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|---|---|---|---|---|
https://chemistry.stackexchange.com/questions/89048/is-glycine-strong-or-weak-field-ligand-if-yes-how
|
Is glycine strong or weak field ligand? If yes, how?
|
The fact that it contains both oxygen and nitrogen atoms is confusing me since N is generally strong ligand field while O has a weak ligand field
| 3 |
[] |
https://chemistry.stackexchange.com/questions/89045/glycine-hydrochloride-salt-how-i-can-get-it
|
glycine hydrochloride salt how I can get it
|
I do not understand what Glycine•HCl salt is (concretely). Is it a mix between glycine and HCl salt (I don't know if HCl salt is existing...) or something else ?
I understand that glycine is an amino acid and in solution glycine HCl is like the following picture
[](https://i.stack.imgur.com/wt0bz.png)
But what happens when the solvent evaporate? What is the meaning of the point • between Glycine and HCl ?
| -1 |
[
[
"\nIn general, the dot in the formula indicates some kind of association compound; the prime examples of such dots are crystal hydrates such as $\\ce{CuSO4 . 5 H2O}$. In that, five molecules of water are present for each pair of copper(II) and sulphate ions.\n\n\nIn organic chemistry, these dots are usually used in a somewhat misleading way including the somewhat nonintuitive description as a ‘hydrochloride salt’. What this means is that the most basic position of the molecule is protonated and this cation forms a salt with a chloride anion. The effect is equivalent to adding hydrochloric acid to a solution of the molecule and then removing the solvent. In glycine’s case, while the compound is often called glycine hydrochloride salt its systematic name would be glycinium chloride; indicating both the fact that the proton is somewhere on the glycine molecule (where depends on the original solvent) and an additional chloride ion forms a salt with this cation.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/88976/does-bet-theory-apply-to-type-h3-and-h4-isotherm-for-calculating-surface-area
|
Does BET theory apply to type H3 and H4 isotherm for calculating surface area? [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/88976/edit).
Closed 5 years ago.
[Improve this question](/posts/88976/edit)
[](https://i.stack.imgur.com/0IliK.png)
BET applies to type 2 and 4. That is basically unanimous.
| 2 |
[
[
"\nThe sample that shows loop of H3 type have **(very likely) no** micropores (small uptake in low p/p0 range). In this case you can apply BET theory in common p/p0 range (0.05-0.3). The sample that shows H4 hysteresis **have** micropores (large uptake in low p/p0 range). The BET theory can be applied here, but in a much lower p/p0 range. For activated carbon, for example, one would look for a linear(!) range between p/p0 = 0.02-0.1. Generally, the BET theory is not suitable for microporous materials. \n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/88973/etymology-of-supernatant
|
etymology of supernatant
|
Why do we call a solution that has been centrifuged a supernatant? It seems to me that a "natant" should be something that floats (from Latin "natare" meaning "to swim"), and a "supernatant" should be something which lies *above* the natant, not below it. Shouldn't the liquid be called a subnatant? Or maybe a superprecipitate?
Edit: fixed Latin (see below)
| 4 |
[
[
"\nIt means \"the liquid that swims above\", from the Latin terms *super* = above and *natare* = to swim.\nSo it is the liquid that remains on top.\n\n\nThe subnatant would be the liquid at the bottom, in case you should centrifuge two liquid phases. \n\n\nThe term you name, *nato*, means son in Latin.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/88971/what-would-affect-the-rate-of-recharge-of-a-lead-acid-battery
|
What would affect the rate of recharge of a lead-acid battery?
|
I am trying to perform an investigation regarding lead-acid batteries and would like to specifically look into the recharging mechanism. However, I realised that this might be quite difficult since the recharge equation is
$$\ce{2PbSO\_4 +2H\_2O->4H^+ +2SO\_4^2- + Pb + PbO\_2}$$
Would anything except for the voltage supplied affect the rate in which the charging takes place (or the current through the electrodes)? For example, would changing the concentration of the electrolyte affect the rate of recharge?
EDIT: I was just looking at the equation and thought that if I increased the concentration of the electrolyte, which is the sulfuric acid, it would actually cause the equilibrium to shift left, meaning that the current would be lower with higher concentrations of sulfuric acid. Is this correct?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/88969/redox-potential-of-a-lead-acid-battery
|
Redox potential of a lead–acid battery
|
In the German Wikipedia there are two reactions on the poles of the battery shown with the following potentials:
>
> $$
> \begin{align}
> \ce{Pb + SO4^2- &-> PbSO4 + 2 e-} &|\pu{-0.36 V}\\
> \ce{PbO2 + SO4^2- + 4 H+ + 2 e- &-> PbSO4 + 2 H2O} &|\pu{+1.68 V}
> \end{align}
> $$
>
>
> $$E\_\mathrm{Ges}^0 = \pu{1.68 V} - (\pu{-0.36 V}) = \pu{2.04 V}$$
>
>
>
I do understand the potential for the second $\pu{1.68 V}$ since for the second reaction the underlying redox pair is $\ce{Pb^4+ + 2 e- -> Pb^2+} $. For this redox pair the electrochemical standard potential is $\pu{1.69 V}$.
But for the first reaction I think that the underlying redox pair has to be $\ce{Pb^2+ + 2 e- -> Pb}$. This redox pair has standard potential of $\pu{-0.1263 V}$.
This result in a voltage of $\approx\pu{1.55 V}$. But Wikipedia and a book of mine tell the the voltage of this battery type is $\pu{2.04 V}$.
What the reason for the $\pu{-0.36 V}$?
**Source:** This is from the [German Wikipedia article on lead-acid batteries.](https://de.wikipedia.org/wiki/Bleiakkumulator) Unfortunately the English version doesn't contain the calculation of the voltage. I took the standard potentials from the book [Elektrochemie by Hamann](https://www.amazon.de/Elektrochemie-Carl-H-Hamann/dp/3527310681).
| 5 |
[
[
"\nThe potentials depend on the form of the compounds. \nIt is true that in solution \n\n\n$\\ce{Pb^{2+} (aq) + 2e^- -> Pb (s)} $ \n\n\nis -0.126 V. \n\n\nBut in the case of a battery we have:\n\n\n$\\ce{PbSO4 (s) + 2e^- -> Pb (s) + SO4^{2-} (aq)}$\n\n\nAnd in this case the $\\ce{Pb^{2+}}$ is in solid form and the potential is -0.356 V. \n\n\nIn a battery the sulphate is insoluble and it is required that it sticks to the electrode, otherwise the reverse reaction can not occur. \n\n\nA table of potentials can be found [here](http://www.chemeddl.org/services/moodle/media/QBank/GenChem/Tables/EStandardTable.htm)\n\n\n",
"3"
],
[
"\nThe underlying redox pair is, as you say, $\\ce{Pb^2+ +2e− -> Pb}$ and has standard potential of $\\pu{−0.1263 V}$.\n\n\nBut standard potential is for $\\pu{1 M}$ concentration. If you look at the \"underlying\" reaction, you must correct for the reduced concentration of $\\ce{Pb^2+}$ due to its insolubility. The correction is made by using the Nernst equation and the solubility (or solubility product) of $\\ce{PbSO4}$, and the half-reaction potential increases because the solubility of $\\ce{Pb^2+}$ is low.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/88963/looking-for-desiccants
|
Looking for Desiccants
|
I am an engineer and not the best at chemistry, I am horrible at it, but I was wondering if it is safe to mix sodium hydroxide and calcium oxide to make a desiccant that also captures carbon monoxide?
I am asking because I am trying to make a high quality zero air generator and need to lower the humidity and carbon monoxide and this seemed like a way to do it.
| 1 |
[] |
https://chemistry.stackexchange.com/questions/88962/any-substance-too-poisonous-to-measure-an-ld50
|
Any substance too poisonous to measure an LD50?
|
A [recent question on poisons](https://chemistry.stackexchange.com/q/88849/11376) was answered in part by bringing up the concept of $\pu{LD\_{50}}$ and animal testing and so on; none of that was new to me, but it did bring to mind a question I don’t know the answer to:
Has there ever been a **substance that scientists have actively tried to determine $\pu{LD\_{50}}$ for**, and could not because they were unable to determine a dosage low enough to kill “just” half of the subjects?
Please note that this question is *not* restricted, strictly, to any technical definition of “poison.” Rather, answers must meet these requirements:
1. There must be research (considering that I am looking for something that might be a negative result and those do not get published as much as they should be, I won’t demand peer-reviewed publication, but the goal should have been to attempt peer-review publication if it had worked out—effectively, I mean serious work), and
2. that research must use the term “$\pu{LD\_{50}}$” to characterize the substance.
Basically, I don’t want to get into debates about what is, or is not, a poison here. If a researcher is willing to call something $\pu{LD\_{50}}$, then I am willing to accept it as a “poison” for the purposes of this question.
I also don’t want speculation, or for a user here to characterize something as $\pu{LD\_{50}}$ when the underlying research doesn’t call it that. It is not enough to say, for example, “well I’m sure even a single atom of antimatter inside your body would be pretty bad,” you need to cite a particular researcher who has performed experiments with the goal of determining what they themselves called $\pu{LD\_{50}}$ for that substance.
I suspect the answer is no, but I have no idea how to research something like this.
| 14 |
[
[
"\nOf course no. [Botulotoxin](https://en.wikipedia.org/wiki/Botulinum_toxin) is probably the strongest known, and still its $\\rm LD\\_{50}$ is counted in nanograms per kilogram, which is pretty manageable. Sure, working with such tiny amounts requires some special measures, but still, it is way greater than *one* molecule. You can divide it again, and again, and again.\n\n\nSo it goes.\n\n\n",
"21"
],
[
"\nSome options that do not work, but have been suggested:\n\n\n1. Prions: looks like LD50 has been successfully determined for the most common Sc237 and 263K prions: [Prions: A Challenge for Science, Medicine, and the Public Health System](https://books.google.fi/books?id=3dGFqapTk84C&pg=PA42&lpg=PA42&dq=ld50%20prions&source=bl&ots=INKvUKTBaH&sig=Isls202jh4Z_RyIetZr7bbEaweY&hl=fi&sa=X&ved=0ahUKEwjb3M2qpt3YAhXDECwKHcwbCNAQ6AEISjAE#v=onepage&q=ld50%20prions&f=false). Though it is also referred to as \"Infectious dose\" as opposed to \"Lethal dose\".\n2. Polonium: While some sources claim that polonium cannot be characterized by LD50, this is not due to high lethality but just because the effects vary depending on what kind of cancers or acute radiation disease the radiation causes. Actual LD50 is around 90 ng for humans for acute radiation disease.\n3. Antimatter: looks like around 1 ng [would be plenty to kill a person](https://steemit.com/science/@renzoarg/what-is-the-most-mass-efficient-material-able-to-kill-us-random-questions-that-i-ve-to-answer-from-time-to-time). LD50 would be much lower, but probably easy enough to measure anyway. Real problem in the experiment would be delivering the antimatter without premature annihilation.\n\n\nBut in reality, it just depends on the means available to the scientist trying to determine the LD50. For example [in this study](http://www.sciencedirect.com/science/article/pii/0041008X71903012), *\"For the remaining solvents LD50 could not be determined due to volume\nlimitations, and have been given a value of <1.0 ml/kg.\"*\n\n\nThe limit of current science would have to be a poison that would kill with a single molecule, as it is quite possible nowadays to manipulate and separate single molecules. Or otherwise single indivisible unit, such as a single cell.\n\n\n",
"6"
]
] |
https://chemistry.stackexchange.com/questions/88960/how-do-i-heat-my-borosilicate-reflux-column
|
How do I heat my borosilicate reflux column?
|
[](https://i.stack.imgur.com/0POWU.jpg)[](https://i.stack.imgur.com/DLhsj.jpg)**I am gonna use a ceramic tape heater, but cannot wrap it around evenly. Is the glass column gonna be okay?**
[](https://i.stack.imgur.com/ys3KU.jpg)
I am running some tests on plastic pyrolysis. I am using the reflux condenser to make the heavier hydrocarbons flow back into the boiling flask. I am also trying out various catalyst in the reflux condenser which will facilitate catalytic cracking.
I have decided to maintain the temperature at top of the reflux column to be of 150C and it will allow the lighter fraction to be removed off and the heavier fraction to be condensed back into the boiling flask.
The boiling flask is made up of quartz and the rest of the glassware is borosilicate.
I am gonna use the glass setup to test catalysts. If results are satisfactory, will get a microreactor made up of steel for further testing. I can better monitor and control temperature in a metal setup I think. Gonna try on only PE and PP types of plastic.
I am a structural engineer by profession and am new to organic chemistry. Thanks for you help in advance.
| 6 |
[
[
"\n[The first answer](https://chemistry.stackexchange.com/a/88961/41328) is great and the borosilicate glass can indeed survive harsh conditions. Here I'd like to suggest a DIY-project to improve the efficiency of the heater and make the column be warmed more evenly. Take a sheet of fiberglass cloth, sew the heating element in a serpentine pattern with the fiberglass threads, then wrap the crafted heating mantle around the column. The exact size can be estimated based on the height and the diameter of the wide part of the column. I sketched an approximate scheme:\n\n\n[](https://i.stack.imgur.com/Zd3uN.png)\n\n\nAlternatively, you can leave a line of column uncovered (1-2 cm) for visual inspection. Fiberglass can withstand the same temperatures as your heating mantle, but I suppose [you've already discovered it the hard way](https://chemistry.stackexchange.com/q/88966/41328).\n\n\n",
"9"
],
[
"\nThat should pose no issue regarding the glass\n---------------------------------------------\n\n\n~~Without images to see how unevenly you will apply the heat, I have to modify my instinctual reaction \"That cannot possibly be an issue\" to \"Most likely it will be OK\".~~ I'd be much more worried about variation in the distilling if you are at some point drawing a distillate. If you cannot control the heat flux, the yield and temperature profile will be offset from the theoretical and can (if the process is delicate) lead to variation in the results.\n\n\nBut borosilicate glass is **sturdy**, you can probably heat it with a naked propane flame - it is designed to have a low thermal expansion coefficient. Directly from that, it will have a high thermal shock resistance.\n\n\n",
"9"
]
] |
https://chemistry.stackexchange.com/questions/88958/what-is-the-driving-force-for-imine-formation
|
What is the driving force for imine formation?
|
I know this that when a primary amine is reacted with a carbonyl compound, an imine is formed:
[](https://i.stack.imgur.com/mP4t1.png)
I'm wondering why the position of this equilibrium lies to the product side / right-hand side. Why is the formation of the C=N double bond favoured?
| 9 |
[
[
"\nIn practice, a drying agent is sometimes added to the reaction: I've used molecular sieves before, but a cursory database search brings up a few other options, e.g. magnesium sulfate, silica gel, or Montmorillonite (this list is probably not exhaustive). These remove water from the system and pull the equilibrium position over to the right-hand side.\n\n\nThat's one possibility, but I don't think that's all there is to it - I think a lot of imine formations proceed perfectly ok even without dehydrating agents being added to them. However, you'll have to hear about these from somebody else, as I don't know what would drive those reactions. It could well be that simply adding an excess of either amine or carbonyl drives the reaction to completion.\n\n\nInterestingly, a 2009 paper by Saggiomo and Lüning1 describes their investigations into supposed imine formation *in water*. Apparently, both the carbonyl compound (benzaldehyde or salicylaldehyde) and amine (aniline) do not dissolve in water - not particularly surprising. The imine did not form appreciably in water and only started to form when water was removed during workup and purification.\n\n\nP/S This is just the top hit I found on Google, it's hardly a complete literature search.\n\n\n\n\n---\n\n\n1. Saggiomo, V.; Lüning, U. On the formation of imines in water—a comparison. *Tetrahedron Lett.* **2009,** *50* (32), 4663–4665. [DOI: 10.1016/j.tetlet.2009.05.117](https://doi.org/10.1016/j.tetlet.2009.05.117).\n\n\n",
"3"
],
[
"\nSee,We compare any reaction thermodanimaclly and kinetically\nLet me put the thermodynamical aspect first which is that the formation of water is the very good sign of thermal stability for a reaction and a energy of 572 KJ/mol is realesed and hence helps to meet with the enrgy demands for formation of carbon and nitrogen bond \nAnd suprisingly carbon and nitrogen bond is stable due to presence of lone pair on nitrogen and high density of electrons around it.\nNow coming back to thermodynamics the carbon nitrogen energy is comparatively very high 614KJ/mol for this carbon and nitrogen bond. **Also important thing note oxygen is more electronegative but yet as wr see oxygen is ready to leave with hydrogen to form carbon nitrogen bond** double bond.\n\n\nNow Kinetics \nThe product hence formed is by first order Kinetic nad hence also it is easy for hydrogen to migrate to alchol group and hence give water and with it leave it as to provide a negative charge to nitrogen and henceforth the water is left to make the way for carbon and nitrogen bond to form.Also **notice** there will be a positive charge on oxygen so would leave easilyand hence there will a negative charge on \"NZ\" leaving formation of the double bond.\nSee this for data <https://chem.libretexts.org/Core/Physical_and_Theoretical_Chemistry/Chemical_Bonding/Fundamentals_of_Chemical_Bonding/Bond_Energies> and \n<http://witcombe.sbc.edu/water/chemistryelectrolysis.html>\n\n\n",
"-2"
]
] |
https://chemistry.stackexchange.com/questions/88956/what-facts-are-being-used-in-this-atomic-structure-question
|
What facts are being used in this atomic structure question?
|
I wish to understand how to solve this atomic structure question:
>
> For the ground state, the electron in the H-atom has an angular
> momentum = $\hslash$ , according to the simple Bohr model. Angular
> momentum is a vector and hence there will be infinitely many
> orbits with the vector pointing in all possible directions. In
> actuality, this is not true, because:
>
>
>
1. Bohr model gives incorrect values of angular
momentum.
2. only one of these would have a minimum energy.
3. angular momentum must be in the direction of spin of electron.
4. electrons go around only in horizontal orbits.
Actually, we have been taught Bohr orbits a lot in class, with formulae for everything (atomic radius/electron orbital speed/various energy levels/etc.) and also basics of the quantum mechanical model. However, we didn't go into such a great depth of Bohr orbits as asked in the above question.
Specifically, I don't know of any reasonable arguments against points 1,2,3. Point 4 is evidently wrong based on the quantum mech model.
>
> How do I develop any claim for/against the points 1,2,3? What are the facts involved here?
>
>
>
| 2 |
[
[
"\nThe only correct answer is #1. \n\n\nIn actuality, the ground state of the hydrogen atom has zero angular momentum. \n\n\nRegarding #2 and #4, there is no preferred plane in which one point should orbit another point. By symmetry, they are all equivalent. \n\n\nRegarding #3, there is no angular momentum in the ground state, and in a state whether there is angular momentum, it doesn't need to be in the same direction as the electron spin angular momentum. \n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/88949/how-can-organocuprates-generated-with-catalytic-cui-be-competitive-in-1-4-addi
|
How can organocuprates generated with catalytic Cu(I) be competitive in 1,4-addition to an enone?
|
Organocopper compounds (e.g. $\ce{LiCuR\_2}$) are often prepared from an organolithium or Grignard reagent when a soften nucleophile is needed, such as for a conjugate addition to an $\alpha,\beta$-unsaturated carbonyl. Clayden et al. (p.499) recommend generating it *in situ* catalytically via transmetallation with 1% $\ce{CuI}$.
My question: if at the start, there is 1% of organocopper reagent and 99% Grignard reagent, why don't those 99% do direct addition? Why is the organocopper reagent so much more reactive that conjugate addition is the main outcome of the reaction?
| 7 |
[] |
https://chemistry.stackexchange.com/questions/88946/iupac-naming-of-1-2-disubstituted-cyclohexane-derivative
|
IUPAC naming of 1,2-disubstituted cyclohexane derivative
|
I'm facing the challenge of naming this molecule:
[](https://i.stack.imgur.com/zT8eD.png)
We have to name it as an alkyl-substituted cycloalkane, since the ring has more carbons than the longest substituent (isopropyl). My questions:
1. When numbering carbons on the ring, we can start counting from either the methyl group (1-methyl...), or the isopropyl group (1-isopropyl...). Which one should I start with? Or do I need to start numbering with the substituent which name comes first, alphabetically?
2. I set the carbon number 1 on the methyl substituent, giving me the following name: *1-methyl-2-isopropylcyclohexane*. However, my book says the correct name is: *1-isopropyl-2-methylcyclohexane*. So the solution managed to sort the name both numerically and alphabetically. Is this what I need to do? Is the name I gave incorrect?
3. Which prefixes count towards alphabetically-naming a compound? Which prefixes don't? All I know is that the numeric prefixes like *di-* or *tri-* don't count. Are there any others which don't count either?
| 5 |
[
[
"\nSimple prefixes (simple substituent groups such as methyl and isopropyl) are arranged alphabetically disregarding any multiplicative prefixes. Any multiplicative prefixes are inserted later and do not alter the alphabetical order. \n\nFor example, ‘methyl’ is considered to begin with ‘m’; ‘isopropyl’ is considered to begin with ‘i’.\n\n\nOn this matter, the current version of *[Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)](http://dx.doi.org/10.1039/9781849733069)* reads as follows:\n\n\n\n> \n> **P-14.5** ALPHANUMERICAL ORDER\n> \n> \n> Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved\n> \n> \n> Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.\n> \n> \n> (…)\n> \n> \n> **P-14.5.1** Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.\n> \n> \n> \n\n\nThus, the correct alphanumerical order in your example corresponds to the name ***x*-isopropyl-*y*-methylcyclohexane.**\n\n\nFurthermore,\n\n\n\n> \n> **P-14.4** NUMBERING\n> \n> \n> When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:\n> \n> \n> (…)\n> \n> \n> (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;\n> \n> \n> (g) lowest locants for the substituent cited first as a prefix in the name;\n> \n> \n> (…)\n> \n> \n> \n\n\nNote that Rule f takes precedence over Rule g; however, Rule f does not permit a decision to be reached between ‘1-isopropyl-2-methylcyclohexane’ and ‘2-isopropyl-1-methylcyclohexane’ since both names correspond to the locant set ‘1,2’.\n\n\nAccording to Rule g, your example is named as **1-isopropyl-2-methylcyclohexane** rather than 2-isopropyl-1-methylcyclohexane since isopropyl is cited first as a prefix in the name.\n\n\nNote, however, that the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’.\n\n\nTherefore, the preferred IUPAC name for your example is actually **1-methyl-2-(propan-2-yl)cyclohexane** because ‘methyl’ is cited first in the name (since alphabetical order is used to establish the order of citation of substituent prefixes in the name).\n\n\n\n\n\n",
"7"
],
[
"\nLet me first quote some rules given by IUPAC (1) and then come to your question.\n\n\nref.1\n\n\n<http://www.acdlabs.com/iupac/nomenclature/79/r79_36.htm#a_2__25>\n\n\nAcyclic Hydrocarbons\n\n\nRule A-2. Saturated Branched-chain Compounds and Univalent Radicals\n\n\n2.25 - Univalent branched radicals derived from alkanes are named by prefixing the designation of the side chains to the name of the unbranched alkyl radical possessing the longest possible chain starting from the carbon atom with the free valence, the said atom being numbered as 1.\nExamples to Rule A-2.25\n\n\n[](https://i.stack.imgur.com/GbRYm.jpg)\n\n\nThe following names may be used for the unsubstituted radicals only:\n[](https://i.stack.imgur.com/Du3Gw.jpg)\n\n\n2.3 - If two or more side chains of different nature are present, they are cited in alphabetical order .\n\n\nThe alphabetical order is decided as follows:\n\n\n(i) The names of simple radicals are first alphabetized and the multiplying prefixes are then inserted.\n\n\n[](https://i.stack.imgur.com/tdGw4.jpg)\n\n\n(ii) The name of a complex radical is considered to begin with the first letter of its complete name.\n[](https://i.stack.imgur.com/sMh7h.jpg)\n\n\n2.4 - If two or more side chains are in equivalent positions, the one to be assigned the lower number is that cited first in the name.\n\n\n[](https://i.stack.imgur.com/iszSM.jpg)\n\n\nFor arranging substituents , which are treated as prefixes the following rule \nis followed.\n\n\nR-0.1.8 Order of prefixes\n\n\n(a) Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplying affixes, if necessary, are then inserted and do not alter the alphabetical order already established.\n\n\n[](https://i.stack.imgur.com/7tamI.jpg)\n\n\nNow comming to your question , it has two substituents methyl and isopropyl at equivalent positions. Using rule 2.4 and R-0.1.8 , numbering is started from isopropyl end and named.\n\n\n[](https://i.stack.imgur.com/PFDQ8.png)\n\n\n**References:**\n\n\n1.<http://www.acdlabs.com/iupac/nomenclature/79/r79_36.htm#a_2__25>\n\n\n2.<http://www.acdlabs.com/iupac/nomenclature/93/r93_93.htm>\n\n\n",
"2"
]
] |
https://chemistry.stackexchange.com/questions/88941/how-does-dry-air-get-saturated-in-the-ostwald-walker-method
|
How does dry air get saturated in the Ostwald-Walker Method?
|
This question was already asked and this was the answer that was posted:
"So basically, before you start the experiment the vapors have already formed, meaning that thermodynamic equilibrium has been achieved. Then you start bubbling through dry gas, which then mixes with the vapor gas already in the bubbles. If you set the gas flow appropriately, you get a near-equilibrium situation (see the graphic below) and the gas that exits the bulb assemply has exactly the same amount of vapor in it as the vapor "layer" (which can not be thought of as a layer anymore because of the bubbling) above the liquid surface.
We do not have an equilibrium situation, which is why so much care has to be taken that we get as close as possible to the equilibrium situation. (Also, we do not have a closed system, so everything starts to get slightly problematic at this point. But to a first degree of approximation, this will hold up fine.)"
(For more, visit: [How does Ostwald-Walker method work?](https://chemistry.stackexchange.com/questions/16575/how-does-ostwald-walker-method-work/18765) )
My questions are:
1) What is this 'near' equilibrium situation that is created?
2) If dry air gets saturated when it passes through the solution bulbs, then how is it able to absorb more solvent vapours when it is passed through the solvent bulbs?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/88938/how-much-calcium-hydroxide-will-precipitate-after-addition-of-sodium-hydroxide-i
|
How much calcium hydroxide will precipitate after addition of sodium hydroxide into saturated calcium hydroxide solution?
|
Below was question 34 in the USNCO 2017 exam:
>
> If $\pu{0.10 mol}$ of solid $\ce{NaOH}$ is added to $\pu{1.00 L}$ of a saturated solution of $\ce{Ca(OH)2}$ $(K\_\mathrm{sp} = \pu{8.0 \times 10^-6})$, what percentage of the calcium hydroxide will precipitate at equilibrium?
>
>
> (A) Roughly 50%
>
> (B) Roughly 75%
>
> (C) Roughly 95%
>
> (D) Over 99%
>
>
>
My solution is as follows:
1. Find concentration of $\ce{Ca^2+}$ $(\pu{0.02 M})$ and $\ce{OH-}$ $(\pu{0.04 M})$ ions from dissolved calcium hydroxide using $K\_\mathrm{sp}$.
2. Add hydroxide ion concentration from sodium hydroxide (assuming full dissolution) to get total hydroxide concentration of $\pu{0.14 M}$
3. Find reaction quotient $Q = 0.02 \times 0.14^2 = 3.92 \times 10^{-4}$
4. Find amount of calcium $(x)$ and hydroxide ions $(2x)$ that will precipitate at equilibrium by using algebraic equation:
$$
(0.02 - x)(0.14 - 2x)^2 = 8.0 \times 10^{-6}, x = \pu{0.019 M}$$
5. Find percentage of calcium hydroxide precipitated:
$$\frac{0.019}{0.02} \times 100\% = 95\%,$$
hence (C)
I am unsure about step 4, where a cubic equation appears, and would not be able to be solved in exam conditions (use of graphing calculator is not permitted).
Is there a simpler method?
| 7 |
[
[
"\nA less analytic aproach:\n\n\nInitial concentrations: \n\n\n$K\\_{\\mathrm{sp}}=\\ce{[Ca^{2+}][OH^-]^2}=x\\cdot (2x)^2=8\\cdot10^{-6}$\n\n\n$\\ce{[Ca^{2+}]=0.0126}$ M; $\\ce{[OH^-]=0.0252}$ M.\n\n\nNow 0.1 mol NaOH is added. Let $p$ be the fraction that precipitates.\nThe concentrations that remain in solution are:\n\n\n$\\ce{[Ca^{2+}}]=0.0126(1-p)$\n\n\n$\\ce{[OH^-]}=(0.10 + 0.0252 -2\\cdot0.0126p)^2$\n\n\nSince the solubility product remains the same:\n\n\n$K\\_{\\mathrm{sp}}=0.0126(1-p)(0.1252 -2\\cdot0.0126p)^2$\n\n\nNow, since there are only four scenarios, try them out substituting every value of $p$ in the equation and take the closest one to $8\\cdot10^{-6}$:\n\n\n\\begin{array}{|c|c|}\\hline\n p&K\\_\\mathrm{sp}\\\\\\hline\n 0.50&\\pu{7.99E-05}\\\\\n 0.75&\\pu{3.56E-05}\\\\\n 0.95&\\pu{6.46E-06}\\\\\n 0.99&\\pu{1.27E-06}\\\\\\hline\n\\end{array}\n\n\nThe closest one is for *roughly* 95% precipitation, without solving any equation.\n\n\n",
"4"
],
[
"\nWhen you run into a cubic in evaluating concentrations, one approach you can use to solve it is the method of successive approximations. \n$$(0.0126-x)(0.1252-2x)^2=8\\cdot10^{-6}$$\n$$(0.0126-x\\_0)(0.1252)^2=8\\cdot10^{-6}$$\n$$(0.0126-x\\_1)(0.1252-2x\\_0)^2=8\\cdot10^{-6}$$\nFor example, moving from the first line to the 2nd line, I guess that $2x=0$ and solve for $x\\_0$. This first guess is not great, but I can use the solution from that approximate equation to generate a better guess. So in the 3rd line, I approximate $2x=x\\_0$ and solve for $x\\_1$. You can continue this procedure until $x\\_{n+1}\\approx x\\_n$, which in this case occurs very quickly. I obtain $x\\_0=0.01209$, $x\\_1=0.0119747$, and $x\\_2=0.011976$ after which the value doesn't change. Checking the ratio, we see that it gives just about $95\\ \\%$. $$\\frac{0.011976}{0.0126}\\times100=95.047\\ \\%$$\n\n\nAn important side note for this case is why I approximated $2x$ rather than $x$. This is because approximating $x$ will cause you to converge towards the two complex solutions to the cubic equation.\n\n\n",
"0"
],
[
"\nNote what the question actually asks you, ***\"roughly\"***. Approximation is an important tool in ionic equilibrium calculations.\n\n\nHere, you have $0.0126M \\text{ }\\ce{Ca}^{+2}$ and $0.0252M\\text{ } \\ce{OH-}$ in the original solution. By addition of $0.1M \\text{ }\\ce{NaOH}$, which is a strong electrolyte, you'll not only add $0.1M\\text{ } \\ce{OH-}$ to the solution, but also cause $Q$ to exceed $K\\_{sp}$, hence, the salt will be precipitated out.\n\n\nNow, the concentration of our ions before precipitation is $0.0126M \\text{ }\\ce{Ca}^{+2}$ and $0.1252M\\text{ } \\ce{OH-}$. Let $y$ be the fraction of existing ions that are *not* precipated out. So, we'll be left with $0.0126 \\cdot y M \\text{ }\\ce{Ca}^{+2}$ and $0.1252 \\cdot (2y-1) M\\text{ } \\ce{OH-}$ ions. Equating their ionic product to the salt's $K\\_{sp}$, we get:\n\n\n$$0.0126 \\cdot y\\cdot(0.1252\\cdot (2y-1))^2=8\\times10^{-6}$$\n\n\n$$y\\cdot(2y-1)^2\\approx0.0405$$\n$$y\\cdot(1+4y^2-4y)\\approx0.0405$$\n\n\n**Magic trick$^\\text{TM}$:** **Neglect $y^3$ at this step.** If $y$ comes out be $\\le5\\%$, we'll assume this neglection to be correct.\n\n\n$$4y^2-y+0.0405\\approx0$$\n$$y\\approx0.050838$$\n\n\nThat means nearly $(100-5)\\% = 95\\%$ of the salt was precipitated out. Hence, your answer. Without a graphing calculator.\n\n\nPS: If you actually solve the [original equation](http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=0.0126+x(0.1252(2x-1))%5E2%3D8*10%5E-6), you'll get $y=0.05$. Hence, our approximate answer is very reasonably close.\n\n\n\n\n---\n\n\n**How did I get $(2y-1)$ as fraction of final concentration of $\\ce{OH-}$ ions?** \n\n\nFairly easy. Observe that $y$ is the fraction of concentration of ions *not* precipitated out, while $x$ is the fraction of concentration of ions precipitated out. Then $x+y=1$. So, $1-x=y=\\text{ concentration of }\\ce{Ca}^{+2}$ ions. And $1-2x=1-2\\cdot(1-y)=2y-1=\\text{ concentration of }\\ce{OH-}$ ions.\n\n\n**But why did I take $y$ anyway? What was wrong with the original $x$?** \n\n\nOur aim was to reduce this problem into an equation which can be solved without a calculator, i.e., by approximation. Taking $y$ ensured that in the end we could neglect higher powers of $y$ (note that we cannot neglect higher powers of $x$) and solve the question easily.\n\n\n\n\n---\n\n\n**Crux of my answer:** if while solving an ionic equilibrium question you're stuck in a tough calculation/unsolveable equation, consider modifying your approach to allow the use of neglection.\n\n\nHope it helps!\n\n\n\n\n---\n\n\nOk, I admit, it wasn't any magic trick :P But just a clever observation that if $x\\le 0.05$, then $x^3\\le0.000125$, which is too small to affect our final result significantly.\n\n\n",
"-1"
]
] |
https://chemistry.stackexchange.com/questions/88933/why-must-polymers-have-a-repeating-unit
|
Why must polymers have a repeating unit?
|
In organic chemistry, we learned that small molecules can form a polymer via a process called polymerization. For example, $\ce{CH2=CH-Br}$ molecules can form the polymer
\begin{align}
\ce{nCH2=CH-Br-> -[-CH2 -}&\ce{CH -]\_n -}.\\
&\;\ce{|}\\
&\ce{Br}
\end{align}
So the polymer is a periodic chain $\ce{-CH2-CHBr-CH2-CHBr-\cdots}$. But since every monomer can have two orientations ($\ce{-CH2-CHBr -}$ or $\ce{-CHBr-CH2 -}\!$), there is no requirement that all monomers must be in the same orientation, does the polymer have to be a peroidic chain? Can it be a random chain with a structure that looks like
$$\ce{-CH2-CHBr-CHBr-CH2-CHBr-CH2-CH2-CHBr-CH2-\cdots}\,?$$
Most textbooks emphasize that $n$ is random, but still assume that the unit repeats.
| 6 |
[
[
"\nEntropy would favour a random orientation of each monomer unit when building up the chain, however whenever you have two different substituents at the ethylene unit, then electrostatics and sterical hindrance will make one side be preferred to make the connection to the growing chain end. \n\n\nPolymers that do grow via chain growth therefore have a regular structure, but there can be defects. The defect density depends on how strong the preference for one side is during the polymerisation reaction. You can have as little as one in ten thousand. That's actually a bit of a problem, because the regularity often allows polymers to crystallise, which is important for the technical application. You want the number of defects always to be the same, because too high crystalinity makes your material brittle, too low makes it soft.\n\n\nPolymers that make step growth (any two molecules randomly connect with their end groups, until you have a long chain) could be different, but also there often a regular structure occurs, because of A-B + A-B , either the A+B like to react, or A+A and B+B. \n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/88923/meaning-of-carbon-hybridization
|
Meaning of carbon hybridization [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/88923/edit)
What do we mean when we talk about carbon hybridization?
I'm trying to understand how it works, what is the difference between non-hybridized and hybridized carbon?
| -3 |
[
[
"\nActually hybridization refers to the electron orbitals around the atom not the atom itself. It actually follows the model called VSPER of electron distribution. So it's not limited to carbon, it applies to all atoms.\n\n\nAtoms depending on the number of electrons they have in their elemental form, has a set number of orbitals. And in order they are as follows, S orbital 2e-, P orbital 6e-, and some other you don't really encounter in intro ochem. \n\n\nThe hybridization is how the orbitals combine to create bonds. You should really read a book about this, I really recommend organic chemistry by Klein chapter 1.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/88922/dicumyl-synthesis
|
Dicumyl synthesis
|
>
> 2,3-Dimethyl-2,3-diphenylbutane (Dicumyl) (63). A solution of 2-bromo-2-phenylpropane (10.0 g, 50 mmol) in anhydrous diethyl ether (25 mL) was stirred with magnesium turnings (0.60 g, 0.025 g atom) overnight. The reaction mixture was poured into aqueous $\ce{NH4Cl}$ solution (100 mL, 5%) and extracted with dichloromethane (100 mL). The solvent was evaporated, and the solid residue was recrystallized from 95%
> ethanol to give 2,3-dimethyl-2,3-diphenylbutane (2.4 g, 47%).
>
>
>
Does anyone have any intuition as to how this reaction is working mechanistically? Is it a Grignard formation and then homocoupling event? Can anyone see a reason for why an *ortho* methyl substituent on the phenyl ring might interfere with this reaction?
| 1 |
[
[
"\nWell, that's interesting as nothing is added here. Today Palladium-catalyzed cross-couplings are quite common, but on a historical side, they took their idea from the very first reactions of sodium and organo-chlorides (Wurtz-coupling) and the magnesium mediated formation of biaryls. Back then transition metal salts were added which lead to the idea to use $\\ce{Pd}$ which is way more carbophilic and can easily enter and leave the catalytic cycle. I have heard that the idea for the cross-coupling derived from the Schlenk-equilibrium, so your Grignard-reagent dimerizes and then equilibrates to from a $\\ce{MgBr2}$ and a $\\ce{MgR2}$. I'd love to say now that this simply reductively eliminates to form $\\ce{Mg}$ and your product, but I have my doubts (although I have heard people talk about it before). \n\n\nSo in your case, we probably have to go the Wurtz path and say that the negatively polarized carbon on your Grignard-reagent nucleophilically attacks the $\\ce{C-Br}$ bond in an $\\ce{S\\_{N}2}$ reaction. If you used sodium the bond would be much more ionic and you'd probably also end up in simply deprotonating the second equivalent of starting material forming a 1-propene derivate. \n\n\n",
"1"
],
[
"\nThe ammount of magnesium was 25 mmol, so only half of the 2-bromo-2-phenylpropane was turned into magnesium bromide.\n\n\nAfter a molecule of magnesium bromide formed, I would guess it undergo a elimination of $\\ce{MgBr^+}$ moity since it is on a tertiary (and also benzyl) carbon and form a relatively stable carboanion. The carboanion then attacks one of the 2-bromo-2-phenylpropane molecules forming the product. But I also wouldn't rule out $\\ce{S\\_{N}2}$ type substitution.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/88911/molar-mass-and-molar-volume-of-monochloroethylene
|
Molar mass and molar volume of monochloroethylene
|
>
> Monochloroethylene gas is used to make polyvinylchloride (PVC). It has a density of $\pu{2.56 g L-1}$ at $\pu{22.8 ^\circ C}$ and $\pu{101 kPa}$. What is the molar mass of monochloroethylene? What is the molar volume under these conditions?
>
>
>
$$PM=dRT$$
$$M=\frac{\pu{2.56 g L-1}\cdot\pu{295.8 K}\cdot\pu{0.0821 atm L mol-1 K-1}}{\pu{0.9968 atm}} = \pu{62.37 g mol-1}$$
After this I can't find molar volume. Which formula should I use?
I know $PV=nRT$, but without knowing the gas mass I can't use this formula, I believe.
| 0 |
[
[
"\nYou don't need to know the volume to determine the [molar volume $V\\_\\mathrm{m}$](https://en.wikipedia.org/wiki/Molar_volume) of a substance. In your notations:\n\n\n$$V\\_\\mathrm{m} = \\frac{M}{d} = \\frac{\\pu{62.37 g mol-1}}{\\pu{2.56 g L-1}} = \\pu{24.36 L mol-1}$$\n\n\nAlternatively, you could've just used ideal gas law:\n\n\n$$V\\_\\mathrm{m} = \\frac{V}{n} = \\frac{RT}{P} = \\frac{\\pu{0.0821 atm L mol-1 K-1}\\cdot\\pu{295.8 K}}{\\pu{0.9968 atm}} = \\pu{24.36 L mol-1}$$\n\n\nP. S. You determined molecular mass correctly, but, as [MaxW pointed out](https://chemistry.stackexchange.com/questions/88911/molar-mass-and-molar-volume-of-monochloroethylene/88916#comment161816_88911), the units are wrong: you probably made a typo and put $\\pu{g L-1}$ instead of $\\pu{g mol-1}$.\n\n\n",
"0"
]
] |
https://chemistry.stackexchange.com/questions/88906/using-cryptands-to-make-a-negatively-charged-solution
|
Using cryptands to make a negatively-charged solution?
|
In our discussion group, my friend mentioned that metal cations can be separated from other cations in solution using cryptands. After addition of the cryptands to the solution of the cations, chelation of the cations takes place and since the complexes formed are more soluble in organic solvent, addition of an organic solvent to this resultant solution would allow these complexes to be migrate to the organic layer. The organic layer can then be decanted and the metal ions have now been separated successfully from the aqueous solution. This was my friend's proposal.
From hearing this, I quickly thought of the possibility of creating a negatively-charged solution. By applying the method of using cryptands which my friend has proposed to separating sodium ions from a solution of sodium chloride, wouldn't the remaining chloride ions in solution give the resultant solution an overall negative charge? This led me to think that there must be something wrong with his idea as this should not be able to occur, based on my instinct. However, I do not know exactly how to refute my friend, to say his proposal does not work. Thus, I would like to ask if anyone knows the reason why my friend's method would not work.
| 4 |
[
[
"\nFor some reason this question reminds me of [Maxwell's demon paradox](https://en.wikipedia.org/wiki/Maxwell%27s_demon). You don't really separate anions and cations here; in reality the resulting metal ($\\ce{M+}$) cryptand ($\\ce{L}$) complex is going to be associated with anionic part ($\\ce{X-}$) in organic solvent ($\\ce{s}$) after it's being extracted from aqueous phase ($\\ce{w}$). For the detailed process, refer to [1], for example.\n\n\nThere are two main processes to consider:\n\n\n\n> \n> $$\n> \\begin{align}\n> \\ce{L\\_\\mathrm{s} + M^+\\_\\mathrm{s} &<=> LM^+\\_\\mathrm{s}}\\label{rxn:1}\\tag{1}\\\\\n> \\ce{\\overline{\\ce{L}} + M^+\\_\\mathrm{w} + X^-\\_\\mathrm{w} &<=> \\overline{\\ce{LMX}}}\\tag{2}\n> \\end{align}$$\n> \n> \n> \n\n\nYou would be correct if only \\eqref{rxn:1} occurs, however, this is not the case.\n\n\n### Reference\n\n\n1. Fyles, T. M. Can. J. Chem. 1987, 65 (4), 884–891. [DOI 10.1139/v87-149](https://doi.org/10.1139/v87-149) (Open Access).\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/88905/tip-of-my-tongue-what-is-the-name-of-this-process
|
Tip of my tongue: what is the name of this process?
|
I remember back in Chemistry class, we did this thing where we would drop some liquid into a solvent. The moment it touched the solvent, it would turn into a gel sphere with the solvent inside it, and you could take these little spheres out and they'd essentially be capsules of solvent. What is the name of this process? The closest thing it reminds me of is Prince Rupert's Drop.
| 2 |
[
[
"\nThe general term for the process would be \"encapsulation\".\n\n\nThe Wikipedia article on [micro-encapsulation](https://en.wikipedia.org/wiki/Micro-encapsulation) lists different processes by which such encapsulation is achieved. Two which sound similar to your described process are interfacial polycondensation and interfacial cross-linking:\n\n\n\n> \n> In interfacial polycondensation, the two reactants in a polycondensation meet at an interface and react rapidly. ... Condensed polymer walls form instantaneously at the interface of the emulsion droplets.\n> \n> \n> \n\n\nAlternatively, the process your class did might have been ionotropic gelation:\n\n\n\n> \n> Ionotropic gelation occurs when units of uric acid in the chains of the polymer alginate crosslink with multivalent cations.\n> \n> \n> \n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/88903/can-naphthenic-acid-be-ionized-in-n-dodecane
|
Can naphthenic acid be ionized in n-dodecane?
|
I am not very clear about the difference between ionization and dissolution. I know that naphthenic acid can dissolve in n-dodecane, but I do not know if it is ionized in n-dodecane.
Thank you!
| 0 |
[
[
"\nIt would not ionize (though the correct term is that it won't dissociate, ie. the O-H bond in the acid won't cleave giving carboxylate anion).\n\n\nIt's because n-dodecane has no lone electron pairs, so it can't complex the hydrogen cation. Unless we're talking about extreme vacuum or plasma, there are no free $H^+$, it always exist in a form of a complex cation with a molecule of solvent or the like. But such solvent must have some electron donor: lone pair, double bond, etc. to form the complex.\n\n\nThe only other possible way here would be self-dissociation, where one of the substance molecules accept the hydrogen carion from the other one, but here it is also borderline impossible as the additionally protonated acid molecule would be very unstable and immidetely such auto-dissotiation would reverse.\n\n\nThe fact that the naphtalenic acid dissolves in n-dodecane is due to low polarity of both, that is acid molecules don't bind strongly to each other and n-dodecane molecules don't bind strongly to each other.\n\n\n",
"1"
]
] |
https://chemistry.stackexchange.com/questions/88899/hexanoic-acid-and-copper-brass-toxicity
|
Hexanoic acid and copper/brass - toxicity?
|
We have a sensor made of brass through which we are flowing organic solvent vapours mixed with air, typically hexane and various alcohols. After a year or so the interior is covered in a greenish-black slime. The smell is definitely that of hexanoic acid, so I am assuming much of the deposit will be the copper salts of the various acids.
Is there likely to be any significant toxicity with the probable compounds that may require special cleaning procedures?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/88893/what-are-the-cip-rules-for-cyclic-substituents
|
What are the CIP rules for cyclic substituents?
|
Here's what I read on [Wikipedia](https://en.wikipedia.org/wiki/Cahn%E2%80%93Ingold%E2%80%93Prelog_priority_rules) (section "Cycles"):
>
> To handle a molecule containing one or more cycles, one must first expand it into a tree (called a hierarchical digraph by the authors) by traversing bonds in all possible paths starting at the stereocenter. When the traversal encounters an atom through which the current path has already passed, a ghost atom is generated in order to keep the tree finite. A single atom of the original molecule may appear in many places (some as ghosts, some not) in the tree.
>
>
>
I don't understand how to assign CIP priorities to cyclic systems, for example, consider this:
How do we decide the CIP priorities? What does that paragraph I found on Wikipedia really mean? Could someone please give a detailed explanation?
| 20 |
[
[
"\nI presume you want to know the E/Z geometry of the double bond. You need to determine on each end of the double bond which ring has priority. Intuitively one may argue that cyclohexane is larger than cyclopentane and cyclobutane is larger than cyclopropane. In this simple case, this is true and the double bond has the Z-configuration. However, digraphs are used in more complex cases. Your structure below has four colored dots. Each attached ring bond is \"cut\" and stretched out until one returns to the original atom. The three red atoms from the cyclohexane ring are identical. The same is true of the other three rings. Now treat the digraph as you would any acyclic alkene. The digraph has the Z-configuration. \n \n\n**Addendum:** For two more complex cases, go [here](https://chemistry.stackexchange.com/questions/6387/how-to-check-for-geometrical-isomerism-in-cyclic-compounds/134639#134639) and [here](https://chemistry.stackexchange.com/questions/139735/assigning-r-s-configuration/139741#139741).\n\n\n[](https://i.stack.imgur.com/ftM94.jpg)\n\n\n",
"27"
]
] |
https://chemistry.stackexchange.com/questions/88884/calculating-average-rate
|
Calculating Average Rate
|
>
> A $\pu{3.45 g}$ piece of marble ($\ce{CaCO3}$) is weighed and dropped into a beaker containing $\pu{1.00 L}$ of hydrochloric acid. The marble is completely gone $\pu{4.50 min}$ later. Calculate the average rate of reaction of $\ce{HCl}$ in $\pu{mol/L/s}$. Note that the volume of the system remains at $\pu{1.00 L}$ through the entire reaction.
>
>
>
I'm not very sure why the units are in $\pu{mol/L/s}$ instead of just $\pu{mol/s}$. Here is what I did:
$$\text{Rate} = \frac{\pu{3.45 g}~\ce{(CaCO3)}}{\pu{4.5 min}} = \pu{0.0128 g/s}$$
$$\pu{0.128 g}~\ce{CaCO3} = \pu{1.28e-3 mol}$$
$$n(\ce{HCl}) = \pu{2.58e-3 mol}$$
$$\text{Rate} = \pu{2.58e-3 mol/L/s}$$
However, the answer given is $\pu{2.55e-4 mol/L/s}$. What am I doing wrong?
| 0 |
[
[
"\nYou haven't accounted for the stoichiometry of the reaction, and I suppose you wrongly converted minutes to seconds. Always start solving problems like this with writing down the chemical reaction:\n\n\n$$\\ce{CaCO3 + 2HCl -> CaCl2 + H2O + CO2}$$\n\n\nBy definition rate of consumption of hydrochloric acid over time $\\Delta t$ is:\n\n\n$$r=\\frac{\\Delta c(\\ce{HCl})}{\\Delta t}$$\n\n\nSince all calcium carbonate reacted completely:\n\n\n$$\\Delta c(\\ce{HCl}) = \\frac{\\Delta n(\\ce{HCl})}{V} = \\frac{2n(\\ce{CaCO3})}{V} = \\frac{2m(\\ce{CaCO3})}{V M(\\ce{CaCO3})}$$\n\n\nwhere $m$ is mass, $M$ - molar mass, $V$ - volume.\n\n\nAnd the average rate is:\n\n\n$$r = \\frac{2m(\\ce{CaCO3})}{V M(\\ce{CaCO3})\\Delta t} = \\frac{2\\cdot\\pu{3.45 g}}{\\pu{1 L}\\cdot\\pu{100.09 g mol-1}\\cdot\\pu{4.50 min}\\cdot\\pu{60 s min-1}} = \\pu{2.55e-4 mol L-1 s-1}$$\n\n\nAlso, be careful with notations. Use proper capitalization, and don't equate moles to grams! This is not tolerable in natural sciences.\n\n\n",
"4"
]
] |
https://chemistry.stackexchange.com/questions/88882/polarity-of-acetic-acid-ethanol-and-1-propanol
|
Polarity of Acetic Acid, Ethanol and 1-Propanol
|
In order to compare acetic acid, ethanol and 1-propanol in terms of solubility in hexane, I would like to order them in terms of (a)polarity.
By looking up the dipole moments, I found that these molecules can be ordered in terms of increasing polarity like this.
1-Propanol (1.68D) < Ethanol (1.69D) < Acetic acid (1.74D)
Without looking up the values for the dipole moments, how can I predict which one of these molecules will be the most/least polar?
| 1 |
[] |
https://chemistry.stackexchange.com/questions/88881/structure-of-ph4br
|
Structure of PH4Br
|
What is the structure of $\ce{PH4Br}$? I would like to know which atom is bonded to which. Are all five atoms directly bonded to phosphorus?
| 1 |
[
[
"\nNo, not in the solid state. The most recent crystal structure analysis suggests there are nearly ideal tetrahedra of phosphonium cations $\\ce{PH4+}$ with bromine anions $\\ce{Br-}$ in between bound via H-bonds [1, ICSD#23691]:\n\n\n[](https://i.stack.imgur.com/w5UJJ.png)\n\n\n### Reference\n\n\n1. Schroeder, L. W.; Rush, J. J. The Journal of Chemical Physics 1971, 54 (5), 1968–1973. [DOI 10.1063/1.1675127](https://doi.org/10.1063%2F1.1675127).\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/88880/why-do-bonding-and-anti-bonding-states-split-in-energy-mathematically
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Why do bonding and anti-bonding states split in energy (mathematically)?
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I am a physicist, and when I hear about the splitting of energy levels for MO antibonding and bonding states, I have a rough intuition that tells me this is reasonable to expect, but I can't think of (nor find) a good formal explanation.
My best shot at understanding this is borrowing from [Avoided Crossing](https://en.wikipedia.org/wiki/Avoided_crossing) and saying that if I have a 2 atom system with a total Hamiltonian $ H = H\_1 + H\_2 $ that is the sum of each individual atom Hamiltonians, then I can take the Eigenstates for each individual Hamiltonian as a basis.
$$\begin{align}
H\_1 | \psi\_1 \rangle &= E\_1 | \psi\_1 \rangle \\
H\_2 | \psi\_2 \rangle &= E\_2 | \psi\_2 \rangle
\end{align}$$
Then the total hamiltonian in this basis becomes,
$$\begin{align}
H &= \left(\begin{matrix}
E\_1 & \langle \psi\_2 | H\_1 | \psi\_2 \rangle \\
\langle \psi\_1 | H\_2 | \psi\_1 \rangle & E\_2
\end{matrix}\right)
\end{align}$$
In the limit of the atoms being really far apart, I would assume the off diagonal terms would vanish, and then the eigenstates are just the atomic orbitals. But if the atoms come closer together, then we get off diagonal terms which causes the splitting as described in [avoided crossing](https://en.wikipedia.org/wiki/Avoided_crossing).
Is this correct?
| 9 |
[
[
"\nYou are (essentially) correct and the off-diagonal terms are related to so-called \"overlap\" between atomic orbitals (AOs). It is common to talk of the AOs \"overlapping\" to form molecular orbitals (MOs), and as you rightly said, the off-diagonal terms (or \"overlap\") goes to zero in the limit of infinite bond length. The term \"avoided crossing\" is not commonly seen in introductory MO theory, but I think it is fair to describe it as such.\n\n\nOf course there are some slight subtleties. Let's stick to a one-electron system such as $\\ce{H2+}$ for now. $\\psi\\_i$ refers to the 1s orbital on hydrogen atom $i$ and $E$ is the energy of a 1s orbital on hydrogen. The minor issue is that $\\hat{H} \\neq \\hat{H}\\_1 + \\hat{H}\\_2$ (following your notation where $\\hat{H}$ is the molecular Hamiltonian and $\\hat{H}\\_i$ is the Hamiltonian for hydrogen atom $i$), because:\n\n\n$$\\begin{align}\n\\hat{H} &= -\\frac{\\nabla^2}{2} - \\frac{1}{r\\_1} - \\frac{1}{r\\_2} + \\frac{1}{R} \\\\\n\\hat{H}\\_1 &= -\\frac{\\nabla^2}{2} - \\frac{1}{r\\_1} \\\\\n\\hat{H}\\_2 &= -\\frac{\\nabla^2}{2} - \\frac{1}{r\\_2} \\\\\n\\end{align}$$\n\n\n(I used atomic units. $r\\_i$ is the distance between the electron and nucleus $i$, and $R$ is the distance between the two nuclei). So actually, it is more like\n\n\n$$\\hat{H} = \\hat{H}\\_1 - \\frac{1}{r\\_2} + \\frac{1}{R}$$\n\n\nand therefore the first term in the matrix\n\n\n$$\\begin{align}\n\\mathbf{H}\\_{11} &= \\left<\\psi\\_1\\middle|\\hat{H}\\_1 - \\frac{1}{r\\_2} + \\frac{1}{R}\\middle|\\psi\\_1\\right> \\\\\n&= E + \\left<\\psi\\_1\\middle|-\\frac{1}{r\\_2}\\middle|\\psi\\_1\\right> + \\frac{1}{R} \\neq E\n\\end{align}$$\n\n\n(since $\\langle\\psi\\_1|\\hat{H}\\_1|\\psi\\_1\\rangle = E$, and also $\\langle\\psi\\_1|(1/R)|\\psi\\_1\\rangle = (1/R)\\langle\\psi\\_1|\\psi\\_1\\rangle = 1/R$).\n\n\nFor more information I refer you to Atkins' *Molecular Quantum Mechanics* 5th ed., pp 262–266. The terminology can be a bit confusing and also Atkins does not use atomic units (in his text he uses $j\\_0 = e^2/4\\pi\\varepsilon\\_0$ which is simply equal to $1$ in atomic units). Adjusting for this, the final quoted formulae for the matrix elements are:\n\n\n$$\\alpha = \\mathbf{H}\\_{11} = \\mathbf{H}\\_{22} = E - j' + \\frac{1}{R}; \\qquad j' = \\left<\\psi\\_1\\middle|\\frac{1}{r\\_2}\\middle|\\psi\\_1\\right>$$\n\n\n(as found earlier), and\n\n\n$$\\beta = \\mathbf{H}\\_{12} = \\mathbf{H}\\_{12} = \\left(E + \\frac{1}{R}\\right)S - k'; \\qquad S = \\langle\\psi\\_1|\\psi\\_2\\rangle;\\quad k' = \\left<\\psi\\_1\\middle|\\frac{1}{r\\_2}\\middle|\\psi\\_2\\right>$$\n\n\nIf you now solve the generalised eigenvalue equation $\\mathbf{Hc} = E\\mathbf{Sc}$ (the simple $\\mathbf{Hc} = E\\mathbf{c}$ only works when your basis set is orthonormal, but in this case it is not) you obtain the two eigenvalues\n\n\n$$E\\_\\pm = E + \\frac{1}{R} - \\frac{j' \\pm k'}{1 \\pm S}$$\n\n\nNote that when the atoms are infinitely far apart both $S$ and $k'$ vanish, so (1) the off-diagonal element $\\beta$ vanishes and (2) the \"bonding\" and \"antibonding\" orbitals are of the same energy, i.e. $E\\_+ = E\\_-$.\n\n\n",
"5"
]
] |
https://chemistry.stackexchange.com/questions/88875/does-alum-react-with-zinc
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Does alum react with zinc?
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I want to remove a hardened steel bolt in a block of zinc (there is a through-hole, so I have two sides to work with). I have done this previously with aluminum and steel before, using a solution of alum to loosen the steel bolt.
Will I have as smooth a time using the alum method in the present case, or will the alum react with the zinc?
| 1 |
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