The mass fraction of potassium nitrate saturated solution is changed by a, increasing temperature b, decreasing temperature c, adding solute d at constant temperature and evaporating solvent at constant temperature Why not d?

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• 1. Saturated solution at a certain temperature. The remaining solution is precipitated by solute crystal after evaporation of part of water at a constant temperature Saturated solution at a certain temperature. The remaining solution is precipitated by solute the water at a constant temperature
• 2. The mass fraction of solute in the saturated solution will decrease after the crystal is precipitated without evaporation of solvent. The mass fraction of solute in the saturated solution decreases with the precipitation of crystals without evaporation of solvent. Is that right? Why? But solute mass fraction = m solute/m solution Since the solute mass is reduced and the solution mass is reduced, how do we compare? The mass fraction of solute in saturated solution must be reduced after the crystal is precipitated without evaporation of solvent. The mass fraction of solute in the saturated solution decreases with the precipitation of crystals without evaporation of solvent. Is that right? Why? But solute mass fraction = m solute/m solution Since the solute mass is reduced and the solution mass is reduced, how do we compare?
• 3. 100G of salt saturated solution at 20°C, the mass fraction of solute is 26.5% the solute mass fraction is () A.26.5% B.>26.5% C.
• 4. At 20.degree. C., the unsaturated solution is divided into two equal parts by mass, one part evaporates 10 g of water and precipitates 4 g of, and the other part evaporates 20 g of water to precipitate 10 g of crystal The solubility of the substance at 20°C is () The answer is 60 g per 100 g of water, Didn't the title say two different copies? Why is it that 10 g evaporates first and 10 g later? And what is 10-4=6g? How did you get here?
• 5. Judgement: If the laboratory evaporates a sodium chloride solution to precipitate crystals, the mass fraction of the solute must be reduced. Judgement: The laboratory evaporates a sodium chloride solution to precipitate crystals, and the mass fraction of the solute must be reduced.
• 6. When the solvent degree of sodium chloride is 36g at 20°C, the mass fraction of solute of saturated sodium chloride solution at this temperature is _______; when 10g water is evaporated from the modified solution and then cooled to 20°C, the mass fraction of solute of the remaining solution _________
• 7. At 10 degrees,131 grams of saturated KCl solution, after evaporation of 10 grams of water, at 10 degrees, the mass of KCl crystals can be precipitated At 10°C,131 g of saturated KCl solution, after evaporation of 10 g of water, the mass of KCl crystals can be precipitated at 10°C
• 8. 200 G of silver nitrate solution at T degree,10 g of precipitated crystal 4 g evaporated,10 g of water evaporated and 6 g washed out,20 g of water evaporated to precipitate crystal mass 200 G of silver nitrate solution at T degree,4 g of precipitated crystals evaporated off 10 g,6 g of precipitated crystals after 10 g of water evaporated, and what is the mass of precipitated crystals after 20 g of water evaporated? 200 G of silver nitrate solution at T degree,10 g of precipitated crystals 4 g evaporated,10 g of water evaporated and 6 g washed out,20 g of water evaporated to precipitate crystal mass 200 G of silver nitrate solution at T degree,4 g of precipitated crystals evaporated off 10 g,6 g of precipitated crystals after 10 g of water evaporated, and what is the mass of precipitated crystals after 20 g of water evaporated? 200 G of silver nitrate solution at T degree,10 g of precipitated crystals 4 g evaporated,10 g of water evaporated and 6 g washed out,20 g of water evaporated to precipitate crystal mass 200 G of silver nitrate solution at T-degree,4 g of precipitated crystals evaporated out of 10 g,6 g of precipitated crystals after 10 g of water evaporated, and what is the mass of precipitated crystals after 20 g of water evaporated?
• 9. At 50.degree. C., the solution of substance A was divided into two equal parts by mass, one part evaporated off 10 g of water to precipitate 4 g of crystal, and the other part evaporated off 20 g of water to precipitate 10 g of crystal Solubility of substance A at 50°C
• 10. 100 G of the solution of the substance A at 50 degrees centigrade was divided into two parts, one part evaporated at constant temperature 10 g of water with 4 g of crystals precipitated, the other part evaporated 20 g of crystals precipitated 20 g, Solubility in 50 degrees Celsius. Must have a process, thank you very much!
• 11. T°C is the solution of KNO3(potassium nitrate). When A g of water is evaporated at constant temperature, a g of crystal is precipitated; To re-evaporate A g of water and precipitate b g of crystal (a not equal to b), answer the following questions: (1) The size of a and b. (2) What is the solubility of potassium nitrate at T°C? T°C is the solution of KNO3(potassium nitrate). When A g of water is evaporated at constant temperature, a g of crystal is precipitated; To re-evaporate A g of water and precipitate b g of crystal (a is not equal to b), answer the following questions: (1) Size of a and b. (2) What is the solubility of potassium nitrate at T°C?
• 12. A solution of substance A was evaporated to 20 g of water at a constant temperature to precipitate 2 g of crystal A. (1) At t°C, the solution of substance A evaporates 20 g of water at a constant temperature, precipitates 2 g of crystal A, and evaporates 20 g of water again at a second time at a constant temperature resulting in the precipitation of 8 g of crystal A, and the solubility of substance A at t°C. (2) To obtain crystals from lime water, the () method is usually used because () A solution of substance A was evaporated to 20 g of water at a constant temperature to precipitate 2 g of crystal A. (1) At t°C, the solution of substance A evaporates 20 g of water at a constant temperature, precipitates 2 g of crystal A, and evaporates 20 g of water at a second time at a constant temperature, resulting in the precipitation of 8 g of crystal A, and the solubility of substance A at t°C. (2) To obtain crystals from lime water, the () method is usually used because ()
• 13. At t.degree. C.,500 g of potassium nitrate solution,40 g of water evaporated,20 g of crystals precipitated, and 40 g of crystals precipitated after further evaporation of 40 g of water How much is the solubility at this temperature?
• 14. Why does the solubility of a gas solid vary with temperature, since it diffuses uniformly in molecular or ionic form after dissolution? Should we consider micro rather than macro at this point? Why does the solubility of a gas solid vary with temperature, since it diffuses uniformly in molecular or ionic form after dissolution? Should microcosmic, not macrocosmic, be considered at this point?
• 15. What is the characteristic of the solubility of various solid substances affected by temperature?
• 16. What is the relationship between the solubility curve and evaporation crystallization or cooling crystallization? Is, for example, a solubility that increases with temperature Many, the other little change, this and other types of what crystal removal methods are needed
• 17. Will the solubility of the substance change while the solvent is being evaporated?
• 18. How to purify potassium nitrate if potassium nitrate contains a small amount of sodium chloride (1 dissolution 2 filtration 3 cooling crystallization 4 heating and concentration) and explain the principle
• 19. Why the potassium nitrate solution contains a small amount of sodium chloride should be purified by cooling crystallization method Why not use evaporative solvents? Is sodium chloride or potassium nitrate separated out? After meeting the question about purification, what kind of separation is it?
• 20. What does cooling crystallization purify?