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## Hot answers tagged ionic-compounds

13

Calcium hydroxide has solubility about 1.9 g/L. This is enough to create pH above 11, i.e. a strongly basic solution. $\ce{CaCl2}$ solutions should be very slightly acidic if they were made from pure $\ce{CaCl2}$. This might not be the case. In industry calcium chloride is produces by reaction of calcium hydroxide with ammonium chloride, so industrial-grade ...

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With enough effort, Born–Haber cycle can be extended to polyatomic ionic solids, however it's practically never done in practice due to the lack of experimental data or because it's impossible to obtain any. From [1, p. 117–118] (emphasis mine): Lattice energies cannot be measured experimentally since they represent hypothetical processes: $$\ce{M^n+... 7 Two sources: water and natural flavors According to the Wisconsin department of healt natural sodium levels in water is \pu{40-60 mg/L} meaning a \pu{20 fl oz (\! 591 mL)} Coke has about \pu{25 mg} of sodium from water. Thus the other \pu{50 mg} of sodium must come from natural flavors. This makes sense since Coca-Cola uses plant products like coca ... 6 They can, and they do. The secret is to have a chiral cation and separately a chiral anion. Letting D^+ be the dextrorotatory form of the cation, L^+ be the levorotatory form, and anologously for the anion, we then have four isomeric salts: D^+D^- D^+L^- L^+D^- L^+L^- The first and second are diastereomers because only the anion is mirror-... 5 London dispersion forces are instantaneous dipole-induced dipole interactions. All compounds can have an instantaneous dipole moment, including ions, so London dispersion forces exist for all compounds. Unlike molecules, ions are charged. The electrostatic interactions between charges are far stronger than the London dispersion forces between ions, so ... 5 In most of these type problems chemists look for ways to simply the mathematical expression by using approximations based on a knowledge of the chemistry. The gist is that because of the significant figures you assume that the reaction essentially goes to completion. So \ce{Ag(CN)_2^− \approx 0.0300} and \ce{CN− \approx 0.0400}. Then you solve for [Ag+]. ... 4 Basically, the Sodium could come from any or all of the ingredients used to manufacture Coca Cola. The ingredients list is based on the raw-materials used in the product, but these ingredients are not necessarily pure, even if they are Food Grade. As an example: The Phosphoric acid used is probably a 75% H3PO4, a common commodity in the industry, but the ... 4 The commonly used method of obtaining ionic radii for higher coordination numbers (C.N.) is to extrapolate values from the Shannon's scale [1] using the relationship between ionic radius and coordination number proposed by Zachariasen [2]: ... the bond lengths D(N_1) and D(N_2) for cation coordination numbers N_1 and N_2 were related as follows ... 4 First and hopefully obvious things first: sulphate ions do form (weak) coordinate bonds to metal centres and thus by definition cause some d orbital splitting. In a simplified crystal field model, you might imagine them as weaker negative charges but nonetheless the do interact or we would not see the structures we see. Comparing the structures of anhydrous ... 3 If you truly believe that the ionic model is a good representation of the species of interest, given a structure you can calculate it analytically, or on a computer with an appropriate piece of software. Madelung energies and Ewald sums are the magic phrases here, you might like to look at https://en.wikipedia.org/wiki/Madelung_constant https://en.... 3 There are a number of ways to explain this. In words, the entropic gain with dissociation overwhelms any attraction between counterions as the electrolyte is diluted. Mathematically: for a m:n electrolyte undergoing dissociation according to the reaction$$\ce{A_mB_n -> mA^{\nu_A -} + nB^{\nu_B +}} if the initial concentration of electrolyte $\ce{... 3 For infinite dilution, water acts as a buffer with$\mathrm{pH}=7$. Therefore,$\alpha=f(\mathrm{p}K_\mathrm{a})$. For$\mathrm{p}K_\mathrm{a}=9$is$\alpha=\frac{1}{101}\overset{cca}=0.01$. For$\mathrm{p}K_\mathrm{a}=8$is$\alpha=\frac1{11}$. For$\mathrm{p}K_\mathrm{a}=7$is$\alpha=0.5$, as only a half of the acid is dissociated. For$\mathrm{p}K_\...

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If the weak electrolyte is a weak acid, we have to reckon with the autoionization of water. With the autoionization, the solvated ions formed from the acid do not reach infinite dilution and therefore the dissociation is limited to below 100%. Let's look at acetic acid, for which (if I remember and can type correctly) $K_a=1.8×10^{-5}$. Thereby* $\dfrac{[... 2 Please be aware that adding salt changes the ionic strength of your solution. Therefore, the activity$a_\mathrm H$of$\ce{H+}$is changed. This changes$\mathrm{pH}=-\log_{10}(a_\mathrm H)$. In a 1 molar$\ce{NaCl}$solution ($I\approx 1\ \mathrm{mol/l}$) the Debye-Hueckel (DH) approximation gives a shift of the activity by around 0.5 pH units (compared to ... 2 Adding acid to the water will not stop the auto-ionization of the water, so: the auto-ionization will continue to give equal concentrations of hydroxide and hydronium. But the addition of acid to the water increases the concentration of the hydronium ions in the solution, pushing water balance with its ions to the left to resist the increase in the ... 2 Interesting question with no generic answer. Boiling the salts, or ionic compounds, makes them a plasma or freely moving ions is nothing but a misconception. The temperature is too low to convert a substance into a plasma. Let us start with ionic liquids which are room temperature cations and anions but made up of organic molecules. You heat them, to the ... 2 They certainly can. Put some water in the bottom of a beaker, fill the beaker with chlorine gas, and drop in bits of calcium carbide. The carbide will produce bubbles of acetylene, which will spontaneously burst into flame when they hit the chlorine. Flame colors depend on many things, even in oxygen. In general, flames in a chlorine atmosphere will ... 2 First, a little bit of background. I assume you're already familiar with atomic orbitals — functions that describe the wave-like behavior of an electron in an atom. There's also molecular orbitals (MO), that describe its behavior in a molecule. You can ignore the odd-looking labels in images and such, since they're not entirely relevant to this question. ... 2 Many elements have multiple valencies. The most common oxidation state of Bi is +3. Bismuth (V) is also possible. Read more here on Bismuth oxides. 2 The dielectric constant$\epsilon$(known also as permittivity) is a measure of the extent to which a substance is polarized under an applied (external) electric field. Polarization amounts to net separation of charge across the substance. A "dielectric" is the common name used to refer to a substance placed between the plates of a capacitor and used to ... 2 Tl, dr -- upon further review we can't really tell whether potassium sulfide or beryllium fluoride is more ionic without some quantitative details. There is more involved in the balance between ionic and covalent bonding than electronegativity differential. You have to consider molecular orbital structure, too. One of the answers to this question ... 2 All of these elements can form compounds in other oxidation states. Aluminium forms some compounds in the +1 state (e.g. see the section in https://en.wikipedia.org/wiki/Aluminium_iodide), as does Zinc (see the section in https://en.wikipedia.org/wiki/Compounds_of_zinc) and Cadmium (e.g. https://en.wikipedia.org/wiki/Cadmium(I)_tetrachloroaluminate). But ... 2 Aspirin has a solubility of ~3 g/L. In addition from a patent by Galat [1]: Sodium aspirin is almost 1000 times as soluble in water as aspirin itself. All of the$\ce{NaOH}$can be assumed to react with acetylsalicylic acid (aspirin) to form the sodium salt. This means the amount of solubilized sodium aspirin is roughly equal to the amount of added$\...

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A traditional softener has a zeolite resin that "holds" positive ions. When it is charged the Na ions fill the resin. Then in service , the positive ions - Ca , Mg , etc , exchange places with the Na ions, putting Na in the water . In theory the back flush of the softener after charging, removes the Cl ions. So that in service the only Cl present would be ...

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If fuel is oxidized at high temperature, the conductivity of the fuel itself is irrelevant -- only the conductivity of the combustion products, which are ionized because of the high temperature and are therefor conductive. Brittanica states that temperatures greater than ~2,500 K are needed for sufficient ionization. Alkali metals can be used to increase ...

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It seems the examples were thrown together without considering that someone would try to understand them. The rule does make sense, but there are several factors involved here, and they are not mentioned. The problem is not even that you are comparing apples and oranges; it's more like fruit salad vs vegetable stew! Let's simplify: Keep one of the ...

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It is true that bismuth burns with a bluish flame in the presence of air to form bismuth trioxide($\ce{Bi2O3}$) as @M.Farooq mentioned. Bismuth pentoxide is not a pure compound but a mixture of various compounds like water, bismuth trioxide and bismuth tetroxide. It is also not stable at strong heat, decomposing at 393 K. Also, there is no direct preparation ...

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Fluorine gas definitely supports flaming combustion. And does so more than oxygen. Very dramatic, actually. Chlorine, less so than oxygen or air, but still possible in right setting. See below for examples of fluorine flames. Examples are not in $100~\%\ \ce{F2}$ environment, but should be obvious that $\ce{F2}$ makes more flames than air. In many ...

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Actually, this only applies for metals. This trend does not apply for non-metals: non-metals form covalent bonds, which are due to the sharing of electrons - thus, there are no free electrons that can carry charge. Additionally, atoms are held tightly by covalent bonds and thus cannot dissociate into ions that carry electric charge. With regards to metals, ...

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You are going wrong in your choice of $x$. You should have recognized in advance that $\ce{CN-}$ is in excess, and hence $\ce{Ag+}$ would go to the complex almost completely, so the equilibrium $\ce{[Ag+]}$ is going to be awfully low, while the concentration of $\ce{[Ag(CN)2]-}$ (which you have unfortunately defined as $x$) is going to be really very close ...

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