13

There is a strong reason. But it is the chemical reaction, not just a better dissolution. $\ce{Mg(OH)2}$ is a base with the limited solubility, defined by $K_\mathrm{sp}=[\ce{Mg^2+}][\ce{OH-}]^2$ $\ce{NH4+}$ ion, created by $\ce{NH4Cl}$ dissolution, acts as a weak acid: $$\ce{NH4+ + H2O <<=> NH3 + H3O+}$$ with $\mathrm{p}K_\mathrm{a}=9.25$ $\...


5

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 ...


3

This is a well known problem in qualitative analysis. When you add $\ce{NH4Cl}$ to a solution containing $\ce{OH-}$ ions, you produce the reaction: $$\ce{NH4+ + OH- -> NH3 + H2O}$$ The result is that the concentration of $\ce{OH-}$ decreases. If this operation was done in a saturated solution of $\ce{Mg(OH)2}$, the solubility product is no more obtained. ...


3

As a general rule, transition metal nitrides such as $\ce{TiN}$ do not react with water. As with hydrogen, nitrogen has only modest electronegativity and cannot capture all the valence electrons available from transition metals into ionic or polar covalent bonds. So the transition metal nitrides retain some metallic character and this inhibits reaction ...


3

$\ce{CaF2}$ 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 ...


2

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 $\...


2

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 ...


2

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 ...


2

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(...


1

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 ...


1

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 ...


1

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 ...


1

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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