According to the meaning of the question, adding sugar to 7% syrup changed the original concentration of syrup, and the quality of syrup increased, but the water quality did not change. Therefore, the quality of water can be calculated according to the concentration of original syrup, and then the quality of current syrup can be calculated according to the concentration of later syrup, and the quality of added sugar can be obtained by subtracting the quality of original syrup from the quality of current syrup.
Mass of water in raw sugar water: 600× (1-7%) = 558 (g)
Now the mass of sugar water: 558 ÷ (1-10%) = 620 (g).
Mass of added sugar: 620-600 = 20 (g)
You need to add 20 grams of sugar
Example 2. A 35% new pesticide is most effective when diluted to 1.75%. 800 kg 1.75% pesticide How many kg of 35% pesticide and how many kg of water are used?
The process of changing a high concentration solution into a low concentration solution by adding a solvent is called dilution. In this dilution process, the mass of solute is constant. This is the key to solve this kind of problem.
800 kg 1.75% of pesticides contain pure pesticides, which are as follows by mass
800× 1.75% = 14 (kg)
The mass of 35% pesticides containing 14kg pure pesticides is
14 ÷ 35% = 40 (kg)
The quality of water diluted from 40 kg of pesticides to 800 kg of pesticides is as follows
800-40 = 760 kg
Answer: 40 kg of 35% pesticide and 760 kg of water can make 800 kg of pesticide with a concentration of 1.75%.
Example 3. The existing concentration of 10% brine is 20kg. How many kilograms of 30% brine can be added to get 22% brine?
This is a problem of solution mixing. The concentration of solution changed before and after mixing, but the total mass of solute and solution did not change as a whole. Therefore, the sum of solutes in the two solutions before mixing is equal to the amount of solute in the solution after mixing.
Quality of salt in 20k g 10% brine.
20× 10% = 2 (kg)
When mixed to 22%, the mass of salt in 20 kg solution.
20× 22% = 404 (kg)
It is necessary to add 30% physiological saline.
(4.4-2) ÷ (30%-22%) = 30 (kg)
A: You need to add 30 kilograms of 30% brine to get 22% brine.
Example 4. 600g of15% brine was prepared by mixing 20% brine with 5% brine. How many grams do 20% saline and 5% saline need?
According to the meaning of the question, 20% brine and 5% brine are mixed to make 15% brine, which shows that the quality of salt in the two kinds of brine before mixing is equal to that in the mixed brine. According to the equal relationship between these quantities, the equation can be solved.
Solution: Suppose that 20% salt water needs X grams and 5% salt water needs 600-X grams, then
20%x+(600-x)×5%=600× 15%
X =400
600-400 = 200 (gram)
Answer: 400g of 20% salt water and 200g of 5% salt water are needed.
Example 5. Three test tubes A, B and C each contain 10g, 20g and 30g of water. Pour a certain mass fraction of saline 10g into the test tube, pour the mixture of 10g into the test tube B, and then take the mixture of 10g out of the test tube B and pour it into the test tube C. At present, the mass fraction of saline in the tube C is 0.5%. What is the mass fraction of brine poured into the nail tube for the first time?
After the idea navigation is mixed, the saline in the three test tubes A, B and C should be 20g, 30g and 40g respectively. According to the meaning of the question, we can find out the quality of salt in C tube. Because there is only 30 grams of water in the C tube, its salt is taken from the B tube 10 grams of salt water. Thus, the quality of salt in 30 grams of brine in the second pipe can be obtained. The salt in the second tube was taken out from the first tube 10g brine, so the quality of salt in 20g brine in the first tube could be obtained. The salt in the nail tube is the salt in a certain concentration of brine, so that the mass fraction of brine initially injected into the nail tube can be obtained.
Mass of salt in C-tube: (30+10) × 0.5% = 0.2 (g)
After pouring into tube B, the mass of salt in tube B is 0.2× (20+10) ÷10 = 0.6 (g).
Pour it into the nail tube, and the mass of salt in the nail tube is 0.6× (10+10) ÷10 =1.2 (g).
1.2÷ 10= 12%
Answer: The mass fraction of brine poured into the A pipe first is 12%.
Reflections on quadrilateral teaching 1
This lesson is based on the fact that students have learned triangles and known squares and rectangles. The
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