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If 32 grams of kerosene with a density of 0.8 g/cm⁻³ are mixed with 8 grams of water, what is the density of the resulting mixture?

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Final answer:

The density of the mixture of 32 grams of kerosene and 8 grams of water, with respective densities of 0.8 g/cm³ and 1 g/cm³, is calculated to be 0.833 g/cm³ after combining the masses and volumes of both substances.

Step-by-step explanation:

To calculate the density of a mixture of kerosene and water, we can use the formula for density, which is mass divided by volume. Kerosene has a density of 0.8 g/cm3, and if we assume we have 32 grams of it, we can calculate its volume using the formula V = m/ρ (where V is volume, m is mass, and ρ is density). The volume of kerosene thus would be 32 g / 0.8 g/cm3 = 40 cm3. Now, for water, it has a density of 1 g/cm3, and 8 grams would also have a volume of 8 cm3, since its density is 1 g/cm3.

The total volume of the mixture would be the sum of the volumes of kerosene and water, which is 40 cm3 + 8 cm3 = 48 cm3. The total mass is 32 grams of kerosene plus 8 grams of water, totaling 40 grams.

To find the density of the mixture, we use the total mass divided by the total volume, which gives us 40 g / 48 cm3 = 0.833 g/cm3. Therefore, the density of the mixture is 0.833 g/cm3.

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