In the experimental class, a junior high school student should prepare 2000 grams of 80% alcohol solution into 60% alcohol solution. Without consideration, a student added 500 grams of water. (1) try to explain whether the student added too much water through calculation; (2) if the amount of water can not be added, how many grams of 20% alcohol solution should be added? If the amount of water is excessive, how many grams of 95% alcohol solution should be added?

In the experimental class, a junior high school student should prepare 2000 grams of 80% alcohol solution into 60% alcohol solution. Without consideration, a student added 500 grams of water. (1) try to explain whether the student added too much water through calculation; (2) if the amount of water can not be added, how many grams of 20% alcohol solution should be added? If the amount of water is excessive, how many grams of 95% alcohol solution should be added?

(1) After adding 500 grams of water, the mass fraction of alcohol solution is: 2000 × 80% 2000 + 500 = 64% > 60%. Therefore, the student does not add too much water. (2) suppose that 20% alcohol solution x g should be added

500 g of 75% disinfectant alcohol solution was prepared from 90% and 60% alcohol solution of a and B, respectively
How much each?
Please list the equations and write the process

Let: we need XG for type A and YG for type B, then 0.9x + 0.6y = 500 * 0.750.1x + 0.4y = 500 * 0.25 to get: x = y = 250g a

How many ml 75% alcohol can be prepared with 500 ml 95% alcohol?
The calculation process is also necessary, thank you!

We're used to saying that
95% and 75% alcohol generally refers to the volume fraction (if the mass fraction needs to know the density)
Set gram to prepare V ml 75% alcohol
According to the solute unchanged
be
500 * 95% = V * 75%
have to
V = 633.33 ml
therefore
633.33 ml 75% alcohol can be prepared with 500 ml 95% alcohol

500ml 95% alcohol, how many ml of water do you need to add to prepare 75% alcohol?

Original concentration multiplied by volume = diluted concentration multiplied by volume
C1v1 = c2v2, ml to L
V2=C1V1/C2=0.95X0.5/0.75=0.6333...
Add 633.33 ml water