The solubility of iodide ions (I-) in water at 30 degrees Celsius is generally around 140 grams per 100 grams of water, depending on the specific iodide salt being considered (e.g., potassium iodide). This indicates that a significant amount of iodide can be dissolved in water at this temperature. However, for precise values, refer to solubility tables specific to the iodide compound of interest.
Solubility, 11.1 g per 100 g H2O at 20°C
AgI (silver iodide) has very low solubility in water, with only about 0.0013 g/100 mL at 20°C. This makes it practically insoluble in water.
This solubility is 103,4 g KNO3/100 g H2O.
Stir the mixture.
The solubility of salt is lower in heavy water (D2O) because deuterium atoms in heavy water are heavier than regular hydrogen atoms in H2O, leading to weaker hydrogen bonding forces between the water molecules and salt ions. This weaker interaction affects the ability of heavy water to dissolve and separate the salt ions.
Solubility, 11.1 g per 100 g H2O at 20°C
AgI (silver iodide) has very low solubility in water, with only about 0.0013 g/100 mL at 20°C. This makes it practically insoluble in water.
When H2O (water) reacts with AgI (silver iodide), it forms a mixture known as a silver iodide suspension. Silver iodide is poorly soluble in water, so it will not dissolve completely but rather will form a suspension with water molecules surrounding the silver iodide particles.
This solubility is 103,4 g KNO3/100 g H2O.
The solubility of sodium nitrate in benzene is likely extremely low. I know that its solubility in dry acetonitrile (<40 ug/mL H2O) is less than 1 mg/mL. So I would guess it would be even worse in a non-polar solvent like benzene.
To find the solubility of a solute, you would typically dissolve a known amount of the solute in a solvent at a specific temperature. Then you would measure the concentration of the solute in the resulting solution. The solubility of the solute at that temperature is the maximum amount that can dissolve in the solvent under those conditions.
Stannous iodide can be converted to stannic iodide by reacting it with an oxidizing agent, such as hydrogen peroxide. The stannous iodide is oxidized to form stannic iodide in the reaction. The chemical equation for this reaction is: 2 SnI2 + H2O2 → 2 SnI4 + 2 H2O.
The solubility of copper sulfate pentahydrate at 100 0C is 114 g/100 g water.
The line on the graph on The Chemistry Reference Table G indicates a saturated solution of NH4Cl has a concentration of 60. g NH4Cl/100. g H2O at 66 degrees celcius.
Solid sodium iodide reacts with water to form sodium ions and iodide ions. No further reactions occur because this compound is a salt of a strong acid and a strong base. Thus, the overall reaction is NaI(s)=> Na+ + I-
Stir the mixture.
The reaction between ammonium iodide (NH4I) and sodium hydroxide (NaOH) will yield ammonia (NH3), sodium iodide (NaI), and water (H2O). The balanced chemical equation for this reaction is: 2NH4I + 2NaOH → 2NH3 + 2NaI + 2H2O