first you need to know the number of liters and moles and the equation. you do someting then multiply the liters times the moles. first you need to know the number of liters and moles and the equation. you do someting then multiply the liters times the moles.
If it is saturated with a solid solute, you would expect some of the solid to precipitate out - as long as the solid could find a surface to nucleate on. If it is saturated with a gas, you would expect more gas to dissolve into it as long as it was still in contact with the saturating gas in the gas phase.
It depends on the amount of salt and the amount of water. If there is only a little salt, it is probably unsaturated. That means more salt could be dissolved into the solution. If there is quite a bit of salt, it is more than likely saturated. If you add more salt and it just floats to the bottom, it is saturated. Unless it is supersaturated of course. For the solution to be supersaturated, you would have had to boil the water, add salt to the point where it stops dissolving into the boiling hot solution, then let the solution cool down. So, it can be any of the three.
Saturated is 36 - 40 g / 100 mL : Divide grams by the molar mass 58.5 g/mol NaCl and you'll find mol/100 mL Multiplying this value by 10 (= dL/L) and you find mol/L. Doing an estimation gives me approx. 7 mol/L, my calculator is tilted, so I hope your's doing better.
That is called a saturated solution. It has reached a point where it can no longer dissolve any more solute (thing to be dissolved) in the solvent.This is more info you might find interesting:Keep in mind compounds dissolve at different rates in solvent. This rate is dependent on heat; goes faster with more. To be sure its not the dissolve rate but saturation you'll need to give it time. Of course a quick fix for a saturated solution is to add more solvent.Unless the goal is to be saturated, like in hand warmers. An example of a super saturated solution. Crystal sodium acetate trihydrate is dissolved in water until the saturation point. It is heated to force more to dissolve and then is allowed to cool down. Once it has cooled the sodium acetate (NaOAc) is still dissolved in solution but is now unstable and easily recrystallized. If even a single crystal more NaOAc is added the entire solution will recrystallize. This recrystallization actually has a surprising effect. It produces heat as an exothermic reaction. That is why this property is used in hand warmers.
To find the amount of pure acid to add, set up an equation based on the amount of acid in the original solution and the final solution. Let x be the amount of pure acid to add. The amount of acid in the original solution is 0.4 × 12 = 4.8 ounces. The amount of acid in the final solution is 0.6 × (12 + x) = 4.8 + 0.6x ounces. Therefore, to get a 60% acid solution, you would need to add 10.67 ounces of pure acid.
The fastest way is to add more salt - if the additional salt falls out of solution and forms a precipitate on the bottom of the container, the solution is saturated.
If it is saturated with a solid solute, you would expect some of the solid to precipitate out - as long as the solid could find a surface to nucleate on. If it is saturated with a gas, you would expect more gas to dissolve into it as long as it was still in contact with the saturating gas in the gas phase.
It depends on the amount of salt and the amount of water. If there is only a little salt, it is probably unsaturated. That means more salt could be dissolved into the solution. If there is quite a bit of salt, it is more than likely saturated. If you add more salt and it just floats to the bottom, it is saturated. Unless it is supersaturated of course. For the solution to be supersaturated, you would have had to boil the water, add salt to the point where it stops dissolving into the boiling hot solution, then let the solution cool down. So, it can be any of the three.
To find the mass of the precipitate that forms when 100.0mL of 0.887M AgNO3 is added to a Na3PO4 solution, you need to determine the limiting reactant. Since Na3PO4 is in excess, AgNO3 is the limiting reactant. Calculate the moles of AgNO3 using its molarity and volume, then use the mole ratio between AgNO3 and the precipitate to find the moles of the precipitate. Finally, convert the moles of the precipitate to mass using its molar mass.
To make a saturated solution with any certain concentration, you simply need to find out the temperature at which saturation occurs for that concentration.Saturation for sugar, assuming you mean sucrose, at 0.6 molL-1 is below the freezing point of water (even at 0oC, it has a solubility of around 5.25molL-1); therefore, creating a saturated solution at this point is impossible.>.actually. 2.4 moles
Saturated is 36 - 40 g / 100 mL : Divide grams by the molar mass 58.5 g/mol NaCl and you'll find mol/100 mL Multiplying this value by 10 (= dL/L) and you find mol/L. Doing an estimation gives me approx. 7 mol/L, my calculator is tilted, so I hope your's doing better.
No more solute can be dissolved in the solvent. If you have ever added a lot of sugar to a drink, you would find that no matter how much you stirred, there would still be undissolved sugar at the bottom of the glass. The drink is a saturated solution- no more sugar will dissolve in it.
Add potassium nitrate to distilled water while stirring continuously; stop adding potassium nitrate once the chemical becomes saturated and stops dissolving in the water. Filter the solution into an evaporating dish and measure the temperature of the solution. Measure the temperature and carefully heat the solution under low heat. Weigh the potassium nitrate to determine its solubility. You should find that approximately 36 g of potassium nitrate can be dissolved in 100 ml of water at room temperature (25 degrees Celsius). santosh kumar.batchu santoshkumar@gmail.com
You need to find a graph of solubility in your textbook. It should have temperature on the x-axis and "grams dissolved per 100 grams H2O" on the y-axis (or some such.) Sodium nitrate is less soluble in cool water than warm. Use the graph to determine how many grams are dissolved at 50°C, multiplying by 2 since there are 200 grams H2O. Then determine how many grams are dissolved at 20°C, again multiplying by 2. Subtract the two numbers to find the grams that will precipitate
Well, isn't that a lovely question! To find the concentration of baking soda in a saturated solution at room temperature, you simply need to divide the solubility (9.30 g) by the volume of the solution (100 ml). This will give you the concentration of baking soda in the saturated solution. Just remember to keep calm and enjoy the process of solving this little mystery!
To find the volume from molarity in a solution, use the formula: volume (in liters) amount of solute (in moles) / molarity (in mol/L). This equation helps determine the volume of a solution based on its molarity and the amount of solute present.
silver chloride (AgCl) AgNO3(aq) + KCl(aq) = AgCl (s) + KNO3 (aq) The only way to find out for your own if silver chloride or potassium nitrate is precipitated out is to look at a solubility chart.