In salt water selection, salt water with density of 1.1x10 ^ 3kg / m ^ 3 is needed In the selection of salt water, salt water with a density of 1.1x10 ^ 3kg / m ^ 3 is needed. A student prepared 350ml salt water, and the quality of the salt water is 0.6kg. Ask whether the salt water meets the requirements? If not, how much water or salt should be added? Especially the process of how much to add!

In salt water selection, salt water with density of 1.1x10 ^ 3kg / m ^ 3 is needed In the selection of salt water, salt water with a density of 1.1x10 ^ 3kg / m ^ 3 is needed. A student prepared 350ml salt water, and the quality of the salt water is 0.6kg. Ask whether the salt water meets the requirements? If not, how much water or salt should be added? Especially the process of how much to add!

35 = 1.71 > 1.1 X10 ^ 3kg / m ^ 3
Suppose that the amount of water added for successful preparation is x kg
There is 0.6 + X / 0.35 + x = 1.1
The solution is x = 2.15
So add 2.15kg water

When selecting seeds with salt water, the density of salt water is 1.15 × 10 ^ 3kg / m ^ 3, 4m ^ 3
The mass of 500cm ^ 3 sample is 600g. Ask (1) the density of concentrated brine? (2) the mass of 4m ^ 3 selected brine? (3) how many m ^ 3 concentrated brine and clean water need to be mixed together to meet the requirements? (regardless of the change of total volume after mixing)

(1) Calculate the density directly according to the density formula, and note that the unit transformation should be equal to 1.2 * 10 cubic kilogram per cubic meter
(2) Use (1) to get the density and the given mass, according to the density formula deformation to get the mass equal to 4.8 * 10 cubic kilogram
(3) According to the calculated density is greater than the given density, so water is needed. The total mass is equal to 600g + m, and the total volume is equal to 500cm3 + m / 1. 1 is the density of water 1g / cm3
When the total mass / volume is equal to 1.15g/cm3, M = 166.7g can be obtained

In the selection of salt water, the density of salt water is 1.1x10 ^ 3kg / m ^ 3. A student prepared 0.5dm cubic salt water, and the weight is 0.6kg. Does the salt water meet the requirements? If not, should we add salt or water? How much?
There should be a process
Be right
Thank you

V salt 1 = 0.5dm & sup3; = 0.5 * 10 ^ (- 3) M & sup3;
P salt 1 = m salt 1 / V salt 1 = 0.6kg/0.5 * 10 ^ - 3M ^ 3 = 1.2 * 10 ^ 3kg / M & sup3;
∵ P salt 1 ∵ P salt ∵ does not meet the requirements, water should be added
Let V M & sup 3 be added to the water
6 kg + V water * 1 * 10 & sup3; kg / M & sup3; = 1.1 * 10 & sup3; kg / M & sup3;
0.5*10^（-3）m³+V
The solution is: V water = 5 * 10 ^ (- 4) M & sup3;
A: this kind of brine does not meet the requirements, and 5 * 10 ^ (- 4) M & sup 3; of water is needed