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marianna13

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chemistry-stackexchange

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[ [ "\nLet the number of atoms be n.\n\n\nIf the number of skeletal p electrons is 2n + 2, the structure is closo (a [deltahedron](http://en.wikipedia.org/wiki/Deltahedron) with n vertices).\n\n\nIf the number of skeletal p electrons is 2n + 4, the structure is nido (an n+1 vertex deltahedron, with one vertex missing).\n\n\nIf the number of skeletal p electrons is 2n + 6, the structure is arachno (an n+2 vertex deltahedron with two vertices missing).\n\n\nIf the number of skeletal p electrons is 2n + 8, the structure is hypo (an n+3 vertex deltahedron with three vertices missing).\n\n\nsee [Excursions Beyond the Zintl Border](http://www.chemistry.illinois.edu/research/inorganic/seminar_abstracts/2002-2003/Allen.Abstract.LitSeminar.pdf) and [Polyanionic Clusters and Networks of the Early p-Element Metals in the Solid State: Beyond the Zintl Boundary](http://www.ncbi.nlm.nih.gov/pubmed/10760845)\n\n\nFor example:\n\n\nNeutral Sn, Pb and Ge each contribute 2 valence p electrons, while Bi and Sb each contribute 3 valence p electrons.\n\n\n$\\ce{Sn\\_5^{2-}}$, $\\ce{Pb\\_5^{2-}}$, $\\ce{Ge\\_5^{2-}}$ and $\\ce{Ge\\_9^{2-}}$ have 2n + 2 skeletal p electrons and are closo (the first three are trigonal bipyrimidal and the last is appoximately a triangular prism with square pyramid caps for the three sides).\n\n\n$\\ce{Sn\\_9^{4-}}$ and $\\ce{Ge\\_9^{4-}}$ have 2n + 4 skeletal p electrons and are nido. ([gyroelongated square bipyramid](http://en.wikipedia.org/wiki/Gyroelongated_square_bipyramid) with one square pyramid apex missing)\n\n\n$\\ce{Bi\\_4^{2-}}$ and $\\ce{Sb\\_4^{2-}}$ have 2n + 6 skeletal p electrons and are arachno (these two are square planar, an octahedron with two vertices missing).\n\n\nAlso, quoting Farley and Castleman's [Observation of Gas-Phase Anionic Bismuth Zintl Ions](http://pubs.acs.org/doi/abs/10.1021/ja00189a068) J. Am. Chem. Soc. 1989, 111, 2734-2735\n\n\n\n> \n> Since bismuth and antimony are typically trivalent, anionic clusters\n> of these elements very quickly exceed the maximum 2n + 8 electrons\n> accountable under Wades rules for “hypo” compounds. In particular, the\n> previously mentioned $\\ce{Bi7^{3-}}$ possesses 2n + 8 valence electrons, and\n> $\\ce{Bi14^{6-}}$ has 2n + 20! Hence, anionic Zintl ions of these species are\n> particularly intriguing due to the inability of current theories to\n> account for them and lack of other corresponding metallic compounds\n> with which they may be compared.\n> \n> \n> \n\n\nNow going back to the example $\\ce{Bi\\_3^{3-}}$ in the orginal question, the consensus in the peer-reviewed literature is that the sturcture is bent (not a triangular ring and not linear), although the structure of the isolated ion has not been experimentally determined. The reasons for considering it to be bent are:\n\n\n1. The structure is isoelectronic with species known to be bent in the ground state including $\\ce{O3}$, $\\ce{S3}$, $\\ce{S2O}$, $\\ce{SO2}$ and allyl anion.\n2. Experimental determination published in [Stabilization of Ozone-like [Bi3]3-](http://pubs.acs.org/doi/pdf/10.1021/ja0038732) and in [a more recent article](http://www3.nd.edu/%7Esevovlab/articles/HeteroatomicDeltahedralClusters.pdf) that it is bent in transition metal complexes. (Similar experiments for isoelectronic [P3]3- and[As3]3-also find bent structures).\n3. Theoretical work in [Zintl Anions Analogous to O3](http://pubs.acs.org/doi/pdf/10.1021/jp980004e).\n\n\nThough ozone and like species are found to be bent, it is also theorized that [cyclic (equilateral triangular ring) ozone](http://en.wikipedia.org/wiki/Cyclic_ozone) represents another higher local potential energy minimum, and this would be the same for $\\ce{Bi\\_3^{3-}}$ too.\n\n\n$\\ce{Bi\\_3^{3-}}$ has 12 p electrons, which is 2n + 6, which implies arachno with respect to a trigonal bipyramid. A equilateral triangular ring structure would be consisent with removing two axial vertices or removing one axial vertex and one equatorial vertex, while the bent structure would be consistent with removing two equatorial vertices.\n\n\n\n", "6" ], [ "\nThe wiki article contains a note\n\n\n\n> \n> To predict the structure of an arachno cluster, the closo polyhedron\n> with n + 2 vertices is used as the starting point, and the n+1 vertex\n> nido complex is generated by following the rule above; a second vertex\n> adjacent to the first is removed if the cluster is composed of mostly\n> small atoms, **a second vertex not adjacent to the first is removed if\n> the cluster is composed mostly of large atoms**.\n> \n> \n> \n\n\n$\\ce{Bi}$ atom is definitely a large atom, so we should remove out non-adjacent vertexes. The only way to choose non-adjacent vertexes in trigonal bipyramid is to take axial vertexes, gaining a triangle.\n\n\nAlso, please not, that removing two adjacent vertexes in trigonal bipyramid may also lead to triangle, if one removes one equatorial and one axial position. removing two equatorial positions will lead not to triangle or line, but to an angle. \n\n\nHowever, for $\\ce{[Bi3]^{3-}}$ the amount of electrons per atom is exactly six. This means, that [6n rules](http://en.wikipedia.org/wiki/Polyhedral_skeletal_electron_pair_theory#6n_rules) suite situation better. 6n rule leads to ring in the case. \n\n\nI recommend to consider Wade's rules as general guidelines, but not a final truth, sometimes they fail spectacularly. For example, consider $\\ce{[CH]6}$ cluster. Wade's rule predicts it to be a Trigonal prism. As we know, IRL this molecule prefers to be a ring, with prismane being an esoteric and a quite rare form.\n\n\n", "3" ], [ "\nSo-called **2n**, **4n**, **14n** and so forth rules are not necessary at all. Simply, Wade's rule in its original form i.e. '**n**' rule is equally applicable to Zintl ions and sufficient to predict their structural type *e.g.*, *closo*, *nido*, *arachno*, *hypho* and so on. For example, the Skeleton Electron Pair Count becomes six [0.5(3 x 5 - 3)] for [Bi5]3+; indicating (*n + 1*) formulation and hence it must adopt *closo* structure. Similarly, the SEPC becomes nine [0.5(3 x 5 + 3)] for [As5]3-, implying its *hypho* structure from (*n + 4*) formulation.\n\n\nMoreover, the contribution of the vertex atom (here it is Bi/As) can be easily calculated from the **'v + x - 2'** rule, where **v** is the # of valance electron(s). The value of **x** (contribution of *exo* ligand towards vertex atom) must be zero since there is no *exo* ligand bonded with the vertex atoms in Zintl ions, Naked Clusters. Jadavpur University, India.\n\n\n", "1" ] ]

[ [ "\nIn general, there are a variety of amino acids that are not chiral (achiral). In physiological proteins, the amino acids are generally alpha-amino acids (one carbon between the amino and carboxylic acid groups), but there are also, [beta-amino acids](https://www.princeton.edu/chemistry/macmillan/group-meetings/b-aminoacids-spb.pdf), gamma-amino acids, etc. \n\n\nJust limiting ourselves to alpha-amino acids, other than glycine, the most interesting example of an achiral amino acid is [amino malonic acid](http://www.hmdb.ca/metabolites/hmdb01147), which has been given its own three-letter abbreviation (Ama) along with the more-common alpha amino acids. Ama has being found in the hydrolysates of E. Coli and human proteins. See [Aminomalonic acid: identification in Escherichia coli and atherosclerotic plaque](http://www.pnas.org/content/81/3/722) PNAS vol. 81 page 722.\n\n\nAlso, the amino acids of protein orgin generally have an alpha H, but if this H is substituted the compound is still an alpha amino acid. If the two side chains on the alpha carbon are the same, the compound is achiral. For example [alpha-methyl alanine](http://en.wikipedia.org/wiki/2-Aminoisobutyric_acid). An example that occurs naturally in cranberries is [1-aminocyclopropane 1-carboxylic acid](http://en.wikipedia.org/wiki/1-Aminocyclopropane-1-carboxylic_acid). The alpha amino acids with two side chains are sometimes referred to as alpha branched amino acids, and have been [found in meteorites](http://www.ncbi.nlm.nih.gov/pubmed/7277509). \n\n\nAnother class of achiral alpha amino acids are those with a side chain double-bonded to the alpha carbon such as [dehydroalanine](http://en.wikipedia.org/wiki/Dehydroalanine).\n\n\n", "6" ], [ "\nThe **alpha** carbon is a chiral carbon atom, with the exception of glycine which has two indistinguishable hydrogen atoms on the alpha carbon. Therefore, all alpha amino acids but glycine can exist in either of two enantiomers, called L or D amino acids, which are mirror images of each other.\n\n\n", "4" ] ]

[ [ "\nOf course, we can't display every pathway there is. We have to select a wide variety of rate constants to get a \"good map\" of the parameter-space. The approach I took was this:\n\n\n1. Construct the differential equations\n\n\n$$\\frac{\\mathrm dA}{\\mathrm dt} = -k\\_1 A + k\\_3 C$$\n$$\\frac{\\mathrm dB}{\\mathrm dt} = -k\\_2 B + k\\_1 A$$\n$$\\frac{\\mathrm dC}{\\mathrm dt} = -k\\_3 C + k\\_2 B$$\n\n\n2. Solve them numerically for every possible combination of 0.1, 1 and 10 for each rate constant and the above stated initial conditions\n3. This leads to 27 pathways, which are plotted in a ternary diagram\n\n\nThe title of the ternary diagram is $k\\_1:k\\_2:k\\_3$, which is 0.1 : 0.1 : 0.1 in the first one.\n\n\n\n[](https://i.stack.imgur.com/kvFnq.png)\n*(click for larger image)*\n\n\n", "12" ], [ "\nWe have a **cyclic** cascade of unimolecular chemical reactions\n\n\n$$\\ce{A} \\xrightarrow{\\kappa\\_1} \\ce{B} \\xrightarrow{\\kappa\\_2} \\ce{C} \\xrightarrow{\\kappa\\_3} \\ce{A}$$\n\n\nwhere $\\kappa\\_1, \\kappa\\_2, \\kappa\\_3 > 0$ are the rate constants. Assuming that each chemical reaction in this cascade has [mass action](https://en.wikipedia.org/wiki/Law_of_mass_action) kinetics, then we have the following system of linear ODEs\n\n\n$$\\begin{bmatrix} \\dot a\\\\ \\dot b\\\\ \\dot c\\end{bmatrix} = \\underbrace{\\begin{bmatrix} - \\kappa\\_1 & 0 & \\kappa\\_3\\\\ \\kappa\\_1 & - \\kappa\\_2 & 0\\\\ 0 & \\kappa\\_2 & - \\kappa\\_3\\end{bmatrix}}\\_{=: \\mathrm M} \\begin{bmatrix} a\\\\ b\\\\ c\\end{bmatrix}$$\n\n\nwhere $a := [\\ce{A}]$, $b := [\\ce{B}]$ and $c := [\\ce{C}]$ are the concentrations of the species. Since $\\dot a + \\dot b + \\dot c = 0$, integrating we obtain the following conservation law\n\n\n$$a (t) + b (t) + c (t) = a\\_0 + b\\_0 + c\\_0$$\n\n\nwhere $a\\_0, b\\_0, c\\_0 \\geq 0$ are the initial concentrations.\n\n\n\n\n---\n\n\n### Stability\n\n\nThe characteristic polynomial of matrix $\\mathrm M$ is\n\n\n$$\\det (s \\mathrm I\\_3 - \\mathrm M) = s \\underbrace{\\left( s^2 + (\\kappa\\_1 + \\kappa\\_2 + \\kappa\\_3) s + (\\kappa\\_1 \\kappa\\_2 + \\kappa\\_1 \\kappa\\_3 + \\kappa\\_2 \\kappa\\_3) \\right)}\\_{=: q (s)} = s \\, q(s)$$\n\n\nSince $\\kappa\\_1, \\kappa\\_2, \\kappa\\_3 > 0$, all three coefficients of the (monic) quadratic polynomial $q$ are positive. Hence, both roots of $q$ have negative real parts and, thus, the cyclic cascade is (internally) marginally **stable**. Given nonzero initial concentrations, the concentrations will not grow unbounded. However, due to the zero eigenvalue, the concentrations will not converge to zero either.\n\n\n\n\n---\n\n\n### Steady state\n\n\nIf the concentrations of the species do not converge to zero, then to what values do they converge? At steady state, $\\dot a = \\dot b = \\dot c = 0$. Hence, we have the following homogeneous linear system\n\n\n$$\\begin{bmatrix} - \\kappa\\_1 & 0 & \\kappa\\_3\\\\ \\kappa\\_1 & - \\kappa\\_2 & 0\\\\ 0 & \\kappa\\_2 & - \\kappa\\_3\\end{bmatrix} \\begin{bmatrix} \\bar a\\\\ \\bar b\\\\ \\bar c\\end{bmatrix} = \\begin{bmatrix} 0\\\\ 0\\\\ 0\\end{bmatrix}$$\n\n\nwhere $\\bar a, \\bar b, \\bar c$ are the steady state concentrations. Using Gaussian elimination, we eventually obtain a **line** passing through the origin that is parameterized as follows\n\n\n$$\\begin{bmatrix} \\bar a\\\\ \\bar b\\\\ \\bar c\\end{bmatrix} \\in \\left\\{ \\gamma \\begin{bmatrix} \\frac{1}{\\kappa\\_1}\\\\ \\frac{1}{\\kappa\\_2}\\\\ \\frac{1}{\\kappa\\_3}\\end{bmatrix} : \\gamma \\in \\mathbb R \\right\\}$$\n\n\nwhere\n\n\n$$\\tau\\_1 := \\frac{1}{\\kappa\\_1} \\qquad\\qquad\\qquad \\tau\\_2 := \\frac{1}{\\kappa\\_2} \\qquad\\qquad\\qquad \\tau\\_3 := \\frac{1}{\\kappa\\_3}$$\n\n\nare **time constants**. Intersecting this line with the **plane** $\\bar a + \\bar b + \\bar c = a\\_0 + b\\_0 + c\\_0$, we obtain\n\n\n$$\\gamma = \\left( \\dfrac{1}{\\tau\\_{1} + \\tau\\_{2} + \\tau\\_{3}}\\right) \\left( a\\_0 + b\\_0 +c\\_0 \\right)$$\n\n\nand, thus, the **steady state** concentrations are\n\n\n$$\\boxed{\\begin{array}{rl} & \\\\ \\qquad\\bar a &= \\left( \\dfrac{\\tau\\_{1}}{\\tau\\_{1} + \\tau\\_{2} + \\tau\\_{3}}\\right) \\left( a\\_0 + b\\_0 +c\\_0 \\right) \\qquad \\\\\\\\ \\qquad\\bar b &= \\left( \\dfrac{\\tau\\_{2}}{\\tau\\_{1} + \\tau\\_{2} + \\tau\\_{3}}\\right) \\left( a\\_0 + b\\_0 +c\\_0 \\right) \\qquad\\\\\\\\ \\qquad\\bar c &= \\left( \\dfrac{\\tau\\_{3}}{\\tau\\_{1} + \\tau\\_{2} + \\tau\\_{3}}\\right) \\left( a\\_0 + b\\_0 +c\\_0 \\right)\\\\ & \\end{array}}$$\n\n\nIf, say, $\\kappa\\_1 \\gg \\kappa\\_2, \\kappa\\_3$, i.e., if $\\tau\\_1 \\ll \\tau\\_2, \\tau\\_3$, then $\\bar a \\ll \\bar b, \\bar c$. If $\\kappa\\_1 = \\kappa\\_2 = \\kappa\\_3$, then $\\bar a = \\bar b = \\bar c$.\n\n\nAssuming that at least one of $a\\_0, b\\_0, c\\_0$ is nonzero, the **ratios** of the steady state concentrations are\n\n\n$$\\frac{\\bar a}{\\bar b} = \\frac{\\kappa\\_2}{\\kappa\\_1} = \\frac{\\tau\\_1}{\\tau\\_2} \\qquad\\qquad\\qquad \\frac{\\bar a}{\\bar c} = \\frac{\\kappa\\_3}{\\kappa\\_1} = \\frac{\\tau\\_1}{\\tau\\_3} \\qquad\\qquad\\qquad \\frac{\\bar b}{\\bar c} = \\frac{\\kappa\\_3}{\\kappa\\_2} = \\frac{\\tau\\_2}{\\tau\\_3}$$\n\n\nHence, the ratio of the steady state concentrations of two species is the ratio of their time constants.\n\n\n\n\n---\n\n\n### Oscillations\n\n\nLet\n\n\n$$\\Delta (\\kappa\\_1, \\kappa\\_2, \\kappa\\_3) := (\\kappa\\_1 + \\kappa\\_2 + \\kappa\\_3)^2 - 4 (\\kappa\\_1 \\kappa\\_2 + \\kappa\\_1 \\kappa\\_3 + \\kappa\\_2 \\kappa\\_3) \\\\ = (\\kappa\\_1 - \\kappa\\_2)^2 + (\\kappa\\_1 - \\kappa\\_3)^2 + (\\kappa\\_2 - \\kappa\\_3)^2 - (\\kappa\\_1^2 + \\kappa\\_2^2 + \\kappa\\_3^2)$$\n\n\nbe the discriminant of quadratic polynomial $q$. When $\\Delta < 0$, the roots of $q$ have nonzero imaginary parts and, thus, the cyclic cascade will exhibit **damped oscillations**. Hence, when\n\n\n$$\\kappa\\_1^2 + \\kappa\\_2^2 + \\kappa\\_3^2 > (\\kappa\\_1 - \\kappa\\_2)^2 + (\\kappa\\_1 - \\kappa\\_3)^2 + (\\kappa\\_2 - \\kappa\\_3)^2$$\n\n\nthe cyclic cascade will exhibit oscillatory behavior. Provided that the inequality above is satisfied, the concentrations will oscillate as follows\n\n\n$$\\begin{array}{rl} a (t) &= \\bar a + \\eta\\_A \\, e^{- \\kappa\\_0 t} \\cos( \\omega\\_0 t - \\phi\\_A)\\\\ b (t) &= \\bar b + \\eta\\_B \\, e^{- \\kappa\\_0 t} \\cos( \\omega\\_0 t - \\phi\\_B)\\\\ c (t) &= \\bar c + \\eta\\_C \\, e^{- \\kappa\\_0 t} \\cos( \\omega\\_0 t - \\phi\\_C)\\end{array}$$\n\n\nwhere\n\n\n$$\\begin{array}{rl} \\kappa\\_0 &:= \\frac{1}{2} \\left( \\kappa\\_1 + \\kappa\\_2 + \\kappa\\_3 \\right)\\\\ \\omega\\_0 &:= \\frac 12 \\sqrt{\\kappa\\_1^2 + \\kappa\\_2^2 + \\kappa\\_3^2 - (\\kappa\\_1 - \\kappa\\_2)^2 - (\\kappa\\_1 - \\kappa\\_3)^2 - (\\kappa\\_2 - \\kappa\\_3)^2}\\end{array}$$\n\n\nEach species will have its own magnitude and phase lag that are dependent on the initial conditions.\n\n\n", "11" ] ]

[ [ "\nI think you are confusing some things.\n\n\nStability is only regarding to individual molecule. \n\n\nInstead, boiling point depends on intermolecular bonding. You can have 2 unstable molecules strongly bonded together via hydrogen bonds for instance as you can have 2 stable compounds without hydrogen so they will not have a very high boiling point because of the absence of such hydrogen bonding.\n\n\nFor the melting point, it's another kind of force sticking molecules or atoms together. So there is no space for comparison beteween individual stability and boiling or even melting point of a whole bunch of molecules. \n\n\nEdit : \n\n\n \n\n\nTrans on the left and Cis on the right. \nWhere can you have hydrogen bonds ? You need some Oxygen or chemicaly equivalent atom to make it !\n\n\n", "2" ] ]

[ [ "\nAccording [Kinetics, Mechanism, and Spectroscopy of the Reversible Binding of\nNitric Oxide to Aquated Iron(II). An Undergraduate Text Book Reaction\nRevisited](http://pubs.acs.org/doi/pdf/10.1021/ic010628q)\n\n\nThe correct structure is $\\ce{ [Fe^{III}(H\\_2O)\\_5(NO^{-})]^{2+} }$\n\n\nFor many years it was thought that iron was reduced to Fe(I) and NO oxidized to NO+, based upon an observed magnetic moment suggestive of three unpaired electrons, however, the current thinking is that high spin Fe(III) (S=5/2) antiferromagnetically couples with NO- (S=1) for an observed spin of S=3/2.\n\n\nOther than the above, I agree with Dhanajay Gupta that the equations in the book are correct, that some of the Fe2+ is oxidized to Fe3+ in the first equation forming NO, and NO coordinates to remaining Fe2+ according to the second equation, but the second equation doesn't specify the charge on Fe or NO, just the overall charge of the complex.\n\n\nIn the first equation in the book, it should be understood that the ions are aqueous. For example H+ doesn't mean there are bare protons, it is understood to mean $\\ce{H\\_3O+}$ and Fe2+ is $\\ce{[Fe(H2O)6]^{2+}}$ and Fe3+ is $\\ce{[Fe(H2O)6]^{3+}}$.\n\n\nIn the second equation in the book, the inner coordination sphere is explicitly shown.\n\n\nIn the article cited above the mechanism of the second equation is explained. Neutral NO attacks hexaaquairon(II), forming a 7-coordinate transition state, a water leaves, and \"This rate-determining displacement of coordinated water is followed by a rapid intramolecular charge-redistribution process to lead to the final $\\ce{ [Fe^{III}(H\\_2O)\\_5(NO^{-})]^{2+}}$ product\".\n\n\n\n> \n> but why won't NO take electrons from the oxidation of Iron ($Fe^{2+}\n> \\rightarrow Fe^{3+} + e^-$), make $NO^-$ and attack the above and\n> substitute one water molecule and give more of the brown ring complex?\n> \n> \n> \n\n\nTo get the electron from Fe2+, NO would enter the inner coordination sphere, and when the electron transfer occurs, the brown ring complex is formed at that Fe. If the NO- then left that Fe, it would destroy one instance of the brown ring complex. Hypothetically, the Fe2+ could reduce NO via an [outer sphere electron transfer](http://en.wikipedia.org/wiki/Outer_sphere_electron_transfer), and then NO- coordinate Fe3+, but this is not what the article reports occurring. \n\n\n", "11" ], [ "\nSince this reaction takes place in aqueous solution, is is implied, that the reaction equations are involving solvated ions. In general both of the given equations are correct. However, I would prefer the first equation to include the highly charged metal ions as aqua complexes: \n$$\\ce{NO3- + 3[Fe(H2O)\\_6]^{2+} + 4H+ -> NO + 3[Fe(H2O)\\_6]^{3+} + 2H2O}$$\n\n\nAlso note that this is a test for nitrate, and nitrite ions will interfere accordingly.\n$$\\ce{NO2- + [Fe(H2O)\\_6]^{2+} + 2H+ -> NO + [Fe(H2O)\\_6]^{3+} + H2O}$$\n\n\nIt is important to realise, that the Iron (II) solution is given in massive excess for this test to work. Hence the newly reduced nitrogen monoxide will react with the yet unreacted excess of $\\ce{Fe^{2+}}$:\n$$\\ce{NO + [Fe(H2O)\\_6]^{2+} -> [Fe(NO)(H2O)\\_5]^{2+} + H2O}$$\n\n\nSince the iron hexaaqua complexes are nearly colourless, they do not interfere with this visibility of this reaction.\n\n\n", "6" ], [ "\nThe reactions given are correct. The nitric acid oxidizes the ferrous ions into ferric ions, and itself get's reduced to NO. But this reaction doesn't happen all at once. It's not as if all the ferrous ions are oxidized to ferric in one go. The remaining ferrous ions then form the brown ring complex with NO. Furthermore, the formation of the brown ring complex itself is a redox reaction. Iron is in the rare +1 oxidation state in the complex, whereas NO is actually NO+. \n\n\nSo yes, ferric ions are present in the solution. They were produced as a result of the first reaction, but are not required in the second reaction.\n\n\n", "2" ] ]

[ [ "\nFor the given structure, the principal chain could be a 3-hydroxyhexanoic acid or a 3-methylhexanoic acid; i.e. the name of the structure could be 3-hydroxy-2-(pentan-2-yl)hexanoic acid or 2-(1-hydroxybutyl)-3-methylhexanoic acid.\n\n\n\n\n\nThe IUPAC recommendations on this matter have been changed.\n\n\nAccording to the old IUPAC recommendations (1979, retained in 1993), the principal chain was the 3-hydroxyhexanoic acid. Thus, the preferred name was 3-hydroxy-2-(pentan-2-yl)hexanoic acid.\n\n\nHowever, the relevant rule in the current version of *[Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)](http://dx.doi.org/10.1039/9781849733069-fp001)* reads as follows:\n\n\n\n> \n> **P-45 SELECTION OF PREFERRED IUPAC NAMES**\n> \n> \n> (…)\n> \n> \n> **P-45.5** CRITERIA RELATED TO ALPHANUMERICAL ORDER OF NAMES\n> \n> \n> **P-45.5.1** The preferred IUPAC name is the name that is earlier in alphanumerical order (…). Alphabetic letters are considered first in the order that they appear in the name; (…)\n> \n> \n> \n\n\nTherefore, the preferred name is **2-(1-hydroxybutyl)-3-methylhexanoic acid** (not 3-hydroxy-2-(pentan-2-yl)hexanoic acid) since ‘hydroxybutyl…’ is earlier alphabetically than ‘hydroxy-pentan-yl…’.\n\n\n", "5" ], [ "\nFor point 3 carbon-carbon single bonds count. Heteroatoms would build up a functional group. However, I could not think of a scenario where point 4 is needed. It should be a scenario where two chains with an equal amount of multiple bonds and single bonds are present but different in length. \n \n\nThe rule, which you mention, in which the number of functional groups count is point 1. \n \n\nAccording to what I have learned the name \"2-(1-Hydroxybutyl)-3-methylhexanoic acid\" is wrong. There are two equal chains, one has a methyl and one a hydroxyl group. According to IUPAC nomenclature of organic chemistry the group with the higher precedence is hydroxyl. Therefore, chemdoodle gave you the wrong name. \n\nThis is a case where this sentence applies which they put above: \"Please give it a try and let us know if you encounter any issues.\"\n\n\n", "-1" ] ]

[ [ "\nThe reason d-orbitals make a difference is that electrons in d-orbitals do not screen nuclear charge as effectively as those in s and p orbitals. This is because of something called **penetration** .\n\n\nThe mathematical shapes of d-orbitals prevent them from allowing electrons to penetrate very closely to the nucleus, compared with electrons in s or p-orbitals. In gallium, you have $10$ electrons in the filled $3$d-subshell, and each of these electrons is doing a slightly worse job (relatively speaking) of screening the nuclear charge than the electrons in the s and p orbitals. Therefore, the effective nuclear charge in gallium is slightly higher than that in aluminum, so the *increase* in the radius is a quite a bit smaller than would be expected based on the difference between boron and aluminum, or gallium and indium. \n\nThe trend goes:\n\n\n* $\\pu{82 pm}$ ($\\ce{B}$)\n* $\\pu{118 pm}$ $\\ce{(Al)}$\n* $\\pu{126 pm}$ $\\ce{(Ga)}$\n* $\\pu{144 pm \\ce{(In)}}$\n\n\n[covalent radii from [www.webelements.com]](https://www.webelements.com).\n\n\nThis effect is generally known as the **d-block contraction**. (It can be more or less pronounced depending on how you define the atomic radii.)\n\n\nA similar thing happens (in principle) when you go from indium to thallium; except in this case you are now dealing with adding a filled f-subshell to the valence shell. \n\n\n*Electrons in f-orbitals are even worse at screening nuclear charge than those in d-orbitals*, therefore again, the effective nuclear charge in thallium is a bit larger than it is in indium, so again the jump in radius is fairly small (from $144$ to $148$ pm). This effect (of the filled f-subshell) is generally known as the **lanthanide contraction**.\n\n\n", "17" ], [ "\nNaively, one should expect the atomic radii to increase in a proportional fashion from boron to aluminium, gallium, indium and thallium, because each of these atoms has one more populated shell.\n\n\nHowever, the difference is not proportional because there is not a constant proton difference between the elements. From boron to aluminium 8 protons are added to the nucleus; from aluminium to gallium and from gallium to indium 18; and finally, from indium to thallium 32. Wherever there are more protons than expected, orbitals contract; thus, both gallium and thallium are smaller than one naively might assume.\n\n\nOf course, one should not forget the additional electrons. However, these are added to the d orbitals (and in thallium’s case, f orbitals). These orbitals are much higher in energy than the corresponding shell’s s and p subshells (as it is evident when these orbitals are populated in the periodic table). Furthermore, these orbitals are associated with two (d) or three (f) nodal planes through the nucleus, meaning that the corresponding electrons are further away from the nucleus on average. Thus, their shielding is less effective than that of s and p orbitals.\n\n\n", "4" ], [ "\nHere's an explanation with the help of [Slater's rule](https://en.m.wikipedia.org/wiki/Slater%27s_rules):\n\n\n\n\n\n| | Group 1 | Group 2 | Rest |\n| --- | --- | --- | --- |\n| s, p | 0.35 | 0.85 | 1 |\n| d, f | 0.35 | 1 | 1 |\n\n\n\nAs you can see from the table, electrons in the same subshell/group contribute only 0.35 screening while the inner electrons do more screening.\n\n\nThe number of electrons that can be accommodated in a subshell increases as s<p<d<f, thus the screening effect decreases as: s>p>d>f.\n\n\n", "0" ], [ "\nd and f orbitals experience more attraction towards nucleus due to lesser shielding effect than s and p orbitals. This is why the ionization energies of d and f block elements are high and so their electronegativities.\n\n\n", "-2" ] ]

[ [ "\nMany organic reactions are unreasonably slow and can take an extended period of time to achieve any noticeable effect so heating is often used to increase the rate of reaction. However, many organic compounds have low boiling points and will vaporise upon exposure to such high heat, preventing the reaction from proceeding in full.\n\n\nTo address this, heating under reflux is used. This refers to heating a solution with an attached condenser to prevent reagents from escaping.\n\n\n\n\n\nAs seen above, any vapor will condense on the cool surface of the attached condenser and flow back into the flask.\n\n\nThe hot water bath pictured is an optional component of heating under reflux and is usually only used for particularly sensitive reactions. Also, using it limits the reaction temperature to 100 degrees Celsius.\n\n\n", "21" ], [ "\nTemperature control is important for reactions, especially in organic chemistry. Some reactions are strongly exothermic or have notable side-reactions that can be suppressed at a low temperature. For others, assuming all reactants survive the temperatures in question, van ’t Hoff’s rule dictates that an increase of the temperature by $10~\\mathrm{^\\circ C}$ increases the reaction rate by a factor of $2$ to $4$. Thus, increasing the temperature is often favourable.\n\n\nAlmost all organic reaction are conducted in a solvent. The choice of solvent dictates the temperature range you can reach; e.g. tetrahydrofurane solidifies at $-108.4~\\mathrm{^\\circ C}$ and boils at $65.8~\\mathrm{^\\circ C}$, so any reactions will have to take place at in-between temperatures.\n\n\nOften, a published reaction will have a set of conditions most likely to work; they typically come with a preferred solvent and a preferred temperature. A Dess-Martin oxidation is typically conducted at $0~\\mathrm{^\\circ C}$ in dichloromethane. For many reactions, the preferred temperature coincides with the solvent’s boiling point — it means that maximum heating is required to conduct the reaction in that solvent. When heated to the boiling point, the solvent will partially evaporate and recondense on colder surfaces. But since also the concentration of reactants is important, one typically wants to recollect the evaporating solvent.\n\n\nThis is where *heating under reflux* comes into play. *Reflux* is the term used to mean ‘letting a solvent boil and collecting its vapour in some kind of condenser to let it drip back into the reaction vessel.’ The most common type of condenser I have encountered for refluxing is the *Dimroth condenser* as shown in the image below (taken from [Wikipedia](https://en.wikipedia.org/wiki/File:Dimroth_mirr_stdzd.png), where a full list of authors is available).\n\n\n[](https://i.stack.imgur.com/9RiHO.png)\n\n\nIt is important to connect the cooling water circuit correctly. **For some reason, most images found on the internet, including the one in the other answer, suggest a suboptimal cooling.** The most optimal cooling efficiency is given in a countercurrent setup. To quote [Wikipedia](https://en.wikipedia.org/wiki/Countercurrent_exchange):\n\n\n\n> \n> The maximum amount of heat or mass transfer that can be obtained is higher with countercurrent than co-current (parallel) exchange because countercurrent maintains a slowly declining difference or gradient (usually temperature or concentration difference). In cocurrent exchange the initial gradient is higher but falls off quickly, leading to wasted potential.\n> \n> \n> \n\n\nThus, in the image above, the water supply should be connected to the top connector while the bottom one should be used as water outlet. This allows the strongest cooling efficiency to be at the top of the condensor, which is important, because if vapour manages to get that high up it needs a quick and efficient cooling.\n\n\n", "4" ] ]

[ [ "\nMolar conductivity is defined as the conductivity of an electrolyte solution divided by the molar concentration of the electrolyte, and so measures the efficiency with which a given electrolyte conducts electricity in solution.\nIt's unit in S.I. is: $$\\ce{S.m^2.mol^{-1}}$$\n\n\n", "8" ], [ "\nunit of conductivity of an electrolyte solution = \n\n\n(unit of length)/(unit of area) \\* (unit of Conductance)\n\n\n= (m/m2). S \n= S . m-1\n\n\nunit of molar conductivity = (unit of conductivity of an electrolyte solution ) / (unit of molar concentration of the electrolyte)\n = ( S m-1) / (mol m-3)\n = S . m . 2 mol-1\n\n\n", "1" ], [ "\nIt does look odd that the numerator is expressed in units of conductance times units of area. Conductivity in Siemens per meter divided by concentration in, say, moles per liter should give you something in units of $S \\cdot l \\cdot mol^{-1} \\cdot m^{-1}$ . But since a liter is simply a volume measurement equal to 1000 $cm^3$ or 0.001 $m^3$, the volume/distance can be cancelled out to just area. So you end up with that $S \\cdot m^2 \\cdot mol^{-1}$.\n\n\nSometimes you see tables that contain molar conductivity values in units of $ 10^{-4} S \\cdot m^2 \\cdot mol^{-1}$. This makes it handy because the $10^{-4}$ factor gives you the same value as when expressed in units of \n$ \\mu S/cm $ per $ mmol/l $. \n\n\nE.g, the specific conductance for NaCl is:\n\n\n$ 0.0126 \\space S \\cdot m^2 \\cdot mol^{-1}$\n\n\n$ 126 \\times 10^{-4} \\space S \\cdot m^2 \\cdot mol^{-1}$\n\n\n$126$ $(\\mu S / cm) \\space \\over \\space (mmol / l) $\n\n\n", "1" ] ]

[ [ "\n1- 3,3-dichloropropanoic acid. Please see <http://www.masterorganicchemistry.com/2011/02/14/table-of-functional-group-priorities-for-nomenclature/>\n\n\n2- The first one is more acidic, as the inductive effect, due to chlorine atom decrease fast with the distance. (So, your answer is correct!).\n\n\n", "3" ], [ "\nWith regard to numbering of locants, the principal characteristic group (the characteristic group chosen for citation at the end of a name by means of a suffix or a class name, or implied by a trivial name, e.g. propanoic acid) has seniority over simple substituent groups (e.g. chloro). On this matter, the current version of *[Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)](http://dx.doi.org/10.1039/9781849733069-fp001)* reads as follows:\n\n\n\n> \n> **P-14.4** NUMBERING\n> \n> \n> When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:\n> \n> \n> (…)\n> \n> \n> (c) principal characteristic groups and free valences (suffixes);\n> \n> \n> (…)\n> \n> \n> (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;\n> \n> \n> (…)\n> \n> \n> \n\n\nNote that Rule c takes precedence over Rule f.\n\n\nTherefore, the preferred IUPAC name for the given structure is 3,3-dichloropropanoic acid (not 1,1-dichloro-3-propanoic acid).\n\n\n\n\n\n\n\n---\n\n\nAs you have correctly assumed, additional chloro substituents increase the acidity of the carboxylic acid due to their electron-withdrawing inductive effect (−I effect):\n\n\n  acetic acid: $\\mathrm{p}K\\_\\mathrm{a} = 4.756$ \n\n  2-chloroacetic acid: $\\mathrm{p}K\\_\\mathrm{a} = 2.87$ \n\n  2,2-dichloroacetic acid: $\\mathrm{p}K\\_\\mathrm{a} = 1.35$ \n\n  2,2,2-trichloroacetic acid: $\\mathrm{p}K\\_\\mathrm{a} = 0.66$\n\n\nThe acidity of chlorocarboxylic acids is hardly directly affected by the length of the carbon chain:\n\n\n  2-chloroacetic acid: $\\mathrm{p}K\\_\\mathrm{a} = 2.87$ \n\n  2-chloropropanoic acid: $\\mathrm{p}K\\_\\mathrm{a} = 2.83$ \n\n  2-chlorobutanoic acid: $\\mathrm{p}K\\_\\mathrm{a} = 2.86$\n\n\nHowever, the effect decreases as the distance between the chloro substituent and the carboxyl group increases:\n\n\n  2-chloroacetic acid: $\\mathrm{p}K\\_\\mathrm{a} = 2.87$ \n\n  3-chloropropanoic acid: $\\mathrm{p}K\\_\\mathrm{a} = 3.98$ \n\n  4-chlorobutanoic acid: $\\mathrm{p}K\\_\\mathrm{a} = 4.52$\n\n\n($\\mathrm{p}K\\_\\mathrm{a}$ values taken from “Dissociation Constants of Organic Acids and Bases”, in *CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010),* David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.)\n\n\n", "2" ] ]

[ [ "\nLet's explain this with your example of silicon that every atom is sp3 hybridised. when you merge two of these orbitals, then it yields bonding and antibonding MO. But when a third atom bounds, it yields 3 MO (not four as you mentioned) bonding, antibonding and non bonding MO. The number of MO equals the number of atomic orbitals constituting them, according to LCAO theory.\nThe energy of the third MO is in between the nonbonding and anti-bonding MOs. When you add a forth, a fifth and so on atoms, you will have more and more MOs very close in energy and you will form a band.\nThis is a qualitative description about the formation of a band.\n\n\n", "4" ], [ "\nI'll give a quick answer first and come back tomorrow for a more complete version.\n\n\nIn your question, you start with two $\\ce{Si}$ atoms and then a third. As you combine these atoms into pairs and molecules, chemistry stresses that the atomic orbitals combine in different ways.\n\n\nSo let's take $\\ce{Si2}$. We know that for the $3p$ orbitals, some will become more stable and some will become less stable. There are 3 for each $\\ce{Si}$ atom, so a total of 6 molecular orbitals due to combinations of $3p$. (Actually, the $3s$ are also involved, and there's some amount of s-p mixing, but let's ignore that for now.)\n\n\n**Take-home message: when we combine atomic orbitals into molecules, some orbitals become more stable and some orbitals become less stable.**\n\n\nOK, now let's add a third $\\ce{Si}$ atom as you said. I'm not going to derive the MO diagram right now. (I'll do that tomorrow and insert the picture.) The main point is that we'll have new combinations. If I take $\\ce{H3}$ as another example, I should have 3 molecular orbitals.\n\n\nAs I grow into a linear chain, a 2D sheet, or a 3D solid, I don't have some small number of atoms like 2, 3, 4, 20, etc. Instead, I have 1,000 or 10,000 or many, many more.\n\n\nWell, those atomic orbitals combine just like they do when you create molecular orbitals. Some become more stable, some become less stable. Yet, rather than having 3 molecular orbitals in $\\ce{H3}$ from 3 atomic orbitals, we have 1,000 or 10,000 or many, many more.\n\n\nSince there are so many orbitals, the energy spacing between them is infinitesimal. These are now called ***bands***.\n\n\nThe type of material (and the atomic composition) will dictate the energy levels of the bands and the so-called \"band structure\" (i.e., the solid-state equivalent of molecular orbital diagrams). But simply having hundreds and thousands of atoms will give you bands.\n\n\n", "1" ], [ "\nThis is explained by the Pauli exclusion principle, as applied to *conductors*: quantum mechanics forbids that fermions have exactly the same spin-state and energy level. In a single atom, that means, for example that H has one s electron in the \"up\" state and anther in thew \"down\". In conductors (and semiconductors), electrons from one atom can exchange with those of another, so they must have (slightly) different energy levels, if all spin-states are occupied, effectively splitting the single level into a \"band\" of energies. See <http://en.wikipedia.org/wiki/Pauli_exclusion_principle> for more detail.\n\n\nBTW, this is not true for \"solids\" in general, e.g. sulfur or window glass, but is true for metals, alloys (e.g. brass) and conductive compounds.\n\n\n", "0" ] ]

[ [ "\nYour guess is correct. Polyamides are polymers where the repeating units are linked by amide bonds. Polypeptides are a specific type of polyamide where the repeating units are amino acids.\n\n\n", "6" ], [ "\nPeptides (or polypeptides) are polymers of amino acids; usually created by cell enzymes.\n\n\nPolyamides usually have a carbon chain between the amides, such as nylon 6 or nylon 66. Peptides could be classified as polyamides, but are not usually included since they have their own group.\n\n\n", "1" ], [ "\nThe idea of a \"mer\" unit is polymer chemistry and polyamides (e.g. kevlar) are common. They are also liquid crystals. A polyamide is anything that has repeating amide bonds. Any amide (diamide to be precise) can be made into a polyer. \n\n\n[](https://i.stack.imgur.com/uydSj.png) \n\n\nFrom Organic-ese.com\n\n\n", "1" ] ]

[ [ "\nThe more particles in a system, the more modes of vibration (and rotation), so one gram of $\\ce{H2}$ has about $235 \\times$ more particles than one gram of U. The heat is \"stored\" in the motions of the atoms and electrons, until reaching millions of kelvin, at which point dissociation of the nucleus would be significant in specific heat. See [these](http://farside.ph.utexas.edu/teaching/sm1/lectures/node71.html) [links](http://nptel.ac.in/courses/115106076/Module%209/Module%209.pdf) and for more information.\n\n\n", "3" ], [ "\nAs the amount of a substance increases the Heat capacity of that substance does go up. However, Specific Heat is a correction for this because it is the Heat Capacity divided by the amount of the substance. \n\n\nFor Specific Heat the amount of a substance is irrelevant and its value is largely determined by the degrees of freedom for the constituent particles. The more degree of freedom, other than transnational, the higher the Specific Heat. An example of another degree being rotational energy which does not contribute to Temperature changes.\n\n\n", "-4" ] ]

[ [ "\nYour energy conservation equation is $$q\\_{rxn} + q\\_{soln} = 0$$ so $$q\\_{rxn} = -q\\_{soln} = -m\\_{soln}c\\_{soln}\\Delta T\\_{soln}$$ Note the minus sign; if $\\Delta T$ is negative, the energy will be positive, and vice versa. And yes, you need the mass of the solution here, and the specific heat capacity of the solution too. \n\n\nIf you only need a couple of significant figures and the solution isn't too concentrated, you can assume $c\\_{soln} \\approx c\\_{\\rm H\\_2O}$. \n\n\nIf you need more precision, though, you'll need to measure or look up $c\\_{soln}$. And you'll need to include the heat absorbed by the calorimeter in your energy conservation equation, too.\n\n\n", "1" ], [ "\nYou have to take the mass of the mixture. So, the second expression is true.\n\n\n", "0" ] ]

[ [ "\nConsidering the number of other atoms a central atom can be bonded to, \"hybridization [is] irrelevant\" and \"the coordination number ... is not restricted by the number of\nhybrid orbitals that the atom can form\", quoting [On the role of d orbitals in sulfur hexafluoride](http://pubs.acs.org/doi/abs/10.1021/ja00273a006?journalCode=jacsat&quickLinkVolume=108&quickLinkPage=3586&selectedTab=citation&volume=108) J. Am. Chem. Soc., 1986, 108 (13), pp 3586–3593. \n\n\nExamples such as $\\ce{CLi6}$ and $\\ce{TeF8^{2-}}$ are cited and emphasis is placed upon the ionic character of the bonding by the cited reference.\n\n\nIn my opinion, the main limiting factor is how many atoms can simultaneously be close to the central atom, this increasing with increasing radius of the central atom and with decreasing radius of the surrounding atoms. \n\n\n", "5" ] ]

[ [ "\nYou have only three isomers with the configurations (1R,2R), (1S,2S) and (1R,2S). The one with the configuration (1R,2S) is achiral, as it has a symmetry plan, and it's superimposable on its image by a mirror (i.e. it is identical to its enentiomer (1S,2R)). \n\n\nThis isomer is known as \"meso compound\". \n\n\nMeso compounds are a special case of stereoisomers which occur when a molecule has two chiral centers and each chiral center has identical substituants.\n\n\n\n\n\n", "4" ], [ "\nLike Lighthart already mentioned in the comments, it is important to know which compound was used in the addition reaction. Because of the [reaction mechanism](http://en.wikipedia.org/wiki/Halogen_addition_reaction#Reaction_mechanism), only certain products are formed, this is very specific. LDC3 also explained that in the comments, but let me dissect this a little further.\n\n\n\n> \n> The first structure is from geminal,\n> \n> \n> \n\n\nIn the scheme below, this is reaction $(1)$ and it forms an achiral product.\n\n\n\n> \n> the second is from cis \n> \n> \n> \n\n\nThis is depicted in $(2)$ and it forms the meso compound described in Yomen Atassis's answer.\n\n\n\n> \n> and the third is trans (or do I have cis and trans backwards). \n> \n> \n> \n\n\n*(No you have not.)* This is depicted in $(3)$, but in this case a racemic mixture of the (*S,S*)- and (*R,R*)-product is formed, depending on which carbon atom in the bromonium ion the nucleophilic attack takes place.\n\n\n\n> \n> The only other isomer I can think of is 1,1-dibromo-2,2-dichloroethane, but it is not made from adding $\\ce{Br2}$ to dichloroethene. \n> \n> \n> \n\n\nFor completeness, this is depicted in $3$.\n\n\nI could further think of another isomer, which is like the above not formed via addition, it is depicted in $(5)$. Note that these last two reactions also yield achiral products.\n\n\n\n\n\n", "1" ] ]

[ [ "\nWhat you have to understand about resonance structures is that they do not actually exist in the molecule. The separate resonance structures are never formed on their own but they can be used to represent a delocalised bonding situation, more accurately described using molecular orbitals.\n\n\nIn the case of propenal the $\\pi$ MOs result from the overlap of four $p$ orbitals on the four atoms in the chain. The two occupied $\\pi$ MOs show how most of the electron density is between C1 and C2, and C3 and the oxygen but there is also significant electron density between C2 and C3. This reflects the different positions of the double bonds shown in the resonance contributors of propenal.\n\n\n\n\n\n\n\n\n<http://www.chemtube3d.com/orbitalsacrolein.htm>\n\n\n", "3" ] ]

[ [ "\nRegardless of what it takes to remove any traces of hydrochloric acid from the drum, the container is probably not certified as food-grade. So in addition to any risks from the acid itself (including things like heavy metals, which are common contaminants in mineral acids), the plastic is not certified to be free of toxic substances left over from manufacturing or that the plastic will not degrade to release such substances over its lifetime.\n\n\n", "9" ], [ "\nI imagine the container is made of polyethylene or polypropylene (it should be marked with a numeric code). If that is the case, I have a hard time accepting that there is any downside to using it for drinking water. Chemists don't want heavy metals contaminating their hydrochloric acid, so I think you are OK on that count as well.\n\n\n", "1" ] ]

[ [ "\nExplosives chemistry is a rather complex topic. I've heard that [this book](https://books.google.ca/books?id=ATiYCfo1VcEC&dq=isbn:3527316566&hl=en&sa=X&ei=w2anVJL-M4GyogT7tYCIBg&ved=0CB4Q6AEwAA) is a good source of information about it (I haven't read it).\n\n\nIn a nutshell, your intuition about the nitro group is accurate. Formation of $\\ce{N2}$ is highly energetically favourable. To get an explosive, what we need is a *rapid* reaction that produces a lot of heat and gas to both cause the obvious effects of an explosion and to propagate the reaction to other molecules of explosive. The rapid part is what separates something like TNT from toluene. The combustion of toluene is also energetically favourable, but in TNT, to create $\\ce{N2}$ is not dependant on mass transport of any other species and can thus happen very rapidly, whereas the combustion of toluene is limited by how quickly oxygen is transported to it. If one were to vaporize toluene in the correct concentration in air, this transport problem goes away and an explosion can occur. TNT also has an advantage in this regard because the nitro groups provide a source of oxygen to react with the carbon and nitrogen remaining (not enough for all of it, but it helps). Explosives are often mixed with fuels or oxidizing agents to produce a more oxygen-balanced mixture for a more efficient explosion.\n\n\nAs for what happens first in an explosion, many different reactions can occur during an explosion, but it may be helpful to consider what it takes to actually detonate TNT and think about the timescale of the reactions. TNT is a solid at room temperature and has a flash point of 163 °C making it difficult to even ignite, and while it will burn in a fire, there is no risk of explosion. For an explosion to occur, enough gas and heat has to be produced to propagate the reaction through the bulk of the material. In practice this is done using a much more sensitive explosive (other explosives like lead azide or nitroglycerine are unstable enough to be set off by heat or pressure) to produce a small shockwave that provides the activation energy to initiate a reaction as it travels through the explosive which then sustains the shockwave through the rest of the material. In a normal explosive (not a fuel-air explosive or the like), the reactions that contribute to the bulk of the explosion are limited to what the explosive is made of because the speed of the explosion is too fast for air to play much of a role initially. \n\n\nIn the case of TNT, the experimentally-measured time it takes for a shockwave to pass through is 100–200 fs (*no idea how one measures that*), so any oxidation of the carbons seem unlikely to contribute much to the the initial explosion, given the only readily available source of oxygen is from the nitro groups which must presumably decompose first. [This group](http://pubs.acs.org/doi/abs/10.1021/jp072121s) proposed a few decomposition pathways for TNT, including homolytic cleavage of the $\\ce{C-NO2}$ bond, rearrangement from $\\ce{C-NO2}$ to $\\ce{C-ONO}$ followed by homolytic $\\ce{O-NO}$ cleavage, and $\\ce{C-H}$ attack from an adjacent nitro group to the methyl ring substituent, but they found that only the first was fast enough to occur during detonation, the others possible only for lower temperature thermal decomposition. This initial decomposition step is the only thing fast enough to contribute to the shockwave that sets off the rest of the TNT, while the reactions that produce the final products occur (relatively) long after the initial blast has initiated the rest of the explosive.\n\n\n", "8" ], [ "\n**What may make things go kaboom without true explosion** \n\n\nFirst of all, whenever you have a mixture of an oxidizer and a fuel, whatever nature, there is always risk of fast combustion. Common gunpowder is a mixture of solids, and it can go kaboom when confined. Since nitrogroups are oxidizing, while the rest of the molecule is usually reducing, many nitrocompounds are capable of combustion without external oxidizer. The process may proceed without true explosion, for example, TNT in small portions may be safely ignited. It burns with bright yellow flame producing a lot of black thick smoke. Anyway, to produce a kaboom, the compound or mixture must have a way of decomposition producing a lot of energy.\n\n\nSelf-sustaining thermal decomposition, say of hydrogen peroxide, may produce large volumes \nof gases, and in case the reaction mixture is confined, the build-up of pressure may eventually lead for the reaction container to... fragment violently.\n\n\n**Detonation wave**\n\n\nA true detonation is described as a process, where instead of burning front, proceeding with subsonic speed, a supersonic detonation front is observed. The speed the front travels with may be above 10 km/s in some solids and over 3 km/s in gases. This implies that the process does not involve diffusion of active species or thermal energy, that travel usually with subsonic speed. So decomposition in detonation wave is pressure-induced, either from adiabatic heating or from direct mechanical stress. To achieve this, the compound must be able of exothermic decomposition, producing enough energy to sustain the pressure wave. \n\n\n**Nitrogroup role**\n\n\nCommonly utilized nitro-explosives are relatively stable and often require a significant primary explosive charge to induce detonation wave. Nitrogroups, despite having positive energy of formation, are relatively durable groups, with high barrier or bond dissociation energies, double so for nitroaromatics. This means, that in detonation wave decomposition of nitrogroups mostly consumes energy, while formation of other molecules, say, water and carbon monooxide, produces energy. So, the role of nitrogroup is mostly to provide a readily available oxygen for oxidation of the rest of the molecule.\n\n\n**Other cases**\n\n\nHowever, there is a lot of explosives (most commonly with high amount of nitrogen), that have comparatively low decomposition energy, but are extremely sensitive. Probably, ones of the most infamous would be heavy metal azides, like lead azide. There in addition to readily available exothermic decomposition route, usually some weak bonds are present in the molecule, allowing easy thermal fragmentation, following by rearrangement. \n\n\nFor example, infamous diazomethane have C-N bond energy around 172 kJ/mol, while energy of $\\ce{CH2=CH2}$ bond is 611 kJ/mol, meaning that the process produces roughly twice more energy than consumes - not a lot, but the very low initial energy investment makes diazomethane to explode from a funny look.\n\n\n**TL;DR.**\n\n\nnitrogroups mostly serve as a readily available source of oxygen to oxidize the rest of the molecule\n\n\n", "4" ] ]

[]

[ [ "\nSo, there are many questions, and I'm not sure if I understand you 100 %:\n\n\nWe have [Avogadros Law](http://en.wikipedia.org/wiki/Avogadro%27s_law), which applies for an [ideal gas](http://en.wikipedia.org/wiki/Ideal_gas) at constant pressure and temperature:\n$$\\frac{V}{n} = k \\quad\\Rightarrow\\quad \\frac{V\\_1}{n\\_1}=\\frac{V\\_2}{n\\_2}$$\nWith $k$ as a constant.\n\n\nAnd we have the [ideal gas law](http://en.wikipedia.org/wiki/Ideal_gas_law): \n$$pV = nRT \\quad\\Rightarrow\\quad \\frac{p\\_1V\\_1}{n\\_1T\\_1} = \\frac{p\\_2V\\_2}{n\\_2T\\_2}$$\n\n\nAnd this is probably what your first question is about. There are problems when you have a state $1$ and a state $2$, which differ. \n\n\nFor example: You have a container where everything ($p\\_1,V\\_1,T\\_1,n\\_1$) is given. Now you double the temperature - and the task is to calculate the pressure. Since the volume and amount of substance is fixed ($V\\_1 = V\\_2$, $n\\_1 = n\\_2$), you can just rearrange everything and calculate the pressure $p\\_2$.\n\n\nIn the ideal gas-model, gases do not seperate. I may be misunderstanding you, but if you put 1 mol $\\ce{N2}$ and 2 mol oxygen in a container, you will have different [partial pressure](http://en.wikipedia.org/wiki/Partial_pressure)s, but no demixing.\n\n\nA relevant question was [posted on physics.SE](https://physics.stackexchange.com/questions/96089/mixing-of-ideal-gas-thermodynamic-equilibrium), that may help you if you have to dive deeper into this topic.\n\n\n", "1" ], [ "\nI think some of your confusion arises from the way your questions have been posed. The laws state a relationship among pressure, volume, temperature and the number of molecules of gas (which is what moles of gas tells you).\n\n\nThe only reason we know the volume is because the gas is in a container. You may not be told this volume in the question, but gases expand to fill whatever container they are in (unless you are doing something on a planetary scale where gravity starts to count and you aren't going to get questions involving this in basic chemistry). So if volume is the unknown, you are being asked to calculate the volume of a container. Or the volume may be known and you need to calculate another of the 4 factors.\n\n\nIt is worth understanding the basis of the laws. They are a high level summary of the consequences of some simple behaviour which can be understood from simple principles. One is that all molecules in an ideal gas (which means anything not close to liquefying from cold or pressure) behave like small objects that don't interact. At a constant temperature and pressure a mole of any gas will occupy the *same* volume. This means that, if we can control the pressure and temperature, we can count the molecules (the number of moles of objects making up the gas) by measuring the volume. Since we can also weigh the gas, we know what the objects consist of, or, at least, their mass). \n\n\nBecause the molecules making up a gas don't interact, a mixture of different molecules will simply mix and we can adjust the gas equations to pretend that the components have *partial* pressures depending on their proportions in the mixture. So air (simplifying and rounding to make the numbers easier) is 75% nitrogen and 25% oxygen. 1 atmosphere of air will contain oxygen with a partial pressure of 0.25 atmosphere and nitrogen of .75 atm. You can apply the equations to the components using these numbers.\n\n\nRemember the principles and you will be able to work out how to apply the laws.\n\n\n", "1" ] ]

[ [ "\n**What's the mechanism of cross-linkage for a typical household 2-party epoxy mix?**\n\n\n\n\n\nConsidering the polymeric epoxy backbone, two functional groups can be distinguished that are chemically active.\nThe hydroxyl-groups may react with phenol-/urea-/melamine-formaldehyde resins at elevated temperatures. At room temperature reaction with for instance isocyanates is possible.\nThe epoxy group is very reactive, reaction enthalpy is around 96 kJ/mol typically, with many chemical species. Curing chemistries using phenol(ic) hardeners is for instance employed in powder coating applications whereas anhydride cure is often used in epoxy composite applications.\nAt room temperature however by far the largest groups of chemicals used are amine-functional curing agents capable of cross linking epoxy resins at room temperature.\n\n\n\n\n\nAn epoxy group reacts with ONE amine-hydrogen, i.e. one hydrogen attached directly to a Nitrogen atom (hence also amine-hydrogen equivalent weight). Note that every time that an epoxy group is opened; one Hydroxyl group is generated (adhesion !, internal catalyst !, …). Initial chemical reaction builds linear molecular weight, the molecules get longer and longer giving (s)low viscosity increase.\n\n\nThe next step in the curing process is the linking of the chains with each other; cross linking. In order for this to occur it is necessary that each amine molecule will have more-than-two active amine-hydrogen’s. The inter-linking of larger molecules causes a rapid increase of the observed viscosity.\nSterical hindrance can result in non-fully chemical cure which can be remedied by means of for instance post-cure or the use of plasticisers.\n\n\n\n\n\nNon-reacted epoxy groups, either deliberately generated through off-stoichiometry or caused by non-full cure can undergo “hydroxyl-cure” catalysed by tert.-amine and heat (sun-light).\n\n\n\n\n\n**What are polymercaptan amines? Do these contain both thiol and amine functional groups?**\n\n\nMercaptans have been used commercially for over 30 years to cure epoxies. Despite the numerous epoxy hardeners on the market, the mercaptans provide some unique properties. The main one is their ability to cure epoxies rapidly at ambient temperature. The other key property is that the mercaptans are less sensitive in their curing speed to mass of material than other hardeners. Thus, they can be used in thin-film cures. Additional factors in favor of the polymercaptans are their low toxicity compared to other hardeners, their light color, their excellent adhesion, and the ability to accelerate other amines and polyamides. \n Polymercaptans are used when the speed of cure is important. Even the “fast”-cure amine systems are slow in comparison, especially when the mass of epoxy is small and a thin film is used. This has led to the use of the mercaptans in a wide variety of applications:\nConsumer and industrial adhesives, Coatings, Sealants, Industrial patch and filler kits,Floor and flooring repair, Encapsulation, and Acceleration of other epoxy systems.\n\n\nA number of commercial mercaptan hardeners are available. They are sold with and without amine accelerators. The polymercaptan without the accelerator (Capcure 3-800) allows the formulator to have more control over the cure speed and flexibility of the product. This material is also used to accelerate amine, amidoamine and polyamide curing of epoxies. The standard five-minute cure is obtained with the accelerated Capcure® 3830-81. The fastest polymercaptan (Capcure 40 sec HV) has a gel time of 40 seconds with a 25-gram mass. There is a more water-resistant version (Capcure WR-6) for exterior uses.\n\n\nTo sum up polymercaptan amines contains at least one pair of $\\ce{SH}$ (thiol) groups and\n\n\nan amine\n\n\n**How does cross-linkage work?**\n\n\nThe chemistry of the mercaptan reactions is fairly straightforward. The tertiary amine catalyst forms a salt with the mercaptan to generate a mercaptide anion, which is a strong nucleophile. The mercaptide will readily open the epoxy. Reaction with another mercaptan group can regenerate the mercaptide anion, as illustrated in the Figure below:\n\n\n\n\n\nThe low activation energy for the process allows the reaction to go at or below room temperature. Since heat is not necessary for the reaction to occur, it can cure in thin films even on metal substrates. The cure rate does slow as the mass gets smaller, and the film is thinner, but the decrease is not as dramatic as when using amine-based hardeners. \n\n\n**Wouldn't we then need a 2:1 ratio of epoxide to hardener?**\n\n\nAnother aspect of the mercaptan chemistry is the formulation flexibility. At a 1:1 weight ratio, there is an excess of epoxy compared to mercaptan groups. However, the tertiary-amine accelerator will cause the epoxy to self-cure. Thus, some variation in ratio is self-compensating: \n1. More epoxy will self-cure by the tertiary-amine accelerator; \n2. More mercaptan will still have epoxy groups with which to react;\n3. More or less tertiary amine changes the rate of cure, but not significantly the final degree of cure.\nOf course, deviating greatly from the 1:1 ratio will cause problems of incomplete cure. \n\n\n", "5" ], [ "\n\"Epoxy\" here refers to an epoxy resin, which is comprised of oligomers that contain at least two epoxide functional groups. Hardening agents, like polyamines and polysufides, react with the resin via nucleophilic attack on the epoxide (as you desribed). Because the hardening agent has multiple nucleophilic groups, it will react with multiple epoxy oligomers. Likewise, the epoxy oligomer reacts with multiple hardening agents. This allows a complex crosslinked network to form.\n\n\nLooking at the MSDS for a typical epoxy hardener, \"polymercaptan amines\" is listed as a \"blend.\" Likely this is just a mix of polysulfides and polyamines that was given an ambiguous name by the manufacturer. These molecules likely contain a varying number of nucleophilic functional groups.\n\n\n\n> \n> Stoichiometry at its finest.\n> \n> \n> \n\n\nThe stoichiometry is actually not straightforward here at all. It is possible that the total number of epoxide groups in the resin equals the total number of nucleophilic groups in the hardener, but that isn't necessarily the case. The reason you get to mix the parts 1:1 is because the manufacturer provides an approprite blend of hardeners to provide the desired stoichiometry, whatever it may be.\n\n\n\n> \n> But then does the resulting oxide anion attack another epoxide ring?\n> \n> \n> \n\n\nThis reaction certainly competes with attack by the hardening agent. Whether it is competitive enough to be relevant depends on the identity and available amount of the hardening agent. It may even intentionally utilized in the crosslinking process.\n\n\nAs a side note, it is unlikely that an alkoxide anion is involved here. Often times a Lewis acid catalyst is added to facilitate crosslinking, which would provide an alkoxide-Lewis acid adduct. Alternatively, there are plenty of hydrogen bond donors available in these mixtures, so a proton would likely be associated with the epoxide during the ring opening, providing a neutral alcohol upon completion of the reaction.\n\n\n", "5" ] ]

[ [ "\nWhen you say: \n\n\n\n```\n When I draw the structure of it in such a way so that the 2 CH3 groups are \n adjacent to each other (on paper ) and take its mirror image I see that I got a \n mirror image which is not superimposable on the original structure.\n\n```\n\nand then:\n\n\n\n```\nOn the other hand when I draw the structure such that the 2 CH3 groups are on \nopposite sides of the central carbon and then take a mirror image of it , \nrotate it by 180° in the plane of the paper I get the original structure.\n\n```\n\nThis tells me you are drawing structures where the bonds are at right angles to each other, like this:\n\n\n\n```\n F CH3\n | |\n CH3-C-CH3 and CH3-C-H\n | |\n H F\n\n```\n\nThis will make the seeing of the structures much harder, you can't quite see the 'tetrahedral'. In fact both of the drawings you have indicated are the same, only the connectivity matters -- that is the central carbon is bonded to a fluorine, a hydrogen and two methyl groups.\n\n\nThis is a good place to use models. And if you don't want to buy tinker toys, tooth picks and colored marshmallows (or cheeses) will work just as well.\n\n\nTo resolve your conundrum, expand requirements such that if there exists any arrangement of the groups in your drawing that can form a superimposable mirror image, the molecule is not chiral.\n\n\nAlso consider using dashes and wedges, such as:\n\n\n\n\n\n", "1" ] ]

[ [ "\nThink of it like this. Ammonia and Water are both neutral molecules, therefore they won't contribute to the charge of the two coordination spheres. There are also no subscripts, which indicates that iron and copper both have the same magnitude of charge. Now copper here could be +1 and iron could be in -1, or copper in +2 and iron in -2. (Of course, this isn't possible since this molecule doesn't exist.) I cannot conclusively say which though, without any further data, such as maybe their spin only angular momenta.\n\n\nFor other complex compounds, i suggest you practice, and eventually you'll get the hang of what oxidation states are common for different metals. Also use the concepts you use for normal covalent compounds. Trust me, it's very easy! \n\n\n", "1" ] ]

[ [ "\nSee [Integration by substitution](http://en.wikipedia.org/wiki/Integration_by_substitution), see also [this question](https://math.stackexchange.com/questions/216132/integrating-each-side-of-an-equation-w-r-t-to-a-different-variable) and [another question](https://math.stackexchange.com/questions/182344/can-anyone-explain-the-intuitive-meaning-of-integrating-on-both-sides-of-the-eq) at math.SE. This is the rule we have to apply here:\n\n\n$$\\int\\_{x\\_0}^{x\\_1} f(u(x))u'(x)\\,\\mathrm dx = \\int\\_{u(x\\_0)}^{u(x\\_1)} f(u)\\,\\mathrm du $$\n\n\nSo we go from this equation:\n$$\\frac1p \\frac{\\mathrm dp}{\\mathrm dT}=\\frac{\\mathrm d\\ln p}{\\mathrm dT}=\\frac{\\Delta H}{RT^2},$$\nand integrate over temperature:\n$$\\int\\_{T\\_1}^{T\\_2}\\frac1p \\frac{\\mathrm dp}{\\mathrm dT}\\,\\mathrm dT = \\int\\_{T\\_1}^{T\\_2} \\frac{\\Delta H}{RT^2}\\,\\mathrm dT,$$\nNow we use the first equation, with $f=\\frac{1}{p}$, $p=u$, and $x=T$:\n$$\\int\\_{T\\_1}^{T\\_2}\\frac1p \\frac{\\mathrm dp}{\\mathrm dT}\\,\\mathrm dT = \\int\\_{p\\_1(T\\_1)}^{p\\_2(T\\_2)}\\frac1p\\,\\mathrm dp=\\int\\_{T\\_1}^{T\\_2}\\frac{\\Delta H}{RT^2}\\,\\mathrm dT$$\n\n\n**Bonus question:** Its just the application of the [chain rule](http://en.wikipedia.org/wiki/Chain_rule), namely\n$$\\frac {\\mathrm d}{\\mathrm dx}z = \\frac {\\mathrm dz}{\\mathrm dy} \\frac {\\mathrm dy}{\\mathrm dx} $$\nwhich, if applied leads to\n$$\\frac{\\mathrm d}{\\mathrm dT}\\ln p = \\frac{\\mathrm d\\ln(p)}{\\mathrm dp} \\frac{\\mathrm dp}{\\mathrm dT} = \\frac{1}{p}\\frac{\\mathrm dp}{\\mathrm dT} .$$\n\n\n", "6" ] ]

[ [ "\n[Palladium (II) Chloride](http://en.wikipedia.org/wiki/Palladium(II)_chloride) follows neither the 18 electron rule nor the octet rule. \n\n\nInstead, like many Pd(II), Pt(II) and Ni(II) complexes, palldium (II) chloride follows the [16-electron rule](http://www.yorku.ca/stynes/Tolman.pdf).\n\n\nPd(II) generally forms square planar complexes, and in palldium (II) chloride each Cl bridges two Pd, with 4 Cl coordinating each Pd in a roughly square planar arrangement. 8 d-electrons plus 2 electrons from each of 4 sigma bonds equals 16 electrons.\n\n\nI have never heard of \"Cl3Pd\", which would involve the rare oxidation state Pd(III), but the 16 and 18 electron rules should only be applied to diamagnetic complexes. Pd(III) complexes are paramagnetic. \n\n\n", "1" ] ]

[ [ "\n*Note that I can only comment on the English translations!*\n\n\n**Cyclohexanol** is correct, cyclohexol is just wrong. This is not the only error in the abstract of the thesis. \n\n\n\n> \n> Cyclohexone is the most [...]\n> \n> \n> \n\n\nisn't correct either. The name is **cyclohexanone**!\n\n\n", "6" ] ]

[ [ "\nHere is the Walsh diagram depicting all the valence molecular orbitals (a diagram showing how individual molecular orbitals change in energy due to bending around the central atom). Oxygen has 6 valence electrons, so ozone has 18 electrons in total. If we start on the right where ozone would be linear, we can see that all the orbitals up to the $2\\pi\\_\\mathrm u$ orbitals (don't worry about why they are named this) are doubly occupied and the two $2\\pi\\_\\mathrm u$ orbitals are both singly occupied.\n\n\n\n\n\nNote: 6a$\\_1$ and 2b$\\_1$ have the incorrect sign on the outer p-orbitals.\n\n\nIf we bend the molecule slightly (moving to the left on the diagram), we can see that there is a favorable interaction between the p-type orbitals on the end as well as between the black of the central p-type orbital and the black of the outer p-type orbitals (Someone drew the $\\mathrm{6a\\_1}$ and $\\mathrm{2b\\_1}$ orbitals wrong, flip the central orbital). This lowers the energy of the molecule.\n\n\nSo why doesn't it keep bending? As a good first-order approximation, we can estimate the relative energy of configurations by the highest energy orbital (provided the other orbitals don't change too much). We can see that if we keep bending, the $\\mathrm{1a\\_2}$ orbital and the $\\mathrm{4b\\_2}$ orbital start rising in energy, eventually rising above the $\\mathrm{6a\\_1}$ orbital. Thus too much bending will be unfavorable, and thus ozone prefers a bond angle of around $117^\\circ$.\n\n\nThis same diagram can be used for other molecules, such as $\\ce{CO2}$. Try using it to figure out why $\\ce{CO2}$ is linear.\n\n\n", "21" ], [ "\nYour guess about electronic repulsion is right: the repulsion between the lone pairs of electrons on the Oxygen atoms leads to intense repulsion at such close ranges leading to instability. In comparison to this resonance is a far far more stable configuration.\n\n\nStrain at the joints in the cyclic structure does its part as well because experimentally it is found to be abnormal. The strain causes instability in bond and eventually the molecule will break into constituents. The ideal angle for stability is 109 degrees and 28 minutes (I don't know why, but experiments show this is true). Angular strain can be found in water as well: to decrease that strain water collapses to an angle of 104 degrees.\n\n\n**Ring strain or angular strain is a type of instability that exists when bonds in a molecule form angles that are abnormal (*i.e., not generally found in nature*).**\n\n\n", "9" ], [ "\nPossible Reasons:\n\n\n* **Electronic Repulsion**: In order to form a 3-member ring your electron pairs would have to bend at 60 degrees; the energy required to push naturally repulsive electrons so close together would make the molecule very unstable.\n* **Unstability of three membered rings (angular strain):**\n\n\n", "1" ] ]

[ [ "\nAmmonia is basic so:\n$$\\ce{NH3 +H+<=>NH4+}$$\nSulphuric acid is acidic(since it's name is itself so acidic!):\n$$\\ce{H2SO4<=>H+ +HSO4- <=>2H+ +SO4^2-}$$\nNow the final reaction would be like:\n$$\\ce{2NH3 +H2SO4<=>NH3 +NH4+ +HSO4- <=>2NH4+ +SO4^2- <=>(NH4)2SO4}$$\nAnd remeber since proteins are polymers of peptides which itself are molecules consisiting of both a carboxylic ($\\ce{-COOH/-COO-}$) and amino ($\\ce{-NH3/-NH4+}$) group, therefore with strong heating ammonium sulfate is expected.\n\n\n", "3" ] ]

[ [ "\nAn acid dissociation constant is a particular example of an equilibrium constant.\nFor the reaction:\n$$\\ce{CH3COOH +H2O <=>CH3COO- +H3O+}$$\nThe equilibrium constant would be:\n$$\\newcommand{\\ac}[1]{\\text{activity of }\\ce{#1}}{\\bf K\\_{eq}}=\\frac{\\ac{CH3COO-}\\cdot\\ac{H3O+}}{\\ac{CH3COOH}\\cdot\\ac{H2O}}$$\nAnd the acid dissociation constant would be:\n$${\\bf K\\_a}=\\ce{\\frac{[CH3COO-][H3O+]}{[CH3COOH]}}$$\nNow the relation between both would be:\n$$\\newcommand{\\acc}[1]{\\gamma\\_{\\ce{#1}}}{\\bf K\\_{eq}}={\\bf K\\_a}\\underbrace{\\frac{\\acc{CH3COO-}\\acc{H3O+}}{\\acc{CH3COOH}\\acc{H2O}}}\\_{\\Gamma\\_0}\\frac1{\\ce{[H2O]}}$$\nNow:\n$${\\bf K\\_a}={\\bf K\\_{eq}}[\\ce{H2O}]\\Gamma\\_0^{-1}$$ \nSo you're missing some factors.They're not exactly the same.\n\n\n", "3" ] ]

[ [ "\nThe random displacements in your suggested method are uncorrelated from step to step. What you are suggesting is essentially heavily damped molecular dynamics at the Brownian limit. The limitations are obvious, you won't sample correlated, large-scale motions efficiently. (That might be a student project right there: turn up the Langevin parameter, see when you lose rotations of phenyl rings, and ask yourself what this means.)\n\n\n", "8" ], [ "\nI use Molecular Workbench (<http://mw.concord.org/modeler/>) when I want to employ MD for teaching. Free of charge and easy to use. No need to reinvent the wheel.\n\n\n", "4" ] ]

[ [ "\nChEMBL database provides API access: <https://www.ebi.ac.uk/chembl/ws>\nproperties can be retrieved from the 'molecule' endpoint, for example aspirin:\n\n\n<https://www.ebi.ac.uk/chembl/api/data/molecule/CHEMBL25.json>\n\n\nDisclaimer: I'm the author of this API.\n\n\n", "7" ], [ "\nI am not sure this is what you are looking for, but there is the Padel descriptor software that calculates a whole slew of properties for compounds you throw at it. Nothing empirical though, all calculated values.\n\n\nSee here: <http://www.yapcwsoft.com/dd/padeldescriptor/>\n\n\nIf you are willing to use a tool like KNIME, there is a Padel KNIME extension available on de Padel website. It is somewhat fiddly to get working properly in KNIME without conflicts, but once you have that going you can pump millions of molecules through this pipeline.\n\n\n", "5" ], [ "\nPubChem now have a [REST API serving properties](https://pubchemdocs.ncbi.nlm.nih.gov/pug-rest-tutorial$_Toc458584424) as part of their [PUG REST](https://pubchemdocs.ncbi.nlm.nih.gov/pug-rest) service.\n\n\nThere's a [table of available 2D and 3D properties that can be selected](https://pubchemdocs.ncbi.nlm.nih.gov/pug-rest$_Toc494865567), and a [variety of input and output formats supported](https://pubchemdocs.ncbi.nlm.nih.gov/pug-rest$_Toc494865555), including database ID, SMILES and SDF input, and JSON/XML/CSV outputs.\n\n\nFor example, to `GET` a set of properties for a SMILES string for benzene, in JSON output format:\n\n\n\n```\nhttps://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/smiles/c1ccccc1/property/MolecularFormula,MolecularWeight,InChIKey/JSON\n\n```\n\nResult:\n\n\n\n```\n{\n \"PropertyTable\": {\n \"Properties\": [\n {\n \"CID\": 241,\n \"MolecularFormula\": \"C6H6\",\n \"MolecularWeight\": 78.11,\n \"InChIKey\": \"UHOVQNZJYSORNB-UHFFFAOYSA-N\"\n }\n ]\n }\n}\n\n```\n\nYou can also [move request parameters into a `POST` request body](https://pubchemdocs.ncbi.nlm.nih.gov/pug-rest$_Toc494865561), allowing `POST`ing of SDF format molecules.\n\n\nFor example, using `curl` on the command line to determine the 3D volume of methane:\n\n\n\n```\n$ curl --url 'https://pubchem.ncbi.nlm.nih.gov/rest/pug/compound/sdf/property/MolecularFormula,Volume3D/JSON' \\\n --form 'sdf=Untitled\n xed2sdf_05210316543D\n\n 1 0 0 0 0 0 0 0 0 0999 V2000\n -5.1559 1.5833 1.7170 C 0 0 0 0 0 0 0 0 0 0 0 0\nM END\n\n$$$$'\n\n```\n\nResult:\n\n\n\n```\n{\n \"PropertyTable\": {\n \"Properties\": [\n {\n \"CID\": 297,\n \"MolecularFormula\": \"CH4\",\n \"Volume3D\": 20.6\n }\n ]\n }\n}\n\n```\n\nFrom a quick test, it appears that properties are pre-calculated for CID database entries matching the query, so it appears that any 3D properties are not calculated on the input conformation, but rather the PubChem database's conformation. The docs suggest this, saying 3D properties are for the \"first diverse conformer (default conformer)\".\n\n\n", "1" ], [ "\nYour question is rather broad. So the answers can't be too narrow. If you look for material properties like DoS and phase diagrams, try [The Materials Project](https://materialsproject.org/). Working with [Pymatgen](http://pymatgen.org/) to brigde to programs like VASP and Gaussian «to compute further», the API is briefly described [here](https://materialsproject.org/docs/api).\n\n\n", "0" ] ]

[ [ "\nOne of the Factors Influencing reaction rate is given below ([Source](http://en.wikipedia.org/wiki/Reaction_rate#Factors_influencing_rate_of_reaction))\n\n\n\n> \n> Temperature: Usually conducting a reaction at a higher temperature delivers more energy into the system and increases the reaction rate by causing more collisions between particles, as explained by collision theory. However, the main reason that temperature increases the rate of reaction is that more of the colliding particles will have the necessary activation energy resulting in more successful collisions (when bonds are formed between reactants). The influence of temperature is described by the Arrhenius equation. As a rule of thumb, reaction rates for many reactions double for every 10 degrees Celsius increase in temperature, though the effect of temperature may be very much larger or smaller than this.\n> \n> \n> \n\n\nReaction rate , Activation energy are connected to temperature through Arrhenius equation. We are interested in [Activation Energy](http://en.wikipedia.org/wiki/Activation_energy) as it gives an idea what temperature the reaction starts. We want to know $E\\_{a}$ in order to manipulate reaction rate essentially. [Catalysts](http://en.wikipedia.org/wiki/Catalysis) or Inhibitors work essentially on this principle.\n\n\nJust looking at rate wont help us in varying the rates to our interest. \n\n\n", "1" ] ]

[ [ "\nThe ions combine while in solution, and since the resulting compound is not soluble, it precipitates. This happens so quickly because as the precipitate forms, it takes those individual ions out of solution, thus pulling the reaction toward the products side, according to **Le Chatelier's principle**.\n\n\n", "5" ] ]

[ [ "\nThe relative strengths of oxidising and reducing agents can be inferred from their standard electrode potentials. The strongest oxidizing agents are shown from the standard electrode table.\n\n\n", "1" ] ]

[ [ "\n2.3:10, 23:100 and (23/123):(100/123) are all correct values for the ratio of the relative abundance of B-10 to B-11. However, only the last value, (23/123):(100/123) or 18.7:81.3 shows the **normalized** ratio of relative abundance of these isotopes. It is this normalized value that adds up to 100.\n\n\n", "7" ] ]

[ [ "\nHere I'll only provide the reasoning for $\\ce{O2}$ and $\\ce{H2}$ diatomic structure, I think similar reasoning was to deduce that of $\\ce{Cl2}$.\n\n\nThe divalency of oxygen, shown by Avogadro's hypothesis.\n--------------------------------------------------------\n\n\nIt is known that two volumes of hydrogen gas and one volume of oxygen gas combine to form water. From Avogadro's hypothesis (equal volumes of gas contain equal numbers of molecules) it can be concluded that water must have two atoms of hydrogen and one atom of oxygen (or conceivably some multiple). Hence, oxygen is divalent (valence of 2) while hydrogen has a valence of 1:\n\n\n$$\\ce{2H + O \\to H2O}$$\n\n\nThis would indicate that $\\frac12$ gram of hydrogen atoms reacted with 8 grams of oxygen atoms. Since the\natomic weight of hydrogen was defined as $H = 1$, then the respective atomic weights of hydrogen and oxygen would be $H = 1$ and $O = 16$ (the values accepted today).\n\n\nThe diatomic nature of hydrogen and oxygen, as shown by Avogadro's hypothesis.\n------------------------------------------------------------------------------\n\n\nIt is observed that when two volumes of hydrogen gas and one volume of oxygen gas combine, two volumes of water gas are formed. If hydrogen and oxygen were monatomic (as shown in the equation above), then only one volume of water would be formed. The only explanation is that hydrogen and oxygen are diatomic:\n$$\\ce{2H2 + O2 \\to 2H2O}.$$\n\n\n(Copied almost verbatim from [this page](http://www.chem.unt.edu/%7Ejimm/CHEM%201360%20Chapters%20and%20Problems/PROOF%20THAT%20OXYGEN%20IS%20DIVALENT%20and%20that%20HYDROGEN%20AND%20OXYGEN%20ARE.pdf))\n\n\n", "3" ], [ "\nThe reason to form a bond is usually for stability, and so that is why some elements are diatomic.\n\n\nIn the case of chlorine, it is has -effectively- 7 electrons in its outer shell. The atom will be most stable when it has 8 electrons, and so it can share an electron with another chlorine atom, which will result in them both having 8 electrons in their outer shell.\n\n\nSimilarly, hydrogen has one electron in its outer shell. At this energy level, it needs 2 electrons in its outer shell, and so can share (in what's called a covalent bond) an electron with another hydrogen atom.\n\n\nOxygen has 6 electrons in its outer shell, and so needs two more electrons to achieve stability. To do this, it forms a double covalent bond with another oxygen molecule, and so it achieves a full outer shell and becomes stable.\n\n\nMetals, such as sodium and potassium, are not actually single atoms, but atoms that are joined together in a lattice. It is made up of positive metal ions surrounded by a 'sea' of delocalised electrons. This means that they are as stable as they want to be, and so do not need to form covalent bonds with other atoms.\n\n\n**Edit After Comment**\nEvidence for diatomic molecules can be seen in several different ways. One example is by looking at an electron density map of, for example, chlorine:\n\n\n\nWe can see that there are two atoms and these are sharing electrons.\n\n\nAlso, we can look at the mass spectrum of a chlorine molecule:\n\n\n\nHere, the two peaks around 35 are the two isotopes of chlorine: chlorine-35 and chlorine-37. Further on, near 70, we see that there are three different peaks. These peaks are where a diatomic molecule of two chlorine isotopes has been formed. The peaks are of a 35-35 molecule, a 35-37 molecule, and a 37-37 molecule, respectively. This shows that chlorine forms a diatomic molecule with two of its isotopes. We can then extend this to saying that because it would be theoretically more stable, and that it is a gas at room temperature, it is likely that gases like chlorine are diatomic in standard conditions and so a covalent compound. \n\n\nThe evidence for the metallic structure which I described is effectively its characteristics. The fact that metals can conduct electricity implies that the delocalised electrons are free to move through the lattice when a potential difference is applied. Similarly, the way metals conduct heat can also be explained using this model, as the electrons move easily and so can transmit kinetic energy rapidly through the lattice.\n\n\nThe high melting and boiling temperatures suggest that the forces between metal atoms must be large because it takes a lot of energy to separate them. The metallic structure suggests that the lattice of positive ions are held tightly together by this electron 'glue'.\n\n\nLastly, metals are malleable and ductile, which indicates that the positive metal nuclei have room to move within the sea of electrons.\n\n\n", "1" ], [ "\nHydrogen, oxygen, chlorine, etc, prefer to fill their valence electron shell by sharing electrons. Energy is released in the sharing a pair of electrons then individual atoms.\n\n\nSodium, potassium, etc, would rather give up an electron (to have a full valance electron shell), then to share electrons in an attempt to fill up the electron shell. In the case of bonding, a little energy is used to release an electron and a lot of energy is released when another atom pulls it into its valance shell. \n\n\nSince the metal atom would use energy to take an extra electron, and the release uses energy, metal atoms will not take an extra electron.\n\n\n", "-1" ] ]

[ [ "\nFrom [Wikipedia](http://en.wikipedia.org/wiki/Hydroxylamine):\n\n\n\n> \n> The reaction of $\\ce{NH2OH}$ with an aldehyde or ketone produces an oxime. \n> \n> $$\\ce {R2C=O + NH2OH·HCl + NaOH → R2C=NOH + NaCl + H2O}$$\n> \n> \n> \n\n\nFor the penta-acetate ester of glucose not to react with hydroxylamine would indicate that the ester is more stable, or there is steric hindrance to the reaction. Since hydroxylamine is heat sensitive, I am guessing that you did not heat the reaction. Given time, you should see the oxime form.\n\n\nFrom [Wikipedia](http://en.wikipedia.org/wiki/2,4-Dinitrophenylhydrazine):\n\n\n\n> \n> A positive test is signaled by a yellow, orange or red precipitate (known as a dinitrophenylhydrazone).\n> $$\\ce {RR'C=O + C6H3(NO2)2NHNH2 → C6H3(NO2)2NHNCRR' + H2O}$$\n> \n> \n> \n\n\n2,4-DNP also reacts with an aldehyde (or ketone), so both glucose and the penta-acetate ester of glucose should react. If you get a different color for the penta-acetate ester, it is due to the substituents on the glucose.\n\n\n", "4" ] ]

[ [ "\nThe H-C-H bond angle in ethene is *ca.* 117 degrees and the H-C-C angle is *ca.* 121.5 degrees. There are two reasons that combine to explain this angular deformation in ethene.\n\n\n**First**, from these bond angles and Coulson's Theorem ([ref\\_1](https://chemistry.stackexchange.com/questions/10653/why-does-cyclopropane-give-bromine-water-test/10666#10666), [ref\\_2](https://chemistry.stackexchange.com/questions/19977/determine-the-bond-angle-in-a-compound/19987#19987)) we can determine that the C-H sigma bonds are $\\ce{sp^{2.2}}$ hybridized and the C-C sigma bond is $\\ce{sp^{1.7}}$ hybridized. \n\n\nFrom these hybridization indices (the index is the exponent \"n\" in the $\\ce{sp^{n}}$ expression) we see that the C-C sigma bond has higher s-character content (1 part s to 1.7 parts p - 37% s) than the C-H bonds (1 part s to 2.2 parts p - 31% s). Since there is more s character in the C-C bond, it is lower in energy and the carbon sigma electrons will tend to flow towards this lower energy C-C bond. Consequently, the C-C sigma bond will contain more electron density than the C-H bonds. Therefore, the electron repulsion between the C-C sigma bond and C-H sigma bonds will be greater than the electron repulsion between the two C-H bonds. Hence the H-C-C bond angle will open up slightly from the $\\ce{sp^{2}}$ ideal of 120 degrees and the H-C-H angle will close down slightly in order to minimize the bond-bond electrostatic repulsions. \n\n\n**Second**, steric factors (which are also really just another way of describing electron-electron repulsion) may also come into play. To whatever extent the cis H-C-C-H hydrogen-hydrogen repulsion is more destabilizing than the geminal H-C-H hydrogen-hydrogen repulsion, it will also serve to increase the C-C-H bond angle and shrink the H-C-H bond angle.\n\n\n", "7" ], [ "\nIn terms of the strongly simplifying VSEPR model, you could arrive at the same conclusion by noting that a carbon atom is larger than a hydrogen atom, so the $\\ce{CH2}$ group will request more space than each hydrogen. You can compare this to ethane $\\ce{H3C-CH3}$, which has $\\angle(\\ce{H-C-C}) = 111.17^\\circ$, slightly deviating from the ideal tetrahedral angle of $109.5^\\circ$ but no π system to explain the deviation. If I did the maths correctly, that means that $\\angle(\\ce{H-C-H}) = 107.72^\\circ$. The deviation in ethane is actually larger than that in ethene, suggesting that steric factors predominate.\n\n\n[](https://i.stack.imgur.com/EVFj1.png) \n\n**Figure 1:** Ethane’s structure including bond lengths and angles. Taken from [Wikipedia](https://commons.wikimedia.org/wiki/File:Ethane-staggered-CRC-MW-dimensions-2D.png?uselang=en), where a full list of authors is available.\n\n\n", "2" ] ]

[ [ "\nWith the concept [Resonance](http://en.wikipedia.org/wiki/Resonance_%28chemistry%29) you can describe delocalized electrons in some molecules. To go with your example:\n\n\n\n([source](http://en.wikipedia.org/wiki/Resonance_%28chemistry%29))\n\n\nImagine the ozone molecule to be frozen in one of those two states. Positive and negative charges are localized, which is generally connected to a higher energy, i.e. a more reactive species. This would be an extreme case for ozone - in reality those electrons will be delocalized all over the molecule. So, what is the reason that molecules with localized electrons/charges have a higher total energy and are, hence, less stable?\n\n\nMaybe you remember the [Particle in a box](http://en.wikipedia.org/wiki/Particle_in_a_box). I wont go into the quantum chemistry here, but one of the take-away-messages is, that the energy for particles in a infinite square well, which is a very rough approximation of the potential in which the outer electrons reside in, does decrease with $ L^{-2} $. $L$ is the size of the box in which the particle can exist. So, the more delocalized the electrons are, the less total energy in a system.\n\n\nWith this, to correct you question a little bit: Resonance is not the reason why molecules are more stable, it is a concept which describes the bonding in a molecule, in which electrons are delocalized, better than a single [Lewis Structure](http://en.wikipedia.org/wiki/Lewis_structure).\n\n\nI would like to have a feedback if this answer is somewhat understandable.\n\n\n", "12" ], [ "\nStrictly speaking resonance *doesn't* stabilise molecules. This is because resonance is merely a descriptive way for chemists to extend simple pictures of bonding to molecules where the simplest model of bonding doesn't work. We *observe* that some molecules are more stable than the simple picture, or a different shape than the simple picture would suggest and we *explain* that difference by adding the idea of resonance.\n\n\nThe problem arises because, when we draw *simple* pictures of the bonding in molecules, we assume each bond consists of two shared electrons. So we draw benzene like this:\n\n\n\n\n\nThis picture neatly accounts for all the electrons in the C-C bonds and is often good enough for understanding and tracking the electrons and bonds in reaction mechanisms. But the picture is clearly **wrong** as it doesn't match the known properties of benzene: in particular *all the bonds in benzene are the same length*. The stability of benzene is also understated compared to the (hypothetical) molecule with isolated double bond. The real reasons for this are pretty complex and involve all sorts of quantum-mechanical molecular orbital gobbledygook. But this is way too complex to be useful when drawing simple molecular pictures or describing reactions or bonding.\n\n\nTo avoid this problem and to help retain the simple picture we add the idea of *resonance*. This allows us to use the simple picture where a line representing a bond still means two shared electrons but we have to draw more than one structure to capture the real molecular structure. We say the multiple structures \"resonate\" but this is really just shorthand for saying that the real structure is an in-between mix of the resonance structures we draw. For benzene we draw like this:\n\n\n\n\n\nIn the first structure there is a single bond between atoms 1 and 2; in the second structure there is a double bond. Real benzene is a mix of the two resonance structures where all the bonds are a sort of mix of a single and double bond. In fact benzene is sometimes drawn like this:\n\n\n\n\n\nThe problem with this structure is that it is is less useful most of the time than the (incorrect) structures with single and double bonds as it makes counting electrons in bonds less clear and this is annoying when trying to follow reaction mechanisms. \n\n\nSo resonance is basically a pragmatic way to retain simplistic pictures of bonding in molecules while recognising that that simplistic pictures is often a poor representation of reality. And thereby avoiding having to do complicated quantum stuff with orbitals every time you want to understand a molecular structure or reaction. \n\n\n", "6" ], [ "\nLet us first discuss the more fundamental phenomenon of covalent bonding. How do you think a covalent bond \"works\"? Let us say we have two hydrogen atoms. We know that the nuclei of these atoms are positively charged and the electrons negatively charged. Coulombic repulsions act between the two nuclei, and between the electrons from the two atoms. Coulombic attractions act between the electrons and nuclei within the same molecule and betweem those of the two different molecules. If the two atoms are far apart from each other, attractions between the electrons of one atom and the nucleus of the other are greater than the repulsions. If the atoms get too close to each other, repulsions are greater than the attractions. At exactly some seperation, attraction balances repulsion and the atoms settle for a seperation called the \"bond-length\", and the atoms are said to be \"covalent-bonded\".\n\n\nAfter that ridiculously long explanation of what a covalent bond is, let us get to the essence of it. The reason why a covalent bond \"works\" is because electrons are holding atoms 'together'. Without the electrons, the atoms would not have anything to 'bind' them. In essence, two electrons are holding two atoms in place, thereby bringing stability, and causing a lowering of potential energy of the system. \n\n\nIn a benzene molecule, the same thing happens. The p-orbitals of the induvidual carbon atoms in benzene overlap with each other, thereby allowing the electrons in these p-orbitals to de-localise, meaning that an electron in the p-orbital of one carbon atom no longer belongs to only one atom, but instead acts to bind all other carbon atoms in the benzene ring, just as the two electrons in the $H\\_2$ molecule helps bind the two atoms of hydrogen together.\n\n\n \n\n\n* Image credits: chemwiki.ucdavis.edu\n\n\nSince the electron belongs to the orbitals of many other atoms, and not only one or two (as in the case of the H-H covalent bond), it holds all these six carbon atoms in place, and thereby acts as a bond between the six carbon atoms, which in turn brings stability to the system and lowers its potential energy. \n\n\nAppendix:\n\n\n**Why doesn't an electron in an atom fall into the nucleus ?**\n\n\nThe picture we often have of electrons as small objects circling a nucleus in well defined \"orbits\" is actually quite wrong. The positions of these electrons at any given time are not well-defined, but we CAN figure out the volume of space where we are likely to find a given electron if we do an experiment to look. For example, the electron in a hydrogen atom likes to occupy a spherical volume surrounding the proton. If you think of the proton as a grain of salt, then the electron is about equally likely to be found anywhere inside a ten foot radius sphere surrounding this grain, kind of like a cloud. \n\n\nThe weird thing about that cloud is that its spread in space is related to the spread of possible momenta (or velocities) of the electron. So here's the key point, which we won't pretend to explain here. The more squashed in the cloud gets, the more spread out the range of momenta has to get. That's called Heisenberg's uncertainty principle. Big momenta mean big kinetic energies. So the cloud can lower its potential energy by squishing in closer to the nucleus, but when it squishes in too far its kinetic energy goes up more than its potential energy goes down. So it settles at a happy medium, and that gives the cloud and thus the atom its size. (from <https://van.physics.illinois.edu/qa/listing.php?id=1226>)\n\n\n", "4" ] ]

[ [ "\nThe formula of $\\ce{Fe3O4}$ is a shorthand for $\\ce{Fe^{II}Fe^{III}2O4}$, so there are two sorts of atoms of iron here in +2 and +3 oxidation state.\n\n\nHowever, it is still possible to gain truly fractional oxidation states. The simplest example probably would be potassium superoxide, $\\ce{KO2}$. Here, the anion is symmetrical $\\ce{O2^{-}}$ anion, so obviously both oxygen atoms are in $\\ce{-1/2}$ oxidation state. There are more complicated examples as well. \n\n\nOften fractional oxidation state comes from symmetrical structures with charge, that would have unpaired electrons otherwise. However, systems with truly fractional oxidation states usually have bonding, hard to describe in terms of common Lewis structures (and with little relevance of idea of oxidation state), and mostly are considered in semi-advanced organic chemistry and advanced inorganic chemistry.\n\n\n", "16" ], [ "\nYour assumption that all iron atoms in magnetite ($\\ce{Fe3O4}$) have the same oxidation state is wrong.\n\n\nIt is a mixed $\\ce{Fe(II,III)}$ oxide.\n\n\n", "8" ], [ "\nThe iron atoms in [magnetite](https://en.wikipedia.org/wiki/Magnetite \"magnetite\"), $\\ce{Fe3O4}$, are not all in the same oxidation state: per Michael's comment, while it is not a [spinel](https://en.wikipedia.org/wiki/Spinel \"spinel\")-type material, it has a spinel structure (strictly, an inverse spinel structure), with one Fe(II) center and two Fe(III) centers.\n\n\n", "6" ], [ "\nAnalyzing the structure:(All other answers tell you why $\\ce{Fe3O4}$ has a fractional oxidation state, but without having a look at the structure you wont be able to tell if there are two $\\ce{Fe}$ atoms in +3 state or one, thus it makes it important to analyze the structure of $\\ce{Fe3O4}$)\n\n\n[](https://i.stack.imgur.com/kJnrv.png)\n\n\nHere the two Iron atoms that are bonded to three oxygen atoms have an oxidation state of +3 each, and the Iron atom that is bonded to two oxygen atom has an oxidation state of +1.\n\n\nThus taking the average (3+3+2)/3 = 8/3;\n\n\nYou would have got the same if you used the fact that $\\ce{Fe3O4}$ is neutral.If you assume the oxidation state of Iron to be x, then 3x+4(-2)=0. Solving you would get x as 8/3(However this oxidation state would be the average oxidation state of Iron).\n\n\n", "1" ] ]

[ [ "\nNature always try to acheive least potential energy for stability . Any organic reaction which makes the product less stable is not favourable . \n\n\nIf the C-H bonds are cleaved homolytically in all cases , the resulting stability of the free radicals will tell us about the feasibility of the reaction . \n\n\nThe primary free radical is the most unstable and to bring about its C-H bond breaking would mean increasing its energy comparitively . Hence its bond energy is the most and so its the stronger bond of the three . \n\n\nTertiary free radical is the most stable while secondary lies in between them . The same explanation can be used to draw out their comparitive strength of the C-H bonds .\n\n\n", "3" ] ]

[ [ "\nResonance structures are a useful way of visualizing and representing the bonding in a molecule - they don't actually exist, but they are a useful tool to see what the bonding and structure of a compound is.\n\n\nBonding in a compound can sometimes be nicely represented using **two-center, two-electron** bonds - for example methane. The structure and reactivity of methane can be rationalised by representing it as a carbon with four hydrogens. Each \"bond\" can be represented by having two electrons, one from carbon, and one from hydrogen, in a **bonding pair**. You might draw this as a combination between an sp3 hybrid and a 1s orbital. In reality these hybrid orbitals don't exist either - they are merely a tool to simplify the bonding to **localised bonding** which is easier to understand. In reality there a multiple interactions between the carbons 2s and 2p orbitals, with the 1s orbital of the hydrogen. However, it is easier, and an accurate enough **model** to model methane with four localised bonds.\n\n\nResonanace is useful where this idea of **two-center, two electron** bonds breaks down - for example we know from experiment that the structure of benzene is a regular hexagon, with **identical** carbon-carbon bonds. However the idea of **two-center two-electron** bonds breaks down here - we can't assign a structure using the ideas we know - carbon makes four bonds, the shape of benzene is a regular hexagon with identical bonds. In this case we can represent benzene as an in-between structure between two structures that **can** be represented as nice **two-center, two-electron** bonds. When we draw benzene as a resonance structure of two rings with alternating double bonds, what we are trying to represent is that the real structure of benzene is somewhere between these two structures - it is neither of the structure, and it doesn't **resonate** - ie. flick really fast between the two structures - instead it is a mixture of the two structures. In benzene the structure we are trying to represent can be understood by considering each carbon to be sp2 hybridized - again these hybrids aren't real, but a tool to understand bonding more easily. Then we have the 6 p-orbitals in the plane of the molecule. In reality these combine two give 6 molecular orbitals. These give a final structure where the bonding is identical between all six carbon atoms, a kind of one and a half bond - but the shape and idea of this delocalised bonding is very important to the reactivity of benzene and other aromatics - which is where this idea of resonance starts to break down in some ways, but the structures can still be useful, but with a proper understanding of the bonding behind the structure.\n\n\n", "5" ], [ "\n\n> \n> How can we use them if we know that they don't actually exist?\n> \n> \n> \n\n\nResonance contributors don't independently exist but rather a composite of the resonance contributors exists.\n\n\n\n> \n> Are they some sort of calculation tool?\n> \n> \n> \n\n\nSo no, resonance contributors don't exist. However, examination of resonance structures can lend insight into reactivity. Why might phenol's conjugate base be unusually stable? We can rationalize it by drawing multiple resonance contributors. These contributors are stabilizing because they show the delocalization of electron density from the oxygen to various sp2 ring carbons. \n\n\nThis, however, we cannot do for something such as cyclohexanol, which has no resonance contributors. \n\n\n", "1" ] ]

[ [ "\nThe reactivity of amines is generally based on how nucleophilic they are. In turn, their nucleophilicity is dependent upon how much of the nitrogen lone pair of electrons really exists on the nitrogen atom (*e.g.* the lone pair electron density on nitrogen). In ammonia ($\\ce{NH3}$), the lone pair electron density on nitrogen is very high making ammonia quite nucleophilic, quite reactive. In the case of nitrogen trifluoride ($\\ce{NF3}$), the **three** extremely electronegative fluorine atoms inductively remove much of the electron density from nitrogen; therefore $\\ce{NF3}$ is less reactive than ammonia in a nucleophilic sense.\n\n\n", "2" ], [ "\nYashbhatt - interesting question, but I think that this question is too vague. Reactivity is a *very* broad term. \n\n\nAre we talking about reactivity in the context of proton transfer? In that case NH3 will be more reactive than NF3 as a proton donor (Brønsted acid) since it actually has protons to donate. \n\n\nIn the context of Brønsted basicity, again, NH3 will likely be more reactive than NF3 because in NF3 the electron-withdrawing fluorines reduces affinity for positive charges. \n\n\nIn the context of nucleophilicity, a lot of factors come into play. In a protic solvent, it's hard to tell *a priori* whether NH3 is more or less nucleophilic than NF3. NH3 has a greater partial negative charge on its central nitrogen atom. But this may also mean greater hydrogen bonding and therefore reduced nucleophilicity versus NF3, which doesn't exhibit hydrogen bonding (due to its lack of hydrogens). \n\n\n", "1" ] ]

[ [ "\nLike user137 has already written in the comment section it should not be harmful unless exposed for a long time. The main charateristic is its hygroscopy, so skin may feel dry.\n\n\nTo cite from a safety data sheet ([source](http://www.sigmaaldrich.com/MSDS/MSDS/DisplayMSDSPage.do?country=DE&language=EN-generic&productNumber=391484&brand=SIAL&PageToGoToURL=http%3A%2F%2Fwww.sigmaaldrich.com%2Fcatalog%2Fproduct%2Fsial%2F391484%3Flang%3Dde)):\n\n\n\n```\n4 FIRST AID MEASURES\n4.1 Description of first aid measures\nIf inhaled\n If breathed in, move person into fresh air. If not breathing, give artificial respiration.\nIn case of skin contact\n Wash off with soap and plenty of water.\nIn case of eye contact\n Flush eyes with water as a precaution.\nIf swallowed\n Never give anything by mouth to an unconscious person. Rinse mouth with water.\n\n```\n\nNevertheless: Always use the safety data sheets your distributor provided and read them before dealing with the chemicals, if the safety hazards are not clear.\n\n\n", "6" ] ]

[ [ "\nI don't think the neutral protomer of an amino acid can be generated in Avogadro. What *can* be done is adjusting the protonation state for different values of pH. That option is in the **Build** menu - **Build**-> **Add Hydrogens for pH**.\n\n\n", "1" ] ]

[ [ "\nThe naming conventions for atoms in PDB file require that at the C-terminus of a protein one of the carboxylate oxygens is called `O` and the other is called `OXT`. But since it's a deprotonated carboxylate group they are both chemically equivalent, and you'd expect that bond lengths are equal.\n\n\n", "2" ] ]

[ [ "\n*If you don't mind, I'll start with a more colloquial and hopfully not to lengthy explanation*:\n\n\nPrecious metals, such gold, do not corrode under \"household conditions\", others, such as iron or aluminium do. Rusting of iron is a typical example. Here, the corroded surface often is very porous and the corrosion carries on until the whole piece of metal is turned to rust (oxidized).\n\n\nIn the case of aluminium, the initial corrosive layer is very tight and serves as a protection for the rest of the aluminium foil. Such a very thin and transparent protection layer often is artificially applied. The item are sold as *Eloxal* (= electrolytically oxidized aluminium).\n\n\nNow comes the table salt, which is sodium chloride ($\\ce{NaCl}$). Sodium chloride is hygroscopic, it \"attracts\" the water and that's why you sprinkled it on the turkey. It is also very soluble in water. When it dissolves, sodium $\\ce{Na+}$ and chloride $\\ce{Cl-}$ ions are formed.\n\n\nThe chloride ions attach and move into the protective oxide layer of the aluminium foil and finally induce the corrosion of the aluminium metal. Aluminium ions are released to the water (= the wet surface of the turkey).\n\n\nFor a more scientific, in-depth explanation of this corrosive processes known as **pitting**, you might want to have a look at this [article](http://www.sciencemadness.org/talk/files.php?pid=237561&aid=17616) or this [review](http://jes.ecsdl.org/content/161/9/C421.full).\n\n\n\n> \n> Was my turkey poisoned with aluminum at this point?\n> \n> \n> \n\n\nThere seem to be some crazy myths concerning the \"toxicity\" of aluminium (ions), often spread by opponents of immunization and other quacks. \n\n\nFact is that \n\n\n* a healthy human body is capable to excrete aluminium ions though the kidneys\n* the [British Alzheimer Society](http://www.alzheimers.org.uk/site/scripts/services_info.php?serviceID=6) does not see any convincing relation between the uptake of aluminium and Alzheimer's Disease\n* the [cancer information service](http://www.krebsinformationsdienst.de/vorbeugung/risiken/mythen.php) of the **dkfz** (Deutsches Krebsforschungszentrum, German Cancer Research Centre) refutes the aluminium induced breast cancer myth\n* many antacids do contain aluminium hydroxide as one ingredient. The effect of \"poisoning\" is not known.\n\n\nIn summary:\n\n\n1. Yes, aluminium ions were released to the wet surface of the turkey.\n2. There's no reason to assume that the turkey is \"poisoned\".\n\n\n", "21" ], [ "\nYou asked:\n\n\n\n> \n> Was my turkey poisoned with aluminum at this point?\n> \n> \n> \n\n\nI don't know. But I do know that it's not recommended for you to combine salty foods and aluminum foil in the future.\n\n\nIn Germany, the Federal Institute for Risk Assessment (BfR) wrote an article named [\"FAQs about aluminium in food and products intended for consumers\"](http://www.bfr.bund.de/en/faqs_about_aluminium_in_food_and_products_intended_for_consumers-191148.html). I shall quote a bunch from that article; the discussion of aluminum and salt is at the very end of my post.\n\n\n\n> \n> ### What health risks does aluminium absorption pose?\n> \n> \n> Any assessment of the hazardous potential of aluminium focuses is on its effects on the nervous system and the fact that it is toxic to reproduction (effects on fertility and unborn life) as well as the effects of aluminium on bone development.\n> \n> \n> When aluminium is ingested with food, its acute toxicity is low. ... But even in healthy individuals, the light metal accumulates in the body in the course of a lifetime, especially in the lungs and the skeletal system.\n> \n> \n> ...\n> \n> \n> ### What quantities of aluminium can be absorbed ... without any health risks?\n> \n> \n> For oral intake from food, the European Food Safety Authority (EFSA) has derived a tolerable weekly intake (TWI) of 1 milligramme (mg) of aluminium per kilogramme of bodyweight.\n> \n> \n> \n\n\nFinally, later in their article, they touch upon mixing aluminum with salty foods:\n\n\n\n> \n> Why can aluminium be transferred to food from packaging or tableware?\n> =====================================================================\n> \n> \n> Aluminium is soluble under the influence of acids or salt. For this reason, packaging and containers used for food such as beverage cans, yoghurt cup lids and aluminium containers for fruit juice are coated on the inside to prevent transfer of aluminium ions to the food or drink.\n> \n> \n> Aluminium from tin foil can be transferred to foods containing acid and salt. For this reason, the BfR recommends that acidic and salty foods are not wrapped in aluminium foil.\n> \n> \n> \n\n\n", "1" ], [ "\nYou might want to look at what happened to people involved in the mass Aluminium Sulphate poisoning event in Camelford, UK.\n<http://www.bbc.co.uk/news/uk-england-cornwall-17367243>\nAlzheimer's-like death and illness.\n\n\n", "0" ] ]

[ [ "\nThere is a journal article almost exactly about the title question, [Mechanisms of Skin Tanning in Different Racial/Ethnic Groups in Response to Ultraviolet Radiation](http://www.nature.com/jid/journal/v124/n6/full/5602865a.html) Journal of Investigative Dermatology (2005) 124, 1326–1332. \n\n\nHowever, the body of the question assumes that eumelanin changes color. Instead, the article finds that melanin gets closer to the surface of the skin, and the degree to which this effect occurs varies with race/ethnicity.\n\n\n", "3" ] ]

[ [ "\nWell, you asked for an example, I'll give you one. \n\n\"Quantum\" is undeniably the most effective keyword our science has seen, it's changed our 'understanding of chemistry and physics' drastically over the past century. \n\nFor example, between 1900 to 1925 (years vary from source to source, but this era exists, that's what that is important), in which Bohr's model was the main focus of study (as wikipedia puts it), is known to be the era of **\"old quantum theory\"**. \n There were series of theories, which started by Max Plank, and partially came to results with the final modifications of Sommerfeld; and ended in Erwin Schrodinger's famous model. New efforts led to new results and therefore the \"old quantum theory\" was never consistent. \n\nWhat was all of this about? Scientists are always eager to find out the differential behavior of matter and energy. Sub-atomic particles (in first level, n, p & e) were discovered in order to explain the phenomenons that Roentgen saw (radioactivity of Uranium and other heavy elements in 'pitchblende'); Faraday experimented (the $\\ce{SnCl2}$ solution experiment); Greek Leucippus and Democritos tried to explain; and so on. After every discovery, since the 'rules' that were discovered usually were up to a point in explaining what's around us, we change them and make new discoveries. And that was the reason of what Einstein, Bell and people like those did. They corrected the quantum for it to be what we know today. \n\nSo you see, the science was always about understanding; so why wouldn't a big discovery like quantization that led to quantum mechanics be able to change our direction of thought, either we are chemists, physicists or simply *humans*? \n\nLinks you might find helpful: [About the old quantum theory in wikipedia](https://en.wikipedia.org/wiki/Old_quantum_theory); [About the old quantum theory 2](http://arxiv.org/abs/0802.1366); [Bell's theorem](http://en.wikipedia.org/wiki/Bell%27s_theorem)(which linked classical quantum to quantum mechanics) and finally [quantum mechanics according to wikipedia](http://en.wikipedia.org/wiki/Quantum_mechanics).\n\n\n", "3" ], [ "\nI find this a strange question, because all contemporary chemical bond theory is quantum mechanical. So quantum mechanics already influences chemistry a great deal (look [here](http://en.wikipedia.org/wiki/Quantum_chemistry))! I suppose you could ask whether deeper quantum theory gives us more knowledge, and I'd guess yes! For example, look at this Periodic Videos analysis of the chemistry of seaborgium ([link](http://www.periodicvideos.com/videos/seaborgium_compound.htm)). We need more experimental data to test all theories, including the quantum mechanical ones.\n\n\n", "0" ] ]

[ [ "\nGenerally, the complete information about a structure is explicitly given by the name and does not rely on any implied information. Nevertheless, the practice of omitting locants when there is no ambiguity is widespread in general nomenclature. For preferred IUPAC names, however, locants are omitted only in a few exceptional cases described in Subsection P-14.3.4 of the current version of *[Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)](http://dx.doi.org/10.1039/9781849733069).*\n\n\nIgnoring the rules for omission of locants, the complete systematic name of the parent compound would be ‘prop-1-ene’. According to Rule P-14.3.4.2, however, the locant ‘1’ is omitted. (Note that no other isomer can be generated by moving the double bond from its position to another.)\n\n\n\n> \n> **P-14.3.4.2** The locant ‘1’ is omitted:\n> \n> \n> (…)\n> \n> \n> (d) in unsubstituted dinuclear and trinuclear alkenes and alkynes and monounsaturated cycloalkenes and cycloalkynes; (…)\n> \n> \n> \n\n\nThus, the preferred IUPAC name for the unsubstituted parent structure is **propene.**\n\n\nIgnoring cis/trans isomerism, the substitutable hydrogen atoms of propene have three different locants (‘1’, ‘2’, and ‘3’). Therefore, the locants for substituent prefixes (e.g. ‘1-chloro’, ‘2-chloro’, or ‘3-chloro’) cannot be omitted in substituted propene.\n\n\nSince, the locants for the substituent prefixes cannot be omitted, also the locant ‘1’ of ‘prop-1-ene’ can no longer be omitted for preferred IUPAC names.\n\n\n\n> \n> **P-14.3.3** Citation of locants\n> \n> \n> In preferred IUPAC names, if any locants are essential for defining the structure of the\n> parent structure or of a unit of structure as defined by its appropriate enclosing marks, then all locants must be cited for the parent structure or that structural unit. (…)\n> \n> \n> \n\n\nFor example, (1*E*)-1-chloroprop-1-ene, (1*Z*)-1-chloroprop-1-ene, 2-chloroprop-1-ene, and 3-chloroprop-1-ene are the preferred IUPAC names for some substituteded propenes.\n\n\nThis principle applies to any substituent. The special situation that replacing the chloro substituent in (1*E*)-1-chloroprop-1-ene, (1*Z*)-1-chloroprop-1-ene, and 3-chloroprop-1-ene with a methyl substituent would actually lead to the names (2*E*)-but-2-ene, (2*Z*)-but-2-ene, and but-1-ene instead of ‘(1*E*)-1-methylprop-1-ene’, ‘(1*Z*)-1-methylprop-1-ene’, and ‘3-methylprop-1-ene’, respectively, does not change the preferred name for 2-methylprop-1-ene. (Note that the substitutable hydrogen atoms of propene have three different locants irrespective of the kind of the substituent.)\n\n\nTherefore, the preferred IUPAC name is **2-methylprop-1-ene.**\n\n\n\n", "2" ], [ "\nWhen considering whether numbering is necessary you must see whether there are alternative structures which have the same basic name but different numbering.\n\n\nIn this example putting the methyl group on the 1 or the 3 position of the longest carbon chain changes its name from prop- to but-. Since the only other position that the methyl group can go is the 2 position then both numbers are unnecessary and so the shortest unambiguous name is methylpropene.\n\n\n", "1" ] ]

[ [ "\nThe second ionization energy of any element will be greater than its first ionization energy. One way to rationalize this phenomenon is by considering that the first electron is being removed from a neutral atom and the 2nd electron is being removed from a positively charged ion that is already electron deficient. Additionally, Slater's Rules for calculating effective nuclear charge can be used to provide a semiquantitative reasoning for this observation.\n\n\n", "5" ] ]

[ [ "\nI'm not an expert in thermodynamics and gas physics so there might be some errors in this calculation but here goes:\n\n\nAccording to several sources (wikipedia included) TNT decomposes according to this equation:\n$$\\ce{2C7H5N3O6 -> 3N2 + 5H2O + 7CO + 7C}$$\n\n\nIf 100kg of TNT was placed in an indestructible $1m^3$ container containing an inert gas (we will say helium) at standard temperature and pressure here is a rough estimate of what might happen if it was detonated.\n\n\nUsing a Hess's Law cycle we can calculate the enthalpy of decomposition of TNT from formation enthalpy data: $$\\Delta\\_{decomp}H = 63.2\\*2 - 285.8\\*5 - 110.5\\*7 = -2076.1~ kJmol^{-1}$$\n\n\nThen we can calculate the energy released by the decomposition: $$M\\_r~of~TNT = 227.13~gmol^{-1}$$ $$n = \\frac{100}{0.22713} = 440.3~mol$$ $$Energy~released = 440.3 \\* 2076.1 \\*1000 = 914.1~MJ$$\n\n\nI'm not sure how to go about specific heat capacities for gas mixtures so I won't put a figure on the temperature rise that this will cause but I suspect it will be in the thousands of degrees due to the amount of energy and so the resultant pressure rise will also be very significant, probably enough to liquefy the gases. Also as @Klaus Warzecha hinted at in his comment these kind of temperatures and pressures will probably cause some interesting chemical reactions to take place.\n\n\n<http://courses.chem.indiana.edu/c360/documents/thermodynamicdata.pdf>\n\n\n", "5" ], [ "\nI did a calculation and came up with an increase in pressure of about 280 atm. You can see my calculations [here](https://drive.google.com/file/d/0B80IPeLPxBV3STM1bFFYSWdtc2s/view?usp=sharing). Most gas cylinders can handle 100 to 150 atm (one is as high as 400 atm), so getting a container to hold 300 atm should not be a problem.\n\n\nAdded: \n\nOn this [page](http://www.alspecialtygases.com/Prd_high-pressure_steel.aspx) is list of cylinders with their working pressure (which is at least 1/2 the maximum pressure it can handle). 2000 psig is about 130 atm and 6000 psig is 400 atm.\n\n\n", "1" ] ]

[ [ "\nYou are going to want to check out Szabo and Ostlund page 143, they explain all the nitty-gritty for this matrix. The math itself is on page 22 and 23. But if you don't have a copy...\n\n\nThe $\\mathbf{S}^{-1/2}$ comes from the need to orthogonalize your basis so that you get your equation (5) instead of the $\\mathbf{FC}=\\epsilon\\mathbf{SC}$. It's one method of several to achieve this orthogonalization, and it's called symmetric orthogonalization. \n\n\nSo how do you compute this function of a Hermitian matrix? Well, functions on diagonal matrices are super easy; you just apply the function to each diagonal element. For non-diagonal matrices, this is not true. More to your question, you can't just take one over the square root of each element of $\\mathbf{S}$. But you can if it's diagonal. And you can make any Hermitian matrix diagonal via a unitary transformation.\n\n\nThe trick, then, is to diagonalize $\\mathbf{S}$ first, apply the inverse square root to each eigenvalue, then transform back using the eigenvectors. So you're going to want to\n\n\n1. Solve $\\mathbf{SU} = \\mathbf{sU}$. Where $\\mathbf{s}$ are the eigenvalues of $\\mathbf{S}$, and $\\mathbf{U}$ are the eigenvectors.\n2. For each $s$ in $\\mathbf{s}$ do $s^{-1/2}$.\n3. Transform back to original basis by $\\mathbf{S}^{-1/2} = \\mathbf{Us}^{-1/2}\\mathbf{U}^{\\dagger}$\n\n\n", "7" ] ]

[ [ "\n$$\\ce{N1V1 = N2V2}$$ The equation is used basically for dilution of solution and determining one of the parameters of acid or base.\n\n\nIn the case of dilution of solution, the equation holds true because the no of gm equivalents of the solute remains constant in the solution even the solution is diluted. \n\n\nInitial no of gram equivalents of solute = Final no of gram equivalents of solute ....................eq (1)\n\n\nAnd we know, $$\\text{Normality} = \\frac{\\text{No of gram equivalents}}{\\text{Volume in Litre}}$$\n$$\\text{No of gram equivalents}=\\text{Normality}\\times\\text{Volume in Litre}$$\nNow lets consider,\n\n\nInitial Normality =$\\ce{N1}$\n\n\nInitial Volume=$\\ce{V1}$\n\n\nFinal Normality=$\\ce{N2}$\n\n\nFinal Volume = $\\ce{V2}$\n\n\nThen , eq (1) becomes\n\n\n$$\\text{Initial Normality}\\times\\text{Initial Volume}=\\text{Final Normality}\\times\\text{Final Volume}$$\n$$\\ce{N1V1 = N2V2}$$\n\n\nIn case of acid and base reaction,\nNo of gram equivalents of acid will always react or consume equal no of gram equivalents of base and vice versa\n\n\nNo of gram equivalents of acid = No of gram equivalent of base .......eq (2)\n\n\nNormality of acid = $\\ce{N1}$\n\n\nVolume of acid =$\\ce{V1}$\n\n\nNormality of base =$\\ce{N2}$\n\n\nVolume of base =$\\ce{V2}$\n\n\nThen eq (2) becomes,\n$$\\text{Normality of acid}\\times\\text{Volume of acid}=\\text{Normality of base}\\times\\text{Volume of base}$$\n\n\n$$\\ce{N1V1 = N2V2}$$\n\n\nThe number of grams of a chemical element or compound equal to the weight of its chemical equivalent. The gram equivalent of a chemical element is numerically equal to the quotient resulting from dividing the element’s atomic weight by its valence. For example, the gram equivalent of iron (atomic weight 55.847) with a valence of 2 is 27.92 g and with a valence of 3, 18.62 g. The gram equivalent of acid is that amount of the acid in grams containing one gram equivalent of hydrogen capable of being displaced by metal with the formation of salt. The gram equivalent of a base is that amount of the base in grams that is necessary for its complete reaction with one gram equivalent of acid. The gram equivalent of salt is that amount of the salt in grams containing one gram equivalent of metal. The concept of gram equivalent is widely used in volumetric analysis.\n\n\n<http://encyclopedia2.thefreedictionary.com/Gram+Equivalent>\n\n\n", "1" ], [ "\nFirst, you need to know when you can use this equation. It is used to find the concentration of the solution when the volume of solvent is changed (when diluting the solute). Since the only thing that changed is the volume of the solvent (the amount of the solute is the same before and after), you use the equation to find the new concentration.\n\n\nE.g. 100 mL of a 0.2 N solution of sodium chloride is diluted to 0.1 N. The new volume is:\n$$\\ce {N\\_{f} \\times V\\_{f} = N\\_{i} \\times V\\_{i}}$$\n$$0.1\\ce {N\\times V\\_{f} = } 0.2\\ce {N \\times 100mL}$$\n$$\\ce {V\\_{f} = 200mL}$$\n\n\n", "0" ] ]

[ [ "\n\n> \n> What reaction can lead to this elimination when it seems that no\n> proton is leaving the molecule (instead of bromide) and acetate is\n> leaving?\n> \n> \n> \n\n\nThis is a reductive elimination reaction. The classic textbook example is the reductive elimination of trans-1,2-dibromocyclohexane to cyclohexene in the presence of zinc dust and acetic acid.\n\n\n\n\n\nIt is basically a redox reaction where the zinc metal is oxidized and the dibromocyclohexane is reduced. As with many reactions involving electron transfer from metal surfaces, the mechanism of this reaction is not fully understood. Perhaps the zinc inserts itself into the carbon-bromine bond as a first step, but in any case, electron flow proceeds in the direction shown above.\n\n\nThe reaction has found many uses and has been widely modified to include cases such as yours where a bromine and another group are eliminated to form a double bond. [Here is a link to a paper](http://www.scielo.cl/pdf/jcchems/v54n1/art21.pdf) where one of the bromines is replaced by an alkoxy group and the same zinc-mediated reductive elimination occurs. [Here is a link to another paper](https://www.jstage.jst.go.jp/article/bcsj1926/50/1/50_1_186/_pdf) where zinc is used to perform a reductive elimination in a series of compounds where both bromines have been replaced, one by acetate and the other by hydroxyl.\n\n\n", "8" ] ]

[ [ "\nYour view of bonding is a little simplified and confused. For the details it is worth reading the great answer referred to in some comments already: [Fundamental forces behind covalent bonding](https://chemistry.stackexchange.com/questions/710/fundamental-forces-behind-covalent-bonding) .\n\n\nI'll stick with a simpler view that focusses just on the energy involved and not the mechanism of chemical bonding. Firstly, you are wrong that boding two hydrogen atoms needs an input of energy. It actually releases energy. This is a thermodynamic view of the reaction. This tells us whether there is more or less energy in the product of the reaction being described than in the chemicals being combined to create the product.\n\n\nBut a lot of chemical reactions are thermodynamically favourable (that is the reaction releases energy) don't happen easily (this is probably not true for atomic hydrogen which should react very easily to form H2). But there are plenty of other reactions that release energy according to thermodynamics but that don't happen easily. For example, elemental carbon plus oxygen will release energy if converted to CO2. But neither the diamond in jewellery or the coal in your fireplace (which is basically impure carbon) reacts easily to give carbon dioxide despite being in an environment with a lot of Oxygen gas in it. Here the problem is that room temperature oxygen molecules and carbon just don't have enough energy to kick off the process that creates the product. This is where we have to look at the kinetic view of the reaction rather than just the thermodynamics. Although the overall reaction is favourable according to thermodynamics, we need to add energy (firelighters for coal fires, a blowtorch or diamond rings) to the mixture to give the atoms in the reactants enough energy to overcome the kinetic barriers to the reaction. Think of having to climb a tall mountain to get to a lower valley on the other side. Or think of giving oxygen molecules enough energy to break carbon atoms apart from their normal solid structure.\n\n\nOnce coal or diamond are warm enough, the reaction will happen spontaneously and you have either destroyed your engagement ring or you have a nice coal fire. In this case you can feel the energy released from the reaction as a pleasant warming of your room. \n\n\nBack to the question. The energy involved in a reaction is an accounting of the amount of energy used in the bonding of the products versus the reactants. The energy released in reactions such as burning carbon goes into the environment as heat. Other reactions happen spontaneously (try dropping a small lump of sodium into water). Some reactions require a tiny amount of encouragement (oxygen and hydrogen in certain concentrations will react explosively to form water when encouraged with the tiniest spark). \n\n\nWe can drive some reactions the other way using inputs of energy. For example we can electrolyse water into a mixture of hydrogen and oxygen. This requires us to add energy to the system (about 286kJ/mol) because the products are in a higher energy state then the reactants.\n\n\nIf you just focus on the thermodynamics of a reaction the numbers (kcal or kJ/mol) tell you how much the energy difference between the products and reactants is. But many reactions that should release energy require some input to kick off the reaction. This is confusing, but chemistry wouldn't be interesting and we wouldn't exist if thermodynamics were the only thing driving reactions. \n\n\nAlso energy is not created or destroyed here, just moved around. You can put energy in to a system to drive a reaction or the environment gets energy back from reactions that release it.\n\n\nSo, some reactions require energy to drive them because the products have higher energy, some release energy because the products have lower energy but many release energy but, confusingly, require some energy input to kick off the actual reaction process.\n\n\n", "7" ], [ "\nAs others have mentioned energy is released on bonding, which is why $H\\_2$ is more stable that two *H* atoms. To answer your last question first, *H* atoms in a box would eventually form $H\\_2$ but only if some third body (another H atom or wall of vessel) can take away some of the energy released before the *H* atoms fly apart again. This leaves $H\\_2$ in a highly vibrational state, further collisions will remove the remaining energy until it is at equilibrium with its surroundings.\n\n\nNow as to bond formation, its easier to examine $H\\_2^+$, this has two protons and one electron; the simplest possible bond. The calculation involves evaluating the interaction between the two protons and either proton and the electron. These are of the coulomb type, that is product of charges divided by separation. (No complications from Pauli principle here as there is only one electron.) \n\n\nSolving Schroedinger's equation involves terms with kinetic and potential energy (this is the Hamiltonian $H\\_0$). The terms for electron proton interaction are negative ($-e^2/r\\_1 - e^2/r\\_2$, where *e* is the electronic charge and $r\\_1$ the distance of proton 1 to the electron and similarly for $r\\_2$)) and the potential energy between protons is $+e^2/R$ where *R* is the internuclear separation). \n\n\nThe calculation proceeds by technical steps, say by using the variational theorem, and integrating over all distances and angles and distances using a basis set of atomic wavefunctions, call them $\\phi\\_1$ and $\\phi\\_2$.(for details see Cohen-Tannoudji, Diu & Laloe, Quantum Mechanics vol II for a full calculation). \n\n\nWe find that there are Coulomb integrals of the form $\\int \\phi\\_1\\frac{e^2}{r\\_2} \\phi\\_1 d\\tau$, and $\\int \\phi\\_2\\frac{e^2}{r\\_1} \\phi\\_2 d\\tau$. ($d \\tau$ simply means integrate over all coordinates). These give the energy of the electrostatic interaction between proton 2 and charge distribution of the electron were it in the 1s orbital of proton 1 and *vice versa*. \n\n\nIn addition there are integrals of the form $A=\\int \\phi\\_1\\frac{e^2}{r\\_1} \\phi\\_2 d\\tau$ which have no classical equivalent, i.e. they would not arise in a classical calculation. These are sometimes called 'resonance integrals' or 'exchange integrals' although this latter term can mean something else to physicists. The integral *A* is difficult to explain, other than to say that it arises out of the mathematics used when making a linear combination of atomic orbitals. The fact that integral *A* is not zero expresses the possibility of the electron ' jumping' as it were from the neighbourhood of one proton to that of the other and hence the possibility of 'quantum resonance' (The term 'resonance' arises by analogy with oscillating pendulums, see Coulson, Valence chapter 4 but does not imply actual motion). \n\n\nThe combination of these integrals (and the overlap integral $S=\\int \\phi\\_1\\phi\\_2 d\\tau$) produces a minimum in the potential energy profile as the calculation is repeated at different internuclear separations *R* and hence describes the energy of the $H\\_2^+$ molecule vs. separation of the nuclei.\n\n\nThe total energy comprises kinetic and potential energy terms, so is it the decrease in kinetic energy or potential energy that is responsible for the chemical bond? The overwhelming effect is due to the decrease in potential energy. This is also confirmed by Virial Theorem calculations.\n\n\nIn the $H\\_2$ molecule we can carry over what has been done for $H\\_2^+$. Now we have two valence electrons instead of one, they will both occupy an orbital similar to that in $H\\_2^+$ and the electrons will have opposite spins by Pauli Principle. The main difference is that there are now two electron contributing to binding and the bond if stronger. The essential principles outlined here can be carried through to forming bonds between other atoms although the calculations become tremendously more involved as there are many more electrons.\n\n\n", "1" ] ]

[ [ "\nIf you don't want to use the classical solid etchants, such as\n\n\n* Iron(III)chloride ($\\ce{FeCl3}$) or\n* Ammonium persulfate ($\\ce{(NH4)2S2O8}$) or\n* Sodium persulfate ($\\ce{Na2S2O8}$)\n\n\nthere's the **mixture of hydrogen peroxide ($\\ce{H2O2}$) and hydrochloric acid ($\\ce{HCl}$)** in various ratios.\n\n\n\n\n---\n\n\nTypical recipes use \n\n\n* water\n* 30% $\\ce{H2O2}$\n* 37% $\\ce{HCl}$\n\n\nin a volume ratio of 1:1:1. \n\n\n\n\n---\n\n\nHydrochloric acid can be considered a \"household chemical\", it was/is available at DIY stores as **muriatic acid** and typically used to clean brickwork.\n\n\nI would advise against mixing concentrated (glacial) acetic acid and hydrogen peroxide solutions of higher concentrations. You might end up forming peracetic acid, which is unhealthy and tends to decompose spontaneously (explode) at higher concentrations.\n\n\nThere's another method worth to be mentioned, although the etchant is usually not a household item.\n\n\nCupric chloride $\\ce{CuCl2}$ in aqueous solution can be used at room temperature to remove metallic copper from the PCB too:\n\n\n$$\\ce{CuCl2 + Cu -> CuCl}$$\n\n\nThinking in terms of transition metal waste management, this seems to be pretty nice, since the resulting copper(I) solution can be regenerated (oxidized) to copper(II) (= the etchant) using hydrochlorid acid and hydrogen peroxide. This [conference proceeding](http://www.www.journalamme.org/papers_amme03/1229.pdf) provides further information.\n\n\nAs far as the permanganate is concerned, it seems that it has been used to **desmear** PCBs in industry, but it has been superseeded by plasma treatment to avoid damage to the boards.\n\n\n", "3" ], [ "\nI’m using $\\ce{CuCl2}$ for three years now. I like it a whole lot better than $\\ce{FeCl2}$ – it’s cleaner, uses domestically available chemicals, and has less of a disposal problem.\n\n\nIt’s related to “mixture of hydrogen peroxide ($\\ce{H2O2}$) and hydrochloric acid ($\\ce{HCl}$)” because that produces the same result, ultimately, though there is more than one way to get there. I have in the past tried $\\ce{HCl/H2O2}$ and ended up stripping the entire board in a few seconds: but those same chemicals can be used to make $\\ce{CuCl2}$ which is far more controlled. I’m not much of a chemist, but my understanding is that $\\ce{Cu + H2O2 + HCl -> CuCl2 + H2O}$: the initial reaction eats copper fast but the result is $\\ce{CuCl2}$ and the reaction with that is slower.\n\n\nHere is a practical guide to making and using $\\ce{CuCl2}$:\n\n\n<http://techref.massmind.org/techref/pcb/etch/CuCl2.htm>\n\n\n", "1" ] ]

[ [ "\nI don't think you can know, *a pr*iori, whether or not s-p mixing occurs without first having some experimental data. I'll use water, a triatomic molecule, as my case in point. \n\n\nFor over 50 years students have been told that water is roughly $\\ce{sp^3}$ hybridized. The general description is that there are two equivalent O-H $\\ce{sp^3}$ sigma bonds and two equivalent lone pairs, also in $\\ce{sp^3}$ orbitals. The lone pair - lone pair repulsion is greater than the sigma bond - sigma bond repulsion, so the lone pair-O-lone pair angle opens up slightly and the $\\ce{H-O-H}$ angle closes down to the observed 104.5 degrees.\n\n\nWith the advent of photoelectron spectroscopy it was found that the two lone pairs in water were not equivalent (2 signals were observed for the lone pairs). Now, the hybridization of water is described as follows:\n\n\n* 2 $\\ce{sp^3}$ O-H sigma bonds\n* one lone pair in a p orbital\n* and the second lone pair in an $\\ce{sp}$ orbital\n\n\nIf chemists knew how to predict s-p mixing, it wouldn't have taken them over 50 years to understand how they mix in water.\n\n\n", "9" ], [ "\n**Preliminary.** \n\n\nHybridization is a questionable model. It was originally introduced to consider organic molecules, and there it works brilliantly. However, it is hard to use outside organic chemistry with molecules with simple bounding. For example, adequate description of cyclopropane requires fractional indices in orbitals. \n\n\nMathematically, hybridization process is transformation of several orbitals using linear transform. Since molecular orbitals are described as linear combinations of atomic orbitals, there is no problem to use any imaginable set of atomic orbitals as long as their linear combinations can produce all 'original' atomic orbitals. Basically, hybridization is similar to change of basis vectors in geometry: it is OK as long as the transformation follows simple rules, and it may be built to preserve values of some scalar vector expressions.\n\n\nSince from mathematical viewpoint hybridization is nothing more than a special transformation of basis set, it does no change 'actual' molecular orbitals, meaning that calculations using hybridized and unhybridized orbitals must provide same result (actually, the very transformation in question is purposefully built this way). So, your question does not have a proper answer as the question itself makes wrong assumption. Hybridization does not occur, it is a model people use, it does not correspond to any real-world phenomenon \n\n\nNow, the proper question would be: **when hybridization model is applicable and provides valuable insight into electronic structure of considered molecule?**\n\n\nWhen you are considering electronic structure of isolated 2-atom molecules it is usually better to not use hybridization. It does not provide any simple conclusions for inferring preferred geometry, but leads away from more 'accurate' MO LKAO model. Considering interaction of such molecule as a ligand with metal center, however, returns some use to hybridization for simple molecules, because it simplifies geometry consideration. For example, using electron count rules it is possible to some extent to predict geometry of $\\ce{NO}$ fragment, inferring preferable hybridization state of the nitrogen atom. It is also much easier to consider interactions with central atom for $\\sigma sp$ lone pair than several dinuclear molecular orbitals. \n\n\nWhen considering geometry of organic and some inorganic molecules ($\\ce{BCHNOF}$) with little angular stress and no hypercoordinated atoms, hybridization works well enough as long as you do not try to use it to predict one-electron properties, like ionisation potentials. However, stressed molecules like [1,1,1]-propellane or dehydroadamantane and hypercoordinated atoms, like methanium cation, require separate, case-by-case analysis to be integrated into framework of hybridization model. For most other atoms hybridization model cannot provide valuable insights.\n\n\n**Again, hybridization does not work for 1-electron properties.** For example, methane, a perfect example of ideal $sp^3$ hybridization has TWO peaks in photoionization spectrum, implying it has 2 unequal sets of valence orbitals (roughly -2.25 and -1.5 eV energy levels) despite hybridization model predicting it to have 4 perfectly equal bonds. MO LKAO model, on the other hand, predicts several bonding orbitals, [one significantly lower in energy than three others](http://www.meta-synthesis.com/webbook/40_polyatomics/polyatomics.html).\n\n\n", "8" ] ]

[ [ "\nThe *heat* generated increases directly with the amount of sodium acetate trihydrate used, but the maximum *temperature*, ignoring surface effects, *should not change at all*. After all, if the heat in 300 ml is used to raise the temperature of 300 ml of substance, or if 300 kg is used to raise the temperature of 300 kg, it's being spread out proportionately. No matter how much you use, it won't exceed ~58 C\n\n\nHowever, do not try this in a container that would not allow room for expansion, e.g. a glass bottle, as in crystallizing, sodium acetate trihydrate might take up more room. See warning at <http://chemmovies.unl.edu/chemistry/dochem/DoChem058.html>. You *could* safely do this on a polyethylene cutting board, though, which melts at ~100 C\n\n\n", "5" ] ]

[ [ "\nSince you want to remove the tarnished silver ($\\ce {Ag2O}$), you would convert it into silver metal (or reduce it). Therefore, you would oxidize the aluminum ($\\ce {Al -> Al^3+}$). The silver and aluminum need to be in physical contact and the sodium chloride or the carbonated water is just providing the medium to transfer the ions; the aluminum does not dissolve first.\n\n\n", "1" ] ]

[ [ "\nThe purpose of hydrogenation is to convert liquid oils into semi-solids at room temperature. Palm \"oil\" is unusual among natural vegetable oils/fats in that it is naturally fairly unsaturated and is thus a semi-solid at room temperature. \n\n\nEconomics is probably the driving factor preventing significant research into the solution (pun intended) you propose with mixing saturated and unsaturated fats to get the desired consistency. Hydrogenation techniques are cheap, well established, agency approved, etc. Same idea with palm oil.\n\n\nI hope I hit the points of your question, comment if I missed anything.\n\n\n", "2" ] ]

[ [ "\nYour teacher is correct, some of the resonance structures in your figure are not aromatic. Still though, as resonance structures they contribute to the overall description of the molecule, just less than the structures that are aromatic. The resonance structures at the right and left end of your figure are aromatic. The fourth resonance structure (the one in the middle) is referred to as being a \"quinone\" rather than an aromatic.\n\n\n*o*- and *p*-benzoquinone are examples of quinones; they are **not** aromatic compounds ([ref](http://en.wikipedia.org/wiki/Quinone), note the sentence \"Quinones are conjugated but not aromatic\").\n\n\n\n\n\nLikewise, their carbon analogues, *o*- and *p*-[xylylene](http://en.wikipedia.org/wiki/Xylylene) are also non-aromatic quinoid-like compounds. \n\n\nFor a compound to be aromatic it must be\n\n\n* cyclic\n* contain 4n+2 pi electrons\n* have a continuous loop of p-orbitals\n* and you must be able to draw double bonds - **internal to the ring** - from each ring carbon atom.\n\n\nIt is this last point that serves to differentiate quinones and quinone-like molecules from aromatic compounds. \n\n\nFinally, aromatic compounds need not be planar, deviations in excess of 15° from planarity can be tolerated and the molecule can remain aromatic (see, for example, [[6]-paracyclophane](http://en.wikipedia.org/wiki/Cyclophane))\n\n\n\n\n\n", "11" ], [ "\nAromaticity is not a very well defined concept (there are no strict rules) and I also believe that it is hard to tell apart from overall resonance stabilisation. The IUPAC goldbook states for [aromatic](http://goldbook.iupac.org/A00441.html):\n\n\n\n> \n> **aromatic** \n> \n> \n> 1. In the traditional sense, 'having a chemistry typified by benzene'.\n> 2. A cyclically conjugated molecular entity with a stability (due to delocalization ) significantly greater than that of a hypothetical localized structure (e.g. Kekulé structure ) is said to possess aromatic character. If the structure is of higher energy (less stable) than such a hypothetical classical structure, the molecular entity is 'antiaromatic'. The most widely used method for determining aromaticity is the observation of diatropicity in the 1H NMR spectrum. \n> \n> See also: Hückel (4n + 2) rule, Möbius aromaticity\n> 3. The terms aromatic and antiaromatic have been extended to describe the stabilization or destabilization of transition states of pericyclic reactions The hypothetical reference structure is here less clearly defined, and use of the term is based on application of the Hückel (4n + 2) rule and on consideration of the topology of orbital overlap in the transition state. Reactions of molecules in the ground state involving antiaromatic transition states proceed, if at all, much less easily than those involving aromatic transition states.\n> \n> \n> \n\n\nHückel's set of rules are often used as nice guidelines, but they are not the definition of aromaticity. In fact they are very, very rigid and strict and most of the compounds you'd usually consider being aromatic in this context, do not comply with these rules. The IUPAC goldbook defines:\n\n\n\n> \n> **[Hückel (4n + 2) rule](http://goldbook.iupac.org/H02867.html)** \n> \n> Monocyclic planar (or almost planar) systems of trigonally (or sometimes digonally) hybridized atoms that contain (4n + 2) π-electrons (where n is a non-negative integer) will exhibit aromatic character. The rule is generally limited to n = 0–5. This rule is derived from the Hückel MO calculation on planar monocyclic conjugated hydrocarbons (CH)m where m is an integer equal to or greater than 3 according to which (4n + 2) π-electrons are contained in a closed-shell system.\n> \n> \n> \n\n\nIn this framework only the first and the last structure obey Hückel's rule. The [proton](http://www.chemicalbook.com/SpectrumEN_100-01-6_1HNMR.htm) and [carbon](http://www.chemicalbook.com/SpectrumEN_100-01-6_13CNMR.htm) NMR clearly proof that *p*-nitroanniline is an aromatic compound, the spectrum is very close to the [data of benzene](https://en.wikipedia.org/wiki/Benzene_(data_page)).\n\n\nResonance structures are just a crutch to explain a very complicated and complex electronic structure in simple and understandable terms, i.e. Lewis structures. In highly delocalised systems a superposition of all resonance structures has to be considered to paint a somewhat conclusive picture of the bonding situation. There are contributions of different structures and not all of them may be equal in the overall description, however, they do not change the fact, that the experimental property is observed. This is again a case, where you can see the limitations of the resonance concept.\n\n\n", "7" ], [ "\nAll the resonating structures follow Huckel's Rule and so they are not losing their aromaticity. \n\n\n* Planar Cyclic Ring\n* 4n+2 pi-electrons (the exocyclic pi electrons are not considered)[1](https://chemistry.stackexchange.com/questions/22100/what-is-the-resonance-structure-of-nitro-aminobenzene-with-a-broken-aromaticity#comment41055_27951),[2](https://chemistry.stackexchange.com/questions/22100/what-is-the-resonance-structure-of-nitro-aminobenzene-with-a-broken-aromaticity#comment41280_27951).\n* All atoms of the ring in conjugation\n\n\nA classic example of an aromatic cabanion is [cyclopentadiene](http://en.m.wikipedia.org/wiki/Cyclopentadiene#Deprotonation) ion. \n\n\nThe negative charge or the lone pair is in resonance and is in an unhybridised orbital. Hence it gives $sp^2$ and not $sp^3$. So its planar. \n\n\n", "4" ] ]

[ [ "\nFrom [ChemGuide](http://www.chemguide.co.uk/organicprops/alkenes/halogenation.html):\n\n\n\n> \n> **Using bromine water as a test for alkenes**\n> \n> \n> If you shake an alkene with bromine water (or bubble a gaseous alkene\n> through bromine water), the solution becomes colourless. Alkenes\n> decolourise bromine water.\n> \n> \n> **The chemistry of the test**\n> \n> \n> This is complicated by the fact that the major product isn't\n> 1,2-dibromoethane. The water also gets involved in the reaction, and\n> most of the product is 2-bromoethanol.\n> \n> \n> \n\n\nIn essence, the electrophilic pi bond in the presence of bromine *and* water attacks diatomic bromine as usual, forming the usual bromonium ion intermediate. \n\n\nAfter this step, however, we have two competing nucleophiles which can open the ring: bromide ion and water. The water is likely present in great excess, making it the more likely attacker (statistically). \n\n\n[](https://i.stack.imgur.com/sj4je.png)\n\n\nOf course, the desired reaction below to form a dibromide also occurs. But it is in competition with other reactions such as the one listed above. \n\n\n[](https://i.stack.imgur.com/p3XNN.png)\n\n\nSo usually electrophilic halogenation of alkenes takes place in an inert medium such as methylene chloride or carbon tetrachloride. These aren't inclined to act as nucleophiles, so they don't participate in ring opening, leaving the $\\ce{Br-}$ ion to do its thing unhindered. \n\n\n", "11" ] ]

[]

[ [ "\nShort answer: Yes.\n\n\nLong answer: Any alloy containing 'passivating' elements such as aluminum, which have the tendency to form a protective oxide as you describe, has the potential to also form the same sort of passivating oxide layer as the pure material. It's not guaranteed, though.\n\n\nPure zirconium also [forms a passivating oxide](http://www.mtialbany.com/metals/zirconium/ \"forms a passivating oxide\") $(\\ce{ZrO2})$$^{(1)}$, so the AlZr alloy you mention is almost certain to form a similar protective oxide.\n\n\nMore broadly, a major area of research at present is in the 'alumina-forming' alloys (\"alumina\" is another word for aluminum oxide, $\\ce{Al2O3}$), where aluminum is included in, e.g., steels or nickel superalloys, to try to achieve this passive aluminum oxide layer on the surface in high-temperature, corrosive environments$^{(2-5)}$. With materials like a stainless steel-plus-aluminum, the passive oxide scale only forms if the alloy has the right percent range of aluminum -- otherwise, the aluminum oxide ends up staying spread throughout the metal and not concentrating at the surface. If the alumina doesn't end up spread evenly over the surface, you don't get nearly as good of a protective effect.\n\n\n(1) <http://www.mtialbany.com/metals/zirconium/> \n\n (2) <http://topaz.ethz.ch/function/web-het-secured/pdfs/Prescott-92.pdf> \n\n (3) <http://www.metalle.uni-bayreuth.de/de/download/publications_downloads/Bensch__Sato__Warnken__Affeldt_______Reed__Glatzel__Acta_Mat_.pdf> \n\n (4) <http://www.netl.doe.gov/File%20Library/Events/2010/ods/Gordon_Tatlock_Corrosion.pdf> \n\n (5) <http://www.ornl.gov/science-discovery/advanced-materials/science-to-energy-solutions/alumina-forming-austenitic-alloys-licensed> \n\n\n", "3" ] ]

[ [ "\nFirst of all, you have it all backwards.\n\n\nYou sound like the $\\ce{AlCl3}$ says, \"Oh, I am in solid state now! Let me turn to crystal (polymer) structure! [a while later] Oh, I am in vapour state now! Let me turn to dimer structure!\"\n\n\nWhere in *reality* it says, \"I don't have enough kinetic energy to break the bonds faster than they form, so I can only be in the crystal structure, in which I cannot move around, so I am in solid state now. [a while later] I have enough kinetic energy now! I can now break bonds faster than they form, but still not enough to break the bonds between the dimer, so let me exist as dimer. Also, I can move around now, since the intermolecular forces are not that strong. I am now a gas.\"\n\n\n\n\n---\n\n\nAlmost everything polymerizes when the kinetic energy is low, forming a solid. The difference is only whether the bond responsible for the polymerization is van der Waals' forces or covalent bond or ionic bond.\n\n\nFor $\\ce{BeH2}$, the bond is dative covalent as that makes the most sense.\n\n\nFor $\\ce{AlCl3}$, the bond is ionic as that makes the most sense.\n\n\n\n\n---\n\n\nWhen the energy is high enough, $\\ce{BeH2}$ decomposes, so all discussions henceforth are meaningless.\n\n\nFor $\\ce{AlCl3}$, dimer has less energy than monomer, so dimer is formed. Remember that things like to be in states of low energy. Dimer enables it to move around, hence it is a gas.\n\n\n\n\n---\n\n\nWhen the energy is even higher, even the dimer bonds break and you have $\\ce{AlCl3}$ monomers.\n\n\n", "2" ] ]

[ [ "\n1 kg for 1 liter is, in fact, liquid water at room temperature. \n\n\nYou need to use the ideal gas law $\\ce {PV=nRT}$.\n\n\n", "1" ], [ "\nIn fact, under NTP conditions, each mole of an ideal gas occupies the volume $22.4\\ce{L }$.\n\n\nIn your case, the volume of $0.45\\ce{mol }$ of $\\ce{O\\_2 }$ is $22.4\\times 0.45=10.08\\ce{L }$.\n\n\n", "0" ], [ "\nAvogadro's law states that \"*equal volumes of all gases, at the same temperature and pressure, have the same number of molecules*\", ie at constant temperature and pressure, 1 mole of any gas will occupy the same volume. At NTP, this volume is ~$24dm^3$. Therefore the volume of $0.45mol$ of $O\\_2$ will be $0.45 × 24 = 10.8dm^3$. $1dm^3 = 1L$ so your answer would be $10.8L$\n\n\n", "0" ] ]

[ [ "\nSee my [answer](https://chemistry.stackexchange.com/questions/22079/draw-a-simplified-mo-diagram-for-the-pi-system-of-methyl-vinyl-ether/22082#22082) to your question on methyl vinyl ether for some background on molecular orbital (MO) theory.\n\n\nIn this case, your molecule is analogous to the all-carbon system, butadiene. In cyanogen you have 4 p orbitals that can align, overlap and mix together to generate 4 molecular orbitals and you have 4 electrons (1 from each p orbital involved in a pi bond) to place in your molecular orbitals. Since the carbons in cyanogen are $\\ce{sp}$ hybridized you actually have a second set of molecular orbitals that are identical to the first in all respects except that the 4 p orbitals in the second set are rotated 90 degrees with respect to the other set of p orbitals.\n\n\nLet's count electrons. We have 4 electrons in each of these orthogonal molecular orbitals that each contains 4 p orbitals, so we have 8 electrons in our two orthogonal sets of molecular orbitals. We also have 2 lone pairs remaining on each nitrogen for a total of 4 electrons. Finally we have the sigma system; we have 2 C-N sigma bonds and 1 C-C sigma bond each containing 2 electrons for a total of 6 sigma electrons. For the whole molecule we have 8+4+6=18 electrons as expected.\n\n\nAs I mentioned in my other answer, we typically focus on the delocalized molecular orbitals formed by combining the p orbitals and disregard the sigma system (usually) when we create our MOs. Disregarding the sigma system usually simplifies matters.\n\n\nThere are a number of ways to create the MO diagram for butadiene or cyanogen, but all of them will have the common theme of mixing 4 p atomic orbitals and generating 4 molecular orbitals. Here is one way to do it, let's start by mixing 2 p orbital to create the MO's for ethylene (or nitrile).\n\n\n\n\n\n[image source](http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch10/ch10-6-1.html)\n\n\nNow we can mix 2 ethylenes (or nitriles) to generate butadiene (or cyanogen)\n\n\n\n\n\n[image source](https://classes.yale.edu/chem225b/studyaids/pericyclic/pericyclic.htm)\n\n\nWe label molecular orbitals $\\ce{\\Psi\\_1 ~- ~\\Psi\\_4}$ in order of increasing energy (so $\\ce{\\Psi\\_1}$ is the molecular orbital at the bottom of the figure and $\\ce{\\Psi\\_4}$ is at the top). Don't worry about the \"S\"'s and \"A\"'s in the diagram, they are just telling us about the symmetry properties of the molecular orbitals which we don't need to focus on right now. \n\n\nNotice how $\\ce{\\Psi\\_1}$ has overlap between the orbitals on the two middle carbons. This indicates overlap and a higher bond order between these 2 carbons, just as your resonance structures II and II indicate. Just like in butadiene, this carbon-carbon bond has some double bond character and, therefore there is a higher barrier to rotation about it then you might otherwise expect.\n\n\n", "4" ], [ "\nThis feature of cyanogen was already mentioned in the comment by Geoff to the Question \"[How many electrons are there in the pi-system of cyanogen?](https://chemistry.stackexchange.com/q/21973/4945)\". (And I was very tempted to vote to close as a duplicate.)\n\n\nYour resonance structures are correct and it is noteworthy, while I might be the predominant one, the other two are still considerably influencing the electronic structure. The problem with this is, that you cannot sufficiently express delocalisation in Lewis terms. A molecular orbitals approach is necessary to understand the bonding situations. Below you can find the complete valence MO scheme of cyanogen, which I repost updated from the linked question. \n\n \n\nYou can clearly see, that the eight $\\pi$ orbitals are delocalised over the complete molecule. Half of them include the bond between the central carbon atoms, i.e. the degenerate $\\pi\\_\\mathrm{u}$ orbitals. \n\nFor a carbon carbon single bond you would expect, a bond length of $\\mathbf{d}(\\ce{C-C})=150~\\mathrm{pm}$ and for a double bond $\\mathbf{d}(\\ce{C=C})=134~\\mathrm{pm}$. The calculated bond length is about $\\mathbf{d}(\\ce{C\\bond{~-}C})=138~\\mathrm{pm}$ (DF-BP86/def2-SVP) and clearly between a single and a double bond. This has of course an effect on the neighbouring carbon nitrogen bonds. You would expect the triple bond to be $\\mathbf{d}(\\ce{C#N})=108~\\mathrm{pm}$ and the double bond $\\mathbf{d}(\\ce{C=N})=120~\\mathrm{pm}$. The calculated bond length is $\\mathbf{d}(\\ce{C\\bond{~=}N})=118~\\mathrm{pm}$ and only little shorter than a double bond. \n\n\nCovalent radii are taken from [Pekka Pyykkö and Michiko Atsumi, *Chem. Eur. J.*, **2009**, *15* (46), 12770-1779.](http://onlinelibrary.wiley.com/doi/10.1002/chem.200901472/abstract)\n\n\n", "3" ] ]

[ [ "\n\n> \n> I am unsure how to use this knowledge to actually draw the MO diagram?\n> \n> \n> \n\n\nFirst, chemists used Lewis structures to explain molecular properties, but eventually this method was replaced with resonance structures (like you've drawn in part iv above). Resonance structures focus primarily upon p orbitals in a molecule and usually do not consider the sigma system. Molecular Orbital (MO) theory was a further improvement on resonance theory. MO theory still focuses on p orbital connectivity and typically disregards the sigma system, but it lets you view the actual patterns that the molecular orbitals generate. It is very similar to the orbital diagram you included above, but we leave out the sigma system and focus on the p orbitals and the electrons they contain.\n\n\nIn part ii) above you conclude that the ether oxygen is $\\ce{sp2}$ hybridized. This means that there is one p orbital on oxygen that contains a lone pair of electrons (there are also 3 $\\ce{sp2}$ orbitals on the oxygen, 2 are used to form sigma bonds to carbon and the third holds the second lone pair of electrons on the oxygen). This p orbital can interact with the 2 p orbitals involved in the carbon-carbon double bond. These 3 p orbitals can interact to produce 3 molecular orbitals (see pictures below). These 3 molecular orbitals are produced by combining (adding and subtracting) a p orbital and the 2 ethylenic molecular orbitals (the bonding and antibonding pi orbitals for the double bond). All of the orbitals shown in the MO diagram below are p orbitals. We will fill the MOs with 4 electrons, 2 from the pi double bond and 2 from the oxygen p orbital that contains one of oxygen's lone pair of electrons.\n\n\nSo, we have determined that we have 3 p orbitals (one on oxygen, 2 on the olefinic carbons) that can overlap and form 3 molecular orbitals. The system we have created with 3 p orbitals from methyl vinyl ether is similar to the all carbon allyl analogue with 3 p orbitals, one p orbital on each of the 3 carbons. Specifically your system is analogous to the ally anion with 4 pi electrons (2 from your double bond and 2 from the oxygen lone pair).\n\n\n\n[Image Source](http://www.chemtube3d.com/orbitalsallylanion.htm)\n\n\nFor more information on how to construct molecular orbitals in organic molecules see [here](http://www.ch.ic.ac.uk/vchemlib/course/mo_theory/main.html) and [here](http://chemwiki.ucdavis.edu/Organic_Chemistry/Organic_Chemistry_With_a_Biological_Emphasis/Chapter_02%3A_Introduction_to_organic_structure_and_bonding_II/Section_2.1%3A_Molecular_orbital_theory%3A_conjugation_and_aromaticity).\n\n\n", "3" ] ]

[ [ "\n\n> \n> Where are they stored? (How does tea plant make them? I'm looking for the reactions that occur in the leaf)\n> \n> \n> \n\n\nBoth citrate and oxalate are associated with ATP production via [the citric acid cycle](http://en.wikipedia.org/wiki/Citric_acid_cycle). As is obvious from the name, citrate is a critical component of the process. Oxalate is a hydrolysis product of oxaloacetic acid, which is also part of the cycle. Both of these ions are present within the plant cells.\n\n\n\n> \n> How and why do they get released from tea leaf? Since you don't usually see any \"big\" differences between tea leaf and a soaked tea leaf? (I'm not looking for the how of solution)\n> \n> \n> \n\n\nIf your tea leaves are crushed or broken in any way, this really is just simple dissolution. If you have whole tea leaves, the oxalate and citrate are migrating into and out of the plant cells. The cell wall is permeable, so the ions (or corresponding acids) can simply flow through it. The cell membrane is semipermeable, and the diffusion of the ions across it is a biologically controlled process that I honestly don't have the background to explain.\n\n\n\n> \n> What makes the color of water change? Why does the color vary with pH alteration?\n> \n> \n> \n\n\nThe pH and color are not related properties. The pH is affected by the citric acid and oxalic acid associated with the anions previously discussed, as well as other minor components that are likely present. The color comes from some of the many other compounds that are extracted into the water, which number in the hundreds, if not thousands. Some of these compounds are inherently colored - no reaction is required to produce color in the solution.\n\n\n", "5" ] ]

[ [ "\nThere are many $\\ce{CrO3}$ oxidation reagents ([Jones](http://en.wikipedia.org/wiki/Jones_oxidation), [Sarett](http://en.wikipedia.org/wiki/Sarett_oxidation), [Collins](http://en.wikipedia.org/wiki/Collins_reagent), [PCC](http://en.wikipedia.org/wiki/Pyridinium_chlorochromate), etc.), however, most are selective and stop the oxidation of a primary alcohol at the aldehyde stage. On the other hand, the Jones reagent is relatively unselective and will oxidize a primary alcohol to the corresponding carboxylic acid - **if** enough reagent is present. It takes two-thirds of an equivalent of Jones reagent to oxidize one equivalent of the alcohol to the aldehyde and an additional two-thirds equivalent to oxidize the aldehyde to the acid. Also the Jones oxidation is run under acidic conditions (typically dilute sulfuric acid).\n\n\nI would have though that A, B (the gem-diol will equilibrate with the aldehyde under acidic conditions) and D were correct answers. It seems like the computer answer is either in error or it is a \"trick\" question (seems unlikely for computer-based learning). The \"trick\" would be that since an excess of reagent was not specified only 1 equivalent is assumed. Therefore there is not enough reagent to carry out two sequential oxidations; as a consequence only the aldehyde or gem-diol could be **completely** oxidized with one equivalent of reagent. But then A or B is not an acceptable answer either.\n\n\n", "3" ] ]

[ [ "\nDissenter is correct about aromatic compounds and that the basic requirement for a compound to be detected by smell is for it to be volatile enough to reach the nose. This is why things like dry salt don't really smell like anything. Once they reach the nose, things become more complex.\n\n\nWhy many esters smell fruity while most thiols smell rather unpleasant is a rather interesting question. Just as most sugars taste sweet, many molecules that have structural similarities also smell similar. Humans have about [350 different olfactory receptors](http://onlinelibrary.wiley.com/doi/10.1111/j.1753-4887.2004.tb00097.x/abstract). These are G protein-coupled receptors that change conformation when they bind a compound and activate a signal pathway that is eventually sent to the brain for processing. However, since we only have 350 different receptors and there are far more than 350 odoriferous compounds, there is not a one-to-one mapping of receptors to compounds—there isn't an \"acetone receptor\". Instead, olfactory receptors have partial and overlapping selectivity. Each receptor type responds to more than one compound, thiols with a certain general shape, for instance, and different receptor types can respond to some of the same compounds. A given compound then has a more or less unique pattern of receptor activation. All this information gets sent to the brain and is processed into the perception of smell. Compounds that have similar structural properties activate many of the same receptors, so they smell similar.\n\n\n", "17" ], [ "\nThe term aromaticity originated with the discovery of unusually stable hydrocarbons that also happened to have strong smells. Many hydrocarbons smell, but not all are aromatic. \n\n\nNowadays, a compound being classified as \"aromatic\" has little to nothing to do with its smell and everything to do with its electron configuration. There are also non-aromatic compounds and anti-aromatic compounds. If we try to relate these to smell as well, what would an anti-aromatic compound smell like? Bad? But isn't that a little too subjective for science? Doesn't benzene smell \"bad\" too? Then why is it aromatic (i.e. unusually resistant to chemical change?) Why not anti-aromatic? \n\n\nEsters smell partly because they exhibit weak intermolecular forces. This allows ester molecules to enter the gas phase and reach your nose. Esters don't exhibit intermolecular hydrogen bonding, unlike alcohols, for example. These are no strongly positively polarized hydrogens in esters to participate in hydrogen bonding. Consider for example ethyl butyrate, which smells like pineapples. \n\n\n\n\n\nMost of the molecule resembles a plain aliphatic hydrocarbon! And we know these only exhibit weak van der Waals intermolecular forces. No wonder esters smell (good *and* bad) - they're volatile and reach our noses easily! \n\n\n", "9" ] ]

[ [ "\n\n> \n> Is any one of these reaction more \"true\" (occurring more often naturally) than the others or is it the case that a little bit of everything is happening?\n> \n> \n> \n\n\nI think the best answer is \"a little bit of everything\". Silver sulfide forms faster but requires exposure of the silver to sulfur-containing materials (like human skin, food, etc.). Silver that isn't exposed to sulfur would still be oxidized by the air, but at slower rates. So depending on the life history of the silver object in question, different amounts of the sulfide and oxide could be present. \n\n\n", "4" ] ]

[]

[ [ "\nThe most likely name would be [naphthenes](http://en.wikipedia.org/wiki/Cycloalkane). It could also be called cycloalkanes.\n\n\nAdded: \n\nSince the reference to the question is about the coal and oil production, the answer would be naphthenes. When the coal and oil were distilled, a layer (called naphthene) was removed which was found to contain mainly cyclopentane and cyclohexane.\n\n\nI would complete the statement as: \n\nIf a **fraction of distillate** contains large amounts of cyclopentane and cyclohexane, it is called **naphthene**.\n\n\n", "1" ] ]

[ [ "\n**It is all about minimizing the energy of a molecule.**\n\n\nIn the case of carbon, the only molecule that adopts a perfect hexagonal geometry in its ground state is benzene (and its derivatives that possess a 6-fold rotational axis). In this case the hexagonal geometry is adopted because all of the carbons are $\\ce{sp^2}$ hybridized. The ideal geometry (lowest energy) for an $\\ce{sp^2}$ hybridized carbon involves three 120 degree bond angles around the carbon atom. This can be accomplished in the case of benzene by placing 6 carbon atoms in a plane and connecting them as shown to produce the benzene structure.\n\n\n\n\n\nOther carbon-based molecules may project as hexagons, but they are not hexagons when viewed in 3 dimensions. For example, cyclohexane is not flat like benzene, but rather typically exists in what is called a \"chair conformation\" in its ground state where the C-C-C angle is around 109.5 degrees.\n\n\n\n\n\n\n> \n> why not octagon?\n> \n> \n> \n\n\nSome other regular polygon shapes can be found in molecules. For example, the cyclopentadiene anion exists as a regular pentagon, and the cyclooctatetraene dianion exists as a regular octagon. Both of these molecules contain only $\\ce{sp^2}$ hybridized carbons.\n\n\n\n\n\nAs mentioned above, these atoms prefer to exist in a geometry with 3 equivalent 120 degree bond angles. However, the internal angle in a regular pentagon and a regular octagon are 108 and 135 degrees respectively. Since these angles deviate from the 120 degree ideal, these molecules will be strained, unlike benzene.\n\n\n\n> \n> What other ring shapes are preferred, low-energy minima?\n> \n> \n> \n\n\nWe've already discussed the $\\ce{sp^2}$ case where the preferred 120 degree angle is found in benzene which exists as a hexagonal structure. $\\ce{sp}$ hybridized carbon is most stable when the X-C($\\ce{sp}$)-X bond angle is 180 degrees. Clearly small or medium size rings comprised solely of $\\ce{sp}$ hybridized carbon cannot exist as stable entities. In the case of $\\ce{sp^3}$ hybridized carbon, the closer we get to the ideal tetrahedral angle (an angle whose cosine = -1/3, *ca.* 109.47 degrees) the lower the energy of the molecule. We've already mentioned cyclohexane and while it's C-C-C angle is close to the ideal tetrahedral angle, it is not exactly the tetrahedral angle (because each of its carbon atoms does not exist in a perfect tetrahedral environment). If we fuse a few more cyclohexane rings onto cyclohexane, we can produce the molecule adamantane with 4 cyclohexane faces all in the chair geometry.\n\n\n\n\n\nThere is less strain energy in adamantane (per carbon) than in cyclohexane. The geometry at each carbon atom is closer to the ideal tetrahedral geometry, but still not quite there. How can we further elaborate adamantane into a ring system where all carbons have the ideal tetrahedral angle? Does adamantane look vaguely familiar? By continuing to add more and more cyclohexane rings we build the following ring structure...\n\n\n\n\n\n... Diamond, where all of the $\\ce{sp^3}$ carbon atoms are in the ideal tetrahedral environment, an extremely stable arrangement.\n\n\n**EDIT:** response to OPs comment\n\n\n\n> \n> So, the lesser energy state must be the reason that after six number\n> of atoms, five is the common one, right?\n> \n> \n> \n\n\nKinda. Here's a table showing estimated ring strains for saturated hydrocarbon rings of various sizes.\n\n\n\n[[Image source](http://ursula.chem.yale.edu/%7Echem220/chem220js/STUDYAIDS/thermo/cycloalkanes/cycloalkanes.html) Also, see this link for a discussion about why the ring strain for cycloheptadecane (-3.4 Kcal/mol) is suspect and probably should be higher.]\n\n\n5- and 7-membered saturated hydrocarbon rings have similar strain energies, yet 5-membered rings are more common. We must consider the free energy of the process, enthalpy **and** entropy changes. In addition to strain (an enthalpic factor) we must also consider entropy. In ring forming reactions it is entropically more difficult for a 7-membered chain to bring both ends together for closure to a ring, then it would be for a smaller 5-membered chain.\n\n\n", "21" ] ]

[ [ "\nIn order of decreasing basicity they are ranked $\\ce{LiCH3}$ > $\\ce{LiNH2}$ > $\\ce{LiOH}$ > $\\ce{LiF}$. We are dealing with the ions $\\ce{F-}$, $\\ce{OH-}$, $\\ce{NH2-}$ and $\\ce{CH3-}$ and so the factor which influences their basicity is the stability of the ion compared to its conjugate acid. \n\n\nThe stability of the ions is determined by the energy of the lone pairs. Carbon has the least effective nuclear charge ($Z\\_{\\text{eff}}$) and so the lone pair on the $\\ce{CH3-}$ ion has the highest energy, making it the least stable ion and so the strongest base. The $Z\\_{\\text{eff}}$ increases as we move across the period until we get to fluorine which has the greatest $Z\\_{\\text{eff}}$ and so its lone pairs have the lowest energy, making $\\ce{F-}$ the least basic ion. \n\n\n", "8" ], [ "\nLook at the conjugate acid of each of the anions (the base). If the conjugate acid is a strong acid (low pKa), then the anion is a weak base. If the conjugate acid is a weak acid (high pKa), then the anion is a strong base.\n\n\nLet's list the anion or base, the conjugate acid and the pKa\n\n\n\\begin{array}{|c|c|c|} \\hline\n\\text{Base }& \\text{Conjugate Acid}& \\mathrm pK\\_\\mathrm a \\\\ \\hline\n\\ce{OH-} & \\ce{H2O} & 15.7\\\\ \\hline\n\\ce{CH3-} & \\ce{CH4} & 48 \\\\ \\hline\n\\ce{NH2-} & \\ce{NH3} & 38 \\\\ \\hline\n\\ce{F-} & \\ce{HF} & 3 \\\\ \\hline\n\\end{array}\n\n\nThe conjugate acid with the highest pKa is methane ($\\ce{CH4}$) it is the weakest acid of the group. Therefore the $\\ce{CH3-}$ in $\\ce{MeLi}$ is the strongest base in the group.\n\n\n", "7" ] ]

[ [ "\nTake a look at [Kelly, P. M., Francis Rose, L.R., The martensitic transformation in ceramics — its role in transformation toughening, Progress in Materials Science, 2002, Vol.47(5):463–557, doi:10.1016/S0079-6425(00)00005-0](http://www.sciencedirect.com/science/article/pii/S0079642500000050).\n\n\nBasically, a transformation is martensitic if it occurs from a collective shear motion of the atoms where each atomic displacement is less than the distance between atoms. There is no diffusion (atoms moving to other lattice sites). The definition is the same as in a metal, but in a ceramic you have multiple atom types in the crystal, that's all.\n\n\n", "1" ] ]

[ [ "\nYour equation $v\\_{rel}=(\\frac{16KT}{\\pi \\frac m2})^{\\frac12}$ is incorrect.\n\n\n$v\\_{mean}=(\\frac{16KT}{\\pi m})^{\\frac12}$ (your last equation) is also incorrect.\n\n\nInstead $v\\_{rel}=(\\frac{16KT}{\\pi m})^{\\frac12}$ and $v\\_{rel}=(\\frac{8KT}{\\pi \\frac m2})^{\\frac12}$ would both be correct.\n\n\nThe need for reduced mass is for when the particles are not the same mass, usually in the context of considering collisions between two different molecules or atoms. \n\n\n$\\mu = m\\_1m\\_2/(m\\_1 + m\\_2)$ by definition. If $m\\_1 = m\\_2$ then $\\mu = m/2$ \n\n\nWhen there are two different masses, there are two different mean velocities. \n\n\nTo see where the equation with the reduced mass comes from, look at appendix 1.4 of [Kinetic Theory of Gases](http://www.cosbkup.gatech.edu/group/chem780/CHAPT1.pdf), along with pages 1-30 to 1-34.\n\n\n", "2" ] ]

[ [ "\nIn general, salt (particularly NaCl) will **increase** the rate of corrosion (rusting).\n\n\nTo understand why, consider metallic iron $\\ce{Fe}$ which rusts (oxidises) to iron(II) oxide $\\ce{Fe2O3}$ in the presence of oxygen $\\ce{O2}$ and water $\\ce{H2O}$.\n\n\n$$\\ce{4Fe +3O2 +6H2O->4Fe(OH)3}$$\n\n\nCorrosion (rust) is a 'redox' reaction, which means it involves reduction of oxygen into hydroxide ions and oxidation of metallic iron to iron cations.\n\n\nFor any redox reaction to take place, electrons are transferred. In this case, from the metallic iron which loses electrons to oxygen. The presence of salt (or any electrolyte) in the water accelerates the reaction because it increases the conductivity of water, effectively increasing the concentration of ions in the water and so increasing the rate of oxidation (corrosion) of the metal.\n\n\nThe situation is complicated by the fact that salt dissolved in water actually reduced the amount of dissolved oxygen present in the water. This is because the water molecules are attracted to the dissolved ions from the salt ([solvation](http://en.wikipedia.org/wiki/Solvation)), which has the tendency to decrease the weak affinity of non-polar oxygen molecules to water, thereby deriving dissolved oxygen out. So if the metal is totally submerged in water, higher salt concentrations can actually reduce the rate of corrosion, if the water is not aerated. \n\n\nSo it depends upon the amount of oxygen getting into the water and onto the metal. For example, underwater plants can release oxygen in salty water which will accelerate the rate of corrosion of the submerged metal. Of course, if the metal is not completely submerged in salt water but only partially submerged, or is affected by salt water 'spray', it will be still be exposed to oxygen and will rust faster.\n\n\n", "15" ] ]

[ [ "\nKT comes in from requiring agreement with the ideal gas law. \n\n\nSee \"section 1.5.3 $\\langle v^2\\rangle$ should agree with the ideal gas law\" and section 1.3 \"The functional form of the dependence of T on $\\langle\\varepsilon\\rangle$ cannot be determined solely from kinetic theory, since the temperature scale may be chosen in many possible ways. In fact, one way to define the temperature is through the ideal gas law: T = pV/(nR)\" in [Kinetic Theory of Gasses](http://www.cosbkup.gatech.edu/group/chem780/CHAPT1.pdf)\n\n\n", "2" ], [ "\nThe Maxwell-Boltzmann distribution of speeds is derived from the Boltzmann distribution, which says that\n\n\n$$\\frac{N\\_i}{N\\_j}=e^{-(\\epsilon\\_i-\\epsilon\\_j)/kT}$$\n\n\nThis formula gives the ratio of the number of particles ($N\\_i$, $N\\_j$) in states with energy $\\epsilon\\_i$ and $\\epsilon\\_j$. From this it is easy to the that $f(v)=Ke^{-\\epsilon/ kT}$ as you noted.\n\n\n", "1" ] ]

[ [ "\nI'll go out on a limb here and say your answer is right and theirs is wrong.\n\n\nNote that the answers differ by a factor of the square root of 3, so I'm guessing they forgot to include the factor of 3. \n\n\nThere would be a small [correction for relativistic effects](http://www.marcelhaas.com/docs/maxrel.pdf), but not enough to explain the differences in the two answers.\n\n\n", "6" ] ]

[ [ "\nThere are two questions being asked here:\n\n\n**Why is aqueous acetic acid a weak electrolyte?**\n\n\nWhen acetic acid is dissolved in water there is an equilibrium reaction:\n$$\\ce{CH3COOH + H2O <=> CH3COO- + H3O+}$$\n\n\nSince acetic acid is a weak acid, the equilibrium position lies well to the left, with only a small fraction of the acetic acid molecules reacting to form ethanoate and hydronium ions. The presence of this small amount of ions results in aqueous acetic acid being a weak electrolyte.\n\n\n**Why is acetic acid highly soluble in water?**\n\n\nSince the vast majority of acetic acid molecules do not dissociate when a sample is dissolved in water, the solubility has to do with the interactions between acetic acid molecules and water molecules. Water has a network of hydrogen bonds between molecules in its liquid phase and so when a substance dissolves in water this bonding is disrupted. However, acetic acid is able to form many new hydrogen bonds to water molecules and so this results in a highly favourable interaction, leading to the high solubility of acetic acid in water.\n\n\n", "9" ], [ "\nI think the Wikipedia page you quote is poorly worded. What on earth does it mean to ‘dissociate into molecules’? Nothing, in my opinion. A better wording is discussed below.\n\n\nBut first, let’s discuss what actually happens when acetic acid is dissolved in water. As you may know, glacial acetic acid consists mainly of $\\ce{H3CCOOH}$ molecules that associate to form hydrogen bonding networks. There is practically no ionisation in glacial acetic acid, i.e. the autoprotonation equilibrium $(1)$ is leaning very strongly to the reactants’ side.\n\n\n$$\\ce{2 H3CCOOH <<=> H3CCOOH2+ + H3CCOO-}\\tag{1}$$\n\n\nSo before dissolution, we are dealing with molecules of acetic acid. If we add these into water, most of them just *stay* being molecules; only a small percentage ionises in water according to reaction $(2)$. The acidity constant shown in the equation is a measure of how many molecules are dissociated; it depends on the concentration.\n\n\n$$\\begin{gathered}\\ce{H3CCOOH <<=> H3CCOO- + H3O+}\\\\\nK\\_\\mathrm{a} = 10^{-4.76} = 1.74 \\times 10^{-5}\\end{gathered}\\tag{2}$$\n\n\nSo upon dissolution we have:\n\n\n* most ($> 99~\\%$) of the acetic acid molecules remaining acetic acid molecules\n* a very small subset deprotonating to form acetate anions\n\n\nOnly the latter are charged compounds and thus only they contribute to the solution’s conductivity. Since they are few in number, conductivity is low.\n\n\nGetting back to the original quote. It seems to imply that dissolving acetic acid in water ‘turns it’ into molecules; which is wrong: the acetic acid molecules remain the same all through. A better wording would emphasise that they *do not change.* For example:\n\n\n\n> \n> Acetic acid is extremely soluble in water, but most of the dissolved compound remains as molecules, rendering it a weak electrolyte.\n> \n> \n> Acetic acid is extremely soluble in water, but only a small fraction is dissociated into ions, rendering it a weak electrolyte.\n> \n> \n> \n\n\n", "3" ] ]

[ [ "\nIt sounds contradictory, but soluble and insoluble are relative terms. Silver chloride's $K\\_\\mathrm{sp}$ is $1.77\\times 10^{-10}$, so one can generally think of it as insoluble, but actually about a milligram will dissolve in a litre of water.\n\n\nMuch like the terms strong acid and weak acid, strong electrolyte and weak electrolyte refer to the dissociation of a substance in a solvent, though they include all electrolytes, not just acids. In the case of silver chloride, though little of it dissolves, what does is present only as $\\ce{Ag+}$ and $\\ce{Cl-}$, not solvated $\\ce{AgCl}$. In water, all salts are strong electrolytes, but in other solvents, things can be different. For example, while perchloric acid is a strong electrolyte in water, it doesn't dissociate completely in acetic so it's a weak electrolyte.\n\n\nSource: (1) Electrochemical Dictionary; Bard, A. J.; Inzelt, G.; Scholz, F., Eds.; Springer Berlin Heidelberg: Berlin, Heidelberg, 2008.\n\n\n", "9" ] ]

[ [ "\n\n> \n> I would like to know how this has a positive charge given that it has 3 bonding pairs of electrons around it, which would give it 6 valence electrons and 8 electrons overall. If carbon has 6 protons, how does this give an ion with a positive charge?\n> \n> \n> \n\n\nYou need to count ALL the protons and electrons in the ion. What about the atoms other than the central carbon?\n\n\n", "4" ], [ "\nWell I get what your problem is. \n\n\nThe thing is that earlier C had 4 electrons in its valence shell. Now it is bonded to 3 other atoms giving it a total of 6 electrons in its shell. The octet rule says that it should have 8. Since the other carbon took away 2 electrons this means that our carbon should have a +2 charge. \n\n\nWell this is where the confusion lies . \n\n\nThe trick is whenever we calculate formal charge we use \n\n\n\n> \n> **Formal Charge= (No.of valence electrons in unbonded state - no of lone\n> pair electrons ) - (no. of bond pair electrons/2)**\n> \n> \n> \n\n\nIn this case the charge comes out to be (4-0) - (6/2) =+1 We usually associate half of the bond pair electrons to each atom while calculating charge. On the other hand When counting the no of electrons in valence shell we count both the electrons in the bond pair. \n\n\nGood Luck :)\n\n\n", "4" ] ]

[ [ "\nWhat actually happens in real life depends on a lot of things. Factors like the shape of the pot can make a big difference to how much of that 2500 W actually goes into vaporizing the water and how much of it is lost to the surroundings. If we assume that all 2500 W is going into the water, it makes things a lot simpler. If the water is already at the boiling point, then all we have to do is calculate the total amount of energy needed to vaporize all the water:\n\n\nThe enthalpy of vaporization of water is 40.65 $\\mathrm{kJ/mol}$, which means 40.65 kJ of heat needs to be added to turn one mole of liquid water to steam. 50 L of water is 2775 moles:\n$$\\mathrm{\\frac{(50\\ L)\\left(1000\\frac{g}{L}\\right)}{\\left(18.02\\frac{g}{mol}\\right)}=2775\\ mol}$$\n\n\nSo the total heat we need is:\n$$\\mathrm{(2775\\ mol)\\left(40.65\\frac{kJ}{mol}\\right)=112.8\\ MJ}$$\n\n\n1 watt is one joule per second, so for a 2500 W heat source:\n$$\\mathrm{\\frac{112.8\\ MJ}{2500\\ J/s}=45\\ ks}$$\nOr, approximately 12.5 hours. In real life, it may be significantly longer for various reasons, but this is the fastest it could go.\n\n\n", "16" ] ]

[ [ "\nIn general, we can derive the expression for the density of any substance as a function of temperature using the definition of [thermal expansion](http://en.wikipedia.org/wiki/Thermal_expansion).\n\n\n$$\\frac{\\mathrm dV}{\\mathrm dT}=\\alpha V\\tag1$$\n\n\nBy differentiating $\\rho=m/V$, we can show that $\\mathrm d\\rho/\\rho = -\\mathrm dV/V$. So using this and equation $\\text{(1)}$,\n\n\n$$\\frac{\\mathrm d\\rho}{\\mathrm dT}=-\\alpha V\\tag2$$\n\n\nIntegrating this above expression and approximating the expression,\n\n\n$$\\rho = \\rho^\\circ\\cdot\\left(1-\\alpha\\Delta T\\right)\\tag3$$\n\n\nwhere $\\rho^\\circ$ is the initial density at the initial temperature, $\\alpha$ is the volumetric thermal expansion coefficient and $\\Delta T$ is the temperature. \n\nFor water, $\\alpha$ is typically [$207\\times10^{-6}\\ \\mathrm{K^{-1}}$](http://en.wikipedia.org/wiki/Thermal_expansion#Thermal_expansion_coefficients_for_various_materials) at $20\\ \\mathrm{^\\circ C}$. Using this, we can observe how temperature affects the density of water.\n\n\n", "1" ] ]

[ [ "\n\n> \n> Is the nitroaldol reaction that much faster?\n> \n> \n> \n\n\nMy guess would be that the nitroaldol is faster than the Cannizzaro. The Cannizzaro reaction involves\n\n\n* attack by hydroxide on the carbonyl to reversibly form the tetrahedral intermediate\n* alignment of the tetrahedral intermediate with a second aldehyde molecule (entropically costly)\n* hydride transfer in a redox reaction (relatively high energy, even complexed hydride is not particularly stable) to form molecules (alcohol and carboxylic acid) that are not that different in stability than the starting material\n\n\nThe nitroaldol involves\n\n\n* generation of the nitroalkyl carbanion\n* attack by the carbanion on the carbonyl to form a tetrahedral intermediate\n* capture of a proton by the negatively charged oxygen to produce a neutral molecule\n* elimination of water to produce a more stable (more conjugated) molecule than we started with (*e.g.* energetically downhill)\n\n\nThe nitroaldol doesn't require any special molecular alignment after the tetrahedral intermediate is formed and is energetically downhill.\n\n\n\n> \n> Can the yields be so different due to the steric hindrance of the\n> additional methyl group?\n> \n> \n> \n\n\nFurther alkylation at the nitroalkyl carbanion center will not help the reaction in a steric sense just as you suggest, but also a secondary carbanion will be less stable (higher energy, more difficult to form) than a primary carbanion.\n\n\n\n> \n> What is the benefit of these nitrogen bases over $\\ce{NaOH}$?\n> \n> \n> \n\n\nThey are somewhat milder reagents than hydroxide and therefore less likely to produce side reactions.\n\n\n", "3" ], [ "\nAlthough it is some time ago already, seeing this question active today made me decide to include the answer given by Prof. Clayden himself over e-mail to me.\n\n\n\n> \n> The reaction you ask about appears in the classic text on practical\n> organic chemistry by Vogel: in fact this preparation is online at\n> <http://www.erowid.org/archive/rhodium/chemistry/nitrostyrene.vogel5.html>\n> \n> \n> It’s not clear to me why amines would be preferable bases in general,\n> although you are right that they avoid the Cannizzaro reaction if the\n> nitro compound is hindered. They could also be preferable if the\n> alcohol were the desired product, rather than the eliminated alkene.\n> \n> \n> \n\n\n", "1" ] ]

[ [ "\nMelatonin is not \"radio-protective\" in the sense that it blocks radiation or anything. As you note, it's an antioxidant. One of the ways in which ionizing radiation can interact with the body is by providing enough energy to break chemical bonds, producing highly reactive radical species that can cause biological damage by reacting with various things in the body. Melatonin can easily be oxidized by many of these reactive species, preventing them from reacting with anything else. [This paper](http://onlinelibrary.wiley.com/doi/10.1111/j.1600-079X.2010.00847.x/abstract) shows the products for the reaction with radio-generated hydroxyl radicals.\n\n\nThe role melatonin plays in sleep is fairly well known (though the effects of exogenous melatonin, less so). Without going into too much detail, there are melatonin receptors in the body, found in cell membranes in the brain and a few other places (G protein-coupled receptors). When these bind melatonin on the outside of the cell, their conformation changes, further activating signalling processes on the inside of the cell. The effects of almost all signalling molecules in the body work in this way. The molecules activate some kind of receptor, which then activates a sequence of other signals, eventually resulting in biological effects.\n\n\n", "2" ] ]

[ [ "\nThe size of a molecule does not correlate with the color of a molecule. \n\n\nColor depends upon which wavelengths of light are absorbed and which are transmitted by a molecule. Visible light typically ranges from around 390 to 700 nm. Light with a wavelength between approximately 635 to 700 nm appears red to the human eye. If we have a molecule that absorbs all visible light below 635 nm, the molecule will appear red; it absorbs all visible light **except** for the red wavelengths, hence only the red wavelengths will be reflected or transmitted and be available for the eye to detect.\n\n\n**What determines which wavelengths of light are absorbed by a molecule?** \n\nThis is controlled by the energy separation (energy gap) between the highest energy orbital that contains an electron (the HOMO or highest occupied molecular orbital) and the lowest energy orbital that an electron can be promoted to (the LUMO or lowest unoccupied molecular orbital). If this energy gap falls in the visible range, then the molecule will appear colored to the eye when the molecule absorbs light at the wavelength that corresponds to the energy gap (*e.g.* when a photon is absorbed that promotes an electron from the HOMO to the LUMO) . \n\n\nMolecules like 2,2-dimethylpropane contain only sigma bonds. The energy gap in typical sigma bonds is very large meaning that it takes a lot of energy to promote an electron from the HOMO to the LUMO. The amount of energy required typically falls in the ultra-violet region, outside the visible range; therefore these molecules are usually colorless since they transmit all visible wavelengths.\n\n\nMolecules containing pi bonds have smaller HOMO-LUMO energy gaps than molecules with just sigma bonds. If there are enough pi bonds and they are conjugated then the gap can be small enough that light in the visible range can be absorbed and make the molecule appear colored. Here is a diagram showing how the energy gap decreases as we increase the length of our conjugated pi system.\n\n\n\n\n\nEven with 1,3,5-hexatriene the light absorption at 258 nm is still outside the visible range, so the molecule will not be colored. But as we continue to extend the conjugation, eventually the molecule will absorb light around 400-500 nm, the blue light range. As the molecule absorbs blue light it will appear yellow. In addition to pi conjugation, certain substituents such as the nitro group and the carbonyl, to just name two, can also serve to extend the conjugation path in a molecule.\n\n\nThe molecule you asked about, penta-1,2,3,4-tetraene, looks roughly like this.\n\n\n\n\n\nIf we examine it closely, we'll notice that the orbitals really look like those in two distinct (orthogonal) 1,3-butadiene molecules. As the chart above shows, 1,3-butadiene does not absorb in the visible range. Nonetheless, because penta-1,2,3,4-tetraene does have an extended pi system it will have a smaller HOMO-LUMO gap than 2,2-dimethylpropane and will absorb light at longer ultra-violet wavelengths (lower energy) than 2,2-dimethylpropane.\n\n\n", "3" ] ]

[ [ "\nSince sound waves are [longitudinal waves](https://en.wikipedia.org/wiki/Longitudinal_wave) (generally in air) there needs to be some kind of interaction between neighboring \"air\" molecules (a mixture consisting mostly of $\\ce{N2}$, $\\ce{O2}$, $\\ce{CO2}$ and $\\ce{Ar}$). You can see that this is just not possible if the mean free path of the \"air\" molecules is larger than the desired wavelength. How would the \"air\" molecules interact on such distance if there are no \"air\" molecules to interact with? This interaction (i.e. sound of a particular wavelength) will only be possible if there are sufficiently many \"air\" molecules nearby to interact with, or formulated otherwise: if the mean free path is sufficiently much smaller than the desired wavelength of the sound. Of course, it is possible to make sound waves with a larger wavelength than the length of the mean free path, just not the converse.\n\n\nOne application I can think of lies more in Physics but whatever. In the days before Einsteins theory of special relativity was developed/accepted, there was a common consensus, which was supported by Maxwell himself, that light was travelling in a special reference frame called \"[The Aether](https://en.wikipedia.org/wiki/Aether_theories)\". In this reference frame the speed of light equaled $c$. We now know of course that light can travel through vacuum and that there exists no such special reference frame as The Aether. There were various problems with this Aether theory, most famously the [Michelson-Morley](https://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment) experiment that suggested that either the Aether did not exist or was moving for a weird reason with the exact same trajectory and speed as earth in space. Another problem, related to your question, was that the speed of waves in a medium is dependent on the particle density of that medium. And since (at least some) electromagnetic waves have an incredible small wavelength (the $\\lambda$ of visible light for example was known), the density of this Aether medium had to be extremely high, yet it had to also weight almost nothing and be able to penetrate trough everything!\n\n\nFrom this and other facts Einstein must have concluded that the Aether theory was in fact invalid. This idea ultimately lead to the theory of special and general relativity, both of which have many spectacular and far reaching consequences.\n\n\n", "5" ] ]

[ [ "\nLiquid crystals are usually rigid rod-like molecules that tend to show some degree of intermolecular order even when notionally liquids (there is a good summary on [Wikipedia](http://www.wikiwand.com/en/Liquid_crystal)). This happens because the rigid polarisable parts of the molecules can have relatively strong directional interactions even in the liquid phase.\n\n\nSome of the structures they form can rotate the plane of polarised light You can get some intuition about why knowing that polarising lenses consist of highly ordered molecules whose interaction with light is highly direction dependent. In normal liquids the molecular orientations are effectively random so, even in each molecule interacts with light depending on its orientation, this effect washes out due to the randomness of the overall structure. Liquid crystals form structures where there is the possibility or large-scale order allowing macro-effects on polarisation (just like the rigid structure in a polaroid film).\n\n\nThe other thing that makes liquid crystals useful is that the details of the ordering in the liquid can be affected by external electric fields, especially in thin films of the compounds. In principle this is possible when any molecule is polar (that is it has a net electrical dipole). Of course, making an effective display requires a lot of fine tuning of the electrical, mechanical and chemical structure, but the basic details are simply that the molecules are affected by electrical fields and this allows some control of the macro-structure which, in turn, alters the way the liquid crystal interacts with light. the effect is not \"turned on\" by the field, just altered and that is enough to make a viable display technology.\n\n\nLiquid crystals have a long history. The first compounds known to show liquid crystal behaviour is cholesterol benzoate, the first molecule shown below, which was dicoveredn in 1888. The other structures show some actual molecules from the early days of LCD displays to modern developments. For more examples of the history there is a great summary from Merck (one of the major manufacturers) in this [slideshow](http://www.slideshare.net/MerckChemicals/merck-chemicals-100-years-of-liquid-crystals-at-merck-3477877).\n\n\n\n\n\n", "4" ] ]

[ [ "\nThe electrical current is carried by **all** ions in the solution, including the $\\ce{SO\\_4^{2-}}$/ $\\ce{HSO\\_4^-}$, $\\ce{NO\\_3^-}$, and $\\ce{PO\\_4^{3-}}$/ $\\ce{HPO\\_4^{2-}}$/ $\\ce{H\\_2PO\\_4^-}$ ions, and not just by $\\ce{H+}$ and $\\ce{OH-}$. So, even though the $\\ce{OH-}$ concentration is decreasing with increasing acid concentration, like you say, it's much more than made up for by the increased concentration of the anions from the acids.\n\n\n**EDIT** (in response to a comment): The contribution of ions that don't participate in the electrochemistry differs depending on the system. For example, in $\\ce{NaOH}$ the $\\ce{Na+}$ effectively just distributes to balance the charge from the concentration gradient in $\\ce{OH-}$:\n\n\n* $\\ce{H2O}$ is reduced to $\\ce{H2}$ and $\\ce{OH-}$ at the cathode, increasing the $\\ce{OH-}$ concentration there\n* $\\ce{OH-}$ is oxidized to $\\ce{O2}$ and $\\ce{H+}$ at the anode, decreasing the $\\ce{OH-}$ concentration there\n* Thus, $[\\ce{OH-}]$ changes from high at the cathode to low at the anode\n* In order for the solution to maintain [electroneutrality](http://www.chem1.com/acad/webtext/elchem/ec1.html \"electroneutrality\"), something has to balance the higher concentration of negative charge at the cathode, and in this system $\\ce{Na+}$ is the major contributor. (Water self-ionization can't provide enough $\\ce{H+}$ to maintain the charge balance.)\n\n\nIn other systems such as $\\ce{H3PO4}$, the acid-base equilibration behavior of an electrochemically inert species can contribute to the charge carrying function:\n\n\n* Adjusted to pH 7 with, say, NaOH, phosphoric acid exists mainly as $\\ce{H2PO4-}$ and $\\ce{HPO4^{2-}}$\n* Water is oxidized at the anode, producing $\\ce{H+}$, which reacts with $\\ce{HPO4^{2-}}$ to form $\\ce{H2PO4-}$\n* Water is reduced at the cathode, producing $\\ce{OH-}$, which reacts with $\\ce{H2PO4^{-}}$ to form $\\ce{HPO4^{2-}}$ and $\\ce{H2O}$\n* So, $\\ce{HPO4^{2-}}$ has a higher concentration around the cathode than the anode, and $\\ce{H2PO4^{-}}$ has a higher concentration around the anode than the cathode. These concentration differences result in a net transfer of charge between the two electrodes.\n* All of the negatively charged phosphate ions are balanced out by a corresponding distribution of positively-charged $\\ce{Na+}$, similar to the $\\ce{NaOH}$ case.\n\n\n", "3" ] ]

[ [ "\nThe positive terminal of a battery is where 'conventional current' comes out of the battery and flows through an external circuit, returning to the battery via the negative terminal. Since electrons are (by convention) negatively charged, the negative terminal of a battery is where electrons are 'released' by the battery.\n\n\nNOTE: Of course, if it is a secondary battery which is 'recharging', electrons will 'enter' the battery at the terminal marked negative.\n\n\n", "3" ] ]

[ [ "\n@Dissenter, you really need to stop asking questions like this. Once again you point out that the emperor has no clothes! :)\n\n\nIt turns out that the mechanism of simple proton transfer is an active area of investigation. There is not a single yes or no answer that would adequately address your questions. But if this truth were told in the introductory texts who would continue on when the answer to seemingly simple questions leads to such a morass? Instead, particularly in introductory circumstances, there is a gentlemen's agreement of sorts that we will simply say that the emperor's clothes are the finest we've seen.\n\n\nLet's look at your two questions in reverse order.\n\n\n\n> \n> why do we depict acid base reactions as concerted?\n> \n> \n> \n\n\nIn many introductory texts, and even some that are for advanced courses, proton transfer is presented as a concerted process based largely on Occam's razor (the simplest explanation is often the correct explanation). However, enough experiments in this area have been run that I think it would be fair to say that a stepwise mechanism is now favored, but by no means mandated for all proton transfers (we'll see why some proton transfers proceed in a concerted fashion below). The stepwise mechanism proceeds as follows: \n\n\n* step 1, formation of the encounter complex $\\ce{A-H}$ and $\\ce{B-}$ diffuse through solution until they are close enough to form a hydrogen bond\n\n\n$$\\ce{A-H + B- -> A-H..B-}$$\n\n\n* step 2, the $\\ce{A-H}$ bond elongates while the $\\ce{B-H}$ bond contracts and a transition state is reached.\n\n\n$$\\ce{A-H..B -> [A..H..B]}$$\n\n\nThe transition state is not necessarily linear or symmetric. There could also possibly be an intermediate accompanying the transition state. This depends upon your definition of an intermediate. How long does a species need to survive in order to be called an intermediate; is it 10 vibrations, 1 vibration or what? Obviously it is difficult to detect such short-lived molecules, but ultra-fast spectroscopy has made great strides and we may soon know whether an intermediate exists along this multi-step reaction coordinate.\n\n\n* step 3, the transition state elongates in the opposite direction\n\n\n$$\\ce{[A..H..B] -> A..H-B}$$\n\n\n* step 4, the hydrogen bond is broken and the complex diffuses apart\n\n\n$$\\ce{A..H-B -> A- + B-H}$$\n\n\n\n> \n> Does hydrogen bonding strength correlate with Brønsted basicity in a\n> given medium?\n> \n> \n> \n\n\nIf we throw out cases where geometry interferes with hydrogen bonding (*e.g.* one example would be where steric factors preclude close approach or proper alignment of the species involved), then it would not be unreasonable to expect some correlation between hydrogen bond strength and the pKa difference between the two species involved. There is certainly experimental evidence to support this view. However, it has been found that if we step back and examine a larger range of pKa differences, the correlation breaks down. \n\n\nHere is a diagram showing how the rate of proton transfers might change as a function of the pKa difference between the two species involved.\n\n\n\n\n\nIf we look at a small portion of the plot, the part at low pKa, we could fairly easily convince ourselves that there is a linear correlation. But eventually as the pKa becomes large, the rate of proton transfer becomes diffusion controlled and insensitive to any further increase in the pKa. The same concept applies to hydrogen bonding. There could be reactions (acid-base proton transfers for example) where there is no activated process, the reaction is diffusion controlled; as a consequence we cannot expect to accurately measure or compare hydrogen bond strengths as a function of Brønsted basicity differences in such a regime.\n\n\n**Summary Answers**\n\n\n\n> \n> why do we depict acid base reactions as concerted?\n> \n> \n> \n\n\nTo keep it simple in introductory classes. It turns out that many acid-base proton transfers are not concerted, but rather involve a multi-step process. Some other proton transfers, for example those involved in diffusion controlled reactions, are probably concerted. This is still an active area of research.\n\n\n\n> \n> Does hydrogen bonding strength correlate with Brønsted basicity in a\n> given medium?\n> \n> \n> \n\n\nThe answer is \"yes\" if we 1) examine only a small range of basicities and 2) look in the range of basicities where the hydrogen bonding is described as an activated process.\n\n\n", "12" ], [ "\nThere are several methods which can be used to compare the Brønsted-Lowry (proton-transfer) basicity and the hydrogen bond basicity. \n\n\nOne of the most common defines a parameter $ \\beta\\_{2} = (log K\\_{B}^{H} + 1.1)/4.636 $, where $logK\\_{B}$ is calculated via the solution of $log K^i = L^ilogK\\_{B}^{H} + D^i$ ($K$ being the experimental equilibirum constant, $L$ and $D$ acid-dependent parameters) over 34 reference compounds. The numerical correction results in $ \\beta\\_{2} = 0 $ and $ \\beta\\_{2} = 1 $ being the lower and upper limits for hydrogen bonding strength. \n\n\n$ \\beta\\_{2} $ values are not well correlated with measures of Brønsted-Lowry basicity (e.g. proton affinity); they are well correlated when comparing very divergent systems such as water vesus phenols but not when comparing members of the same family; e.g. different halo-subsituted phenols. \n\n\nWe can rationalize this by considering that hydrogen bonding strength is dependent on the electronegativity of the donor atom as well as the donor lone pair LUMO/H $\\sigma\\*$ HOMO energy-match. Increasing the electronegativity of the donor atom at constant $pK\\_{a}$ results in greater hydrogen bond basicity.\n\n\nAs for your second question, proton transfer is not going to occur step-wise because an unsolvated proton is very unstable; loss of a proton is always associated with formation of a charged complex in solution\n\n\nSources: [1](http://pubs.acs.org/doi/abs/10.1021/ja01045a038), [2](http://pubs.rsc.org/en/content/articlelanding/1990/p2/p29900000521#!divAbstract), [3](http://www.ncbi.nlm.nih.gov/pmc/articles/PMC4004274/)\n\n\n", "4" ] ]

[ [ "\nThe only \"hydroxyl\"-type species whose recombination has been observed to yield hydrogen peroxide are **hydroxyl radicals**.\n\n\n$$\\ce{HO^. + ^.OH -> H2O2}$$\n\n\nIn order to generate these hydroxyl radicals from water, a lot of energy has to be provided, typically by radiolysis (short wave uv, $\\gamma$, electron beam).\n\n\nThe conceivable steps yielding $\\ce{HO^.}$ are either\n\n\n$$\\ce{H2O -> H^. + ^.OH}$$\nor\n$$\\ce{H2O -> H2O^{+.} + e-}$$\n\n\n$$\\ce{H2O^{+.} -> H+ + ^.OH}$$\n\n\n", "5" ], [ "\n\n> \n> Is it possible that $\\ce{2 OH-}$ from water could react to form\n> $\\ce{H2O2}$?\n> \n> \n> \n\n\nThere would be two electrons left unaccounted for. There needs to be a balanced chemical equation, then we could consider the equillbrium constant for the reaction. \n\n\nWe could write $\\ce{2H2O -> H2 + H2O2}$, but it is extremely unfavorable, like water decomposing to hydrogen and oxygen gas. You could calculate the Gibbs free energy of the reaction and an equilibrium constant, but the equilibrium constant will be extremely small. \n\n\n", "3" ] ]

[ [ "\nProbably you are having problems with Le Chatelier's Principle.\n\n\nSuppose you have an equilibrium established between four substances $\\ce{A}$, $\\ce{B}$, $\\ce{C}$ and $\\ce{D}$, such that\n\n\n$$\\ce{A + B <=> C + D}$$\n\n\n \n\n\n> \n> What would happen if you changed the conditions by increasing the\n> concentration of $\\ce{A}$?\n> \n> \n> \n\n\n \n\nAccording to Le Chatelier, the position of equilibrium will move in such a way as to counteract the change. That means that the position of equilibrium will move so that the concentration of $\\ce{A}$ decreases again - by reacting it with $\\ce{B}$ and turning it into $\\ce{C + D}$. The position of equilibrium moves to the right, making more of the products. Of course, this assumes that there is still some amount of $\\ce{B}$ left in the reaction vessel.\nFor better understanding, refer [this](http://www.chemguide.co.uk/physical/equilibria/lechatelier.html).\n\n\n", "18" ], [ "\nSimply it means that the equilibrium will force the conversion of certain substances to reduce the disturbance, to maintain the equilibrium itself. So, for example, $\\ce{A + B <=> C + D}$ (both directions). \n\n\nIf you increase the concentration of the reactants — $\\ce{A}$ or $\\ce{B}$, then you’re increasing the conversion of reactants into products, because the rate of reaction and overall number of product formed increases. Therefore, we need to form more $\\ce{C + D}$, to decrease the concentration of $\\ce{A + B}$ and balance out the rate of change. \n\n\nTherefore, even though the reaction occurs simultaneously as a forward and backward reaction, the movement of equilibrium to the right (to the products) means the conversion of reactants to products is further emphasised, and generally — there is more of it occurring than in a usual equilibrium, in order to balance the change. Same thing would’ve happened if you increase the concentration of products, you want the equilibrium to shift to the left (reactants) so concentrations of both reactants and products balances out, reducing the disturbance (Le Chatelier’s Principle — specifically referred to by Ashu). Simply shifting equilibrium means increased rate of conversion of substances, predicating on the change in the reaction in the first place. \n\n\nSorry, it’s quite long winded and irrelevant at times, I hope you’ve benefitted from this though. I’ve also used the idea of changing concentrations in particular from Ashu, if that’s alright.\n\n\n", "10" ], [ "\nLet’s do it simply. You start from some equilibrium. For example you have simple reaction $\\ce{A <=> B}$, and in the equilibrium state there is $60~\\%$ of $\\ce{A}$ and $40~\\%$ of $\\ce{B}$. If it said that equilibrium is shifted to the right it means that you have more product $\\ce{B}$ (*i.e.* more than $40~\\%$). If you shifting to the left you gaining more substrate. Now how it's done — look on the previous answers.\n\n\n", "3" ], [ "\n“Equilibrium shift” is jargon. The equilibrium constant stays the same when you disturb an equilibrium by adding or removing species. However, the system might not be at equilibrium anymore.\n\n\nWhen the system was at equilibrium, there was no net reaction. Now that the equilibrium is disturbed, there will be a net reaction in the direction of reactants or products (depends on the specifics of how the equilibrium was disturbed). The jargon for this net reaction is to say the equilibrium “shifted to the left” (or to the right).\n\n\nIt would be more formal to say equilibrium was attained again by a net reverse (or forward) reaction. This would also be easier for someone learning these concepts because it better describes what is going on.\n\n\n1. The reaction is at equilibrium\n2. The equilibrium is disturbed\n3. There is a net reaction (in one or the other direction) until we are at equilibrium again.\n\n\nTypically, the concentrations present at the original equilibrium state will be different from those for the second equilibrium state.\n\n\nWhen a change in temperature causes a change in the equilibrium constant, the phrase “equilibrium shift” might be a little bit more intuitive. There will be a net reaction to adjust to the new equilibrium constant.\n\n\n", "2" ] ]

[ [ "\nBatteries are constructed with an internal barrier between the two electrolytes that does **not** conduct electricity (i.e., electrons can't pass), but that **does** allow ions to pass through. See the figure below (taken from [Wikimedia](http://upload.wikimedia.org/wikipedia/commons/thumb/0/04/Alkaline-battery-english.svg/2000px-Alkaline-battery-english.svg.png \"Wikimedia\"), open source), of a simple alkaline dry cell (e.g., an AA battery), in particular the part labeled \"ION CONDUCTING SEPARATOR.\"\n\n\n\n\n\nThe reason batteries are made this way is because in order for the anode electrolyte and the cathode electrolyte to come to thermochemical equilibrium, there **has** to be a transfer of **both** electrons and ions. By putting the separator in there that only conducts ions, you block equilibration of electrons except in the case where you have it connected to an external circuit. If you conduct an imaginary experiment where you \\* **SNAP** \\* your fingers and make that ion conducting separator disappear, the electrons would indeed surge from one electrolyte to the other as you describe, as fast as mass transfer would allow. (You would then having to imagine the battery either exploding or melting, because this would make a ***LOT*** of heat.)\n\n\n", "7" ], [ "\nThe charge doesn't \"move back through the battery\" because in order to do that, it would have to pass directly through the chemical reactions that put it there in the first place. It moves through the external circuit, because that's the only path available to it.\n\n\nTo take it to the next level of detail, the half-reactions that occur at the two plates inside the battery are actually occurring in both directions all of the time. When the battery is open-circuit, both reactions are occurring at the same rate. The forward reactions are slowed down by the electric field created by the separation of charge, and the reverse reactions are accelerated by it, and the battery maintains equilibrium.\n\n\nWhen an external path for current is created across the battery terminals, some of the charge flows through it reducing the voltage and the E-field slightly, and this allows the forward reactions within the battery to occur at a higher rate than the reverse reactions, and this is what eventually dicharges the battery as the reactants are consumed.\n\n\n", "5" ], [ "\nThe answer is: it will! When no salt bridge or similar conducting element is present between the two half cells, the reaction will quickly run to equilibrium. The state in which the potential force of the difference in electron affinity between the two electrodes more or less equals the Coulomb force of the opposite charges that accumulate in them. However, if you make use of a salt bridge this charge difference accumulation can be removed, since an electrolyte, which sort of completes the circuit. Usually salt bridges are constructed in a manner that prevents intermixing of the ions in both half cells, so that there is some certainty about what redox reaction is actually going on, and how. An example of a common salt bridge used in high school chemistry experiments is a filter paper soaked in a solution of potassium chloride.\n\n\n", "1" ] ]

[ [ "\nThis is the square planar splitting diagram\n\n\n\n\n\nMany d8 metal complexes are usually square planar because a very stable electron configuration is achieved if you put the electrons on this splitting diagram -6 electrons stabilizing the compound and only two slightly destabilizing electrons-\n\n\nNow you could ask, why then not d6 and d7 adopt this cinfiguration? Because for 6 or 7 electrons, an even more stable state is achieved with octahedral fields.\n\n\nAbout hybridization... Forget about it when talking about coordination complexes. It's completely useless. It can't explain or predict anything that CFT/LFT can't\n\n\n", "7" ], [ "\nThe use of hybridisation for transition metal complexes is extremely outdated and often leads to incorrect predictions and rationalisations of transition metal chemistry. If you are using it routinely, please stop using it.\n\n\nYou may wish to first refer to this other answer of mine for a discussion of a very similar case:\n\n\n[Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral?](https://chemistry.stackexchange.com/questions/40880/why-is-pdcl42-square-planar-whereas-nicl42-is-tetrahedral)\n\n\nThere, the comparison is between a 3d and a 4d metal. However, the factors that affect the geometry of a $\\mathrm{d^8}$ complex are the same. They are described in much more detail in that link. To summarise:\n\n\n1. Adoption of a square planar geometry is favoured due to occupation of four low-energy orbitals formed.\n2. Adoption of a tetrahedral geometry is favoured due to (a) relief of steric repulsions; and (b) the relief of electron-electron repulsions due to adoption of a high-spin configuration.\n\n\nWhen the ligand is $\\ce{CN-}$, a strong sigma donor and pi acceptor, the corresponding splitting of the orbitals in $D\\_\\mathrm{4h}$ symmetry will be very large. Consequently, the electronic stabilisation derived upon adoption of a square planar geometry in $\\ce{[Ni(CN)4]^2-}$ is much larger than for other complexes, e.g. $\\ce{[NiCl4]^2-}$.\n\n\nIt's quite possible that the linear ligand $\\ce{CN-}$ also introduces less steric repulsions in a square planar geometry than a halide, like $\\ce{Cl-}$ or $\\ce{Br-}$, would. However, the main factor is likely to simply be the large electronic stabilisation, which in turn derives from the fact that $\\ce{CN-}$ is very high in the spectrochemical series.\n\n\n", "7" ], [ "\nIt's tetrahedral; as not 6 lone pairs are coming from $\\ce{CN}$, only $\\ce{CN}$ are attached. Therefore, only 4 lone pairs will affect the configuration.\n\n\nSo by now applying CFT you will get its hybridization to be $\\ce{dsp^2}$ which will make the shape of tetrahedral.\n\n\n", "-4" ] ]