation.
| 0 |
[
[
"\nAs Alchimista points out, two different principles are at work here. Concentration, or how many of the species are available, and are these species able to react when they do meet.\n\n\nThe transcript of the video at 3:32 states:\n\n\n'some constant that takes into account things like temperature and how the molecules are actually configured. Because it's not dependent on them just being there. You have to worry about their kinetic energies...'\n\n\nSo, Molarity will not affect the probability that any two molecules that meet will react, but it does absolutely affect rate of reaction. If the reaction is possible (there is sufficient activation energy) then the more concentrated the substance, the more likelihood of reactable species meeting, and therefore a greater rate of reaction.\n\n\n",
"2"
],
[
"\nIn this video the author is first explaining the origin of the Keq constant based on the rates of the forward and backward reaction and then moves on to explain why the rate law expression is of the form rate=k[A]$^a$[B]$^b$. \n\n\nNo problem with the first part but disagree with the second especially when he says that [A]$^a$[B]$^b$ expresses the probability of having a of A molecules and b of B molecules in a small area at the same time:\n\n\nRate = kProbability(Having a of A molecules and b of B molecules in a small area at the same time)\n\n\nI am no mathematician but the probablity of anything has values between 0 and 1 which is obviously not the case for [A]$^a$[B]$^b$. Only this makes the video wrong.\n\n\nHe also says that:\n\n\n\"The probability of having V molecules in some volume = [Concentration]xvolume\"\n\n\nwhich is incorrect as well. In a solution of constant concentration you will always have (Conc. x volume) amount of molecules evenly spread in that volume. There is no probability involved. If there is it would be the probability of having (Conc.x volume) molecules of compound in that volume which would be 1 since that is how we define concentration.\n\n\nIn order to understand the form of the rate law equation you should look into collition theory (e.g. <http://umich.edu/~elements/03chap/html/collision/index.htm>) where you will be able to see exactly how the probability of two coliding molecules reacting or not is properly used.\n\n\nNote that that the probability of two molecules colliding and reacting is higher with increasing concentration, as Alchimista commented, and this is one of the principles collition theory relies on but not in the way it is presented in the video.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90502/how-could-i-easily-quantify-the-amount-of-egcg-in-a-sample-of-green-tea
|
How could I easily quantify the amount of EGCG in a sample of green tea?
|
I am currently conducting an experiment to study the effects of pH on the rate of thermal decomposition of Epigallocatechin Gallate (EGCG) in green tea.
One of the key steps in my procedure is quantifying this EGCG content in the green tea. What are some of the methods which I could use to determine and quantify this EGCG content?
| 0 |
[
[
"\nTo analyze specifically this component, you will have to use HPLC. \nIn case you can use a more generic approach, you can analyze the total phenolic content with the Folin-Ciocalteu method. \n\n\nThe Folin-Ciocalteu method is based on a reaction with polyphenols that gives a blue color. \nThis color can then be measured in a UV-Vis spectrophotometer and the concentration can be established. \n\n\nThis method has been used to quantify polyphenols in many plants, as for example [here](http://www.mdpi.com/1420-3049/18/6/6852/pdf).\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/90501/can-carbon-be-the-central-atom-of-a-complex-can-it-form-a-quadruple-bond-with-i
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Can carbon be the central atom of a complex? Can it form a quadruple bond with itself? [closed]
|
**Closed**. This question needs to be more [focused](/help/closed-questions). It is not currently accepting answers.
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Why does $\ce{[C(NH3)2]}$ not exist as a complex? Here, $\ce{C}$ is carbon and the configuration of $\ce{C}$ in ground state is $\ce{[He] 2s^2 2p^2}$.
Can't the two $\ce{NH3}$ molecules pair the two $\ce{p}$ electrons of carbon and form coordinate bond with carbon?
Why does $\ce{C2}$ (one carbon atom bonded with other through quadruple bonds) not exist? Two $\ce{2p}$ orbitals of $\ce{C}$ should form π bonds and one $\ce{2p}$ and one $\ce{2s}$ should form σ bonds with the other $\ce{C}$ (without hybridisation).
I'd appreciate an answer for a beginner.
| 3 |
[
[
"\nAsking multiple unrelated questions at once is considered a bad practice, but let's forget that for a moment.\n\n\nYour first molecule has more or less satisfactory electronic configuration, but that's not enough for the long-term survival. N would be more than willing to relinquish a proton, and C to accept it, and thus in no time you will have $\\ce{H2N-CH2-NH2}$, which then may or may not transform into something else.\n\n\nYour second molecule has been discussed here before on multiple occasions. Long story short, it does exist, but doesn't have a quadruple bond. (Such bonds are known, but not in $\\ce{C2}$; they have a [delta component](https://en.wikipedia.org/wiki/Delta_bond), and that requires *d* orbitals). The problem with your description is that the sigma-oriented *2p* orbital of one carbon will interact not just with the similar *2p*, but also with the *2s* orbital of the other. If not for this fact, it would work out pretty much as you expected.\n\n\n",
"3"
],
[
"\nWe may regard oxyanions as complexes with oxide or oxo ligands; indeed such an approach may offer advantages in understanding the electronic structure of some of these ions. In that sense carbonate ion is indeed a complex with carbon as a central atom. \n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90500/how-to-define-equivalents-here
|
How to define equivalents here
|
>
> $3g$ of activated charcial was added to $50 ml$ of acetic acid solution ($0.06N$) and filtred after an hour and it was found to be $0.042 N$ .The amount of acetic acid adsorbed (per gram of charcoal) is
>
>
>
The solution for this question is exactly given as I have written below
No of equivalents before adsorption $=50\times0.06 =3$
No of equivalents after adsorption $=50 \times 0.042=2.1$
No of equivalents left $=0.9/1000 \times 60=54mg$
Acetic acid per gm $= 54mg/3=18mg$
Now I have some doubts regarding the question and it's answer
>
> 1. Why is normality being decreased here?
> 2. Why are they taking volume to be same when it generally increases when we add solutes?
> 3. What does equivalent mean here? According to Wikipedia, it is said to be "the mass of one equivalent, that is the mass of a given substance which will 1) combine or displace directly or indirectly with 1.008 parts by mass of hydrogen or 8 parts by mass of oxygen or 35.5 parts by mass of chlorine – these values correspond to the atomic weight divided by the usual valence; or 2) supply or react with one mole of hydrogen cations (H+) in an acid–base reaction or 3) supply or react with one mole of electrons (e−) in a redox reaction." The equivalent talked in the question doesn't meet any of this criteria so what is its role here. Can somebody please explain me this?
>
>
>
| 3 |
[
[
"\nNumber of equivalents and normality are out-of-date concepts in chemistry. For acids, they are just like number of mols and molar concentration (or molarity), with an extra factor indicating how many hydrogens can be effectively ionized in solution. Because acetic acid ($\\ce{CH3COOH}$) is monoprotic, you can just exchange the terms by number of mols and molarity and you are good.\n\n\n\n> \n> Why is normality being decreased here?\n> \n> \n> \n\n\nSome of the acetic acid adsorbs - meaning it bonds - in the charcoal. So the concentration *in solution* is decreased. Pay attention to the fact that the charcoal remains solid and *does not* dissolve.\n\n\n\n> \n> Why are they taking volume to be same when it generally increases when we add solutes?\n> \n> \n> \n\n\nAs I wrote above, the charcoal stays solid, so it has no effect on the solution volume. The loss of acetid acid in solution might have a *slight* effect on volume, but that is mostly negligible.\n\n\n",
"7"
]
] |
https://chemistry.stackexchange.com/questions/90497/why-are-ammonium-chloride-and-ammonium-hydroxide-used-in-calcium-salt-detection
|
Why are ammonium chloride and ammonium hydroxide used in calcium salt detection?
|
On adding $\ce{ (NH4)2CO3 }$ solution to $\ce{ Ca^{2+} }$ aqueous salt solution, $\ce{ CaCO3 }$ is formed as a white precipitate, which confirms the presence of the cation.
While doing the test in my school laboratory, we were given a list of procedures to follow; for $\ce{ Ca^{2+} }$ it read:
>
> Experiment: To the original solution added a small amount of $\ce{ NH4Cl }$, and an excess of $\ce{ NH4OH }$, followed by solid $\ce{ (NH4)2CO3 }$
>
>
> Observation: white ppt
>
>
>
Why do we need to add $\ce{ NH4Cl }$ and $\ce{ NH4OH }$?
| -1 |
[
[
"\nYou want to fix the carbon in solution so it doesn't volatize as $\\ce{CO2}$, since the equilibrium $$\\ce{H2CO3 <=> CO2 + H2O}$$ favours the product side. By adding a base ($\\ce{NH4OH}$), you shift the water ionization equilibrium, which in turn shifts the carbonic acid dissociation equilibrium. $\\ce{NH4Cl}$ serves to regulate the $pH$ as a buffer. With your carbonate in solution, it is then able to precipitate as a solid by adding calcium. \n\n\nUsing a strong base would hinder the solid precipation by favouring the solubilization of ions, justifying the use of ammonium hydroxide.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/90491/what-metal-is-used-to-back-this-mirror
|
What metal is used to back this mirror?
|
My father is attempting to remove the mirror from a piece of glass. We stripped the black paint from the back and exposed the shiny surface with acetone. In the past he has been able to remove aluminum mirror with lye, but that didn't react. We also attempted bleach in case it was actually silver but that didn't react either. Is there another metal that could be used to back a mirror? Are we doing anything wrong? Also adding hydrogen peroxide to the lye made the mirror bubble but no black silver oxide or white aluminum oxide.
| 7 |
[
[
"\nThe back side of a mirror is typically made of glass, which is coated with a thin layer of metal (such as aluminum or silver) to create the reflective surface. This reflective coating is applied to the back of the glass, and it is this side that is usually polished to achieve a smooth, flat finish.\n\n\nRemoving the reflective coating from the back side of a mirror can be done through a chemical reaction that dissolves the metal layer. One way to achieve this is by using a mixture of nitric acid and hydrochloric acid, also known as aqua regia. This mixture can dissolve many metals, including the ones commonly used in mirror coatings. You can learn more about mirrors [here](https://sarathiacademy.com/sims/geometric-optics-basics_en.html).\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90489/possible-to-calculate-electronic-charge-distributions-densities-for-organic-mole
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Possible to Calculate Electronic Charge Distributions/Densities for Organic Molecules?
|
I was wondering if there are any programs that can calculate the electronic charge distributions around an organic molecule, possibly in the form of a volumetric charge volume density? I am reading up about using DFT programs that (correct me if I'm wrong) I believe may be able to tell us something about finding the charge density.
I'm working on a project that involves obtaining the electric field values surrounding an organic molecule. My current approach involves finding a function for the molecule's electronic volume charge density and doing a volume integral to get the electric field.
Is this a potentially valid method? If not, what else should I look into?
| 3 |
[
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"\nIf I understand correctly, you want the total electric field at some point $\\mathbf{r}$ due to a molecule. This will have at least three components. The first due to the electric field from the nuclei. This can be calculated classically from the field due to point charges. The second due to the dipole moment of a molecule. This can also be calculated completely classically (though you will have to use some electronic structure calculation to get the dipole and direction). Finally, you will need to find the total charge density of the electrons.\n\n\nThe charge density can be found using either DFT or any post-Hartree Fock method such as MP2 or CCSD(T), etc. When using something like MP2, you will get the charge density from the molecular orbitals. You can do this manually, but one code I know of that will give you the relevant information from the HF orbitals is NWChem. You will want to use the [dplot command](http://www.nwchem-sw.org/index.php/DPLOT). This allows you to print out electron densities for each orbital using a grid. You will want to use a fine enough grid to capture the actual spatial changes in density. Notice that you also provide limits i x, y, and z. This is because, in principle, the charge density extends to infinity in all directions. Thus, you will want to choose a cutoff where the density is sufficiently small as to be negligible. Then, you can again calculate the field due to the electron density from classical equations which can be found online or in an E&M textbook. You can either calculate the field discretely from the points on the grid, or fit it to some functional form and then do a true integral over the density.\n\n\nAnother way to do this is to use a code which will give you approximate charges per atom from some type of population analysis such as Mulliken charges. You can do this with Gaussian using [population keyword](http://gaussian.com/population/). This will aggregate the nuclear charge and the electronic charge into one field. Fair warning, however, Mulliken charges (and all population analyses) are approximate and not even necessarily physical.\n\n\nThere may be other ways to do this from DFT, but I surprisingly cannot find any software packages that do this directly.\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/90485/weak-adhesives-has-some-critical-chemical-been-banned
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Weak adhesives, has some critical chemical been banned? [closed]
|
**Closed**. This question is [opinion-based](/help/closed-questions). It is not currently accepting answers.
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I notice that in the last 10 years or so, the adhesives I encounter in everyday life have gotten noticeably weaker, is this because some critical chemical has been banned or regulated out of consumer products?
For example, the typical duct tape, packing tape, masking tape, and scotch tape that I can buy in stores now are all noticeably weaker and less adhesive than the same products 10-15 years ago. For example, duct tape used to a have thick, white adhesive that was extremely sticky, but duct tape that you would find in a hardware store now has a clear adhesive that is much less sticky.
I know that there other consumer products that have become inferior due to chemical banning; for example most plastic wrap is now made of HDPE plastic instead of the original polyvinylidene chloride, so it is greatly inferior to the original Saran wrap. Has the same thing happened to tape adhesives? If so, what is the operative chemical that has been banned?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/90475/how-can-a-reaction-exhibit-first-order-kinetics-when-there-is-more-than-one-reac
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How can a reaction exhibit first order kinetics when there is more than one reactant (the other being water)
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This is for the hydrolysis of p-nitrophenyl phosphate by alkaline phosphatase, whereby the enzyme is in a solution of distilled water. We also add glycine buffer to the solution. It exhibits first-order kinetics, even though there is a second reactant (being water) - how is this possible?
| -2 |
[
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"\nIt is because water is the solvent in the reaction and therefore its concentration remains practically constant throughout the course of the reaction. Hence, it is absorbed in the rate constant: \n\n\nrate=k[p-nitrophenyl phosphate]$^1$[H$\\_2$O]$^x$=k'[p-nitrophenyl phosphate]$^1$\n\n\nHence, we say the reaction is peudo-first order. Also, in general, you could have a reaction with two reactants and it could still be first order overal since it could be 0 order in reactant A and first order in reactant B.\n\n\n",
"0"
]
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https://chemistry.stackexchange.com/questions/90473/how-to-change-the-glass-transition-temperature-tg-of-a-compund
|
How to change the Glass transition Temperature ( Tg ) of a Compund? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
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we wanna Increase our compound damping (loss) , we know that Damping properties of a polymer is dominated by glass transition...
So how can we change Tg In our PP/EPDM Compounds?
| -3 |
[
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"\nYou cant change the Tg of a polymer without changing its composition which means it wont be the same polymer any more. A specific compound or polymer will have a specific Tg. You could vary the Tg of say PE by controlling its degree of branching, but each PE you will get will be a different compound each with its own Tg although all of them will fall under the general category of PEs.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90472/why-can-we-ignore-the-weak-conjugate-base-acid-when-calculating-the-ph-in-a-titr
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Why can we ignore the weak conjugate base/acid when calculating the pH in a titration problem? [closed]
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**Closed.** This question is [off-topic](/help/closed-questions). It is not currently accepting answers.
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**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)
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Take the reaction 350 mL of 0.8M HCN and 0.6M NaOH for example. The problem asks us the calculate the pH when we add 500 mL of 0.6M NaOH into the 350 mL of 0.8M HCN (This is more than enough to neutralize the weak acid). However, the problem states that we can ignore OH contributions from water or any weak base. Can someone explain why we can do this?
| -2 |
[
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"\n$\\ce{NaOH}$ is a strong base. It is assumed that it **dissociates completely** in aqueous solutions, hence, it's contribution to the $\\ce{OH-}$ concentration is the *maximum* as compared to other weak bases or water. Moreover, due to common ion effect, the presence of $\\ce{OH-}$ ions in solution due to $\\ce{NaOH}$ further hinders the dissociation of the *already* weak bases, thereby making their contributions negligible.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90471/carbon-14-in-organic-chemistry
|
What happens to a radioactive carbon dioxide molecule when its carbon-14 atom decays?
|
When carbon-14 decays, the decay products are nitrogen-14 and an electron (and an electron antineutrino, but that's chemically irrelevant\*):
$$\ce{^14\_6C -> ^14\_7N + e- + \overline{v\_e}}$$
Let's assume that the carbon atom in question is part of a carbon dioxide molecule in the atmosphere. What would happen to the molecule when the atom decays into nitrogen? Will it be converted into a $\ce{NO2}$ molecule, or will it split apart? Will the electron created in the decay have sufficient energy to escape the molecule and form a positive ion?
[Here](https://chemistry.stackexchange.com/questions/39458/curious-about-the-chemistry-of-carbon-14-after-its-produced-in-the-atmosphere)'s a somewhat related question dealing with the formation of radioactive carbon dioxide.
\* Of course, not all the energy from the defect will be transferred to the beta particle's kinetic energy, so this is in fact relevant for the rate. See Loong's answer for details.
| 48 |
[
[
"\nAn [article](http://pubs.acs.org/doi/abs/10.1021/j150569a008) by Snell and Pleasanton, 'The Atomic and Molecular Consequenses of Radioactive Decay', (*J. Phys. Chem.*, 62 (11), pp 1377–1382, $1958$) supports Ben Norris's comment.\n\n\n\n> \n> It is clear ... that $\\ce{^{14}CO2}$ remains predominantly bound as $\\ce{NO2+}$, a result that is perhaps not surprising. [This occurs in] $81$% of the decays. In $\\ce{^{14}CO2 -> NO2^+}$ dissociation yielding $\\ce{NO+}$, $\\ce{O+}$ and $\\ce{N+}$ follows [in], respectively, $8.4$, $5.9$, and $3.6$% of the decays.\n> \n> \n> \n\n\nA table summarising the results is given.\n\n\n$$\\begin{array}{|c|c|}\n\\hline\n\\mathbf{Ion} & \\mathbf{\\%\\ abundance} \\\\\n\\hline\n\\ce{NO2+} & 81.4(16) \\\\\n\\ce{NO+} & 8.4(4) \\\\\n\\ce{O+} & 5.9(6) \\\\\n\\ce{N+} & 3.6(4) \\\\\n\\ce{NO2^{2+}} & 0.40(06)\\\\\n\\hline\n\\end{array}$$\n\n\n",
"39"
],
[
"\nThe antineutrino is not completely irrelevant. The decay energy is shared between the beta particle and the neutrino. Therefore, the beta particles appear with an energy distribution that ranges from zero to the maximum beta energy. The maximum beta energy for the decay of $\\ce{^14C}$ is $E\\_\\text{max}=0.1565\\ \\mathrm{MeV}$; the average beta energy is $E\\_\\text{avg}=0.0495\\ \\mathrm{MeV}$.\n\n\n[](https://i.stack.imgur.com/SSbsM.png)\n\n\nNevertheless, even beta particles with a relatively low beta energy have a much larger kinetic energy than any chemical bond dissociation energy or ionization energy. Thus, the beta particle cannot be captured in the electron shell of the affected atom. The beta particle leaves the atom with almost relativistic speed before the rest of the molecule understands what went wrong. This process is too fast for immediate chemical reactions. The daughter nuclide remains in the same chemical structure. Thus, the beta decay of a $\\ce{^14C}$ atom in $\\ce{CO2}$ simply leaves a $\\ce{NO2+}$ ion in an excited state. This primary product is unstable and can dissociate slightly later. The resulting secondary products can be radicals or ions, which tend to react with almost any other molecule or atom.\n\n\nHowever, the energy released in beta decay is not only distributed between the neutrino and the beta particle. A small amount is also allocated to the recoil of the nucleus. Since the mass of the nucleus is much larger than the mass of the beta particle, the recoil energy is much smaller than the energy of the neutrino and the beta particle and can usually be neglected in a first approximation.\n\n\nFor the beta decay of $\\ce{^14C}$, the maximum recoil energy of the $\\ce{^14N}$ nucleus can be calculated as $E\\_\\text{recoil, max}=7.08\\ \\mathrm{eV}$. This value is much smaller than the total decay energy of $0.1565\\ \\mathrm{MeV}$ and could thus be neglected, but it is larger than the energy of a chemical bond. By way of comparison, the bond dissociation energy in $\\ce{NO2}$ is only about $306\\ \\mathrm{kJ/mol}$ or $3.17\\ \\mathrm{eV}$. Therefore, the recoil of the $\\ce{^14N}$ nucleus can be sufficient to break a bond immediately during the beta decay of $\\ce{^14C}$.\n\n\n",
"26"
],
[
"\nThere seem to be three processes to consider here:\n\n\n1. As the high-energy electron exits, it creates a rapidly varying electrical current, which produces intense electromagnetic fields for a short time. This may act on the other charged particles that are present, possibly creating further ionization and/or breaking the chemical bond.\n2. The nitrogen nucleus recoils, and this recoil could break the bond and/or create further ionization.\n3. The charge of the nucleus changes, so even if effects 1 and 2 didn't exist, the state of the electrons is no longer the ground state. It is now a linear combination of different states of the new element, which means there is some probability of excitation or ionization.\n\n\nA paper by [Oksyuk and Gerasimenko](http://www.jetp.ac.ru/cgi-bin/e/index/e/19/1/p176?a=list) says that effect 2 is normally the important one.\n\n\nFrom Linear Christmas's answer, we know that in the case of $\\ce{^14C}$ decay, which has an unusually low energy, there is a fairly small but still appreciable probability of breaking the bond, and this is presumably caused by one or more of the above processes.\n\n\nFor process 1, it certainly isn't safe to assume that it's negligible because the electron leaves the molecule so quickly. If this were true, then beta particles wouldn't produce ionization when they came along from the outside and hit atoms. Although the time-scale for the electron to exit is short, its electromagnetic fields are intense. By the way, the motion of this electron is not insanely fast compared to the velocities of the other electrons. It has a typical velocity in this decay of about $0.5c$, which could be compared to an estimate of $Zc/137\\approx 0.04c$ for an inner-shell electron in carbon.\n\n\nTo estimate process 1, let's use the mean rate of energy loss for beta particles in a solid. For $\\pu{0.1 MeV}$ betas in a solid made of fairly light elements, this is about $(dE/dx)/\\rho\\approx 0.3$ MeV.m2/kg. Taking $\\rho$ to be the density of water, and $\\Delta x=0.2$ nm, we find $\\Delta E=0.06$ eV, which seems to be a couple of orders of magnitude too low to break a bond. However, energy loss of betas is a process that has a lot of random variation about the mean, so it doesn't seem unreasonable to me to imagine that there is something like a 1% chance that it deposits 100 times this energy in the parent atom on its way out. This puts us in the right ballpark for this mechanism to contribute significantly to the observed probability of breaking a bond in NO.\n\n\nSo now let's consider process 2. Let $Q$ be the energy released in the decay and $M$ the mass of the recoiling nitrogen nucleus. We'll see that for kinematical reasons, almost all of the energy goes into the electron and antineutrino, not the nucleus. Suppose we want to find the maximum energy of the recoiling nucleus. This is attained in the case where the electron gets almost 100% of the energy, because for a fixed energy, a more massive particle carries more momentum. If the electron and neutrino were to share the energy, then their momentum vectors could also partially cancel, further reducing the recoil.\n\n\nThe decay energy of $\\ce{^14C}$ is unusually low, but in most beta decays the beta is much more relativistic. Let's do the ultrarelativistic case first, both because the math is simpler and because it's a better guide to our intuition about what happens in general.\n\n\nIn the approximation that the beta is ultrarelativistic, its momentum (in the case where it carries all the energy in this example) is $p\\approx Q/c$, and by conservation of momentum, this is also the momentum of the recoiling nucleus. Since the nucleus is nonrelativistic, its kinetic energy is $K=p^2/2M\\approx Q^2/2Mc^2$.\n\n\nAs a typical example, let's take $\\ce{^40K}$, which is the strongest source of naturally occurring beta radioactivity in our environment. It has an 89% probability of decaying to $\\ce{^40Ca}$ plus an electron and an antineutrino. The energy for this decay mode is $\\pu{1.33 MeV}$, which is almost $10$ times that of $\\ce{^14C}$. The ultrarelativistic approximation for the electron is not too ridiculous; in the case where it gets almost all the energy (none to the neutrino), its velocity is about $0.96c$. The maximum kinetic energy of the recoiling calcium nucleus in this approximation is about $\\pu{24 eV}$, which is clearly plenty of energy to break a chemical bond.\n\n\nWithout the ultrarelativistic approximation, the momentum of the beta, in the maximum-recoil case, isn't $Q/c$ but rather $\\sqrt{(x+Q)^2-x^2}/c$, where $x=mc^2$, and $m$ is the mass of the electron. Even in the case of $\\ce{^40K}$, it turns out that the ultrarelativistic approximation is not so great. The actual maximum recoil energy for the calcium is $\\pu{41.7 eV}$, so although the approximation gives the right order of magnitude, it's off by almost a factor of $2$.\n\n\nIn the example of $\\ce{^14C}$, the result for the maximum energy of the recoiling nitrogen is $\\pu{7.0 eV}$, in agreement with Loong's answer. This is plenty of energy to break the bond. Another interesting case with a very low energy is $\\ce{^3H}$ , in which the recoiling $\\ce{^3He}$ has a maximum energy of only about $\\pu{3 eV}$. People have used this to try to study the chemistry of helium.\n\n\nSo the take-away here is that in almost all cases, beta decay is very likely to break up the molecule in which it occurs.\n\n\n",
"15"
]
] |
https://chemistry.stackexchange.com/questions/90467/what-happens-to-the-leftover-ions-in-the-mass-analyzer
|
What happens to the "leftover" Ions in the Mass Analyzer
|
I am very intrigued by what happens to the ions which do not make it to the detector in mass spectrometers, do they just pile up? I have never seen anyone clean our quadrupole, is this done? or do they just stay there indefinitely? I am talking of a triple quadrupole detector in a GC instrument, which deals with very small quantities, but after thousands of injections there must be a considerable mass of analyte on the analyzer. Even more so considering that most ions which go through the quadrupole do not make it to the detector.
Thanks.
| 3 |
[
[
"\nMost would end up being sucked out though the vacuum pump with the other residual gases. I assume that a lot of them collide with the sides of the vacuum chamber and are neutralized before sublimating in the vacuum.\n\n\nThe quads can become contaminated and require cleaning after several years of heavy use. Because there are such small quantities at that stage of the process, it's not uncommon for an MS to reach the end of it's service life before the quads need to be cleaned.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/90466/use-of-the-box-method-to-determine-lowest-term-symbol-for-lanthanides-but-not-ir
|
Use of the box method to determine lowest term symbol for lanthanides but not iron
|
I am trying to compare the predicted magnetic moment for Fe(CN)6 using the spin-only formula magnetic moment = $\sqrt{4S(S+1) + L(L+1)}$.
However, if I use the box method for Fe(CN)6 to predict the lowest term symbol of Fe3+, I get 6S. If L = 1, however, then the predicted value of the magnetic moment is similar to the experimentally derived value.
Can we not use this method because in the complex the octahedral field changes the energy levels of the d orbitals?
If this is the case, then why can we use the box formula to determine the magnetic moments of lanthanides using the formula magnetic moment = g$\sqrt{J(J+1)}$.
| 1 |
[] |
https://chemistry.stackexchange.com/questions/90462/what-does-the-shaded-area-represent
|
What does the shaded area represent?
|
[](https://i.stack.imgur.com/zNi9L.png)
I'm not sure what the answer is, all of my friends and my chemistry teacher said it is A, I think it is obvious that the graph is talking about 2 catalysts so it could be D. But I have no idea what the E\_a thing is, I've tried googling it to no avail. I got this question on an advanced Chemistry test. One thing for sure, it is not B because the word "additional" isn't there. And another thing, we can't say that one is the graph of the molecule without a catalyst and the other is its graph with a catalyst because according to the graph that would mean that sometimes the molecule without a catalyst is more active than without a catalyst which is obviously wrong.
| 7 |
[
[
"\nShort answer: The two curves are the distribution of kinetic energy of molecules at two different temperatures. A catalyst won't change the distribution; only temperature change will change the energy distribution. The line marked Ea is the activation energy; a molecule has to have more than this to react. A catalyst could change the activation energy; since there is only one Ea, we can eliminate catalytic activity as an issue. All the molecules with less energy than this just bounce around, not reacting.\n\n\nThe curve with the higher peak represents the lower temperature. The one with the lower peak has a longer and higher tail out to higher energies, so the shaded area represents the greater number of high energy (reactive) molecules at the higher temperature. Answer: A.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/90458/what-material-can-be-used-as-scraper-in-a-chemical-reactor
|
What material can be used as scraper in a chemical reactor?
|
I am having a 10L reactor built using SS316 for Plastic Pyrolysis.
TEMPERATURE : 550C
I would like a scraper to clean the char that forms on the reactor's inner layer.
[](https://i.stack.imgur.com/hDEP9.jpg)
Teflon scraper is available but it will not withstand 550C temperature.
So I am thinking about using flexible expanded graphite (Similar to graphite gland packing ropes) material. What do you think?
[](https://i.stack.imgur.com/t6VeL.jpg)
What other options do you suggest that I look into?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/90455/why-is-the-ph-scale-only-from-0-14
|
Why is the pH scale only from 0-14? [duplicate]
|
**This question already has answers here**:
[Is a negative pH level physically possible?](/questions/5737/is-a-negative-ph-level-physically-possible)
(7 answers)
[pH range outside conventional 0-14 [duplicate]](/questions/6997/ph-range-outside-conventional-0-14)
(3 answers)
Closed 5 years ago.
Why can't there be more hydrogen ion concentration in acids or lesser hydrogen ion concentration in bases? Why should there even be any hydrogen ion concentration in alkalis? (I know pH is calculated by the negative log of H+ ion concentration.)
| 1 |
[
[
"\nIt isn't. We're just used to working in the lab where solutions are generally 1 M or less in concentration. So we get no more than 1M (solvated) hydrogen ions where pH is 0, or at the other end no more than 1 M hydroxide ions corresponding to pH 14.\n\n\nThese 1 M limits are, however, just for our convenience and not any law of nature. Right in our own labs we likely have concentrated acid stock solutions, well above 1M, where if we attempted to measure pH we would get less than zero. Acid pickling solutions in the steel industry are concentrated enough to do the same thing, as manufacturers are more interested in getting hot mill scale cleaned off than in a convenient pH scale. Bases like sodium hydroxide also have concentrated stock solutions with pH above 14.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/90451/sulphate-ion-confirmatory-test
|
Sulphate ion- confirmatory test
|
On adding $\ce{BaCl2}$ solution to $\ce{ SO4^{2-}}$ salt aqueous solution $\ce{ BaSO4}$ is formed as a white precipitate, which confirms the presence of the anion.
While doing the test in my school laboratory, we were given a list of procedures to follow; for $\ce{ SO4^{2-}}$ it read:
'Experiment: Aqueous solution of sample + dilute $\ce{ HCl}$ + $\ce{ BaCl2}$ solution
Observation: white ppt'
Why do we need to add $\ce{ HCl}$?
| 5 |
[
[
"\nThe method consists\nof slowly adding a dilute solution of barium chloride to a hot solution of the\nsulphate slightly acidified with hydrochloric acid\n\n\nIt is customary to carry out the precipitation in weakly acid solution in order\nto prevent the possible formation of the **barium salts of such anions as chromate,\ncarbonate, and phosphate, which are insoluble in neutral solutions**; moreover,\nthe precipitate so obtained consists of large crystals, and is therefore more easily filtered\n\n\nReference: *Vogel's Textbook of 'Quantitative Chemical\nAnalysis* 11.72 Sulphate: Determination of sulphate as barium sulphate\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/90448/why-are-ionic-reactions-very-fast-in-aqueous-medium
|
Why are ionic reactions very fast in aqueous medium? [closed]
|
**Closed**. This question needs [details or clarity](/help/closed-questions). It is not currently accepting answers.
---
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Closed 5 years ago.
[Improve this question](/posts/90448/edit)
(No, I am not talking about why Ionic Reactions are faster than Covalent reaction.)
This question has me stumped. I mean, at first it looks obvious; water is *polar* in nature, but what exactly does that help with? Also, can ionic reactions even happen in non-aqueous mediums?
| 1 |
[
[
"\nWater has a permanent dipole character, which means it has a slight positive and negative charge permanently in it (due to electronegativity difference). So, in case of an ionic reaction, it helps in easier solvation of ions by pulling them with its charged centres (positive centre attracts negative ions and negative centre attracts positive ions)\n\n\n*Aqueous* is a term specially for water as a solvent. Any polar solvent can support ionic reactions.\n\n\n",
"1"
]
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https://chemistry.stackexchange.com/questions/90447/resistance-of-electrolytes
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Conductivity of water and chlorine?
|
I am a novice in chemistry and i have a question: I know that water is not a good conductive material (since there is a perfect bond in water I think) but I am thinking that if we add chlorine which has 7 valence electrons this means that there is a hole and this increases conductivity... am I right?
| 1 |
[
[
"\nIn a nutshell, water is very much ***not*** like semiconductors. Its conductivity is due to altogether different mechanism, and follows different trends.\n\n\nTypical semiconductor is a rigid network of atoms held together with bonds. Nearly any \"wrong\" atom would provide either extra electrons or holes (that is, missing electrons). Holes and electrons are the [charge carriers](https://en.wikipedia.org/wiki/Charge_carrier) here. Being very small, they move fast, so you only need a tiny amount of those impurity atoms (say, a few ppm) to create significant conductivity.\n\n\nWater is quite different. You can't have free electrons in water, to begin with; they will instantly react with something and become ions. Ditto for holes. Ions are the only allowed charge carriers in water. Ions are as big as atoms, and each carries a thick coat of water molecules stuck to it, so effectively they are even *bigger* than atoms. An ion would move in an electric field, but with great difficulty, much like a man pushing his way through a busy crowd. (Compared to it, the electrons would be like little sparrows darting overhead.) Therefore you need **a lot** of ions to create some real conductivity.\n\n\nThings that produce ions in water (*dissociate*, as we say) are called electrolytes, and there is plenty of them around. For example, table salt dissociates into $\\ce{Na+}$ and $\\ce{Cl-}$; sulphuric acid dissociates into $\\ce{H+}$ and $\\ce{SO4^2-}$. Elemental chlorine ($\\ce{Cl2}$), being a neutral molecule, does not dissociate; also, you can't add much of it, for it is a gas, and its solubility is not that great. Over time, it would react with water to produce certain compounds which **do** dissociate, so you'll have some conductivity after all, but that's a weird and inefficient way of achieving it. Using any soluble salt is much more straightforward.\n\n\nThe choice of particular electrolyte depends on your ultimate purpose.\n\n\n",
"4"
]
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https://chemistry.stackexchange.com/questions/90446/in-halogenation-of-ethene-using-bromine-for-example-why-does-the-lone-pair-of-b
|
In halogenation of ethene using bromine for example, why does the lone pair of bromine attack the C atom?
|
In the presence of the electron rich (pi bond) of ethene, the bromine molecule becomes a temporary induced dipole, one end being electron deficient and acting as the electrophile. (The pi bond in) ethene attacks the electrophile.
At the same time, one of the lone pair of electrons in bromine attacks one of the C atom. Why does this happen?
[](https://i.stack.imgur.com/IZ1dN.png)
Normally, I'm familiar with the same process, but with the carbocation being formed without the lone pair of electrons of bromine attacking the other C atom:
[](https://i.stack.imgur.com/Y728D.png)
Does this have something to do with resonance as shown by the diagram below?
[](https://i.stack.imgur.com/Xdw78.png)
| 0 |
[
[
"\nThere are too many bond lines to explain the reaction; the question suggests that we have to explain the bond lines!\n\n\nLet's start from scratch, with a different viewpoint. Ethylene isn't the attacker; it doesn't burst into flame in oxygen or do anything aggressive. It may be reactive, but in a \"normal\" range. Bromine, on the other hand, is a potent oxidizer and does go around interacting with relatively inert materials - it's a powerful bleach. \n\n\nSo, when bromine comes near the ethylene pi bond, it attacks the electron pair and grabs it, but the second bromine steals it away and departs as Br-. After all, bromine is more electronegative than carbon; the first bromine is just a link in the chain of events that allows the second bromine to escape with an electron. The first bromine winds up stuck to the C-C double bond; the two carbons and the bromine share the positive charge. The bromine certainly gets the lion's share of electron density, leaving the carbons with most of the positive charge. That's why another bromine can attack the carbons from the rear to give a dibromide. (If a Br- attacked the bromine attached to the double bond, it would just regenerate the reactants.) I bet that the second bromine to bond to the ethylene is not from the same bromine molecule as the first bromine. (But since all bromine molecules look the same, I can't prove it.) Doing the bromination in a solution with an overwhelming abundance of another nucleophile would probably give a mixed product.\n\n\nA C-C double bond is about 1.34 Angstroms long; a bromine atom (in Br2) is about 2.29 Angstroms in diameter. Rather than envisioning the cation as a triangle of three grapefruit, I see it as a cantaloupe sitting on two large oranges. Just consider the empty p orbital of bromine and the pi lobe of the carbons; no need or benefit from making a three-membered ring.\n\n\n",
"-1"
]
] |
https://chemistry.stackexchange.com/questions/90443/why-are-there-no-dihedral-mirror-planes-in-the-d3h-point-group
|
Why are there no dihedral mirror planes in the D3h point group?
|
Consider the molecule $\ce{BH3}$, which belongs to the $D\_\mathrm{3h}$ point group. Why are the three mirror planes in this point group labelled as $\sigma\_\mathrm v$ instead of $\sigma\_\mathrm d$? Don't they bisect the angles between a pair of rotational axis $C\_{2}$ axes as shown in the diagram below?
[](https://i.stack.imgur.com/6HeNg.png)
| 5 |
[
[
"\nThe σ$\\_v$ planes in a molecule are a direct result of the C$\\_n$ main axis i.e. they are interchangeable if you perform the C$\\_n$ operation on them. On the other hand σ$\\_d$, which are called dihedral planes bisect the dihedral angles between members of the σ$\\_v$'s set. Therefore molecules with odd C$\\_n$ such as BH$\\_3$ wont have any σ$\\_d$ planes; only n (3 for BH$\\_3$) σ$\\_v$'s. In contrast a molecule such as benzene which has a C$\\_6$ main axis will have 3 σ$\\_v$'s (the ones including the carbon atoms) and 3 σ$\\_d$'s which are the ones passing between the carbon atoms. Of course it doesnt really matter which set is called dihedral and which is called vertical, it could be the other way round as well. Ref: Chemical applications of group theory by Cotton.\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/90284/ap-chemistry-sig-figs-question
|
AP Chemistry: Sig Figs Question
|
This is the official answer of a problem on the AP Chemistry 2009B Exam:
[](https://i.stack.imgur.com/1Yhpy.png)
The answer has two sig figs, but I think it should have 4 sig figs. I was taught that your end answer should have as many sig figs as that of the number which has the smallest amount that is **used in your calculation.**
When I solved this problem, I used the second approach in the image. The only number I took from the question is 10.93, which has 4 sig figs, so my answer should end up with 4 sig figs.
Is it because 14 has two sig figs? I would think that shouldn't count.
| -1 |
[
[
"\nWhen I read this question, I forgot that the sig figs for pH is different. For pH, the number of sig figs is equal to the number of digits after the decimal.\n\n\nSince the pH given is 10.93, it has two sig figs (since two digits after decimal) which is the lowest, so the answer should have two sig figs which is does.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90281/will-naoh-react-with-ethanol
|
Will NaOH react with ethanol? [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 5 years ago.
[Improve this question](/posts/90281/edit)
I am having a doubt in here that NaOH reacts with methanol but not with ethanol.
Theoretically my instructor on chemistry says that the conjugate base of ethanol would be stronger than NaOH.
But I can see in many websites that the reaction is given(reaction of CH4 and NaOH).
Also a side question to this is CH4,C2H6 will not react with NaOH. Is there any reason other than full saturation CH4 and C2H6.
| 0 |
[
[
"\nJust because one can write down the reaction equation (as you have read in those websites) doesn't mean it takes place to any considerable extent. Yes, ethanol and NaOH do theoretically react as in $\\ce{EtOH + NaOH \\to EtO^-Na+ + H2O}$, but can you observe this reaction in practice? The answer lies in the thermodynamics of the reaction: the equilibrium is severely shifted to the left, so the reaction as written is almost irrelevant. Conversely, methanol ionization does present a measurable equilibrium constant at appropriate conditions. \n\n\n",
"5"
]
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https://chemistry.stackexchange.com/questions/90276/trend-in-reducing-property-of-dioxides
|
Trend in reducing property of dioxides
|
I am not sure of the explanation for the following statement:
>
> The reducing property of dioxides of group 16 elements decreases from $\ce{SO2}$ to $\ce{TeO2}$.
>
>
>
* Is the statement valid? If yes, how do I explain it theoretically?
* If not, what would be the actual trend in that case?
| 0 |
[
[
"\nThe statement is actually valid.If you go from S to Te, the compounds of Se and Te are stable at lower oxidation states due to inert pair effect. \n\nSo, the tendency of these group 16 elements to be in higher oxidation states decreases down the group.That means, tendency of the elements getting oxidised decreases down the group and thus their reducing property(reducing others, getting oxidised itself) decreases frm S to Te.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90274/stability-of-resonance-structure
|
Stability of resonance structure
|
My book states- " the one that has ...less separation of opposite charges ,...., more dispersal of charge is more stable than others"
Aren't 'less separation ' and 'more dispersal' opposite to each other?
Please explain with an example.
| 0 |
[
[
"\nRe-read the statement carefully. It says that the separation of opposite charges (\"charge separation\") destabilizes the resonating structures. The second phrase (\"charge delocalisation\") refers to the delocalisation of the *same* charge around a resonating structure.\n\n\nFor example, here's an example of how \"charge delocalisation\" increases stability:\n\n\n[](https://i.stack.imgur.com/IQhD4.png)\n\n\n(*[source](http://www.chem.ucla.edu/harding/tutorials/resonance/draw_res_str_key.html)*)\nNotice the same positive charge is delocalised throughout the conjugated system. It reduces - the unit positive charge on a single carbon atom - to a partial positive charge over several carbon atoms in the resonance hybrid, thus increasing the stability of the molecule.\n\n\nHere's an example of \"charge separation\":\n\n\n[](https://i.stack.imgur.com/U1BeK.png)\n\n\nwhich destabilises the molecule. Remember that uncharged structures are more stable than charged structures.\n\n\nI hope it helps!\n\n\n",
"3"
]
] |
https://chemistry.stackexchange.com/questions/90269/how-to-make-ammonia-from-urea
|
How to make ammonia from urea?
|
The chemical formula of urea is $\ce{CO(NH2)2}$ or is sometimes written as $\ce{CH4N2O}$. How would I decompose or precipitate the $\ce{CO}$ from urea to get ammonia? I know that in the olden days people used to make gunpowder (potassium nitrate) and ammonium nitrate from human urine, which was made by reacting ammonia and nitric acid, so it should be possible. (Just as a side note: this is for a high school chemistry project, I am not attempting to make explosives out of this).
| 0 |
[
[
"\n**Short answer:** use crushed watermelon seeds (urease)\n\n\nCombine urea in solution with urease, get some ammonia with carbamate intermediary, which rapidly decomposes to ammonia and carbon dioxide.\n\n\n$$\\ce{(NH2)2CO + H2O ->[urease] NH3 + H2NCOOH -> 2NH3(gas) + CO2(gas)}$$\n\n\nOf course the ammonia gas dissolves in the water and increases the pH.\n\n\n**Sources:**\n\n\n* <https://www.hemmaodlat.se/research/citrullus%20lanatus%20seeds%20as%20a%20urine%20catalyst%20for%20anthroponic%20use.pdf>\n* <https://en.wikipedia.org/wiki/Urease>\n* <https://en.wikipedia.org/wiki/Ammonia_volatilization_from_urea>\n\n\n",
"2"
],
[
"\nThe following article (available freely under a creative commons license) outlines two different procedure to decompose urea to ammonia, one via thermal hydrolysis (THU), the other using a Ni catalyst under an applied electrochemical potential:\n\n\n\n> \n> ECS Electrochem. Lett. 2015 volume 4, issue 10, E5-E7\n> \n> \n> <http://eel.ecsdl.org/content/4/10/E5.full>\n> \n> \n> doi: 10.1149/2.0041510eel\n> \n> \n> \n\n\nThe following passage describes the THU process which is carried out at high temperature in water (references and more details can be found in the article):\n\n\n\n> \n> Warner postulated that urea hydrolysis -THU summarized by reaction 1-\n> consists of two steps. The first step is the decomposition of urea to\n> ammonia and isocyanic ion (Eq. 6) which is irreversible at pH less\n> than 5 and greater than 12. The second step is the hydrolysis of\n> isocyanic ion to produce ammonia and CO2 (Eq. 7).\n> \n> \n> $\\ce{ NH2CONH2 -> NH3 + H+ + CNO- ~~} $ [6]\n> \n> \n> $\\ce{ CNO- + H+ + H2O -> NH3 + CO2 ~~~} $ [7]\n> \n> \n> \n\n\n",
"1"
],
[
"\n1. First hydrolyze the urea as follows to form ammonium carbonate.\n$$\\ce{NH2CONH2 + H2O -> (NH4)2CO3}$$\n2. Next decompose this ammonium carbonate to yield our required product, $\\ce{NH3}$.\n$$\\ce{(NH4)2CO3 -> 2NH3 + H2O + CO2}$$\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/90268/can-oxidation-of-hydrogen-happen-in-room-temperature
|
Can oxidation of hydrogen happen in room temperature? [duplicate]
|
**This question already has answers here**:
[Why don't everyday things burn?](/questions/55094/why-dont-everyday-things-burn)
(2 answers)
Closed 5 years ago.
Here is my explanation about the reaction between hydrogen or methane and oxygen.
>
> Hydrogen molecules and oxygen molecules can collide sucessfully and react if they have enough energy. In room temperature, molecules are moving slowly, but **they do have a chance to collide**, and **they do have a chance to have high energy level** to collide successfully. Therefore, even in room temperature, hydrogen can slowly react with oxygen to form carbondioxide and water.
>
>
>
And as temperature increases, the chance to collide and to have enough energy gets bigger, so rate of reaction increases.
PS: the kenetic energy of particles satisfies something like normal distribution, so molecules with high energy must exist even in low temperature.
My chemistry teacher don't agree with me. why?
Which brought me to a second question: what is the definition of **ignition point**(the minimal temperature required to burn) if reaction can happen just at any temperature?
| -1 |
[
[
"\nYour reasoning is correct. When we say a reaction occurs only with sufficient energy (activation energy), it is truly a tail of a single (pair of) molecule. But the ensemble of molecules still provides molecules of high energy, able to react, and molecules of low energy which will collide and not react. That effect is expressed macroscopically by the reaction kinetics, with a higher rate of collisions resulting in a higher reaction rate. The kinetic effect of temperature is thus continuous and there is no single temperature at which a system *has* the activation energy to react (as some people believe). It is indeed true that a reaction may become spontaneous in a given direction only above a certain temperature, but that is a matter of thermodynamics and not kinetics.\n\n\nDespite all of this, it sort of happens that indeed some reactions do seem to begin to take place at a certain temperature while not changing its thermodynamic spontaneity. That is because the kinetic effect of temperature is not linear, it is actually much more intense that a linear model might predict, sometimes doubling the reaction rate at every 10⁰C increase - this is captured by the *Arrhenius equation*. Sometimes you do have some molecules reacting, but it is *so slow* at a the temperature you're looking it's not even noticeable. Bring the temperature up and you will see a rapid increase in reaction rate in a short temperature interval. So it looks like it was an on/off transformation.\n\n\nMore often than not in chemistry, the theory predict really small quantities, but the theory is also naive to assume the molecular distribution is continuous and not made of discrete entities as we know. So it sometimes happens that the idea that there no molecules reacting is in this situations actually closer to reality than the theory suggests.\n\n\nA remark: molecular energy distribution is *not* normal generally, but follows something like a [Maxwell-Boltzmann distribution](https://en.wikipedia.org/wiki/Maxwell%E2%80%93Boltzmann_distribution).\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/90263/small-but-modest-displacements-at-negligible-forces-from-dft
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Small but modest displacements at negligible forces from DFT
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When looking at a Gaussian log file today, I noticed the following information in the convergence criterion section after an analytical frequency calculation was performed:
`Item Value Threshold Converged?
Maximum Force 0.000001 0.000450 YES
RMS Force 0.000000 0.000300 YES
Maximum Displacement 0.000190 0.001800 YES
RMS Displacement 0.000030 0.001200 YES`
This struck me as odd. How could the forces be *so low* but the maximum displacement still be non-negligible? Does anyone have any hypothesis about why this phenomenon would occur? This makes me wonder if the convergence condition for the maximum displacement is unreasonably low if essentially no acting forces can result in a maximum displacement that's merely 1 order of magnitude lower than the default threshold. I believe the force units are Hartrees/Bohr and the distance units are angstroms.
| 1 |
[
[
"\nFor clarity I will assume Gaussian performs a generic Newton-Raphson minimization (NR), which should suffice to explain the phenomenon.\nIn NR, the linear problem\n$$\n\\nabla\\nabla^\\ast E \\Delta + \\nabla E = 0\n$$\nis solved, where $\\Delta$ is the displacement. In order to arrive at \"large\" displacements despite \"small\" (but non-zero) forces ($-\\nabla E$), it suffices for the Hessian ($\\nabla\\nabla^\\ast E$) to have eigenvalues close to 0 itself, because a small number is divided by an even smaller number. This happens when the potential energy surface is very flat.\n\n\n",
"3"
]
] |
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