Salt water seed selection needs 1.1 × 103kg / m3 of salt water. Now 0.05m3 of salt water is prepared, and its mass is 60kg. Does this salt water meet the requirements? If not, should water or salt be added? How much more?
(1) Let the density of brine be ρ, the density of brine be: ρ = MV = 60kg0.05m3 = 1.2 × 103kg / m3, ∵ ρ > ρ 0 = 1.1 × 103kg / m3, ∵ the prepared brine does not meet the requirements; (2) the density of brine is too high, so water needs to be added to reduce the density; (3) if the mass of water to be added is △ m, then m after adding water is always equal to M
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- 1. A piece of paraffin with a density of 0.9 g / cm3 is placed at the bottom of the beaker, and then salt water with a density of 1.1 g / cm3 is slowly injected into the beaker. Why?
- 2. Place a piece of paraffin with a density of 0.9 g / cm3 at the bottom of the beaker, and then add salt with a density of 1.1 g / cm3 A piece of paraffin with a density of 0.9 g / cm3 is placed at the bottom of the beaker, and then salt water with a density of 1.1 g / cm3 is slowly injected into the beaker. The experimental results show that the paraffin does not float up. Why? If you change paraffin into wood? What's the result? Why?
- 3. The density of aluminum is 2.7 * 10 cubic kg / cm3, the mass is 270g, and the volume is 270cm3_____________ Floating and sinking?
- 4. A metal ball made of pure lead is known to have a volume of 0.5dm3, a mass of 0.81kg and a density of 2.7x10 cubic kg / m3 Solid? In two ways
- 5. The volume of hollow aluminum ball with density of 2.7kg/m3 and mass of 270g is 270cm3. What is the hollow volume in the middle of the aluminum ball
- 6. The mass of a hollow aluminum ball is 27kg, and the total mass is 48g after the hollow part is filled with water P aluminum = 3 times of 2.7 * 10 kg / cm3
- 7. The volume of aluminum ball is 400cm cubic and the mass is 810g. Q: is the ball hollow or solid? If it is hollow, what is the volume of the hollow part? (P = 2.7 * 10 ^ 3kg / m ^ 3) (p aluminum = 2.7 * 10 ^ 3kg / m ^ 3)
- 8. The total mass of the hollow part is 59G after alcohol injection. The volume of the aluminum ball is calculated. (the density of aluminum is 2.7 × 103kg / m3)
- 9. The mass of a hollow aluminum sphere is 27g. After the hollow part is filled with alcohol, the total mass of the hollow part is 48g (ρ alcohol = 0.8 × 103kg / m3, ρ aluminum = 2.7 × 103kg / m3)
- 10. The mass of a hollow aluminum sphere is 27g. After the hollow part is filled with alcohol, the total mass of the hollow part is 48g (ρ alcohol = 0.8 × 103kg / m3, ρ aluminum = 2.7 × 103kg / m3)
- 11. When selecting seeds with salt water, the density of salt water is 1.1 * 10 ^ 3kg / m ^ 3. Now 400ml salt water is prepared, and the mass is 0.6kg Does this salt water meet the requirements? If not, how much salt or water should be added?
- 12. Salt water seed selection needs 1.1 × 103kg / m3 of salt water. Now 0.05m3 of salt water is prepared, and its mass is 60kg. Does this salt water meet the requirements? If not, should water or salt be added? How much more?
- 13. Dilute 250 ml of 98% concentrated sulfuric acid with a density of 1.84 g / cm3 to 600 ml. what is the concentration of concentrated sulfuric acid after dilution
- 14. Use 98% concentrated sulfuric acid with a density of 1.84g/cm3 to prepare 1000ml of 2mol/l sulfuric acid solution. How to prepare it? Urgent, thank you
- 15. Dilute concentrated sulfuric acid with a density of 1.84g/cm and a mass fraction of 98% to 1000ml. The mass concentration of the substance is 2mol / L. the density is 1.20g/cm and a mass fraction of 179; Find (1) the volume of concentrated sulfuric acid needed (2) Volume of water required I hope I can answer in detail, because my knowledge of this part is not solid at all
- 16. 250 g 4.9% sulfuric acid was prepared with 98% concentrated sulfuric acid and 1.84 g / cm3 density How many cubic centimetres of water are needed (accurate to 1 cubic centimeter, water density 1 g / cubic centimeter) This concentration is required for a few cubic centimeters (accurate to 0.1 cubic centimeters)
- 17. 50 ml & nbsp; 20% dilute sulfuric acid (density 1.14 g / cm3) was prepared with 98% concentrated sulfuric acid (density 1.84 g / cm3), and water should be poured into the beaker first___ Ml, and then___ Ml & nbsp; 98% concentrated H 2SO 4 was slowly injected into water and continuously stirred
- 18. Dilute the concentrated sulfuric acid with the mass fraction of 250mlh2so4 of 98% and the density of 1.84g/cm3 to 1000ml (1) What is the amount of H2SO4 in the solution? (2) What is the concentration of H + in the solution? The answer is (1) 4.6mol (2)9.2mol/L So you need to give a good answer in detail
- 19. The concentrated sulfuric acid with density of 1.84 g / cm3 and 98% mass fraction was diluted into 100 ml. the mass concentration of the substance was 2 mol / L and the density was 1.20 g / cm3 Find the volume of sulfuric acid needed, the volume of water needed?
- 20. Dilute the concentrated sulfuric acid with density of 1.84 g / cm3 and mass fraction of 98% to dilute sulfuric acid with density of 1.20 g / cm3 and mass concentration of 2 mol / L. calculate: (1) the volume of concentrated sulfuric acid required; (2) the volume of water required