The water content of alcohol produced in industry should not exceed 10%. At present, there are four kinds of samples (a, B, C and D) with densities of 0.81g/cm3, 0.815g/cm3, 0.82g/cm3 and 0.83g/cm3 respectively. What are the qualified products? (the known alcohol density is 0.8g/cm3) Solution steps. Urgent!

The water content of alcohol produced in industry should not exceed 10%. At present, there are four kinds of samples (a, B, C and D) with densities of 0.81g/cm3, 0.815g/cm3, 0.82g/cm3 and 0.83g/cm3 respectively. What are the qualified products? (the known alcohol density is 0.8g/cm3) Solution steps. Urgent!

Suppose the total mass of alcohol is m, in which the mass of pure alcohol is 90% m and the maximum mass of water content is 10% m, then the maximum density of qualified alcohol is as follows:
The calculation results show that if the water content exceeds 10%, the alcohol density will be greater than 0.816g/cm3; otherwise, if the alcohol density is greater than 0.816g/cm3, the water content exceeds 10%, which does not meet the requirements (ρ water > ρ wine), so the qualified products that meet the requirements are 0.81g/cm3 and 0.815g/cm3
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When the water content of alcohol is 10.20% of the total mass, the density of alcohol can be calculated

The density of pure alcohol is 0.8g/cm and 179;
Suppose you have 100g of this alcohol
If the water content is 20%, the mass of water is 20g, the density of water is 1g / cm, and the volume of water is 20cm;
The mass of alcohol is 80g, the density is 0.8g, the volume of alcohol is 100cm3
Then the volume of 100g alcohol is 120cm3, and the average density is 100 / 120 = 5 / 6G / cm & # 179;
Similarly, the water content is 10% and the density is 40 / 49
So it's 40 / 49 to 5 / 6

There are 320G of water with density of 1G / cm3 and 320G of pure alcohol with density of 0.8g/cm3. How many grams of medical alcohol with density of 0.84g/cm3 can be prepared by using them? (the volume change in the mixing process is ignored) is a physics problem in grade two,

The volume of water is 320G / cm ^ 3 = 320cm ^ 3
The volume of pure alcohol is 320G / cm ^ 3 = 400cm ^ 3
Assuming that all 320 grams of water and 320 grams of alcohol are mixed, the density after mixing is (320 + 320) / (320 + 400) ≈ 0.89 g / cm ^ 3
Obviously higher than the required density, so if you want to configure the most medical alcohol, then all pure alcohol should be used, but not all water
According to the meaning of the question: (320 + x) / (400 + x) = 0.84
The solution is x = 100
It needs 100 grams of water and 320 grams of pure alcohol
So the quality of medical alcohol is 320 + 100 = 420
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