# Tag Info

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To acknowledge OP's request, I am putting this as my answer: What OP's initially referring to (US20110008445A1) was a patent application. Since it was a 2011 application, I searched for the patent and found it (Ref.1). It's abstract states that: A Suspension of ascorbic acid in glycerol or in glycerol comprising diglycerol, in which the content of ...

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Many phosphates are not soluble in water at standard temperature and pressure, except for the sodium, potassium, rubidium, cesium, and ammonium phosphates, which are all water-soluble. As a rule, the hydrogen and dihydrogen phosphates are slightly more soluble than the corresponding phosphates (Wikipedia). For example, based on Wikipedia Solubility Table, ...

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Goggle translate from Swedish yields a slightly different translation. You add $\pu{100 ml}$ of $\pu{0.01 M} \; \ce{Na2SO4}$ solution in $\pu{100 ml}$ of $\pu{0.02 M} \; \ce{CaCl2}$ solution. A $\ce{CaSO4}$ precipitate forms. How much of $\pu{0.01 M}\; \ce{Na2SO4}$ solution has yet to be added to completely dissolve the $\ce{CaSO4}$ precipitate formed? $... 3 When all of the$\ce{CaSO4}$is dissolved completely, all the calcium ions originate from the$\pu{100 mL} \; \ce{CaCl2}$solution, so it is easy to determine the amount: $$n_\ce{Ca^2+} = \pu{100 mL \times 0.02 mol/L = 2 mmol}$$ To get the total sulfate amount, we have to consider what is present initially and what we add when we add a volume of$V_\text{...

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As you've already noted, there are 3 substances of interest here: $\ce{Ca^{2+}}, \ce{SO4^{2-}}, \ce{H2O}$. Let $V$ be the additional volume of sodium sulfate solution. Then: The total volume is $V + 0.2\ \mathrm{L}$. The total amount of sulfate is $(V + 0.1\ \mathrm{L}) (0.01\ \mathrm{M})$. The total amount of calcium is $(0.1\ \mathrm{L}) (0.02\ \mathrm{... 1 To answer this kind of problem, it'd be better starting with elimination approach. It is given the solubility of sodium carbonate in$\pu{100 g}$of water as$\pu{25.0 g}$at$\pu{22 ^{\circ}C}$. By that, you know, solubility of its hydrate should be more than$\pu{25.0 g}$at$\pu{22 ^{\circ}C}$because the mass of hydrate is including additional mass of ... 0 Your calculation would be good, if you did not forget that hydrates contain water. The solubility of the anhydrous carbonate is 25 g in 100 g of water. Both carbonate forms, once diluted, are the same compound, forming the same ions plus eventually releasing water. So in case of the hydrate, the ratio carbonate:water must be 1:4 as well.$M_{\ce{Na2CO3}}=...

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This is a confusing question because, while solubilities can be reported in mL/L, there can be ambiguity when choosing a pressure during conversion to this unit, for instance using the following equation to convert from molarity $c$ to volume/volume units: $$\rho = \frac{cRT}{p}$$ In this online data page, for instance , in some columns the solubility is ...

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As noted by Wikipedia, "Henry's Law is a gas law that states that the amount of dissolved gas in a liquid is proportional to its partial pressure above the liquid." So you are absolutely right. 4 times the pressure of oxygen should result in 4 times as much oxygen being dissolved. The answer key for the problem must be wrong. User Delta_G made an ...

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There is no "or" in the first place. An emulsion is a two-phased system, it is just that one phase comes in a form of small droplets. Small as they are, however, they are pretty big from molecular point of view. Now to the point. If you stir the liquids vigorously, you'll have an emulsion. Short of that, they will attempt to minimize the surface, and the ...

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Well, if you are speaking about an equilibrium mixture of two pure substances, then you would not have an emulsion, since an emulsion of two such substances would not be thermodynamically stable. Formation of an emulsion requires an additional energy input to form the unstable dispersion. The stable structure (for instance a sphere or two side-by-side ...

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As the comment above suggests, if there was an obvious way to do this it would already be making someone wealthy. Self heating cans (and gloves) are seen where an exothermic chemical reaction or heat of solution is used to thermally warm food or gloves. The most common application of this method are 'flameless ration heaters' used to heat MRE (military ...

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Iodine forms charge transfer complexes, most notably with starch. In these complexes, a fraction of the electronic charge is transferred to the other molecule. The formation of these complexes increase the solubility of iodine. This has also been called solvation, see for instance this article. One effect of the formation of these complexes is the change ...

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I believe you are asking solubility of diethyl ether and acetanilide in water. If that the case, then diethyl ether is is more soluble in water than acetanilide at $\pu{25 ^{\circ}C}$: Solubility of diethyl ether in water at $\pu{25 ^{\circ}C}$: $6.05\% (w/v)$ (Wikipedia) Solubility of acetanilide in water at $\pu{25 ^{\circ}C}$: $\lt 0.56\% (w/v)$ (...

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Acetanilide is approximately 10x more soluble in diethyl ether (1g dissolves in18ml) than in water (1g dissolves in 185ml). Source here As @Mithoron said in the comments, that phenyl group has a big effect.

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This is a simplistic question which doesn't require a great deal of analysis. When ionic salts dissolve the typical assumption is that they completely ionize. So: $$\ce{Ca(OH)2 -> Ca^2+ + 2OH-}$$ $$\ce{Mg(OH)2 -> Mg^2+ + 2OH-}$$ The assumption is that the calcium hydroxide will dissolve completely. However the magnesium hydroxide is less soluble and ...

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Iodine as a halogen has the strong electron affinity, what is the best met by the free electron pair of diethylether, then by $\pi$ electron density of toluene. Note that iodine is also well soluble in ethanol, commercially used in medical treatment of skin injuries.

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