Salt water with a density of 1.1x10 ^ 3kg / m3 is needed for seed selection. Now 500ml salt water is prepared, and its mass is 0.6kg. Is this salt water suitable for seed selection

Salt water with a density of 1.1x10 ^ 3kg / m3 is needed for seed selection. Now 500ml salt water is prepared, and its mass is 0.6kg. Is this salt water suitable for seed selection


The volume of brine v = 500ml = 500cm, the mass of brine M = 0.6kg = 600g,
The density of brine is ρ = m / v = 600 g / 500 cm, = 1.2 g / cm, = 1.2 × 10 m, kg / m, kg / m, kg / m, kg / m, kg / m, kg / m, kg / m, kg / m, kg / M, kg / m, kg / m, kg / m, kg / m, kg / m;
Such salt water does not meet the requirements



Salt water selection needs 1.1x10 ^ 3kg/m3 of salt water. Now 500ml salt water is prepared, and its mass is 0.6kg. Does this salt water meet the requirements?


The density of the prepared brine is ρ = m / v = 600 g / 500 cm & 179; = 1.2 g / cm & 179; = 1.2 × 10 & 179; kg / M & 179;
∵ 1.2 × 10 & # 179; kg / M & # 179; > 1.1 × 10 & # 179; kg / M & # 179;
Such salt water does not meet the requirements



It is reliable and easy for farmers to select seeds with salt water. The density of salt water is required to be 1.1 × 103kg / m3. Now 500ml salt water is prepared and its mass is 0.6kg. Does this salt water meet the requirements? If not, should salt or water be added? How much more?


The density of the sample: ρ sample = MV = 600g500cm3 = 1.2g/cm3, ∵ 1.2g/cm3 > 1.1 × 103kg / m3, ∵ does not meet the requirements, water should be added; set the amount of water to be added as Ag, because the density of water is 1g / cm3, so the volume of water to be added is acm3, so we can get from the meaning: 600g + ag500cm3 + acm3 = 1.1g/cm3, the solution, the value of a is 500, so 500g of water should be added The water does not meet the requirements; add water, add 500g



Required brine density: 1100kg / m3, now 0.5dm3 brine is configured, the mass is 0.6kg. Should add salt or water? How much?


Since the mass of 0.5dm3 brine is 0.6kg, the density of brine ρ = m / v = 0.6/0.5kg/dm3 = 1.2 * 10 ^ 3kg / m3 is greater than the required density of 1.1 * 10 ^ 3kg / m3, so water should be added
Suppose: the volume of water is xdm3 with XKG water
(0.6+x)/(0.5+x)=1.1
The solution is: x = 0.5
Therefore, 0.5kg of water should be added