{"post_id": "99321", "domain": "chemistry_test", "history": "Why do we still need to know about the Rankine temperature scale? I am learning process principles in chemical engineering and I was taught various temperature units like Fahrenheit, Celsius, Kelvin, and Rankine. I see the first three being used quite commonly, but I rarely see the fourth being used. Where is the Rankine scale used then?", "created_at_utc_A": 1531408130, "created_at_utc_B": 1531355214, "score_A": 38, "score_B": 18, "human_ref_A": "Rankine is commonly used in the aerospace industry in the United States. Rankine is to Fahrenheit what Kelvin is for Celsius. So when people in the United States were creating programs and using equations that needed an absolute temperature, they used Rankine before Celsius became dominate for scientific calculations. The reason people still sometimes use it in the aerospace industry is that there are a lot of programs that were developed using Rankine, so in order to be compatible with those old programs, it's often simpler to just use Rankine in the new programs too. Thus, as an intern at NASA, I was writing simulations that used Rankine in 2009.", "human_ref_B": "Most of us in the world use the Celsius scale to measure temperature for day-to-day purposes. The Kelvin scale has been designed in such a way, it is not only an absolute temperature scale, but also 1C change is equal to a 1K change. This makes conversion from Celsius to Kelvin pretty easy, involving just the addition or subtration of a certain constant (in this case, it turns out to be 273.16). However, our friends who live in the United States prefer to use the Fahrenheit scale. The problem is, converting Fahrenheit to Kelvin is not as easy as it is for people using Celsius. So what do they do? Here steps the Rankine scale. The Rankine scale is sort of like the Kelvin analog for Fahrenheit users. Rankine is an absolute temperature scale, and has the property of having a 1R change equal to a 1F change. This means Fahrenheit users who need to work with absolute temperatures will find it easier to use the Rankine scale instead of the Kelvin scale. Fahrenheit can be easily converted to Rankine by: $$R = F + 459.67$$ This indeed is the case, as engineering systems using the Fahrenheit scale use the Rankine scale for absolute temperatures: (source: Wikipedia)
The Rankine scale (/rkn/) is an absolute scale of thermodynamic temperature named after the Glasgow University engineer and physicist William John Macquorn Rankine, who proposed it in 1859. (The Kelvin scale was first proposed in 1848.) It may be used in engineering systems where heat computations are done using degrees Fahrenheit.
", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/99321, Response URL: https://chemistry.stackexchange.com/questions/99353, Post author username: Pritt says Reinstate Monica, Post author profile: https://chemistry.stackexchange.com/users/42741, Response author username: Rick, Response author profile: https://chemistry.stackexchange.com/users/52575", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/99321, Response URL: https://chemistry.stackexchange.com/questions/99322, Post author username: Pritt says Reinstate Monica, Post author profile: https://chemistry.stackexchange.com/users/42741, Response author username: Pritt says Reinstate Monica, Response author profile: https://chemistry.stackexchange.com/users/42741", "seconds_difference": 52916.0, "score_ratio": 2.111111111111111, "upvote_ratio": -1.0, "c_root_id_A": "99353", "c_root_id_B": "99322"} {"post_id": "84206", "domain": "chemistry_test", "history": "Synthesis Golf VI: Muscarine October's synthesis golf is rather late - apologies. I've chosen a slightly different-looking target this round, with the aim of inspiring a wider variety of routes. Muscarine is a natural product found in some mushrooms. Most notably, it is a good agonist of one class of acetylcholine receptors - these receptors are therefore called muscarinic acetylcholine receptors. The only conditions for this round are that: the synthesis must be enantioselective; and no more than one chiral centre may be purchased in the form of a building block. (Chiral catalysts, ligands, auxiliaries, etc. do not fall under this rule.) Otherwise, any commercially available starting material (as usual, in the SigmaAldrich catalogue) is fair game! There is also no restriction on the counterion.", "created_at_utc_A": 1508355651, "created_at_utc_B": 1508605446, "score_A": 10, "score_B": 14, "human_ref_A": "Allylation of ethyl acetate should provide the unsaturated ester. Sharpless asymmetric dihydroxylation should proceed with the formation of the lactone. The free alcohol was then tosylated, substituted with azide and the azide was reduced by Straudinger procedure and the obtained amine was protected with Bn group. Davis hydroxylation should be reagent controlled to give the hydroxyl group with the correct stereochemistry. Tebbe olefination, was followed by OH-directed hydrogenation and Bn deprotection. Selective methylation of the amino group should provide the target molecule.", "human_ref_B": "1. Retrosynthesis This proposed route relies on being able to make the tetrahydrofuran via some type of ring closing etherification. Through disconnection of the C-O bond, a linear diol is revealed which may be accessed in several ways. Initially, I'd thought about doing a Sharpless dihydroxylation of (Z)-hex-1,4-diene, however the starting material is surprisingly expensive and there are few literature reports of doing this dihydroxylation (and in those instances where it has been carried out, the e.e. and regioselectivity has been moderate to poor). A different disconnection, using the chiral pool (less elegant, perhaps, but the starting material is cheap and ensures good selectivity) allowsthe diol can be disconnected across the C-C bond, revealing an aldehyde which is derived from a lactate ester. 2. Forward synthesis Reagents and conditions: (a) TIPSOTF, 2,6-lutidine (b) DIBAL; allyl-MgBr (c) BnTCA, PPTS (d) Shi epoxidation (e) HF-Py, Py (f) TsCl, Py (g) NaN3 (h) H2, Pd/C (i) MeI The synthesis starts with protection of methyl lactate with a TIPS group - a silyl group is chosen here as upon deprotection under fluoride conditions later in the synthesis, the THF should spontaneously close. The TIPS protected methyl lactate may then be treated with DIBAL to afford the lactaldehyde. This must be used immediately to prevent epimerisation. In all likelihood, this could be achieved by adding a solution of allyl magnesium bromide to the same pot as the DIBAL reduction but failing this, there is good precedence for a reagent controlled boron mediated allylation (Tet. Lett. 2003, 44, 1737). The terminal olefin installed by allylation is now setup to do a Shi epoxidation (Sharpless is terrible at terminal olefins) providing the final stereocentre. Treatment of the substrate with HF-Py should then deprotect the TIPS ether with concomitant formation of the THF ring system. Finally, functional group transformations provide the desired natural product as shown in the scheme. Some issues may be observed when attempting to methylate the amine in the presence of the alcohol however from experience of similar substrates, the amine is likely to quaternise faster than the alcohol methylates and as such careful control of stoichiometry and conditions should allow the desired product to be isolated as the major component. Summary A synthesis of muscarine is proposed in 9 steps (linear) starting from methyl lactate, with the remaining stereocentres being installed using a substrate controlled allylation and a reagent controlled asymmetric epoxidation. Addition steps could likely be cut out by (1) not protecting the alcohol that is currently protected with a benzyl group and (2) making use of an asymmetric aziridination in place of the Shi epoxidation.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/84206, Response URL: https://chemistry.stackexchange.com/questions/84418, Post author username: orthocresol, Post author profile: https://chemistry.stackexchange.com/users/16683, Response author username: EJC, Response author profile: https://chemistry.stackexchange.com/users/7159", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/84206, Response URL: https://chemistry.stackexchange.com/questions/84543, Post author username: orthocresol, Post author profile: https://chemistry.stackexchange.com/users/16683, Response author username: NotEvans., Response author profile: https://chemistry.stackexchange.com/users/21296", "seconds_difference": 249795.0, "score_ratio": 1.4, "upvote_ratio": -1.0, "c_root_id_A": "84418", "c_root_id_B": "84543"} {"post_id": "84206", "domain": "chemistry_test", "history": "Synthesis Golf VI: Muscarine October's synthesis golf is rather late - apologies. I've chosen a slightly different-looking target this round, with the aim of inspiring a wider variety of routes. Muscarine is a natural product found in some mushrooms. Most notably, it is a good agonist of one class of acetylcholine receptors - these receptors are therefore called muscarinic acetylcholine receptors. The only conditions for this round are that: the synthesis must be enantioselective; and no more than one chiral centre may be purchased in the form of a building block. (Chiral catalysts, ligands, auxiliaries, etc. do not fall under this rule.) Otherwise, any commercially available starting material (as usual, in the SigmaAldrich catalogue) is fair game! There is also no restriction on the counterion.", "created_at_utc_A": 1508376323, "created_at_utc_B": 1508605446, "score_A": 8, "score_B": 14, "human_ref_A": "ethylene glycol, pTsOH, reflux mCPBA (S,S)-salen Co (Jacobsen kinetic resolution) $\\ce{NaN3}$. Fortunately, this epoxide will open the right way since it's not a styrene... toluene, pTsOH, water, Dean-Stark reflux. I'm trying to do this one-pot. The conditions for hydrolysis of the acetal are so close to the conditions for dehydation to form the dihydrofuran after intramolecular ring closure. We might need to play with solvent to get enough water in there for hydrolysis. mCPBA. I'm hoping to get some diastereoselectivity here. MeMgBr. We will likely need to optimize reaction conditions here. I think low temperature plus the possible addition of a Lewis acid will help selectivity. $\\ce{PPh3}$. Staudinger reaction 2.9 equivalents of methyl triflate. Slow addition at low temperature.", "human_ref_B": "1. Retrosynthesis This proposed route relies on being able to make the tetrahydrofuran via some type of ring closing etherification. Through disconnection of the C-O bond, a linear diol is revealed which may be accessed in several ways. Initially, I'd thought about doing a Sharpless dihydroxylation of (Z)-hex-1,4-diene, however the starting material is surprisingly expensive and there are few literature reports of doing this dihydroxylation (and in those instances where it has been carried out, the e.e. and regioselectivity has been moderate to poor). A different disconnection, using the chiral pool (less elegant, perhaps, but the starting material is cheap and ensures good selectivity) allowsthe diol can be disconnected across the C-C bond, revealing an aldehyde which is derived from a lactate ester. 2. Forward synthesis Reagents and conditions: (a) TIPSOTF, 2,6-lutidine (b) DIBAL; allyl-MgBr (c) BnTCA, PPTS (d) Shi epoxidation (e) HF-Py, Py (f) TsCl, Py (g) NaN3 (h) H2, Pd/C (i) MeI The synthesis starts with protection of methyl lactate with a TIPS group - a silyl group is chosen here as upon deprotection under fluoride conditions later in the synthesis, the THF should spontaneously close. The TIPS protected methyl lactate may then be treated with DIBAL to afford the lactaldehyde. This must be used immediately to prevent epimerisation. In all likelihood, this could be achieved by adding a solution of allyl magnesium bromide to the same pot as the DIBAL reduction but failing this, there is good precedence for a reagent controlled boron mediated allylation (Tet. Lett. 2003, 44, 1737). The terminal olefin installed by allylation is now setup to do a Shi epoxidation (Sharpless is terrible at terminal olefins) providing the final stereocentre. Treatment of the substrate with HF-Py should then deprotect the TIPS ether with concomitant formation of the THF ring system. Finally, functional group transformations provide the desired natural product as shown in the scheme. Some issues may be observed when attempting to methylate the amine in the presence of the alcohol however from experience of similar substrates, the amine is likely to quaternise faster than the alcohol methylates and as such careful control of stoichiometry and conditions should allow the desired product to be isolated as the major component. Summary A synthesis of muscarine is proposed in 9 steps (linear) starting from methyl lactate, with the remaining stereocentres being installed using a substrate controlled allylation and a reagent controlled asymmetric epoxidation. Addition steps could likely be cut out by (1) not protecting the alcohol that is currently protected with a benzyl group and (2) making use of an asymmetric aziridination in place of the Shi epoxidation.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/84206, Response URL: https://chemistry.stackexchange.com/questions/84430, Post author username: orthocresol, Post author profile: https://chemistry.stackexchange.com/users/16683, Response author username: Zhe, Response author profile: https://chemistry.stackexchange.com/users/35555", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/84206, Response URL: https://chemistry.stackexchange.com/questions/84543, Post author username: orthocresol, Post author profile: https://chemistry.stackexchange.com/users/16683, Response author username: NotEvans., Response author profile: https://chemistry.stackexchange.com/users/21296", "seconds_difference": 229123.0, "score_ratio": 1.75, "upvote_ratio": -1.0, "c_root_id_A": "84430", "c_root_id_B": "84543"} {"post_id": "21709", "domain": "chemistry_test", "history": "Why does coordinate covalent bond form? Coordinate covalent bonds are bonds on which both electrons from one atom. But why does this happen? Some may think it is because one of the bonding atoms have strong electronegativity. But experimental evidence suggests otherwise. For example, carbon monoxide. The EN of carbon is 2.6, and oxygen 3.4. This may lead you to think that carbon will donate a pair of electron, but on Princeton.edu:
Carbon monoxide (CO) can be viewed as containing one coordinate bond and two \"normal\" covalent bonds between the carbon atom and the oxygen atom. This highly unusual description illustrates the flexibility of this bonding description. Thus in CO, carbon is the electron acceptor and oxygen is the electron donor.
Nevertheless, in some cases like NO2, it does follows that the less electronegative atom domates electron. But why??", "created_at_utc_A": 1421591147, "created_at_utc_B": 1418946302, "score_A": 17, "score_B": 6, "human_ref_A": "
We can share the women, we can share the wine We can share what we got of yours, cause we done shared all of mine
\"Jack Straw\" Grateful Dead But seriously now, the most important thing in the Princeton.edu article (which explains at the bottom that its content comes from Wikipedia) is \"The distinction from ordinary covalent bonding is artificial\" Does $\\ce{H2}$ come from a hydride and a proton or from two hydrogen atoms? The reason a molecule is stable is independent of a particular path by which it is made. For $\\ce{CO}$, it is only in your mind that oxygen brought more wine (electrons) to the party, but what really matters is that there is enough wine (electrons) at the party for everyone to be satisfied. $\\ce{CO}$ is isoelectronic with $\\ce{N2}$ and $\\ce{CN-}$; the reason for bonding is the same for each.", "human_ref_B": "Electronegativity affects coordinate covalent bonds; indeed atoms which accept electrons should be \"electron hungry\" enough to take them. However it stems mainly from that the acceptor doesn't have complete electron shell, thus acts as Lewis acid and donor acts as base.", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/21709, Response URL: https://chemistry.stackexchange.com/questions/23898, Post author username: most venerable sir, Post author profile: https://chemistry.stackexchange.com/users/4805, Response author username: DavePhD, Response author profile: https://chemistry.stackexchange.com/users/5160", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/21709, Response URL: https://chemistry.stackexchange.com/questions/21894, Post author username: most venerable sir, Post author profile: https://chemistry.stackexchange.com/users/4805, Response author username: Mithoron, Response author profile: https://chemistry.stackexchange.com/users/9961", "seconds_difference": 2644845.0, "score_ratio": 2.8333333333333335, "upvote_ratio": -1.0, "c_root_id_A": "23898", "c_root_id_B": "21894"} {"post_id": "17089", "domain": "chemistry_test", "history": "Why doesn't calcium carbonate dissolve in water even though it is an ionic compound? I studied the solubility of compounds in water. I have found that calcium carbonate doesn't dissolve in water. The teacher stated that the ionic compounds dissolve in water except some carbonates. What would be a clear elaboration of this phenomenon?", "created_at_utc_A": 1412534640, "created_at_utc_B": 1412529981, "score_A": 16, "score_B": 5, "human_ref_A": "As someone said here, this:
The teacher stated that the ionic compounds dissolve in water except some carbonates.
Is indeed an oversimplification. First of all, the distinction between an \"ionic compound\" to other compounds isn't too defined. What your teacher probably said, or didn't say but wanted to, is that some ionic compounds easily dissolve in water. Salt (halite - NaCl) is the best example. Calcium carbonate, in nature, also commonly dissolves. It's just not as immediate as dissolution of the more soluble ionic compounds. You are probably familiar with this phenomenon: This forms because calcium carbonate dissolves. The rock is limestone, which is usually composed of pure calcium carbonate. Acidic water greatly enhances the solubility of calcium carbonate, and it doesn't even need to be highly acidic. Rain or river water that come into contact with the atmosphere absorb the $\\ce{CO2}$ as $$\\ce{H2O + CO2 <=> H2CO3},$$ which then facilitates calcium carbonate dissolution with $$\\ce{CaCO3 + H2CO3 <=> Ca^2+ + 2HCO3-}.$$", "human_ref_B": "Solubility of salts is not a prediction that I like to make. It all comes down to $\\Delta{G}_{sol} < 0$. That condition is fulfilled when $\\Delta{H}_{sol} < 0$ and $T \\Delta{S}_{sol} > 0$ or one factor outweighs the other. There's two problems. Lattice enthalpies are large, as are solvation enthalpies, and the common observation is that $\\Delta{H}_{sol}$ in most cases is small. We are calculating differences of large numbers, and the one fact we can confidently assert is that the difference is impossible to predict and generally meaningless. Same thing for $\\Delta{S}_{sol}$. Lattice entropies are small, but entropies of solution are lower than one would think thanks to the short-range order the solvated cation imposes on the solvent. You are gaining translational entropy from dissolving the lattice and losing translational entropy from the solvent through the inner coordination shell. You are calculating differences of numbers of equal magnitude again, and just by staring at reagents it's impossible to tell which way the pendulum is going to swing and if the result is positive or negative.", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/17089, Response URL: https://chemistry.stackexchange.com/questions/17102, Post author username: Paul Janson, Post author profile: https://chemistry.stackexchange.com/users/8197, Response author username: Gimelist, Response author profile: https://chemistry.stackexchange.com/users/8083", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/17089, Response URL: https://chemistry.stackexchange.com/questions/17101, Post author username: Paul Janson, Post author profile: https://chemistry.stackexchange.com/users/8197, Response author username: Abel Friedman, Response author profile: https://chemistry.stackexchange.com/users/7920", "seconds_difference": 4659.0, "score_ratio": 3.2, "upvote_ratio": -1.0, "c_root_id_A": "17102", "c_root_id_B": "17101"} {"post_id": "92244", "domain": "chemistry_test", "history": "What is the difference between , ||, radial probability, and radial distribution of electrons? I am very confused regarding these four terms- ,||, radial probability and radial distribution I know that is called the wave function but is it the same as radial probability? I also know that || is called the probability density of finding an electron inside an atom but is it the same as radial distributionwhich is the probability of finding an electron in a volume $\\mathrm dV$ inside a nucleus?", "created_at_utc_A": 1520949787, "created_at_utc_B": 1520949780, "score_A": 3, "score_B": 2, "human_ref_A": "According to the Copenhagen interpretation of quantum mechanics, $|\\Psi|^2$ is the \"probability density\" (the probability per volume of finding a particle, such as an electron, in a given volume, in the limit the volume approaches zero). If $\\Psi(r,\\theta,\\phi)$ is separable as $R(r)Y(\\theta,\\phi)$, such as in the hydrogen atom, then: \"Radial Probability Density\" is $R^2(r)$ (still proportional to probability per volume for s-states) \"Radial Probability Distribution\" is $4\\pi r^2R^2(r)$ (probability per incremental distance from origin, as the incremental distance approaches zero)", "human_ref_B": "I'd say the wave function is your $$. This wavefunction describes your system. When you want to determine something like its energy or other operations you need to describe your system like your atom using a wave function. $||$ is as you already said the probability density. Your wavefunction for example in solving the hydrogen atom using the Schrdinger-equation has this step where you transform to spherical coordinates and then seperate into a radial part and an angular part. From this radial part you can get the radial probability, so to find an electron for example at the distance x. And the radial distribution is this function times the area of a sphere having the size of the radius for that distance you are looking at the moment. So this will give you the probability for the electron to be located somewhere on a sphere at the distance x. But I can't really tell you the actual difference between taking only the radial part or the wave function itself to describe probability.", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/92244, Response URL: https://chemistry.stackexchange.com/questions/92247, Post author username: Aura Sartori, Post author profile: https://chemistry.stackexchange.com/users/59105, Response author username: DavePhD, Response author profile: https://chemistry.stackexchange.com/users/5160", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/92244, Response URL: https://chemistry.stackexchange.com/questions/92246, Post author username: Aura Sartori, Post author profile: https://chemistry.stackexchange.com/users/59105, Response author username: Justanotherchemist, Response author profile: https://chemistry.stackexchange.com/users/41761", "seconds_difference": 7.0, "score_ratio": 1.5, "upvote_ratio": -1.0, "c_root_id_A": "92247", "c_root_id_B": "92246"} {"post_id": "92244", "domain": "chemistry_test", "history": "What is the difference between , ||, radial probability, and radial distribution of electrons? I am very confused regarding these four terms- ,||, radial probability and radial distribution I know that is called the wave function but is it the same as radial probability? I also know that || is called the probability density of finding an electron inside an atom but is it the same as radial distributionwhich is the probability of finding an electron in a volume $\\mathrm dV$ inside a nucleus?", "created_at_utc_A": 1520949780, "created_at_utc_B": 1520950053, "score_A": 2, "score_B": 17, "human_ref_A": "I'd say the wave function is your $$. This wavefunction describes your system. When you want to determine something like its energy or other operations you need to describe your system like your atom using a wave function. $||$ is as you already said the probability density. Your wavefunction for example in solving the hydrogen atom using the Schrdinger-equation has this step where you transform to spherical coordinates and then seperate into a radial part and an angular part. From this radial part you can get the radial probability, so to find an electron for example at the distance x. And the radial distribution is this function times the area of a sphere having the size of the radius for that distance you are looking at the moment. So this will give you the probability for the electron to be located somewhere on a sphere at the distance x. But I can't really tell you the actual difference between taking only the radial part or the wave function itself to describe probability.", "human_ref_B": "All the four terms are different and they represent different concepts in quantum mechanics.Firstly, the term $\\Psi$ represents wave function of a particle which is distributed in a three dimensional space. This wave function is a function of four coordinates ($x$, $y$, $z$, and $t$), and it gives the values which are in complex-space. For a typical example, $$\\Psi(x,y,z,t) = \\sqrt{\\frac{8}{abc}} \\sin\\left(\\frac{n_x\\pi x}{a}\\right) \\sin\\left(\\frac{n_y\\pi y}{b}\\right) \\sin\\left(\\frac{n_z\\pi z}{c}\\right) e^{-2\\pi iEt/h}$$ is an example of a wave-function. But the $|\\Psi|^2$ is mathematically defined as $\\Psi\\cdot\\Psi^*$. Max Born interpreted the value of this real valued function as the probability of finding the particle in 3 dimensional space. If you consider the previous example then $$|\\Psi(x,y,z,t)|^2 = \\frac{8}{abc} \\sin^2\\left(\\frac{n_x\\pi x}{a}\\right) \\sin^2\\left(\\frac{n_y\\pi y}{b}\\right) \\sin^2\\left(\\frac{n_z\\pi z}{c}\\right)$$ Here the complex part will not appear as in the previous example because $\\Psi$ is multiplied with its complex conjugate. The probability of finding the particle in a unit volume element $\\mathrm dV$ is $|\\Psi|^2 \\mathrm dV$. In spherical polar coordinates, it is $|\\Psi|^2 r^2 \\, \\mathrm dr \\, \\sin\\theta \\, \\mathrm d\\theta \\, \\mathrm d\\phi$. When you are only concerned about the radial part, the polar angular integral and azimuthal angular integral are replaced by $4\\pi$ as, $\\int_{0}^{2\\pi} \\int_{0}^{\\pi} \\sin\\theta \\, \\mathrm{d}\\theta \\, \\mathrm{d}\\phi =4\\pi$. thus we are left with only the radial part which is your radial probability distribution function. $$\\Pr(r) = |\\Psi|^2 4 \\pi r^2 \\, \\mathrm dr$$ The radial probability can be thought as the probability of finding the particle within an interval of length $\\text{d}r$ at $r=r_0$. So, the radial distribution is a function but the radial probability as described can be calculated by integrating that function from $0$ to $r_0$.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/92244, Response URL: https://chemistry.stackexchange.com/questions/92246, Post author username: Aura Sartori, Post author profile: https://chemistry.stackexchange.com/users/59105, Response author username: Justanotherchemist, Response author profile: https://chemistry.stackexchange.com/users/41761", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/92244, Response URL: https://chemistry.stackexchange.com/questions/92248, Post author username: Aura Sartori, Post author profile: https://chemistry.stackexchange.com/users/59105, Response author username: Soumik Das, Response author profile: https://chemistry.stackexchange.com/users/58497", "seconds_difference": 273.0, "score_ratio": 8.5, "upvote_ratio": -1.0, "c_root_id_A": "92246", "c_root_id_B": "92248"} {"post_id": "92244", "domain": "chemistry_test", "history": "What is the difference between , ||, radial probability, and radial distribution of electrons? I am very confused regarding these four terms- ,||, radial probability and radial distribution I know that is called the wave function but is it the same as radial probability? I also know that || is called the probability density of finding an electron inside an atom but is it the same as radial distributionwhich is the probability of finding an electron in a volume $\\mathrm dV$ inside a nucleus?", "created_at_utc_A": 1520950053, "created_at_utc_B": 1520949787, "score_A": 17, "score_B": 3, "human_ref_A": "All the four terms are different and they represent different concepts in quantum mechanics.Firstly, the term $\\Psi$ represents wave function of a particle which is distributed in a three dimensional space. This wave function is a function of four coordinates ($x$, $y$, $z$, and $t$), and it gives the values which are in complex-space. For a typical example, $$\\Psi(x,y,z,t) = \\sqrt{\\frac{8}{abc}} \\sin\\left(\\frac{n_x\\pi x}{a}\\right) \\sin\\left(\\frac{n_y\\pi y}{b}\\right) \\sin\\left(\\frac{n_z\\pi z}{c}\\right) e^{-2\\pi iEt/h}$$ is an example of a wave-function. But the $|\\Psi|^2$ is mathematically defined as $\\Psi\\cdot\\Psi^*$. Max Born interpreted the value of this real valued function as the probability of finding the particle in 3 dimensional space. If you consider the previous example then $$|\\Psi(x,y,z,t)|^2 = \\frac{8}{abc} \\sin^2\\left(\\frac{n_x\\pi x}{a}\\right) \\sin^2\\left(\\frac{n_y\\pi y}{b}\\right) \\sin^2\\left(\\frac{n_z\\pi z}{c}\\right)$$ Here the complex part will not appear as in the previous example because $\\Psi$ is multiplied with its complex conjugate. The probability of finding the particle in a unit volume element $\\mathrm dV$ is $|\\Psi|^2 \\mathrm dV$. In spherical polar coordinates, it is $|\\Psi|^2 r^2 \\, \\mathrm dr \\, \\sin\\theta \\, \\mathrm d\\theta \\, \\mathrm d\\phi$. When you are only concerned about the radial part, the polar angular integral and azimuthal angular integral are replaced by $4\\pi$ as, $\\int_{0}^{2\\pi} \\int_{0}^{\\pi} \\sin\\theta \\, \\mathrm{d}\\theta \\, \\mathrm{d}\\phi =4\\pi$. thus we are left with only the radial part which is your radial probability distribution function. $$\\Pr(r) = |\\Psi|^2 4 \\pi r^2 \\, \\mathrm dr$$ The radial probability can be thought as the probability of finding the particle within an interval of length $\\text{d}r$ at $r=r_0$. So, the radial distribution is a function but the radial probability as described can be calculated by integrating that function from $0$ to $r_0$.", "human_ref_B": "According to the Copenhagen interpretation of quantum mechanics, $|\\Psi|^2$ is the \"probability density\" (the probability per volume of finding a particle, such as an electron, in a given volume, in the limit the volume approaches zero). If $\\Psi(r,\\theta,\\phi)$ is separable as $R(r)Y(\\theta,\\phi)$, such as in the hydrogen atom, then: \"Radial Probability Density\" is $R^2(r)$ (still proportional to probability per volume for s-states) \"Radial Probability Distribution\" is $4\\pi r^2R^2(r)$ (probability per incremental distance from origin, as the incremental distance approaches zero)", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/92244, Response URL: https://chemistry.stackexchange.com/questions/92248, Post author username: Aura Sartori, Post author profile: https://chemistry.stackexchange.com/users/59105, Response author username: Soumik Das, Response author profile: https://chemistry.stackexchange.com/users/58497", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/92244, Response URL: https://chemistry.stackexchange.com/questions/92247, Post author username: Aura Sartori, Post author profile: https://chemistry.stackexchange.com/users/59105, Response author username: DavePhD, Response author profile: https://chemistry.stackexchange.com/users/5160", "seconds_difference": 266.0, "score_ratio": 5.666666666666667, "upvote_ratio": -1.0, "c_root_id_A": "92248", "c_root_id_B": "92247"} {"post_id": "19517", "domain": "chemistry_test", "history": "Is there a way to contain fluorine gas for long term so that it can be visually observed? Bromine, chlorine and iodine can all be sealed in a glass container for display without the elements reacting with the glass. But if you try to seal fluorine in glass I believe it will react and fog the surface making it harder to see the gas, right? What's the best way, if at all possible for long term containment of fluorine for display?", "created_at_utc_A": 1616268307, "created_at_utc_B": 1615954860, "score_A": 7, "score_B": 6, "human_ref_A": "This is not something to speculate on...it has been done. Fluorine gas was observed by Moissan, actually, (i.e. over a hundred years ago and by the man who first isolated the element). He used a platinum tube with CaF2 windows at the ends. He actually published a color image--if you have access to a first class academic library or a good file server, may be able to see it there. But here is an image from Wikipedia Commons (fluorine in the middle): Although perhaps looking in the journal hard copy would be better...then again, reproduction of observed color is a tricky deal (even nowadays, don't get me started on three wavelength color matching for car paint!). Journal cite: H. Moissan, Ann. Chim. Phys. 25 (1892) 125 Also, I'm not clear if the Moissan image was a color photo or a drawing. Image on wiki is not clear enough to tell. But in any case, it was a tough experiment then and still would be now. But if you have a good machine shop and the money...it's possible to do it again. I would advise a light source to shine through the tube. But the Pt cost and the seal of the flourite stoppers will be the hardest part. Safety note: Do it in a fume hood, please, and with proper safety precautions (limited amounts of gas, manual transfers). Fluorine gas is no joke. Don't be a \"martyr\".", "human_ref_B": "In addition to the suggestion of using metal fluorides like CaF2, I want to mention the possibility of using carbon-based fluoropolymers like polytetrafluoroethylene (PTFE), perfluoroalkoxy alkanes (PFA), and fluorinated ethylene propylene (FEP), all of which may be sold under the brand name \"Teflon\", which has become synonymous with PTFE for many people. These plastics are some of the most commonly used materials for dealing with highly reactive fluorine compounds like HF, UF6, and HSbF6, as well as other reactive compounds. All these plastics are colorless, but PTFE is more just translucent, and generally looks white, while FEP and PFA are the properly transparent ones. You can see on this chemical resistance chart that PTFE is considered \"resistant\" to dry F2 up to at least 60C, as well as everything else they have data for. PFA is apparently similar. FEP is also apparently compatible with F2 up to at least 60C: https://www.polyfluor.nl/en/chemical-resistance/fep/ That page also mentions \"better gas and vapor permeability\", though, which sounds somewhat worrying if you want to hold gaseous fluorine in it. However, I think \"better\" might actually mean lower, since this pdf of general info on FEP films from Teflon.com mentions it having exceptionally low permeability to gases and non-gases. Some rather vague comparisons of these 3 substances are given here. These polymers may be cheaper than metal fluorides (though I'm not certain of that), but they are not fully fluorinated, and thus can still react with F2, e.g.: (CF2)n(s) + nF2(g) nCF4(g) As this page by a company selling them puts it, \"[t]he extremely potent oxidisers, fluorine (F2) and related compounds (e.g., chlorine trifluoride, ClF3 ), can be handled by PTFE/PFA only with great care and recognition of potential hazards. Fluorine is absorbed into the resins, and with such intimate contact the mixture becomes sensitive to a source of ignition such as impact.\" That quote implies that teflon can burn in fluorine (as can normal oxide-based glass). This makes sense because I know it can burn in oxygen, although it does not normally burn in air, because \"teflon\" fires in the liquid oxygen tanks are believed to have been part of the cause of the Apollo 13 disaster. Also, those chemical resistance sheets I linked to list FEP, PTFE, and PFA all as \"not recommended\" for use with F2 at 100C, and that Teflon.com pdf specifically mentions that FEP reacts with \"molten alkali metals, fluorine at elevated temperatures, and certain complex halogenated compounds, such as chlorine trifluoride, at elevated temperatures and pressures.\" When combined with the fact that these fluoropolymers melt and chemically decompose into (somewhat toxic) gases at temperatures fairly typical for plastics, this means they are not very resistant to fire-like temperatures. (Thus, for example, you probably should avoid trying to close F2 in an FEP ampule by melting the tip, though maybe it would work if the F2 was in the form of a liquid far from the part you were melting. I've seen specific references to FEP being good for cryogenic temperatures and not becoming brittle, so it probably is the best for liquid F2.) (There is a question on this stack-exchange specifically about what can attack PTFE: Is there ANY chemical that can destroy PTFE, or Teflon? .) Metal fluorides, on the other hand, are much more refractory (though not as refractory as oxides). CaF2, in particular, has one of the highest melting points (my best number for the melting point is 1423C from my CRC Handbook), though ScF3, LaF3, and CeF3 at least have higher melting points (but much lower boiling points). There is also no way for CaF2 and many other metal fluorides to be further fluorinated or burned in any way. (Some can react to form higher fluorides, which may even be gases, but there's no reason to use those.) One potential problem with CaF2 and many other metal fluorides is that they have non-trivial solubilities in water. The solubility of CaF2 in water is still very low, though, (similar to CaCO3) so I'm guessing it won't dissolve unless you leave it in a large body of water or use it as plumbing for a really long time, let it be rained on for many years, or dump acid on it. (The latter is how HF is commercially produced.) CaF2 and many other metal fluorides can be very transparent if pure (fluorides and fluoride mixtures like ZBLAN are used for fiberoptics, lenses, and windows on lab equipment), and I'm not sure if that's true of fluoropolymers. SiO2, Al2O3, and some other other non-fluorine compounds are highly transparent, refractory, and insoluble in water, and also quite resistant to oxidation by F2. They can, however, still theoretically be corroded by F2: SiO2(s) + 2F2(g) SiF4(g) + O2(g) G298K,1atm = -172.55 kcal/mol for glass or -171.13 for quartz (reaction is spontaneous) 2Al2O3(s) + 6F2(g) 4AlF3(s) + 3O2(g) G298K,1atm = -303.0 kcal/mol for corrundum to crystalline AlF3 (reaction is spontaneous) Note that SiF4(g) is a gas (boiling point somewhere around -86C or -90.3C), while AlF3 is a colorless refractory solid (sublimation point 1290~1291C) that might form a protective layer on the inside of a container, similar the NiF2 layer that forms on the inside of the nickel containers commonly used to store F2 and other highly fluorinating compounds. Thus, I suspect that Al2O3 is a better option than SiO2. It's possible this advantage could also be gained with \"aluminosilicate\" glasses (made of just Al, Si, and O), and, in any case, these probably work at least as well as fused quartz (pure SiO2 glass). As I mentioned, melting an ampule containing F2 closed is not a very good idea, unless it's made of metal fluorides (and some metal fluoride glasses do have very low glass-transition temperatures), because both fluoropolymers and oxide-glasses react with F2 at high temperature, and may ignite under such conditions. Sealed ampules are also quite difficult to put things into or take things out of without breaking the seal. Other methods of making air-tight seals I know of tend to involve some kind of rubbery material (\"elastomer\", at least if a polymer). This may be used as an O-ring underneath some kind of cap or tube attatchment, or could be used to make a self-sealing membrane. I've often seen that silicones are used as these elastomers, especially for self-sealing membranes. There are fluorosilicones that may be quite resistant to F2, but normal silicone should generally be attacked by F2. Silicones all involve Si-O-Si chains, and at least almost always contain Si-C or O-C bonds, and these bonds can be attacked by fluorine. Some websurfing shows that apparently there are quite rubbery fluorocarbon elastomers, too, though. If you look at page 19 of this pdf, you can see that 3 elastomers they consider (\"PERLAST\", \"FKM (fluorocarbon)\", and \"FVMQ (fluorosilicone)\") are listed as having \"good\" (but not \"excellent\") compatibility with fluorine. (\"Good\" apparently means they swell 10~20% when used with F2, indicating something bad might be happening.) I think the first two are both fluorocarbon-based, with PERLAST being \"FFKM\", which is specifically perfluorocarbon-based, like PTFE or FEP, based on the website selling it. In any case, whoever makes the normal nickel or whatever containers F2 is sold in must have already solved this problem. (Actually, the comments on uhoh's answer point towards Halocarbon 25-5s grease as being fluorine-proof and in-fact specifically recommended for use with gaseous F2. I don't know if that's good enough by itself for a long-term seal, though.)", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/19517, Response URL: https://chemistry.stackexchange.com/questions/147821, Post author username: docscience, Post author profile: https://chemistry.stackexchange.com/users/9411, Response author username: guest, Response author profile: https://chemistry.stackexchange.com/users/106397", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/19517, Response URL: https://chemistry.stackexchange.com/questions/147669, Post author username: docscience, Post author profile: https://chemistry.stackexchange.com/users/9411, Response author username: H. H., Response author profile: https://chemistry.stackexchange.com/users/98246", "seconds_difference": 313447.0, "score_ratio": 1.1666666666666667, "upvote_ratio": -1.0, "c_root_id_A": "147821", "c_root_id_B": "147669"} {"post_id": "41675", "domain": "chemistry_test", "history": "MgCl2 acidic or neutral in water? I'm currently taking chemistry 12. On our test we were asked, when given a $0.1~\\mathrm{M}$ solution of certain compounds whether the resulting solution when added to water would be acidic, basic, or neutral. At first, I thought the solution would be neutral, because we learnt in class that alkali earth metals and chloride are spectator ions and therefore don't interact with water and pH. However, I then realized that $\\ce{Mg(OH)2}$ has low solubility in water, and I assume that adding $\\ce{MgCl2}$, would dissociate and then precipitate $\\ce{Mg(OH)2}$ from the solution, therefore the resulting solution would be acidic. My final answer was acidic. The teacher marked my answer wrong, but when questioned could only say that $\\ce{Mg}$ and $\\ce{Cl}$ are both spectators, but was unable to give direct contradictory proof against my answer. I looked elsewhere online and Wikipedia states that anhydrous $\\ce{MgCl2}$ is a Lewis acid, despite being a weak one. But anhydrous seems to mean without water, so I currently have no idea which answer is correct. I would like an explanation to why it is acidic or neutral and why the contradicting argument (both spectators vs low solubility of $\\ce{Mg(OH)2}$ ) is invalid or flawed.", "created_at_utc_A": 1499902582, "created_at_utc_B": 1449272353, "score_A": 7, "score_B": 5, "human_ref_A": "As a physics teacher turned chemistry teacher, I cannot be 100% sure, but I think the answer is getting at the complex formed with water. $$\\ce{MgCl2 -> Mg^2+ + 2Cl-}$$ $\\ce{Cl-}$ is pretty unreactive with its high electronegativity and completes outer shell of electrons (hence why $\\ce{HCl}$ for example gives almost 100% dissoscation and does not reform). Well, for the $\\ce{Mg^2+}$ on the other hand, water molecules can now use their lone pair of electrons on the oxygen atom to form a dative bond with the metal ion (this seems to happen all the time!) forming $\\ce{[Mg(H2O)6]^2+}$. If it accepts a lone pair, it is a Lewis acid (perhaps this is what your wikipedia research showed). ANYWAY, now the acidic bit... So you have an $\\ce{Mg^2+}$ ion, surrounded by 6 water molecules. Due to the high charge density around the ion (I guess this part of my thinking means that the likes of strontium, further down the period are less acidic?), it can actually help pull the electrons in the $\\ce{O-H}$ covalent bonds in the water molecule (which are already being pulled away from the H atoms due to the high electronegativity on oxygen). If the electron pair is pulled enough, it will be beyond the electrostatic force of the proton in the hydrogen atom and hence the proton will dissociate into the solution leaving behind $\\ce{OH-}$ (no longer water). This $\\ce{H+}$ (no longer $\\ce{H}$, because it loses its electron in its shared electron pair) goes into the solution, thus increasing the $\\ce{H+}$ concentration, thus decreasing the pH. SHORT ANSWER: $\\ce{Mg^2+}$ forms a complex with water - $\\ce{[Mg(H2O)6]^2+}$ The high (positive) charge density on the $\\ce{Mg}$ ion hydrolyses the surround water ligands This releases $\\ce{H+}$ (protons) into the solution, decreasing the pH slightly.", "human_ref_B": "Long story short, the ultimate cause of non-neutral pH in the solutions of some salts is hydrolysis. If a salt is made from a weak acid and strong base, then the acid anion would get partially protonated by water, like $\\ce{CH3COO- + H2O <=> CH3COOH + OH-}$, and that $\\ce{OH-}$ would be responsible for the overall alkaline pH of the resulting solution. If we have a salt of a strong acid and weak base, it is the other way around. Which acids or bases are strong and which are weak is determined by looking up their $pK_a$ or $pK_b$ in the reference books. I don't want to get deeper into this because running far ahead of your syllabus might do more harm than good. Just remember that neither $\\ce{Mg^2+}$ nor $\\ce{Cl-}$ would hydrolyze significantly, so the solution would remain neutral. The precipitation is not all that important, as hydrolysis of most common salts rarely goes all the way up to metal hydroxide. For example, $\\ce{AlCl3}$ would make the solution pretty acidic, but still produce no precipitate. Also, consider $\\ce{NH4+}$ which is prone to hydrolysis despite having soluble hydroxide. (One might argue this does not quite qualify as hydrolysis, but anyway.) If you want an example when the precipitation actually does occur, think of $\\ce{TiCl4}$. As for the spectator ions, that's a lame argument indeed. There is no such thing as \"spectator ion in general\". True, certain ions are spectators in certain reactions, but there is no ion that would always remain a spectator, no matter what the situation is. As for the Lewis acids, that's a different concept altogether, to the point that you shouldn't automatically think of \"Lewis acids\" as acids (though many of them are).", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/41675, Response URL: https://chemistry.stackexchange.com/questions/78730, Post author username: user2804925, Post author profile: https://chemistry.stackexchange.com/users/23538, Response author username: Somewhere in Between, Response author profile: https://chemistry.stackexchange.com/users/48581", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/41675, Response URL: https://chemistry.stackexchange.com/questions/41678, Post author username: user2804925, Post author profile: https://chemistry.stackexchange.com/users/23538, Response author username: Ivan Neretin, Response author profile: https://chemistry.stackexchange.com/users/19383", "seconds_difference": 50630229.0, "score_ratio": 1.4, "upvote_ratio": -1.0, "c_root_id_A": "78730", "c_root_id_B": "41678"} {"post_id": "5544", "domain": "chemistry_test", "history": "Experiment to show that air contains about 20% oxygen I remember that in my primary classed I had an experiment in science to show air contains 20% oxygen. The experiment involved taking a trough of water, lighting a candle in the middle of the trough and inverting a glass jar over the candle so that air supply for the candle is cut of. When candle extinguishes, water level in the glass jar rises. I have seen this happen. My question is why should the water raise in the jar, since burning of candle should produce an equivalent amount of carbon dioxide, so the molar amount of gas in jar is constant? Also the heat from the candle should expand the gasses in the jar, water level should go down I think.", "created_at_utc_A": 1373287586, "created_at_utc_B": 1373189803, "score_A": 7, "score_B": 6, "human_ref_A": "One problem with using combustion reactions to determine the percent of oxygen in air is the additional products, namely $\\ce{CO2}$ and water vapor. If performing a water displacement reaction, these gases add to the uncertainty and imprecision of this method. Water displacement by a combustion reaction can, however, be used if an appropriate combustion reaction is chosen. A more reasonable experiment to determine the percent oxygen in air is to use the oxidation of iron $$\\ce{4Fe(s) + 3O2(g) -> 2Fe2O3(s)}$$ The benefits of this reaction are (a) only one product is formed and (b) the product is a solid so it will not influence the water displacement in a significant way. This experiment has been reported in the Journal of Chemical Education and a brief search of the web found an adaptation of this experiment if you don't have access to the journal. Briefly, one measures the volume of a test tube, adds a known amount of steel wool into the tube and inverts the tube in a beaker of water. The oxidation of the iron is catalyzed by first rinsing it in a dilute acid (vinegar) solution. Knowing the density of iron one can calculate the volume of the test tube occupied by the solid and then determine the volume of air before and after the reaction is completed. There are a number of possible errors that can be introduced into this experiment, but unlike the combustion of a hydrocarbon, these errors are much more manageable for an undergraduate or high school chemistry experiment. Most significantly, the reaction occurs near room temperature so the temperature change of the gas can be ignored.", "human_ref_B": "The paraffin wax is an alkane ($\\ce{C_nH_{2n+2}}$); its combustion reaction is (complete combustion): $$\\ce{2C_nH_{2n+2} + $(3n+1)$ O2 -> $2(n+1)$ H2O + $2n$ CO2}$$ or (incomplete combustion): $$\\ce{2C_nH_{2n+2} + $(2n+1)$ O2 -> $2(n+1)$ H2O + $2n$ CO}$$ In the two cases the quantity (in moles) of $\\ce{O2}$ is greater than the quantity of $\\ce{CO2}$ or $\\ce{CO}$. Moreover the gaseous $\\ce{H2O}$ becomes liquid.", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/5544, Response URL: https://chemistry.stackexchange.com/questions/5548, Post author username: Gautam, Post author profile: https://chemistry.stackexchange.com/users/1946, Response author username: bobthechemist, Response author profile: https://chemistry.stackexchange.com/users/1573", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/5544, Response URL: https://chemistry.stackexchange.com/questions/5545, Post author username: Gautam, Post author profile: https://chemistry.stackexchange.com/users/1946, Response author username: user5402, Response author profile: https://chemistry.stackexchange.com/users/1899", "seconds_difference": 97783.0, "score_ratio": 1.1666666666666667, "upvote_ratio": -1.0, "c_root_id_A": "5548", "c_root_id_B": "5545"} {"post_id": "9067", "domain": "chemistry_test", "history": "What is the carbon dioxide content of a soda can or bottle? I sort of know how carbonated beverages are carbonated: a lot of $\\ce{CO2}$ gets pushed into the liquid, and the container is sealed. There are at least two things I don't know. First, how much carbon dioxide is actually dissolved in the liquid? Second, what is the resulitng partial pressure of $\\ce{CO2}$ in the headspace and the total pressure in the headspace? I'm interested in cans, plastic bottles, and glass bottles. I know from experience that there is some variation among manufacturers even for the same beverage, so I will be happy with general numbers or a good estimate.", "created_at_utc_A": 1394401933, "created_at_utc_B": 1394493952, "score_A": 3, "score_B": 4, "human_ref_A": "It depends from drink and time, according to S. Teerasong, Analytica Chimica Acta 668 (2010) 4753 for a normal Cola is about $3.1 \\frac{g_{CO_2}}{L}$ (take a look even to Glevitzky, Chem. Bull. \"POLITEHNICA\" Univ. (Timioara) Volume 50 (64), 1-2,2005 but his esteem is very high). Regarding the pressure in the headspace this is very difficult to esteem theoretically because you have to take in account not only the temperature but the whole system. :you can shake the bottle and increase the pressure dramatically.", "human_ref_B": "Through an interesting approach, authors estimate $\\ce{CO2}$ pressure inside carbonated beverages, by measuring the freezing point ($fp$) depression caused by $\\ce{CO2}$. In other words, when $\\ce{CO2}$ is dissolved in water, a solution is formed and the freezing point is lowered. The molality of $m_{\\ce{CO2}}$ in water can be obtained, according to: $$ m_{\\ce{CO2}}=\\Delta{T_{fp}}/k_{fp} $$ with $k_{fp}$ for water $=-1.86\\,^{o}C \\, kg \\, mol^{-1}$. For a particular brand of sparkling water, a value of $\\Delta{T_{fp}}$ = $-0.42^oC$ was measured, thus yielding a value for $m_{\\ce{CO2}}$ equal to $0.23$ $mol \\, kg^{-1}$. This first result allows to estimate the mass of ${\\ce{CO2}}$ dissolved in one liter of that kind of beverage. Said mass of ${\\ce{CO2}}$ is approximately equal to $10\\,g$ per liter (or $3.6\\,g$ per can content volume). Applying Henry's law, $$ P_{\\ce{CO2}}=m_{\\ce{CO2}}/k_{H} $$ with $k_{H}$ being Henry's law constant ($~0.077\\,mol\\,kg^{-1}\\,atm^{-1}$ at $0^{o}C$), one finally gets the pressure $P_{\\ce{CO2}}$ exerted by the gas over the liquid, at $0^{o}C$: $$ P_{\\ce{CO2}} = 3.0\\,\\text{atm} $$ that corresponds to a mass of about $0.1\\,g$ of ${\\ce{CO2}}$ inside the headspace volume of a can (about $15\\,ml$), mass calculated via $PV=nRT$. EDIT: for most carbonated beverages, $\\Delta{T_{fp}}$ should fall at around $-0.2^oC$ (as reported by Brooker): that would give a $P_{\\ce{CO2}} = 1.5\\,\\text{atm}$ at $0^{o}C$, that rises at about $3\\,\\text{atm}$ at $25^{o}C$.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/9067, Response URL: https://chemistry.stackexchange.com/questions/9069, Post author username: Colin McFaul, Post author profile: https://chemistry.stackexchange.com/users/323, Response author username: G M, Response author profile: https://chemistry.stackexchange.com/users/2094", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/9067, Response URL: https://chemistry.stackexchange.com/questions/9104, Post author username: Colin McFaul, Post author profile: https://chemistry.stackexchange.com/users/323, Response author username: mannaia, Response author profile: https://chemistry.stackexchange.com/users/3871", "seconds_difference": 92019.0, "score_ratio": 1.3333333333333333, "upvote_ratio": -1.0, "c_root_id_A": "9069", "c_root_id_B": "9104"} {"post_id": "87951", "domain": "chemistry_test", "history": "What kind of chemical used in a chromatography run could be denoted with the letters D.H.? A fellow translator is having trouble translating (into Russian) the mysterious D.H. abbreviation in the following passage:
Slight variations of the ratio of the mobile phase constituents or adjustments of the mobile phase flow-rate can be made occasionally, in order to provide a suitable elution times for Methyl p-hydroxybenzoate and D.H., and to meet the requirements of the system suitability tests.
If by some chance it's a widely known compound, or at least there are several chemicals that could be abbreviated thus, I'd be grateful for answers. A crapshoot, I know, but I became curious. It could be some rare chemical; in that case it would be futile to try figuring it out.", "created_at_utc_A": 1548402913, "created_at_utc_B": 1578435370, "score_A": 2, "score_B": 4, "human_ref_A": "Since the context is about chromatography and elution times it seems probably that D.H. actually stands for an abbreviation of DHBA. Dihydroxybenzoic acids are family of phenolic acids (as it can be seen here) similar to methyl 4-hydroxybenzoate. Another possibility is that D.H. stands also for Dorset-Henley liquid which is a synthetic medium for the culture of tubercoline but this seems improbable.", "human_ref_B": "Recently I had to read up on HPLC-MS technique and I encountered D.H./DH several times as an acronym for dehydrogenase, also, like in the quoted section, in the context of parabens analysis with HPLC. For example, DH is explicitly defined as an acronym for dehydrogenase in Kastner's Protein liquid chromatography [1, p. 5] or in the paper by Zimmerling et al. [2]. Other examples for the exact dehydrogenase composition include acronyms such as ADH (alcohol dehydrogenase), FDH (formate dehydrogenase), $\\small\\text{D}$-LDH ($\\small\\text{D}$-lactate dehydrogenase), 11DH (11-hydroxysteroid dehydrogenase) and many more. References Protein Liquid Chromatography; Kastner, M., Ed.; Journal of chromatography library; Elsevier: Amsterdam; New York, 2000; Vol. 61. ISBN 978-0-444-50210-0. Zimmerling, J.; Tischler, D.; Gromann, C.; Schlmann, M.; Oelschlgel, M. Characterization of Aldehyde Dehydrogenases Applying an Enzyme Assay with In Situ Formation of Phenylacetaldehydes. Appl Biochem Biotechnol 2017, 182 (3), 10951107. DOI: 10.1007/s12010-016-2384-1.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/87951, Response URL: https://chemistry.stackexchange.com/questions/108525, Post author username: CowperKettle, Post author profile: https://chemistry.stackexchange.com/users/7213, Response author username: Jojostack, Response author profile: https://chemistry.stackexchange.com/users/58093", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/87951, Response URL: https://chemistry.stackexchange.com/questions/126251, Post author username: CowperKettle, Post author profile: https://chemistry.stackexchange.com/users/7213, Response author username: andselisk, Response author profile: https://chemistry.stackexchange.com/users/41328", "seconds_difference": 30032457.0, "score_ratio": 2.0, "upvote_ratio": -1.0, "c_root_id_A": "108525", "c_root_id_B": "126251"} {"post_id": "8140", "domain": "chemistry_test", "history": "Defining and testing custom made DFT functionals? What is a good and free software for this endeavor? Any recommended articles on the subject? I want to design dft functionals, trying out different parametrizations and optimizing mixing constants for exchange and correlation contributions for specific data sets. Even add PT2 perturbation energy into the mix if possible. I am aware of Gaussian's user-defined functional options, but would like to have a bit more control in the parameters and contributions, instead of just 6 parameters.", "created_at_utc_A": 1390940025, "created_at_utc_B": 1406751076, "score_A": 3, "score_B": 6, "human_ref_A": "Since you mentioned \"free\", you maybe want to have a look at GAMESS from the Gordon group. I have really no idea whether it's fit for the task, but last time I checked it was free for academic use. There's of course Turbomole, but according to an old thread in a forum from 2007, they weren't fond of custom functionals either at that time. The statement of one of the developers might be worth reading anyway.", "human_ref_B": "If you're trying to modify the source code and run DFT jobs, then you basically have only three options: PSI4 (very powerful, written in both C and Fortran) NWChem (very powerful and fast) GAMESS (kind of fast, fully written in FORTRAN) If you would like to play with codes and tweak it as you wish, not worrying about the speed and optimization level, then you must try PyQuante. They recently added DFT and MP2 codes in it, so you it will be easy for you to play with. Needless to say, it runs slowly because the top layer is written in Python, while the inner layers are written in C and FORTRAN. I guess you will love playing with it!", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/8140, Response URL: https://chemistry.stackexchange.com/questions/8142, Post author username: beangoben, Post author profile: https://chemistry.stackexchange.com/users/4340, Response author username: Klaus-Dieter Warzecha, Response author profile: https://chemistry.stackexchange.com/users/418", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/8140, Response URL: https://chemistry.stackexchange.com/questions/14757, Post author username: beangoben, Post author profile: https://chemistry.stackexchange.com/users/4340, Response author username: user3786990, Response author profile: https://chemistry.stackexchange.com/users/6986", "seconds_difference": 15811051.0, "score_ratio": 2.0, "upvote_ratio": -1.0, "c_root_id_A": "8142", "c_root_id_B": "14757"} {"post_id": "8140", "domain": "chemistry_test", "history": "Defining and testing custom made DFT functionals? What is a good and free software for this endeavor? Any recommended articles on the subject? I want to design dft functionals, trying out different parametrizations and optimizing mixing constants for exchange and correlation contributions for specific data sets. Even add PT2 perturbation energy into the mix if possible. I am aware of Gaussian's user-defined functional options, but would like to have a bit more control in the parameters and contributions, instead of just 6 parameters.", "created_at_utc_A": 1390940025, "created_at_utc_B": 1390941806, "score_A": 3, "score_B": 14, "human_ref_A": "Since you mentioned \"free\", you maybe want to have a look at GAMESS from the Gordon group. I have really no idea whether it's fit for the task, but last time I checked it was free for academic use. There's of course Turbomole, but according to an old thread in a forum from 2007, they weren't fond of custom functionals either at that time. The statement of one of the developers might be worth reading anyway.", "human_ref_B": "Depending on what you're after, you might want to look for programs using libxc, so you can completely define your own functionals. From the libxc section of the Octopus wiki, these include:
Abinit plane-wave code APE an atomic code Atomistix ToolKit numerical orbitals code AtomPAW projector augmented wave functions generator BigDFT wavelet code CP2K - A program to perform atomistic and molecular simulations of solid state, liquid, molecular, and biological systems. DP Dielectric Properties, a linear response TDDFT code Elk FP-LAPW code ERKALE a DFT/HF molecular electronic structure code based on Gaussian orbitals exciting FP-LAPW code GPAW grid-based projector-augmented wave method JDFTx plane-wave code designed for Joint Density Functional Theory MOLGW - a small, but accurate MBPT code for molecules octopus real-space (TD)DFT code Yambo solid state and molecular physics many-body calculations code
On a more personal note, and this is just my opinion, if you're just arbitrarily mixing functionals, you should really make sure you have a scientific justification for doing this, otherwise you're just generating numbers without basis.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/8140, Response URL: https://chemistry.stackexchange.com/questions/8142, Post author username: beangoben, Post author profile: https://chemistry.stackexchange.com/users/4340, Response author username: Klaus-Dieter Warzecha, Response author profile: https://chemistry.stackexchange.com/users/418", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/8140, Response URL: https://chemistry.stackexchange.com/questions/8144, Post author username: beangoben, Post author profile: https://chemistry.stackexchange.com/users/4340, Response author username: Aesin, Response author profile: https://chemistry.stackexchange.com/users/39", "seconds_difference": 1781.0, "score_ratio": 4.666666666666667, "upvote_ratio": -1.0, "c_root_id_A": "8142", "c_root_id_B": "8144"} {"post_id": "133521", "domain": "chemistry_test", "history": "Why do the first and second laws of thermodynamics not contradict each other? We're learning about entropy right now in general chemistry, and I'm trying to understand something. From the fact that entropy can be directly compared with enthalpy, free energy, electromotive force, etc., and its being measured in units of $\\pu{J//mol*K}$ including joules, the units of energy it would seem that entropy is some form of energy. By the first law of thermodynamics, then, entropy can't be created nor destroyed, only transformed and transferred. From Gibbs' equation we see that entropy is being transformed out of the molecular potential energy, and from Clausius' equations we see that entropy can be transferred between a system and its surroundings, with $\\Delta S_\\mathrm{system}=-\\Delta S_\\mathrm{surroundings}$. Since $\\Delta S_\\mathrm{universe}=\\Delta S_\\mathrm{system}+\\Delta S_\\mathrm{surroundings}$, it follows that $\\Delta S_\\mathrm{universe}=0$ as one would expect from the first law. Yet the second law of thermodynamics states that $\\Delta S_\\mathrm{universe}>0$. If entropy is a form of energy, then how can universal entropy tend to increase? If entropy is not a form of energy, then how can it be compared with actual forms of energy and measured in energy units?", "created_at_utc_A": 1589296012, "created_at_utc_B": 1589294121, "score_A": 14, "score_B": 11, "human_ref_A": "It is something of a historical accident that entropy has units of J/K. It came out of the fact that the connection between heat, temperature, and energy was not obvious to early scientists, and so they effectively picked different units for measuring temperature and for measuring energy. In the more modern statistical interpretation of entropy, the entropy of a system is simply a number. Specifically, if the number of microstates associated with a given macrostate is $\\Omega$, then $S = k \\ln \\Omega$. The number of microstates ($\\Omega$) is just a number, without any units, and therefore so is $\\ln \\Omega$. You can see that we actually have to insert Boltzmann's constant, with its units of J/K, to make the units \"come out right\". Arguably, a more natural way to define entropy would be to just make it it a dimensionless quantity: $S = \\ln \\Omega$, without the factor of $k$.1 This would be equivalent to measuring temperature in units where $k$ is equal to 1 exactly, rather than defining our unit of temperature such that $k= 1.380649 \\times 10^{-23}$ J/K exactly. If we did this, we'd be effectively measuring temperature in units of energy as well; for example, in an ideal monatomic gas with a \"temperature of 1 J\", the average KE of each molecule would be $\\frac{3}{2}$ J. Quantities such as Helmholtz free energy would still have units of energy, since we'd still define $F = U - TS$, with $T$ having units of energy and $S$ being dimensionless. Of course, in this parallel universe where entropy is defined as a dimensionless number, another gen. chem. student would be asking why temperature is not the same thing as energy, even though they're measured in the same units. But that's another question and another answer. 1 In fact, entropy is defined exactly this way in information theory, since there's not really a notion of energy (or temperature) to speak of in such contexts.", "human_ref_B": "Incorrect assumptions
[OP] we see that entropy can be transferred between a system and its surroundings, with $\\Delta S_\\mathrm{system}=-\\Delta S_\\mathrm{surroundings}$
This equation is usually not correct, except when you have a reversible process (an ideal situation where something happens even though everything is at equilibrium). For an equilibrium, you are right: The entropy does not increase.
[OP] If entropy is a form of energy, then how can universal entropy tend to increase?
Entropy is not a form of energy. It does not even have the same dimensions. Also, there are forms of energy that increase without breaking the first law. You could have a space heater turning electrical energy into thermal energy. The first law can't be applied separately to electrical energy (\"the electrical energy in the universe is constant\" is not true).
[OP] If entropy is not a form of energy, then how can it be compared with actual forms of energy and measured in energy units?
It is not measured in energy units. The term $T \\Delta S$ is measured in energy units. Consider speed vs time and speed vs distance. They share units, but that does not mean speed is the same thing as time, or as distance. And laws about distances or time do not automatically apply to speed. Simple counter example If two bodies of different temperature are brought into thermal contact, they will reach thermal equilibrium (same temperature). The thermal energy lost by the hotter body is equal to the thermal energy gained by the colder body (first law). The entropy lost by the hotter body is less than the entropy gained by the colder body (entropy increases, second law). Why there is no contradiction Lots of quantities have the same dimensions as energy (work, heat), and there are a lot of forms of energy. The first law does not apply to any of those, just to all energy combined. So applying the first law directly to entropy does not make any sense, and entropy is not a form of energy.", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/133521, Response URL: https://chemistry.stackexchange.com/questions/133562, Post author username: DonielF, Post author profile: https://chemistry.stackexchange.com/users/54511, Response author username: Michael Seifert, Response author profile: https://chemistry.stackexchange.com/users/17515", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/133521, Response URL: https://chemistry.stackexchange.com/questions/133560, Post author username: DonielF, Post author profile: https://chemistry.stackexchange.com/users/54511, Response author username: Karsten, Response author profile: https://chemistry.stackexchange.com/users/72973", "seconds_difference": 1891.0, "score_ratio": 1.2727272727272727, "upvote_ratio": -1.0, "c_root_id_A": "133562", "c_root_id_B": "133560"} {"post_id": "84684", "domain": "chemistry_test", "history": "Magnetic moment approximation I've been reading a bit about the magnetic moment (spin-only) $\\mu_{s.o}$ where they give a formula relating this to the number of unpaired electrons $$\\mu_{s.o}=\\sqrt{n(n+2)}$$ where $n$ is the number of unpaired electrons. However in our lecture today we were using the approximation $\\mu_{s.o} \\approx n+1$. Is this an acceptable approximation for the magnetic moment or should I stick to using the previous one. Obviously using $\\mu_{s.o} \\approx n+1 $ is easier to use for calculations but I would like someone's opinion on this.", "created_at_utc_A": 1508847876, "created_at_utc_B": 1508861036, "score_A": 6, "score_B": 11, "human_ref_A": "To me this seems like a blind usage of the Taylor expansion. Let's say we want to get a Taylor approximation of $\\mu (n) = \\sqrt{n(n+2)}$. Then we know we can expand any function $f(x) = f(x_0 +h)$ if $h/x_0 \\ll 1$ into $f(x_0 + h) \\approx f(x_0) + \\frac{\\partial f}{\\partial x}(x_0) \\cdot h $. If I apply this to the function $\\mu(n) = \\mu(0 + n)$ then I get the approximate result $\\mu(n) \\approx \\frac{n+1}{\\sqrt 2}$. But here of course it is not true that $n/0 \\ll 1$, so the assumption needed for using Taylor is violated. However, if I plot both those functions in Wolframalpha they seem to agree OK enough for large n. So I guess a different approximation technique might have been used deriving this result, and it depends if you're working at large n or small n.", "human_ref_B": "I think you can come at this approximation in two ways. Using more advance methods, the approximation is obtained as a truncation of the Laurent series of $\\sqrt{x(x+2)}$ about $x=\\infty$. This is possible, but I think needlessly complex in this case. Using just algebra, we can note $$\\sqrt{n(n+2)}=\\sqrt{n^2+2n}\\approx\\sqrt{n^2+2n+1}=\\sqrt{(n+1)^2}=n+1$$ By looking at a plot, we can see this approximation is very good, giving essentially the exact result at $n=10$.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/84684, Response URL: https://chemistry.stackexchange.com/questions/84688, Post author username: Patrick Moloney, Post author profile: https://chemistry.stackexchange.com/users/37924, Response author username: AtmosphericPrisonEscape, Response author profile: https://chemistry.stackexchange.com/users/53818", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/84684, Response URL: https://chemistry.stackexchange.com/questions/84702, Post author username: Patrick Moloney, Post author profile: https://chemistry.stackexchange.com/users/37924, Response author username: Tyberius, Response author profile: https://chemistry.stackexchange.com/users/41556", "seconds_difference": 13160.0, "score_ratio": 1.8333333333333333, "upvote_ratio": -1.0, "c_root_id_A": "84688", "c_root_id_B": "84702"} {"post_id": "84684", "domain": "chemistry_test", "history": "Magnetic moment approximation I've been reading a bit about the magnetic moment (spin-only) $\\mu_{s.o}$ where they give a formula relating this to the number of unpaired electrons $$\\mu_{s.o}=\\sqrt{n(n+2)}$$ where $n$ is the number of unpaired electrons. However in our lecture today we were using the approximation $\\mu_{s.o} \\approx n+1$. Is this an acceptable approximation for the magnetic moment or should I stick to using the previous one. Obviously using $\\mu_{s.o} \\approx n+1 $ is easier to use for calculations but I would like someone's opinion on this.", "created_at_utc_A": 1508861036, "created_at_utc_B": 1508858573, "score_A": 11, "score_B": 6, "human_ref_A": "I think you can come at this approximation in two ways. Using more advance methods, the approximation is obtained as a truncation of the Laurent series of $\\sqrt{x(x+2)}$ about $x=\\infty$. This is possible, but I think needlessly complex in this case. Using just algebra, we can note $$\\sqrt{n(n+2)}=\\sqrt{n^2+2n}\\approx\\sqrt{n^2+2n+1}=\\sqrt{(n+1)^2}=n+1$$ By looking at a plot, we can see this approximation is very good, giving essentially the exact result at $n=10$.", "human_ref_B": "Using $\\sqrt{x}-\\sqrt{y} = { x -y \\over \\sqrt{x} + \\sqrt{y}} $ we have $\\sqrt{n(n+2)} - (n+1) = -{ 1\\over n \\left( \\sqrt{1+ {2 \\over n}}+1 + {1\\over n}\\right )} $, so $|\\sqrt{n(n+2)} - (n+1)| \\le {1 \\over 2n}$. The approximation is reasonable for large $n$. Addendum: To see where the approximation comes from, note that $\\sqrt{1+ {2 \\over n}}+1 + {1\\over n} \\ge 2$, hence $ {1\\over n \\left( \\sqrt{1+ {2 \\over n}}+1 + {1\\over n}\\right )} \\le {1 \\over 2n}$.", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/84684, Response URL: https://chemistry.stackexchange.com/questions/84702, Post author username: Patrick Moloney, Post author profile: https://chemistry.stackexchange.com/users/37924, Response author username: Tyberius, Response author profile: https://chemistry.stackexchange.com/users/41556", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/84684, Response URL: https://chemistry.stackexchange.com/questions/84699, Post author username: Patrick Moloney, Post author profile: https://chemistry.stackexchange.com/users/37924, Response author username: copper.hat, Response author profile: https://chemistry.stackexchange.com/users/53834", "seconds_difference": 2463.0, "score_ratio": 1.8333333333333333, "upvote_ratio": -1.0, "c_root_id_A": "84702", "c_root_id_B": "84699"} {"post_id": "158520", "domain": "chemistry_test", "history": "What is the English translation of the German term \"Aufschluss\", a method to get insoluble analytes into solution? In the analysis of solids, often the first step is to get the analyte into solution in some form. Sometimes, solids are heated in the presence of acids (acid digestion) or molten salts (sample decomposition through fusion according to this source). These methods combined are call \"Aufschluss\" in German (roughly translated unlocking). There is a joke in the qualitative analysis teaching lab that for particularly difficult samples, you use the \"Assistenten Aufschluss\", i.e. you have to cheat and beg the T.A. to tell you what is in your sample. Is there a technical term in English or in other languages?", "created_at_utc_A": 1633648971, "created_at_utc_B": 1633646871, "score_A": 10, "score_B": 9, "human_ref_A": "Digest (the result) and digestion (the process) are seen again and again (e.g., already in the abstract here, or revising the state of the art here (first page, right hand column). Speaking of this university, Freiberg digestion is a common term, too (cf. e.g., Sundermayer in ACIE (p. 228/229). If the intermediate melt is alcaline, alkali fusion is used, too. Digest (like in Reader's digest) applies well on ores and biological samples, though the concept applied on cells seems more often called lysis instead. Digestion equally applies well for French, for which Le Petit Noveau Robert (1996, print edition; a one-volume orthography dictionary for the general use, not tied to chemistry) explicitly states the chemical relevance of this noun:
2. (chim) Dissolution d'une substance dans un liquide haute temprature ou extraction de certains lements de cette substance.
which translates into
(chemistry) Dissolution of a substance in a liquid of high temperature, or extraction of certain components of this substance.)
backed nicely by the entry in the dictionary of the Academie Franaise. Voil.", "human_ref_B": "It's not a special case and there is no problem in translating the German term Aufschluss into English once the context is given, i.e. disambiguation is performed. In the absence of one there cannot be a definitive translation since the word has multiple meanings (Duden lists six) allowing for the wordplay as with your example with TA. Staying in the realm of chemistry and biology, der Aufschluss generally refers to a process of transferring to a soluble state that is convenient for processing. The most universal English term in this context would likely be dissolution, followed by more field-specific ones [1, p. 16]:
Aufschluss chem (in Lsung bringen) dissolution, digestion; disintegration, decomposition; dissociation, solubilization; lysis; fractionation; maceration; (paper) cooking, pulping.
Interestingly enough, Aufschluss hasn't been borrowed verbatim into other languages as, for example, Anschluss has been: another example of how important the choice and use of separable prefixes in German are. Reference Cole, T. C. H. Wrterbuch Der Chemie / Dictionary of Chemistry (Deutsch/English), 2., berarbeitete und erweiterte Auflage.; Springer Spektrum: Berlin, 2018. ISBN 978-3-662-56330-4.", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/158520, Response URL: https://chemistry.stackexchange.com/questions/158523, Post author username: Karsten, Post author profile: https://chemistry.stackexchange.com/users/72973, Response author username: Buttonwood, Response author profile: https://chemistry.stackexchange.com/users/1782", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/158520, Response URL: https://chemistry.stackexchange.com/questions/158522, Post author username: Karsten, Post author profile: https://chemistry.stackexchange.com/users/72973, Response author username: andselisk, Response author profile: https://chemistry.stackexchange.com/users/41328", "seconds_difference": 2100.0, "score_ratio": 1.1111111111111112, "upvote_ratio": -1.0, "c_root_id_A": "158523", "c_root_id_B": "158522"} {"post_id": "27631", "domain": "chemistry_test", "history": "Reason for the formation of azeotropes Why is it that some liquid mixtures (that exhibit positive/negative deviation from Raoult's law) form azeotropic mixtures at certain compositions? What is the physical reason behind this; are there any extra forces holding the two components together? If so, please elaborate on this.", "created_at_utc_A": 1550616707, "created_at_utc_B": 1550864830, "score_A": 6, "score_B": 7, "human_ref_A": "Before getting to the azeotropes, it's worth reminding ourselves of the relevant assumption that we make about ideal mixtures: that the intermolecular interactions have constant energy regardless of the mixture. Given the important role that intermolecular interactions have in determining the boiling point of a pure liquid, it's surprising that so many mixtures behave approximately ideally, in line with this assumption. There are two main reasons for this. First, liquids are only miscible if they have somewhat similar types of intermolecular interactions with similar magnitudes of energy. For example, water and hexane have very different intermolecular interactions, so surrounding water with hexane or hexane with water is quite unfavorable, and the two do not mix. Thus, any actual mixture has interactions between the two types of molecules that are not hugely different from the interactions within the pure liquids. Second, substances that are liquid are room temperature generally do not have especially strong intermolecular forces, so the small differences in the intermolecular interactions of two miscible liquids is usually dwarfed by the impact of the difference in molecular weight that is a major determinant of boiling point. Water is quite unusual in having such a low molecular weight and still being a liquid at room temperature, a symptom of the exceptionally strong role of intermolecular interactions in determining its boiling point. With that in mind, we can talk about azeotropes. To get a minimum boiling azeotrope, the deviation from Raoult's Law requires that the vapor pressure of the lower boiling component be reduced from what is predicted for ideal behavior and that the vapor pressure of the higher boiling component be increased. In order for that to happen, we need two liquids with substantially different intermolecular interactions whose mixing changes those interactions in opposite directions. In the example of the water:ethanol azeotrope, the water is able to increase the amount of hydrogen bonding between ethanol molecules by increasing the density of hydrogen bond donors and acceptors per unit volume. At a low water concentration, the water also does not interfere greatly with the hydrophobic interactions between the ethyl ends of the molecules. As a result, the average intermolecular interaction for the ethanol molecules is stronger than in pure ethanol, and we have the necessary decrease in the vapor pressure of ethanol. For the water, when it is at low concentration in mixture, each water molecule is surrounded by ethanol molecules, so it has fewer hydrogen bonding interactions than in pure water. As a result, its average intermolecular interactions are weaker than in pure water, and we have the necessary increase in vapor pressure. Many low molecular weight alcohols form azeotropes with water for the same reasons. When the water concentration gets below what is required for the azeotrope, enough of the alcohol molecules interact only with other alcohol molecules that the deviation from ideality decreases. For the maximum-boiling azeotropes, the requirement is that the vapor pressure of both components is decreased. This is quite common in acid:water mixtures where the degree of ionization of the acid in the mixture is higher than when it is in pure form, which means that there are more ionic interactions, which are quite strong. If the water concentration gets below what is required for the azeotrope, the amount of ionization decreases and the deviation from ideality decreases with it. When the water concentration is higher than in the azeotrope, there may be a great deal of ionization of the acid, but there is enough water that is not interacting with the acid (and conjugate base) molecules that the deviation from ideality is less.", "human_ref_B": "A solution that has a maximum or minimum vapour pressure (vs mole fraction) is called an azeotrope. The liquid is in equilibrium with the vapour and mole fractions in the liquid are the same as in the vapour at a given temperature. As the composition of liquid and vapour are the same the $dp/dx_A=0$ where $p$ is pressure and $x_A$ mole fraction of species A. (The mole fraction of species B is $x_B=1-x_A$). The cause of this effect is intermolecular interaction between species A and B. This may be hydrogen bonding but need not be since benzene/cyclohexane and chloroform/hexane and many others form an azeotrope. Other intermolecular interactions are often generally called van-der-waals interactions and are dipole-dipole and pi-pi interactions among others. In a perfect solution the total pressure $p$ of a binary mixture varies linearly with mole fraction of either component, (Raoult's law)i.e. $\\displaystyle p=x_Ap^0_A+x_Bp^0_B = x_Ap^0_A+(1-x_A)p^0_B$ where $x$ are the mole fractions and $p^0_A,p^0_B$ the partial pressures of the pure liquids. Experimentally this 'law' is found to occur only when the species A and B are very similar, such as ethylene bromine and propylene bromide, or benzene and dichlorobenzene. Clearly such mixtures will not form azeotropes. What is needed is to modify the partial pressures to account for interaction between molecules. The result of allowing pairs of molecules to interact with energy $\\Delta E$ is $$p=p^0_Ax_Ae^{(1-x_A)^2\\Delta E/k_BT}+p^0_B(1-x_A)e^{x_A^2\\Delta E/k_BT}$$ which now depends on the A-B interaction energy and predicts variations from Raoults law for energies comparable or larger than $k_BT$ where $k_B$ is the Boltzmann constant. The detailed calculation is shown below and the figure shows how the vapour pressure deviates from the idea. The numbers show different values of $\\Delta E/k_BT $. A value of zero corresponds to Raoult's law. The total pressure expressed in this way has a maximum or minimum depending on the sign and magnitude of the interaction energy. A figure showing ratio of vapour pressure vs mole fraction for species A. The numbers correspond to $\\Delta E/k_BT$. The curve with $\\Delta E/k_BT=1$ is close to that for ethanol in water. When the total pressure for species A and B is plotted a maximum or minimum is produced just as in an azeotrope. Derivation of vapour pressure curve equation. The aim is to calculate the chemical potential using statistical arguments and the interaction energy between molecules. The chemical potential is given by $\\mu=\\mu^0+RT\\ln(a)$ where $a$ is the activity, and the partial pressure $p=p^0a=p^0\\gamma x$ where $\\gamma$ is the activity coefficient and $x$ the mole fraction. The chemical potential is the partial derivative of the Gibbs energy with number of moles $n$, for species A for example $\\displaystyle\\mu_A= \\left( \\frac{\\partial G}{\\partial n_A}\\right)_{T,p,n_B} $. Notice that $T,p$ and $n_B$ are the temperature, pressure, and the number of moles of species B, all of which should be held constant. The Gibbs energy is $\\Delta G = \\Delta H-T\\Delta S = \\Delta E+p\\Delta V-T\\Delta S$ and as we shall make the volume constant $\\Delta G = \\Delta H-T\\Delta S = \\Delta E-T\\Delta S$ thus the energy and entropy changes have to be determined. A regular solution is considered to be a random mixture of type A and B molecules in which specific interactions, such as hydrogen bonding, are absent but in which there is nevertheless interaction between molecules of either type and between different types, i.e interactions of the form AA, BB and AB and these need not be equal to one another. The entropy of mixing is assumed to be that of an ideal solution. Some simplifying, but not entirely realistic, assumptions are made so that a tractable solution can be obtained and are (i) that the molecules are situated on a lattice (ii) the molecules are of approximately the same size, which will allow random mixing and no change in volume, and (iii) intermolecular forces are only between nearest neighbours. This is called the Bragg-Williams model. As we assume $z$ nearest neighbour interactions only, the intermolecular potential energy of solution A with molecules at their equilibrium positions is $U_{AA}=N_Azw_{AA}/2$ for $N_A$ molecules with average potential energy $w_{AA}$. The 1/2 arises because there are $N_Az/2$ pairs of molecules. There is a similar energy for solution B. In the mixture there are three types of pairs AA, BB and AB and the total energy of the mixture is then $$U_{AB}=N_{AA}w_{AA}+N_{BB}w_{BB}+N_{AB}w_{AB}$$ and the change in energy on mixing is $$\\Delta E_{mix}= U_{AB}-U_{AA}-U_{BB} $$ To continue the calculation $N_{AA}$ and $N_{BB}$ are needed in terms of $N_{AB}$. Each A molecule has $z$ neighbours and so $zN_A$ in total. This number of neighbours is also $N_{AB}+2N_{AA}$ producing $N_{AA}=(zN_A-N_{AB})/2$ and similarly for B, $N_{BB}=(zN_B-N_{AB})/2$. The factor of 2 arises because AA pairs are counted twice and AB pairs only once. Substituting gives $$\\Delta E_{mix} = N_{AB} \\left( w_{AB}-\\frac{ w_{AA} }{2}-\\frac{ w_{BB} }{2} \\right)$$ A statistical argument is now used to find the number $N_{AB}$. As $N = N_A+N_B $ the first random choice to pick any molecule can be made in $N$ ways, the next can be made in $N-1$ ways hence to pick two $N(N-1)/2 \\approx N^2/2$ as $N$ is very large. (The 1/2 arises are there are two ways of choosing, i.e. it does not matter which is chosen first.) This gives the number of all pairs. If, however, we must choose molecule A first and then B the number is $N_AN_B$ then the fraction of these AB pairs is $2N_AN_B/N^2$. As each of the $N$ molecules has $z$ neighbours, the total number of neighbouring pairs is $Nz/2$ thus $$N_{AB}= \\frac{zN}{2}\\frac{2N_AN_B}{N^2}=\\frac{zN_AN_B}{N}$$ The energy is now $$\\Delta E_{mix} = z\\frac{N_AN_B}{N_A+N_B} \\left( w_{AB}-\\frac{ w_{AA} }{2}-\\frac{ w_{BB} }{2} \\right)$$ Letting $n_A,\\,n_B$ be the number of moles of A and B respectively, and $N_0$ Avogadro's number and making $\\displaystyle w= w_{AB}-\\frac{ w_{AA} }{2}-\\frac{ w_{BB} }{2} $ gives $$\\Delta E_{mix} = N_0 z w\\frac{n_An_B}{n_A+n_B}$$ As we assume that the mixing is random the entropy change on mixing is that for an ideal mixture, namely $$\\Delta S_{mix} = -R\\left(n_A\\ln(x_A)+n_B\\ln(x_B)\\right)$$ where $x$ are the mole fractions. The derivation of the entropy change can be found here Deriving the entropy of mixing of a non-ideal solution. Combining equations gives $$\\Delta G_{mix} = N_0 z w\\frac{n_An_B}{n_A+n_B} -RT\\left(n_A\\ln(x_A)+n_B\\ln(x_B)\\right)$$ The free energy is $G=G_A+G_B+\\Delta G_{mix} $ and differentiating to give the chemical potential gives $$\\mu_A=\\mu_A^0 + \\left( \\frac{\\partial \\Delta G_{mix}}{\\partial n_A}\\right)_{T,p,n_B} $$ where $G_B$ differentiates to zero because $n_B$ is a constant when differentiating with $n_A$. By definition $\\partial G/\\partial n_A=\\mu^0$. Performing the differentiation $\\Delta G_{mix}$ with $n_A$ gives $$\\mu_A=\\mu_A^0 +RT\\left(\\ln(x_A)+\\frac{N_0zw}{RT}x_B^2\\right) $$ simplifying by letting $\\beta = N_0zw/RT$ gives $\\displaystyle \\mu_A=\\mu_A^0 +RT\\left(\\ln(x_A)+\\beta x_B^2\\right) $. There is a similar equation for $\\mu_B$. Since $\\mu_A=\\mu^0_A+RT\\ln(a_A)$ then $\\ln(a_A)=\\ln(x_A)+\\beta x_B^2$ and as $a_A=\\gamma_A x_A$ then $\\gamma_A=e^{\\beta x}$ and so $$p_A=p_A^0\\gamma_A x_A = p_A^0x_Ae^{\\beta x_B^2}$$ and also for $p_B$ $$p_B=p_B^0x_Be^{\\beta x_A^2}$$ as $\\beta = N_0zw/RT=zw/k_BT$ then as $zw$ is an energy this partial pressure becomes equal to $p^0_Ax_Ae^{(1-x_A)^2\\Delta E/k_BT}$ which is the equation for the partial pressure we sought. The azeotrope can be determined by calculating the derivative of the total pressure and setting it to zero, $\\left( dp/dx_A \\right)_T =0$ giving $\\displaystyle x_A^{azeo} =\\frac{1}{2} \\left( 1+ \\frac{k_BT}{\\Delta E}\\ln\\left( \\frac{p_A^0}{p_B^0} \\right) \\right)$", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/27631, Response URL: https://chemistry.stackexchange.com/questions/109769, Post author username: AADHI, Post author profile: https://chemistry.stackexchange.com/users/15090, Response author username: Andrew, Response author profile: https://chemistry.stackexchange.com/users/73692", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/27631, Response URL: https://chemistry.stackexchange.com/questions/109930, Post author username: AADHI, Post author profile: https://chemistry.stackexchange.com/users/15090, Response author username: porphyrin, Response author profile: https://chemistry.stackexchange.com/users/30424", "seconds_difference": 248123.0, "score_ratio": 1.1666666666666667, "upvote_ratio": -1.0, "c_root_id_A": "109769", "c_root_id_B": "109930"} {"post_id": "395", "domain": "chemistry_test", "history": "Why are strong acids and bases not suitable as primary standards? This is one of the questions with which I have puzzled over, and can arrive at no definite conclusion. Why are strong acids or bases, such as $\\ce{H2SO4}$,$\\ce{ HNO3}$, $\\ce{HCl}$, and $\\ce{NaOH}$ not suitable primary standards? How will their higher or lower pH affect the accuracy of the results of a titration?", "created_at_utc_A": 1337092356, "created_at_utc_B": 1337127916, "score_A": 3, "score_B": 9, "human_ref_A": "There's nothing particular about strong acids or strong bases that disqualifies them from being a primary standard, but most of them are too unstable or difficult to work with to be useful as a primary standard. A primary standard should have these qualities: High purity Stability in presence of air Absence of any water of hydration which might vary with changing humidity and temperature. Cheap Dissolves readily to produce stable solutions in solvent of choice A larger rather than smaller molar mass Now for the examples you gave: $\\ce{H2SO4}$ is hygroscopic, so it's concentration will change quite readily in air. $\\ce{HNO3}$ is very hygroscopic as well. The $\\ce{NO3-}$ anions can also start redoxing other things which isn't particularly useful for titrations. $\\ce{HCl}$ is a gas, so it will be difficult to determine the concentration of the solutions (you'd have to compare it against another primary standard), and high-concentration solutions are unstable. $\\ce{NaOH}$ is hygroscopic as well. As Georg pointed out below, one generally needs high-purity solid compounds so they can be weighed accurately (and in addition to the cons above, these substances are difficult to find/weigh in solid form). I hope this helps.", "human_ref_B": "strong Bases in general are not suitable as they will react with $\\ce{CO2}$ from the air. This is actually more problematic than hygroscopicity, as it means that you also cannot store diluted $\\ce{NaOH}$ solution (openly) as standard. Instead, $\\ce{Na2CO3}$ or $\\ce{KHCO3}$ are much better alternatives (for strong acids). The acids are either a gas ($\\ce{HCl}$) or are produced from gas. I don't see how you can produce a conentration reliably enough for use as primary standard. even if you would manage to produce saturated $\\ce{HCl}$, it's much more difficult and less reliable than known alternatives (see below). but that wouldn't work e.g. for $\\ce{H2SO4}$ as $\\ce{SO3}$ is soluble in $\\ce{H2SO4}$ (Oleum). $\\ce{HNO3}$: redox reactions are not only not particularly useful, but actually harmful as they use up acid equivalents: $\\ce{NO3- + 2e- + 2 H+ -> NO2- + H2O}$ also here, a solid alternative (KH-Phthalate) is available. If you don't have that, you may get away with oxalic acid which is still better than the acids suggested here.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/395, Response URL: https://chemistry.stackexchange.com/questions/399, Post author username: Bidella, Post author profile: https://chemistry.stackexchange.com/users/94, Response author username: LeakyBattery, Response author profile: https://chemistry.stackexchange.com/users/21", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/395, Response URL: https://chemistry.stackexchange.com/questions/406, Post author username: Bidella, Post author profile: https://chemistry.stackexchange.com/users/94, Response author username: cbeleites unhappy with SX, Response author profile: https://chemistry.stackexchange.com/users/160", "seconds_difference": 35560.0, "score_ratio": 3.0, "upvote_ratio": -1.0, "c_root_id_A": "399", "c_root_id_B": "406"} {"post_id": "10465", "domain": "chemistry_test", "history": "How can water catalyse a reaction between iodine and aluminium? A few drops of water can initiate a reaction between iodine and aluminium. https://www.youtube.com/watch?v=SKSU72-1ERc How can this happen, since iodine is only slightly soluble in water?", "created_at_utc_A": 1399241134, "created_at_utc_B": 1399238945, "score_A": 12, "score_B": 6, "human_ref_A": "A naked aluminum surface (freshly sanded) in air immediately grows about 4nm of adherent aluminum oxide that passivates the surface. Iodine slightly dissolves in water and disproportionates $$\\ce{I2_{(s)} + H2O_{(l)} -> HOI_{(aq)} + I^{-}_{(aq)} + H+_{(aq)}}$$ Iodine is strongly solubilized by forming triiodide. $$\\ce{I2 + I- -> [I3]-}$$ The acids chew through the alumina surface barrier. When bare metal and triiodide meet, everything takes off, also igniting the bare metal in air. http://www2.ucdsb.on.ca/tiss/stretton/database/inorganic_thermo.htm $$\\begin{alignat}{2} &\\ce{AlI3_{(s)}}\\quad &&\\Delta G_\\text{f}^\\circ = -300.8\\ \\mathrm{kJ/mol}\\\\ &\\ce{Al2O3_{(s)}}\\quad &&\\Delta G_\\text{f}^\\circ = -1582.4\\ \\mathrm{kJ/mol}\\\\ \\end{alignat}$$ Absent water the iodine must diffuse through the intact alumina barrier, react with naked aluminum, then swell and disrupt the surface passivation from underneath.", "human_ref_B": "It is very difficult to have a solid react with another solid due to the low contact area and the inability of the molecules to obtain the activation energy. Even though there is little iodine dissolved in the water, the contact area is greatly increased. Once the reaction starts, the heat of the reaction provides the activation energy and the gaseous iodine (which makes the contact area large).", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/10465, Response URL: https://chemistry.stackexchange.com/questions/10467, Post author username: Paul Richards, Post author profile: https://chemistry.stackexchange.com/users/5410, Response author username: Uncle Al, Response author profile: https://chemistry.stackexchange.com/users/4296", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/10465, Response URL: https://chemistry.stackexchange.com/questions/10466, Post author username: Paul Richards, Post author profile: https://chemistry.stackexchange.com/users/5410, Response author username: LDC3, Response author profile: https://chemistry.stackexchange.com/users/4934", "seconds_difference": 2189.0, "score_ratio": 2.0, "upvote_ratio": -1.0, "c_root_id_A": "10467", "c_root_id_B": "10466"} {"post_id": "41140", "domain": "chemistry_test", "history": "Does Gibbs free energy of formation always equal zero for elements in standard state? I was looking through the appendix in my chemistry textbook when I noticed that every compound that had a $\\Delta H_f$ of zero also had a $\\Delta G_f$ of zero. So of course the compounds that have a $\\Delta H_f$ of zero are the elements in their standard states but why does that make their $\\Delta G_f$ zero? Does every compound that has a $\\Delta H_f$ of zero also have a $\\Delta G_f$ of zero? What exactly is $\\Delta G_f$?", "created_at_utc_A": 1448292458, "created_at_utc_B": 1448316342, "score_A": 2, "score_B": 4, "human_ref_A": "If you look more closely through your textbook, you will find it pointed out that the heats of formation and free energies of formation of all elements in their natural states at 298 and 1 bar are taken to be zero by convention. This in no way limits your ability to use these tables to determine the changes in enthalpy and free energy for arbitrary chemical reactions involving compounds comprised of these elements.", "human_ref_B": "
Does Gibbs free energy of formation always equal zero for elements in standard state?
No. Elements occur in different allotropes. $\\Delta H_f^o$ and $\\Delta G_f^o$ are defined to be zero at 298K, 1 bar for the lowest energy allotrope, with the exception that the values for white phosphorous are defined to be zero even though it is not the lowest energy allotrope. The other allotropes do not have zero $\\Delta H_f^o$ and $\\Delta G_f^o$ even though they are still elements.
So of course the compounds that have a $\\Delta H_f$ of zero are the elements in their standard states but why does that make their $\\Delta G_f$ zero?
$\\Delta H_f = 0$ does not imply $\\Delta G_f = 0$ The two values are independently defined to be zero for a particular allotrope as explained above.
Does every compound that has a $\\Delta H_f$ of zero also have a $\\Delta G_f$ of zero?
If a compound (other than by definition) happens to have $\\Delta H_f = 0$, then $\\Delta G_f$ could be any value, it is not necessarily zero.
What exactly is $\\Delta G_f$
It is the change in Gibbs energy upon forming a compound in its standard state from its constituent elements in their standard states at 298K.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/41140, Response URL: https://chemistry.stackexchange.com/questions/41152, Post author username: Lubed Up Slug, Post author profile: https://chemistry.stackexchange.com/users/22517, Response author username: Chet Miller, Response author profile: https://chemistry.stackexchange.com/users/15095", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/41140, Response URL: https://chemistry.stackexchange.com/questions/41175, Post author username: Lubed Up Slug, Post author profile: https://chemistry.stackexchange.com/users/22517, Response author username: DavePhD, Response author profile: https://chemistry.stackexchange.com/users/5160", "seconds_difference": 23884.0, "score_ratio": 2.0, "upvote_ratio": -1.0, "c_root_id_A": "41152", "c_root_id_B": "41175"} {"post_id": "4033", "domain": "chemistry_test", "history": "Can flexible plastics have reflective coating (vacuum metalized)? Is it possible to achieve an optical mirror coating to a flexible plastic part? If vacuum metalization is possible, would the reflective coating layer be durable enough to withstand the bending of a flexible plastic such as polyurethane or a thermoplastic elastomer, without cracking over time?", "created_at_utc_A": 1426060616, "created_at_utc_B": 1426261359, "score_A": 2, "score_B": 6, "human_ref_A": "I would expect that gilding - classic manual gold plating using gold leaf - works. (But as the frequency, amplitude and radius of flexing is not known, it's hard to tell) It would be starting with the polished polymer surface. That is coated with a special glue, the gold leaf is just layed down on it, and the gold surface is polished/burnished to create the mirror surface. The difficult part may be finding the right glue substance, which depends on environment conditions. But a standard product should work. So - why do I propose this ancient method? I see the advantage in using an actual (microscopic) compact layer of sheet metal. Gold leaf is just a thin gold foil. But there is a special property in gold: it's very ductile. It's so ductile, that gold leaf can actually be produced roughly by taking a piece of gold and hammering it flat. Now, I would think that the gold layer will not have much problems with the flexing polymer under it. Another important property of gold here is the innertness of the metal, because we have the mirror layer directly exposed to air - not behind glass like in \"consumer mirrors\". With silver, that would be a problem Gold makes a very good mirror for many wavelengths - you did not say which you are interested in. But if the yellow color tint of reflected visible light is a problem, one could try the same with other metals. No other metal has the level of ductility of gold, so they are possibly less suitable. What makes me think that creating a sheet of metal first, and putting it on the surface in a second step is that any kind of metal plating where the metal is deposited on the surface will follow any roughness of the surface, which will cause irregular thickness and mechanical strength. On the other hand, the layer following the contours has a larger area, when looking closely, which shoud help with mechanical stretching. The typical thikness of deposited metas ist thinner than a foil. It can be made as thick for sure, but this could make coating expensive in terms of processing time.", "human_ref_B": "There are many techniques for metal-coated polymers, as linked by permeakra Generally, vacuum deposition is an issue because the metal is hot. Cold deposition techniques (e.g., sputtering, ink coating) often work better. Common, commercially-available materials include: Pyralux - copper-coated polyimide from DuPont. Mylar - DuPont's name for biaxially-oriented polyethylene terephthalate (BoPET) As mentioned in other answers, adhesion is an issue, and to my knowledge, the processes to bind Pyralux and metabolized Mylar are proprietary. Chemistry.About.com does mention vapor deposition of metals into Mylar, though.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/4033, Response URL: https://chemistry.stackexchange.com/questions/27176, Post author username: Edan, Post author profile: https://chemistry.stackexchange.com/users/1150, Response author username: Volker Siegel, Response author profile: https://chemistry.stackexchange.com/users/5078", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/4033, Response URL: https://chemistry.stackexchange.com/questions/27291, Post author username: Edan, Post author profile: https://chemistry.stackexchange.com/users/1150, Response author username: Geoff Hutchison, Response author profile: https://chemistry.stackexchange.com/users/5017", "seconds_difference": 200743.0, "score_ratio": 3.0, "upvote_ratio": -1.0, "c_root_id_A": "27176", "c_root_id_B": "27291"} {"post_id": "4033", "domain": "chemistry_test", "history": "Can flexible plastics have reflective coating (vacuum metalized)? Is it possible to achieve an optical mirror coating to a flexible plastic part? If vacuum metalization is possible, would the reflective coating layer be durable enough to withstand the bending of a flexible plastic such as polyurethane or a thermoplastic elastomer, without cracking over time?", "created_at_utc_A": 1380408527, "created_at_utc_B": 1426261359, "score_A": 3, "score_B": 6, "human_ref_A": "It might be possible to deposit a metallic film on a sheet of plastic. It is however, a problem with respect to the bonding of that layer. Depending on the plastic used, it might not have a strong bond with the metal, which would then over time develop defects. On the other hand, it might be possible, when carefully manufacturing this, that the plastic has 'functional groups' (parts of the molecule with a specific function) that would bind a metal. Note however, that in this case, the functional groups might also react with other chemicals. Almost no functional group will only under go one reaction, especially not when metals are involved. Therefore, this would not be the most durable of options. Furthermore, because of the interface between, most likely, an amorphous solid (wikipedia, amorphous solids) and a metal structure is bound to create some defects. A last option I can think of, is the use of conductive polymers. These polymers, because in many ways they mimic metals, also have a similar 'shininess' to them. However, I'm not sure about their really being a mirror. Hope this helped.", "human_ref_B": "There are many techniques for metal-coated polymers, as linked by permeakra Generally, vacuum deposition is an issue because the metal is hot. Cold deposition techniques (e.g., sputtering, ink coating) often work better. Common, commercially-available materials include: Pyralux - copper-coated polyimide from DuPont. Mylar - DuPont's name for biaxially-oriented polyethylene terephthalate (BoPET) As mentioned in other answers, adhesion is an issue, and to my knowledge, the processes to bind Pyralux and metabolized Mylar are proprietary. Chemistry.About.com does mention vapor deposition of metals into Mylar, though.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/4033, Response URL: https://chemistry.stackexchange.com/questions/6361, Post author username: Edan, Post author profile: https://chemistry.stackexchange.com/users/1150, Response author username: Eljee, Response author profile: https://chemistry.stackexchange.com/users/2359", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/4033, Response URL: https://chemistry.stackexchange.com/questions/27291, Post author username: Edan, Post author profile: https://chemistry.stackexchange.com/users/1150, Response author username: Geoff Hutchison, Response author profile: https://chemistry.stackexchange.com/users/5017", "seconds_difference": 45852832.0, "score_ratio": 2.0, "upvote_ratio": -1.0, "c_root_id_A": "6361", "c_root_id_B": "27291"} {"post_id": "32699", "domain": "chemistry_test", "history": "After completing an experiment, a student finds that the percentage yield exceeded 100%. What could have occurred to the sample to lead to this? I am looking for a general answer; nothing too detailed. And no. This is not a homework question. It happened to me in lab and I am wondering what error I could have committed.", "created_at_utc_A": 1433956958, "created_at_utc_B": 1433950038, "score_A": 9, "score_B": 4, "human_ref_A": "Preface: The problem is that you have overstated your percentage yield, and the symptom is that it is above 100%. Example: You and your friend each do an experiment where the literature states the yield should be ~65%. You both use poor technique and fail to dry your product, causing you to attribute water mass as product mass. Your friend's yield is 101%, and yours is 99%. Both you and your friend are wrong, for the same reason! Don't make the mistake of stopping your analysis when percentage yield drops below 100%. Don't chastise your friend for having an \"impossible\" yield while patting yourself on the back for your \"99%\"! Overstatement of percentage yield happens for the following reasons: Zero) Your model of the reaction chemistry is just wrong. Unlikely in an undergraduate setting, but worth mentioning. 1) Your math is wrong. Worth double checking and easy to rule out as a source of error. 2) You used more reactant than you thought. Most errors in reactant measurement cause you to use less reactant than recorded (e.g. transfer errors, impure reactants), but you could possibly have used more than you thought. Beyond simple measurement errors, perhaps your limiting reagent was provided to you in solution, and that solution was of higher concentration than reported. Put a bit of time into thinking about this, but [especially in educational experiments] protocols usually minimize the chance of this kind of error. 3) You actually have less product than you think. This is the most likely culprit. You have to be really honest with yourself here. In most experiments, you take the mass of something at the end, and call it the mass of your product. Is it really your product? Or does that mass contain impurities which cause you to misrepresent impurities as product. Likely culprits are solvents (e.g. water from improper drying), side-products, and/or unreacted reagent. Have you really done your best to filter and remove them? If not, your yield will be inflated.", "human_ref_B": "Imagine a reaction where your starting material is cleaved and the product has a lower molecular weight. If the conversion is incomplete and the product contaminated with some starting material, the resulting weight may suggest a yield higher than 100%.", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/32699, Response URL: https://chemistry.stackexchange.com/questions/32711, Post author username: Ordinary Owl, Post author profile: https://chemistry.stackexchange.com/users/16852, Response author username: brian_o, Response author profile: https://chemistry.stackexchange.com/users/16857", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/32699, Response URL: https://chemistry.stackexchange.com/questions/32704, Post author username: Ordinary Owl, Post author profile: https://chemistry.stackexchange.com/users/16852, Response author username: Klaus-Dieter Warzecha, Response author profile: https://chemistry.stackexchange.com/users/418", "seconds_difference": 6920.0, "score_ratio": 2.25, "upvote_ratio": -1.0, "c_root_id_A": "32711", "c_root_id_B": "32704"} {"post_id": "32699", "domain": "chemistry_test", "history": "After completing an experiment, a student finds that the percentage yield exceeded 100%. What could have occurred to the sample to lead to this? I am looking for a general answer; nothing too detailed. And no. This is not a homework question. It happened to me in lab and I am wondering what error I could have committed.", "created_at_utc_A": 1433956958, "created_at_utc_B": 1433946442, "score_A": 9, "score_B": 3, "human_ref_A": "Preface: The problem is that you have overstated your percentage yield, and the symptom is that it is above 100%. Example: You and your friend each do an experiment where the literature states the yield should be ~65%. You both use poor technique and fail to dry your product, causing you to attribute water mass as product mass. Your friend's yield is 101%, and yours is 99%. Both you and your friend are wrong, for the same reason! Don't make the mistake of stopping your analysis when percentage yield drops below 100%. Don't chastise your friend for having an \"impossible\" yield while patting yourself on the back for your \"99%\"! Overstatement of percentage yield happens for the following reasons: Zero) Your model of the reaction chemistry is just wrong. Unlikely in an undergraduate setting, but worth mentioning. 1) Your math is wrong. Worth double checking and easy to rule out as a source of error. 2) You used more reactant than you thought. Most errors in reactant measurement cause you to use less reactant than recorded (e.g. transfer errors, impure reactants), but you could possibly have used more than you thought. Beyond simple measurement errors, perhaps your limiting reagent was provided to you in solution, and that solution was of higher concentration than reported. Put a bit of time into thinking about this, but [especially in educational experiments] protocols usually minimize the chance of this kind of error. 3) You actually have less product than you think. This is the most likely culprit. You have to be really honest with yourself here. In most experiments, you take the mass of something at the end, and call it the mass of your product. Is it really your product? Or does that mass contain impurities which cause you to misrepresent impurities as product. Likely culprits are solvents (e.g. water from improper drying), side-products, and/or unreacted reagent. Have you really done your best to filter and remove them? If not, your yield will be inflated.", "human_ref_B": "Yields above 100% are obviously impossible, so either you used more educts than you thought, or you don't have as much product as you think. An error when measuring the amount of educts for the reaction is one possible reason. A miscalculation at any point of the amounts needed or the final yield is also something you might want to check again. A really common reason is the presence of other material in your product, often solvents like water are still present if you weren't careful enough when drying. This is really the most likely cause, especially in student labs. You should think about the analytics you performed, and check if they show the presence of anything aside from your product. And then think about what kinds of substances could be in there (solvents, educts, other reaction products) that wouldn't be visible in the kind of analytics you performed.", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/32699, Response URL: https://chemistry.stackexchange.com/questions/32711, Post author username: Ordinary Owl, Post author profile: https://chemistry.stackexchange.com/users/16852, Response author username: brian_o, Response author profile: https://chemistry.stackexchange.com/users/16857", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/32699, Response URL: https://chemistry.stackexchange.com/questions/32701, Post author username: Ordinary Owl, Post author profile: https://chemistry.stackexchange.com/users/16852, Response author username: Mad Scientist, Response author profile: https://chemistry.stackexchange.com/users/7", "seconds_difference": 10516.0, "score_ratio": 3.0, "upvote_ratio": -1.0, "c_root_id_A": "32711", "c_root_id_B": "32701"} {"post_id": "32699", "domain": "chemistry_test", "history": "After completing an experiment, a student finds that the percentage yield exceeded 100%. What could have occurred to the sample to lead to this? I am looking for a general answer; nothing too detailed. And no. This is not a homework question. It happened to me in lab and I am wondering what error I could have committed.", "created_at_utc_A": 1433956958, "created_at_utc_B": 1433946748, "score_A": 9, "score_B": 3, "human_ref_A": "Preface: The problem is that you have overstated your percentage yield, and the symptom is that it is above 100%. Example: You and your friend each do an experiment where the literature states the yield should be ~65%. You both use poor technique and fail to dry your product, causing you to attribute water mass as product mass. Your friend's yield is 101%, and yours is 99%. Both you and your friend are wrong, for the same reason! Don't make the mistake of stopping your analysis when percentage yield drops below 100%. Don't chastise your friend for having an \"impossible\" yield while patting yourself on the back for your \"99%\"! Overstatement of percentage yield happens for the following reasons: Zero) Your model of the reaction chemistry is just wrong. Unlikely in an undergraduate setting, but worth mentioning. 1) Your math is wrong. Worth double checking and easy to rule out as a source of error. 2) You used more reactant than you thought. Most errors in reactant measurement cause you to use less reactant than recorded (e.g. transfer errors, impure reactants), but you could possibly have used more than you thought. Beyond simple measurement errors, perhaps your limiting reagent was provided to you in solution, and that solution was of higher concentration than reported. Put a bit of time into thinking about this, but [especially in educational experiments] protocols usually minimize the chance of this kind of error. 3) You actually have less product than you think. This is the most likely culprit. You have to be really honest with yourself here. In most experiments, you take the mass of something at the end, and call it the mass of your product. Is it really your product? Or does that mass contain impurities which cause you to misrepresent impurities as product. Likely culprits are solvents (e.g. water from improper drying), side-products, and/or unreacted reagent. Have you really done your best to filter and remove them? If not, your yield will be inflated.", "human_ref_B": "Percentage yield exceeds 100% if impurities are present due to inadequate purification. Such impurities could be present for many reasons, such as if the vessels used to collect the product are contaminated. Also, it is possible that the product absorbs components from air like water vapor or carbon dioxide.", "labels": 1, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/32699, Response URL: https://chemistry.stackexchange.com/questions/32711, Post author username: Ordinary Owl, Post author profile: https://chemistry.stackexchange.com/users/16852, Response author username: brian_o, Response author profile: https://chemistry.stackexchange.com/users/16857", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/32699, Response URL: https://chemistry.stackexchange.com/questions/32702, Post author username: Ordinary Owl, Post author profile: https://chemistry.stackexchange.com/users/16852, Response author username: Prafulla , Response author profile: https://chemistry.stackexchange.com/users/16804", "seconds_difference": 10210.0, "score_ratio": 3.0, "upvote_ratio": -1.0, "c_root_id_A": "32711", "c_root_id_B": "32702"} {"post_id": "32699", "domain": "chemistry_test", "history": "After completing an experiment, a student finds that the percentage yield exceeded 100%. What could have occurred to the sample to lead to this? I am looking for a general answer; nothing too detailed. And no. This is not a homework question. It happened to me in lab and I am wondering what error I could have committed.", "created_at_utc_A": 1433946442, "created_at_utc_B": 1433950038, "score_A": 3, "score_B": 4, "human_ref_A": "Yields above 100% are obviously impossible, so either you used more educts than you thought, or you don't have as much product as you think. An error when measuring the amount of educts for the reaction is one possible reason. A miscalculation at any point of the amounts needed or the final yield is also something you might want to check again. A really common reason is the presence of other material in your product, often solvents like water are still present if you weren't careful enough when drying. This is really the most likely cause, especially in student labs. You should think about the analytics you performed, and check if they show the presence of anything aside from your product. And then think about what kinds of substances could be in there (solvents, educts, other reaction products) that wouldn't be visible in the kind of analytics you performed.", "human_ref_B": "Imagine a reaction where your starting material is cleaved and the product has a lower molecular weight. If the conversion is incomplete and the product contaminated with some starting material, the resulting weight may suggest a yield higher than 100%.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/32699, Response URL: https://chemistry.stackexchange.com/questions/32701, Post author username: Ordinary Owl, Post author profile: https://chemistry.stackexchange.com/users/16852, Response author username: Mad Scientist, Response author profile: https://chemistry.stackexchange.com/users/7", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/32699, Response URL: https://chemistry.stackexchange.com/questions/32704, Post author username: Ordinary Owl, Post author profile: https://chemistry.stackexchange.com/users/16852, Response author username: Klaus-Dieter Warzecha, Response author profile: https://chemistry.stackexchange.com/users/418", "seconds_difference": 3596.0, "score_ratio": 1.3333333333333333, "upvote_ratio": -1.0, "c_root_id_A": "32701", "c_root_id_B": "32704"} {"post_id": "32699", "domain": "chemistry_test", "history": "After completing an experiment, a student finds that the percentage yield exceeded 100%. What could have occurred to the sample to lead to this? I am looking for a general answer; nothing too detailed. And no. This is not a homework question. It happened to me in lab and I am wondering what error I could have committed.", "created_at_utc_A": 1433946748, "created_at_utc_B": 1433950038, "score_A": 3, "score_B": 4, "human_ref_A": "Percentage yield exceeds 100% if impurities are present due to inadequate purification. Such impurities could be present for many reasons, such as if the vessels used to collect the product are contaminated. Also, it is possible that the product absorbs components from air like water vapor or carbon dioxide.", "human_ref_B": "Imagine a reaction where your starting material is cleaved and the product has a lower molecular weight. If the conversion is incomplete and the product contaminated with some starting material, the resulting weight may suggest a yield higher than 100%.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/32699, Response URL: https://chemistry.stackexchange.com/questions/32702, Post author username: Ordinary Owl, Post author profile: https://chemistry.stackexchange.com/users/16852, Response author username: Prafulla , Response author profile: https://chemistry.stackexchange.com/users/16804", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/32699, Response URL: https://chemistry.stackexchange.com/questions/32704, Post author username: Ordinary Owl, Post author profile: https://chemistry.stackexchange.com/users/16852, Response author username: Klaus-Dieter Warzecha, Response author profile: https://chemistry.stackexchange.com/users/418", "seconds_difference": 3290.0, "score_ratio": 1.3333333333333333, "upvote_ratio": -1.0, "c_root_id_A": "32702", "c_root_id_B": "32704"} {"post_id": "84811", "domain": "chemistry_test", "history": "HOMO Odd, LUMO Even: Why does the affected orbital change depending on the carbon? How are carbons numbered? While looking at how substituents affect the absorption spectra of molecules, I came across this:
Substituents alter the emitted wavelengths of these molecules by affecting the electron densities. Electron Withdrawing Groups (or EWGs) withdraw electron density, thereby making the arrangement more stable. The energy level of the orbital affected subsequently lowers, causing a typically-noticeable red shift in the maximum wavelength. Alternately, Electron Donating Groups (EDGs, for short), add to the electron densities of the molecule. This causes the affected orbital to be less stable, and requires more energy to obtain. The orbital affected depends on the placement of the substituent. A useful mnemonic device in this matter is HOLE: HOMO Odd, LUMO Even. This explains that the affected orbital alternates between odd and even placements.
Why does the orbital affected depend on the placement of the substituent? I thought substituents that caused bathochromic shifts just extended conjugation and hence reduced the HOMO-LUMO gap? Is this odd-even rule only applicable to azulene, or can it be applied to other chromophores? Also, why are the carbons numbered the way they are? Here's a picture with the numbering. Shouldn't the carbons with the question mark also be counted?", "created_at_utc_A": 1509028944, "created_at_utc_B": 1509029961, "score_A": 3, "score_B": 6, "human_ref_A": "This numbering is the one used for azulene alone (see http://www.acdlabs.com/iupac/nomenclature/79/r79_72.htm ). Carbon atoms which are part of a \"bridge\" (as are the carbons denoted by \"???\") are actually numbered \"3a\" (between atoms 3 and 4) and \"8a\" (between atoms 8 and 1). This is the standard numbering for those compounds, aka \"fused polycyclic hydrocarbons\".", "human_ref_B": "To start with, I'll address the frontier orbitals. Lets look at what the HOMO and LUMO look like. Notice the LUMO has no density on carbons $1$ or $3$ and (while it is a little difficult to see in this image) little density on carbons $5$ and $7$. Similarly the HOMO has little to no density on any of the even carbons. So, if we want to affect the energy of these orbitals, we have to add a substituent to a carbon that has electron density from that orbital. The odd-trend is likely not a general rule, but if you know the HOMO and LUMO reside mainly on different carbons, you can assume that selective functionalization of those carbons will affect the orbital of interest. I say \"assume\" because you should always keep in mind that the orbital picture is an approximation and that once you make a substitution you have to check to make sure you haven't perturbed the system enough where your new compound doesn't have frontier orbitals that look like this anymore. In regards to the numbering, this is really just a matter of convenience for a fused ring system like this. We don't number these bridging carbons because they can't be functionalized; that is, we can't add a group to these carbons without turning the molecule into something other than azulene. So the numbering starts from next to these bridging carbons and skips over them as you move around the rings.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/84811, Response URL: https://chemistry.stackexchange.com/questions/84819, Post author username: oswinso, Post author profile: https://chemistry.stackexchange.com/users/53593, Response author username: SteffX, Response author profile: https://chemistry.stackexchange.com/users/32114", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/84811, Response URL: https://chemistry.stackexchange.com/questions/84821, Post author username: oswinso, Post author profile: https://chemistry.stackexchange.com/users/53593, Response author username: Tyberius, Response author profile: https://chemistry.stackexchange.com/users/41556", "seconds_difference": 1017.0, "score_ratio": 2.0, "upvote_ratio": -1.0, "c_root_id_A": "84819", "c_root_id_B": "84821"} {"post_id": "18641", "domain": "chemistry_test", "history": "What is the entropy of mixing of two ideal gases starting out with different pressures? The entropy of mixing of ideal gas is given by this equation: $\\Delta S_{mix} = -nR(x_1\\ln x_1 + x_2\\ln x_2)$ Does this equation works only when the initial conditions of both compartments are at the same pressure? If ideal gas 1 at initial conditions $n_1,P_1,V_1,T$ and ideal gas 2 at $n_2,P_2,V_2,T$ are mixed, which equation should I use to obtain $\\Delta S_{mix}$ given that $P_1 P_2$? Assume $V_1 + V_2 = V_{tot}$, and $T_1 = T_2 = T$.", "created_at_utc_A": 1614062104, "created_at_utc_B": 1614152705, "score_A": 2, "score_B": 3, "human_ref_A": "There's a simple way to solve this: We know that, for an ideal gas at constant T: $$\\Delta S = -nR \\ln \\frac{p_f}{p_i}= nR \\ln \\frac{p_i}{p_f}$$ And since ideal gases ignore each other, we can calculate the entropy change for gas 1 and gas 2 separately, and simply sum the two. [At any given $T$, the chemical potential of an ideal gas is determined only by its own partial pressure, which is independent of the presence of any other gas(es).] Thus: $$\\Delta S = \\Delta S_1+ \\Delta S_2 = n_1R \\ln \\left( \\frac{p_{1, initial}}{p_{1, final}} \\right )+n_2R\\ln \\left( \\frac{p_{2, initial}}{p_{2, final}} \\right),$$ where $p_{1, final}$ and $p_{2, final}$ are the respective final partial pressures of gases 1 and 2. If we don't know the pressures, we can substitute the following into the above expression: $$p_{1, initial} = \\frac{n_1 R T}{V_1}$$ $$p_{1, final} = \\frac{n_1 R T}{(V_1+V_2)}$$ $$p_{2, initial} = \\frac{n_2 R T}{V_2}$$ $$p_{2, final} = \\frac{n_2 R T}{(V_1+V_2)}$$ After simplifying, this gives us the following: $$\\Delta S = n_1R \\ln \\left( \\frac{V_1+V_2}{V_1} \\right )+n_2R\\ln \\left(\\frac{V_1+V_2}{V_2} \\right)=n_1R \\ln \\left( \\frac{V_{total}}{V_1} \\right )+n_2R\\ln \\left(\\frac{V_{total}}{V_2} \\right)$$", "human_ref_B": "I would like to give an alternative derivation for theorist's answer. You can get the same answer with statistical mechanics using a lattice model. Let us discretize the container into $M_1$ respective $M_2$ cells. The volume of a cell is $V_{cell}$. Each gas particle fits exactly into one cell and can hop from cell to cell in a discrete manner. Since we are dealing with an ideal gas, the particles ignore each other. In other words, all particles can in principle occupy the same cell, i.e. all particles can always choose from all the cells. We therefore get the following number of possible states $\\Omega$: $$\\Omega_1 = \\frac{(M_1)^{N_1}}{N_1!} \\\\ \\Omega_2 = \\frac{(M_2)^{N_2}}{N_2!} \\\\ \\Omega_{mix} = \\frac{(M_1+M_2)^{(N_1)}}{N_1!}\\cdot\\frac{(M_1+M_2)^{(N_2)}}{N_2!}$$ where $N_i$ is the respective number of particles. The division by the factorial is the correction factor for the fact that the particles of the type $i$ are indistinguishable. If the states are degenerated, i.e. have the same energy, we can use Boltzmann's entropy formula: $$S = k_B\\ln(\\Omega)$$ $$\\implies\\Delta S=k_B\\ln(\\Omega_{mix})-k_B\\ln(\\Omega_1)-k_B\\ln(\\Omega_2)$$ When we plug in all $\\Omega_i$ and summarize with logarithmic identities, we obtain: $$\\require{cancel}\\Delta S=N_1k_B\\ln\\left( \\frac{M_1+M_2}{M_1}\\cdot\\frac{\\cancel{N_1!}}{\\cancel{N_1!}}\\right)+N_2k_B\\ln\\left(\\frac{M_1+M_2}{M_2}\\cdot\\frac{\\cancel{N_2!}}{\\cancel{N_2!}}\\right)$$ Finally, when we plug in $M_i = V_i/V_{cell}$ and $N_ik_B=n_iR$, we obtain theorist's solution: $$\\Delta S = n_1R\\ln\\left( \\frac{V_1+V_2}{V_1}\\right)+n_2R\\ln\\left(\\frac{V_1+V_2}{V_2}\\right)$$ Bonus: If we consider the constant pressure case for an ideal gas by plugging in $V_i=n_i\\cdot RT/p$: $$\\Delta S = n_1R\\ln\\left( \\frac{n_1+n_2}{n_1}\\right)+n_2R\\ln\\left(\\frac{n_1+n_2}{n_2}\\right)$$ This expression can be reformed to OP's formula by applying logarithmic identities and $x_i=n_i/n_{tot}$: $$\\Delta S = -n_{tot}R(x_1\\ln x_1 + x_2\\ln x_2)$$", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/18641, Response URL: https://chemistry.stackexchange.com/questions/146669, Post author username: t.c, Post author profile: https://chemistry.stackexchange.com/users/8140, Response author username: theorist, Response author profile: https://chemistry.stackexchange.com/users/68027", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/18641, Response URL: https://chemistry.stackexchange.com/questions/146709, Post author username: t.c, Post author profile: https://chemistry.stackexchange.com/users/8140, Response author username: Kexanone, Response author profile: https://chemistry.stackexchange.com/users/104336", "seconds_difference": 90601.0, "score_ratio": 1.5, "upvote_ratio": -1.0, "c_root_id_A": "146669", "c_root_id_B": "146709"} {"post_id": "127951", "domain": "chemistry_test", "history": "Do metals form covalent bonds besides ionic and cordinate bond? My chemistry textbook says that metals form ionic or cordinate bonds whereas non metals form covalent bonds. But in another textbook I read that Lithium, Beryllium, Aluminium, Chromium, Manganese etc, which are metals, form covalent compounds like LiCl, BeH2, AlCl3, H2CrO4, H2MnO4 respectively Now my question is whether metals form covalent bonds or not? And if yes then how to predict which metallic compound contains ionic and which contains covalent bonds?", "created_at_utc_A": 1581858552, "created_at_utc_B": 1581864251, "score_A": 4, "score_B": 14, "human_ref_A": "Your textbook is right. Lots of metals form covalent bonds. In the case of lithium chloride such bonding is one explanation for the solubility of this compound in organic solvents (see this answer). One additional example you might want to know about is Grignard reagents, a class of highly basic compounds in which carbon is covalently bonded to magnesium. These are widely used in organic synthesis processes.", "human_ref_B": "The first thing that should be said is that there's no difference between a coordinate bond (dative bond) and an ordinary covalent bond. Yes, the electrons \"come from different places\"; but the molecule doesn't actually know this, nor does it care. Once a covalent bond is formed, it is a covalent bond, regardless of where the electrons \"come from\". Some of the examples you have given can equally be described as coordinate bonds or covalent bonds. For example, $\\ce{MnO4-}$ could be viewed as four $\\ce{O^2-}$ ligands forming coordinate bonds to a central $\\ce{Mn^7+}$. Or, you could also just view them as plain old Mn=O covalent bonds. It doesn't actually matter. Having established that there is no real difference between coordinate bonds and covalent bonds, the only real question is how can you tell a covalent bond apart from an ionic bond. The answer is mostly to do with polarisability. If you imagine an ionic bond between a strongly polarising cation ($\\ce{Al^3+}$) and a polarisable anion ($\\ce{Cl-}$), the cation will pull over some of the anion's electron density: this is exactly what is meant by \"polarising\". This changes the nature of bonding from an ionic model (where the bonding electrons reside entirely on one atom) to a covalent model (where the bonding electrons are somewhere in between the two atoms). You'll notice that all the cations you've listed: $\\ce{Li+}$, $\\ce{Be^2+}$, $\\ce{Al^3+}$, (formally) $\\ce{Cr^6+}$, and $\\ce{Mn^7+}$, are very highly polarising. The first three have a very small size, and the last two have exceptionally large positive charges. That said, covalent bonds between two transition metals are reasonably common, too: a simple example is $\\ce{Mn2(CO)10}$ (Wikipedia), which is essentially $\\ce{(OC)5Mn-Mn(CO)5}$ with a single bond between the two. The bond order between two metals can go up to slightly ridiculous numbers. CrCr and ReRe quadruple bonds are well-known in $\\ce{Cr2(OAc)4(H2O)2}$ and $\\ce{[Re2Cl8]2-}$ respectively. Even quintuple and sextuple bonds between metals have been described (although some of these accounts are not without controversy). How can you predict this, though? It's not trivial at all, and in truth you probably don't need to care about these exotic examples until you have enough experience to understand the bonding.", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/127951, Response URL: https://chemistry.stackexchange.com/questions/127952, Post author username: dRIFT sPEED, Post author profile: https://chemistry.stackexchange.com/users/68424, Response author username: Oscar Lanzi, Response author profile: https://chemistry.stackexchange.com/users/17175", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/127951, Response URL: https://chemistry.stackexchange.com/questions/127957, Post author username: dRIFT sPEED, Post author profile: https://chemistry.stackexchange.com/users/68424, Response author username: orthocresol, Response author profile: https://chemistry.stackexchange.com/users/16683", "seconds_difference": 5699.0, "score_ratio": 3.5, "upvote_ratio": -1.0, "c_root_id_A": "127952", "c_root_id_B": "127957"} {"post_id": "127951", "domain": "chemistry_test", "history": "Do metals form covalent bonds besides ionic and cordinate bond? My chemistry textbook says that metals form ionic or cordinate bonds whereas non metals form covalent bonds. But in another textbook I read that Lithium, Beryllium, Aluminium, Chromium, Manganese etc, which are metals, form covalent compounds like LiCl, BeH2, AlCl3, H2CrO4, H2MnO4 respectively Now my question is whether metals form covalent bonds or not? And if yes then how to predict which metallic compound contains ionic and which contains covalent bonds?", "created_at_utc_A": 1581858552, "created_at_utc_B": 1581869643, "score_A": 4, "score_B": 6, "human_ref_A": "Your textbook is right. Lots of metals form covalent bonds. In the case of lithium chloride such bonding is one explanation for the solubility of this compound in organic solvents (see this answer). One additional example you might want to know about is Grignard reagents, a class of highly basic compounds in which carbon is covalently bonded to magnesium. These are widely used in organic synthesis processes.", "human_ref_B": "$\\ce{Na2}$ and $\\ce{K2}$ do exist in the sodium and potassium vapors. This can be proved by analyzing the highly diluted flames produced by hot vapors of these metals in contact with diluted $\\ce{Cl2}$ vapor, as proved by Polanyi and co-workers. See the following references: M. Polanyi, Atomic Reactions, Williams and Norgate, London (1932); M.G. Evans, M. Polanyi, Transactions Faraday Society 35, 178 (1935); C. E. H. Bawn, Ann. Repts. Chem. Soc. 39, 36 (1942).", "labels": 0, "metadata_A": "Post URL: https://chemistry.stackexchange.com/questions/127951, Response URL: https://chemistry.stackexchange.com/questions/127952, Post author username: dRIFT sPEED, Post author profile: https://chemistry.stackexchange.com/users/68424, Response author username: Oscar Lanzi, Response author profile: https://chemistry.stackexchange.com/users/17175", "metadata_B": "Post URL: https://chemistry.stackexchange.com/questions/127951, Response URL: https://chemistry.stackexchange.com/questions/127960, Post author username: dRIFT sPEED, Post author profile: https://chemistry.stackexchange.com/users/68424, Response author username: Maurice, Response author profile: https://chemistry.stackexchange.com/users/84629", "seconds_difference": 11091.0, "score_ratio": 1.5, "upvote_ratio": -1.0, "c_root_id_A": "127952", "c_root_id_B": "127960"}