They collected 1000ml salt water and weighed it to be 1.05kg. The density of salt water was 2700kg / m2

From ρ = m / V, M = ρ V is obtained to express the relationship between the mass and volume of salt and water: M1 = ρ 1v1, M2 = ρ 2v2; from M1 + M2 = m and V1 + V2 = V, the equations are: ρ 1v1 + ρ 2v2 = m; V1 + V2 = v. the solution is: V1 = 29.4cm & sup3;, V2 = 970.6cm & sup3;; then the salt content in brine is: M1 / M = ρ 1v1 / M = (2

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1. If you want to prepare 1000 ml, density of 1.1 * 10 ^ 3 kg / m ^ 3, mass fraction of 0.1 brine solution, how many kg of water and salt do you need
2. If you want to prepare 1000 ml brine solution with a density of 1100 kg / m3 and a mass fraction of 10%, how many kg of water and salt do you need?
3. If you want to prepare 1000ml salt water solution with density of 1.1 * 10 ^ 3kg / m ^ 3 and mass fraction of 10%, how many kg of water and salt are needed
4. 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
5. 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?
6. 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
7. 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
8. 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)
9. 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
10. 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
11. The density of salt water is 1.1 * 10 ^ 3kg / m3 when selecting seeds with salt water. Now, 0.5 cubic decimeter salt water is prepared, and its mass is 0.6kg. Does this salt water meet the requirements? (if it does not meet the requirements, should salt or water be added? How much to add?) 1.1×103kg/m³
12. A bottle contains 500 CM & sup3;, 1.2 × 10 & sup3;, kg / M & sup3;, how much more water should be added to dilute the brine to 1.1 × 10 & sup3;, kg / M & sup3?
13. A hollow copper ball with a volume of 30 cubic centimeters and a mass of 178 grams. If the hollow part of the ball is filled with aluminum, what is the mass of the ball? The density of aluminum is equal to 2.7 times 10 cubic kilogram per cubic meter, and the density of copper is equal to 8.9 times 10 cubic kilogram per cubic meter
14. A hollow copper ball with a volume of 30 cubic centimeters and a mass of 178 grams is used to calculate the volume of the hollow part of the copper ball
15. 1. A copper ball with a mass of 178 grams and a volume of 30 cubic centimeters is solid or hollow. What is the hollow volume? If the hollow part is filled with aluminum, what is the total mass? 2. In order to reduce the weight of the aircraft, an engineer changed a steel part into an aluminum part to reduce its mass by 1.56 kg. What is the mass of aluminum required? The density of steel is 7.9 × 10 cubic kg / cubic cm, and the density of aluminum is 2.7 × 10 cubic kg / cubic cm
16. The volume of the hollow copper ball is 30cm3, the mass m = 178g. After the hollow part is filled with some liquid, the total mass m = 314g? (ρ Cu = 8.9 × 103 kg / m3)
17. The volume of the hollow copper ball is 30cm3, the mass m = 178g. After the hollow part is filled with some liquid, the total mass m = 314g? (ρ Cu = 8.9 × 103 kg / m3)
18. The volume is 30 cm3 and the mass is 178 g. if the hollow part is filled with aluminum, what is the mass of aluminum? Density of copper P = 8900 kg / m3, density of aluminum P = 2700 kg / m3 Clear calculation process
19. If the hollow part of a hollow copper ball with a volume of 30 cm3 and a mass of 178 G is filled with aluminum, what is the total mass of the ball filled with aluminum? (known: the density of aluminum is 2.7 × 103kg / m3, the density of copper is 8.9 × 103kg / m3)
20. A copper ball contains 178g of copper. If the total volume of the ball is 40 cubic centimeters, what is the hollow volume of the ball To process, urgent!