a)The element with the greatest density is osmium, which has a density of 22.6g/cm3. Calculate the mass of 32.1 cm3 of osmium.

b)A team of students determined the density of a sample of wood to be 0.52 g/cm3. A handbook of chemistry reported the density of 0.548 g/cm3 for the same type of wood. what is the percent error of the students' value?

I never heard of a chemistry course called "Dobson".

a) density time volume equals mass

b) The students may have made a negligible error. The density of any type of wood depends upon how dry it is. Handbook values are only averages. Seasoned wood thet is ready for firewood or lumber is less dense than the wood in a living or freshly cut tree.

Assuming the Handbook value is correct, the percent error is
(0.548-0.52)/0.548 x 100

a) To calculate the mass of osmium in 32.1 cm3, you can use the formula:

Mass = Density x Volume

Given:
Density of osmium = 22.6 g/cm3
Volume of osmium = 32.1 cm3

Substituting the values into the formula, we get:

Mass = 22.6 g/cm3 x 32.1 cm3
Mass = 725.46 g

Therefore, the mass of 32.1 cm3 of osmium is 725.46 grams.

b) To calculate the percent error, we can use the formula:

Percent Error = |(Experimental Value - Accepted Value) / Accepted Value| x 100

Given:
Experimental Density = 0.52 g/cm3
Accepted Density = 0.548 g/cm3

Substituting the values into the formula, we get:

Percent Error = |(0.52 g/cm3 - 0.548 g/cm3) / 0.548 g/cm3| x 100
Percent Error = |-0.028 g/cm3 / 0.548 g/cm3| x 100
Percent Error = 5.11%

Therefore, the percent error of the students' value is 5.11%.

a) To calculate the mass of osmium, you can use the formula:

Mass = Density * Volume

Given that the density of osmium is 22.6 g/cm3 and the volume is 32.1 cm3, simply multiply these values together:

Mass = 22.6 g/cm3 * 32.1 cm3
Mass = 725.46 g

Therefore, the mass of 32.1 cm3 of osmium is 725.46 g.

b) To calculate the percent error, you can use the following formula:

Percent Error = [(Measured Value - Actual Value) / Actual Value] * 100

In this case, the measured value is the density determined by the students (0.52 g/cm3), and the actual value is the density reported in the handbook (0.548 g/cm3).

Calculate the percent error as follows:

Percent Error = [(0.52 g/cm3 - 0.548 g/cm3) / 0.548 g/cm3] * 100
Percent Error ≈ -5.11%

Therefore, the percent error of the students' value is approximately -5.11%. Note that the negative sign indicates an underestimate in comparison to the actual value.