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You can make the silicates by fusing the hydroxides with silica. Potassium silicate fertilizer grade are successfully produced by direct fusion of silica ($\ce{SiO2}$) and potasium compounds($\ce{KOH}$ and $\ce{K2CO3}$) in furnaces at temperatures up to melting point of mixture. The fusion temperature reaches around 1350 °C. $$\ce{SiO2 + 2 KOH ->[\Delta] ... 1 An analogy might help. Solubilization of a salt in a solvent (say NaCl in water) is similar to condensation of a gas. Imagine you have an amount of gas in a cylinder with a piston to regulate the volume and pressure. Beginning at a pressure below the vapor pressure of the substance (which means that all is gas, none is liquid), the pressure is increased (... 0 For most salts the equilibrium constant for ionization is very large and thus the reaction practically proceeds only in the forward direction and so we generally say it is irreversible. But in reality no chemical reaction is irreversible and some backward reaction always takes place. So, in a strict sense ionization is a reversible equilibrium process but ... 2 Sander (Ref. 1) has compiled a useful review of Henry's law constants in water that includes an introduction showing notation and conversions. There are two types of Henry's law constants: Solubility constants convert from pressure to concentration in solution (solubility):$$c=Hp$$Volatility constants convert from concentration in solution (solubility) ... 3 There are three kind of oxides/hydroxides: basic (e.g., \ce{NaOH}, metal hydroxides), acidic (e.g., \ce{CO2}, non-metal oxides), and amphoteric (e.g., \ce{Al(OH)3}, Group 13 and 14 hydroxides). Amphoteric Hydroxides act either as Bronsted-Lowry bases (accepting protons) or as Lewis acids (accepting an electron pair), depending on reaction conditions. ... 1 Glycerol (glycerin) is polar molecule with three hydroxy groups. Fats are tri-esters of glycerol and very long-chained fatty acids, and hence relatively non polar molecules. Thus it is safe to assume that fats might not be dissolved in glycerol and vise versa. This assumption is based on the fact that glycerol is insoluble in biodiesel (Ref.1). An excerpt ... 0 The OP asked the question again, and proposed the following solution from his book. My textbook says to solve it in the following way: 1) Ignore the 0.10 M KI since it doesn't matter. 2) Since you have excess AgCl(s), you can calculate the Ag+ concentration using its 𝐾sp. 𝐾sp=1.77×10−10=𝑥^2 𝑥=1.3×10^−5 This is the Ag+ concentration 3) Substitute ... 2 I decided to give a help here since OP has given enough effort to solve this problem. First, OP needs to know the molarity of a solution is defined as the amount of solute (in this case \ce{NaBr}) in \pu{mol} per \pu{1.0 L} of solution. The first part of problem states: If \pu{2.60 g} of \ce{NaBr} is dissolved in enough water to make \pu{160.0 mL}... 2 It comes to be \pu{252 mL} precisely:$$V = \frac{\pu{2.6 g}}{\pu{103 g}} \times \frac{1}{0.1} \times \pu{1000 mL} = \pu{252.4 mL}$$\text{Molar Mass of NaBr} = \pu{103 g mol-1}. 2 I think the experts here have muddled the problem by stating "no equilibriums," so I'll solve it. Given the reaction:$$\ce{KI(aq) + AgCl(s) <=>[excess AgCl(s)] KCl(aq) + AgI(s)} there are equilibriums. There are always equilibriums in chemistry. In this case it is much better to think of the reaction as proceeding quantitatively. That is ...