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The water molecules undergo a sort of attraction between them. This attraction is due to $\ce{H}$ atoms being attracted by the oxygen atoms of neighboring water molecules. At the surface of the water this attraction produces a sort of "skin" due to the attraction of the molecules between them. If a solid is touching this surface, it may produce a similar ...


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In the case of o-nitrophenol, the acidic hydrogen is hydrogen-bonded to the nitro group's oxygen atom, making it less acidic. In the case of catechol, one acidic hydrogen is hydrogen-bonded to the adjacent OH group's oxygen atom, but the other, more acidic hydrogen is not hydrogen bonded. This hydrogen is more acidic than that of hydroquinone. The resulting ...


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Adding heat is risky and could lead to overpressure or damage. I'd also expect it to boil more of the water off with the ammonia, which is probably not ideal. Not sure about efficiency per se, but you can definitely improve the effectiveness with fans. It's a heat pump that maintains a given temperature delta not between outside and inside exactly but ...


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Sure they can. Even if most of the ions are in ion pairs or other complexes, they can still move by being transferred from one complex to another. All you need is to get the ionic compound to dissolve, which in most cases requires a polar solvent with which the ions will not react. A rather unexpected case involves magnesocene, whose bonding is a mixture ...


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Consider three compositions: A. 2NaF + CaCO3 B. CaF2 + Na2CO3, and C. NaF + 0.5 CaCO3 + 0.5 CaF2 + 0.5 Na2CO3. Using data from the CRC Handbook (62nd ed), the heats of formation of A and B are respectively 560.47 and 560.6 kcal, so there is little driving force to make a reaction go to completion. Note that A should be near neutral pH, but B ...


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AFAIK, such a reaction has no name. However, if one were so inclined, one could call it a replacement reaction, because the $\ce{MgCO3}$ is being converted into $\ce{Mg(OH)2}$. To calculate the "equilibrium point," which I take to mean the concentration of all the ions at equilibrium, we would need to know the $k_\text{sp}$ of both $\ce{MgCO3}$ and $\ce{Mg(...


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Yes, the calcium ion could lead to precipitation. The solubility of $\ce{CaCO3}$ in distilled water is about 15 mg/L, which is about 0.15 mM calcium ion if there is no other source of carbonate. The solubility constant for $\ce{CaF2}$ is about $4\times 10^{-11}$, which means that we can only have 0.5 mM fluoride ions before precipitation will start. That's ...


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$\ce{CaF2}$ has a relatively low solubility in water, about 15 mg/L, with a $K_{sp}$ of about $4\times 10^{-11}$. According to the paper you linked, the concentrations of fluoride in tea are around 5 mg/L, which corresponds to about $1.3\times 10^{-4}$ M. Thus, the concentration of $\ce{Ca^2+}$ required to precipitate the fluoride out is about 2.5 mM, which ...


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What Atharva and Ericmuch say is sometimes contrary to the truth. Calcium fluoride cannot be dangerous, because it is insoluble in water, and in diluted acidic solutions. It can produce negative effects such as skeletal fluorosis, but this happens only in cases where it is taken in huge amounts. Fluorides have no negative effects, if taken in small amounts. ...


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Plastics Design Library series include tabulated parameters for the variety of polymers. One of such parameters is a PDL number rating from 0 to 9: 0: solvent dissolved disintegrated 1: decomposition 2: severe distortion; oxidizer and plasticizer deteriorated … 9: highest resistance, no change In the table below I assembled data on thermoplastics ...


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Assume: Water has a constant density of 1.000 g/ml from $20 ^\circ\pu{C}$ to $100 ^\circ\pu{C}$. Given: $400\pu{g}\ \ce{Na2SO4}$ contains $400/142.04 = 2.8161$ mole of $\ce{Na2SO4}$. $1000\ \pu{ml}\ \ce{H2O} = 1000\ \pu{g}\ \ce{H2O} = 1000/18.015 = 55.509\ \pu{moles}\ \ce{H2O}$ $19.4\ \pu{grams}$ of $\ce{Na2SO4}$ per $100\pu{mL}$ of water $= 19.4/142.04$ ...


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Calcium bicarbonate $\ce{Ca(HCO3)2}$simply does not exist. It is impossible to fill up any container with pure $\ce{Ca(HCO3)2}$. Nobody has ever been able to produce a powder containing $\ce{Ca(HCO3)2}$. It is possible and easy to produce a solution containing the ions $\ce{Ca^{2+}}$ and the ions $\ce{HCO3^-}$, by bubbling $\ce{CO2}$ into a solution of ...


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Even I am not a chemist, but according to my textbook there is already a counter ion present in hard water to the $\ce{Ca^2+}$ and $\ce{Mg^2+},$ which are mostly bicarbonates (temporary hardness) or fluorides, chlorides and sulfates (permanent hardness). These counter-ions would by themselves prevent any appreciable amount of fluoride forming a bond with the ...


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While solubility, thermal stability and density are pretty hard to explain as @maurice said. But, there is good logic for the solubility of group 2 hydroxides. The lattice energy of a solid is inversely proportional to the radius ratio (as lower the radius ratio greater the packing efficiency). So, hydroxide being a small ion, makes a more stable aqueous ...


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Nobody is able to foresee the solubility of a product. There are some experimental rules, but they all have exceptions, that nobody is able to explain. Just have a look on the Calcium salts made with the halogens (F, Cl, Br, I). There is a nice analogy among Cl, Br and I, but not F. Look ! The Calcium chloride $\ce{CaCl2}$, bromide $\ce{CaBr2}$ and iodide $\...


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