Question

what is the percent abundance (in units of percent) of zinc in a sample whose density is 7.843 g/ml and the only other component is copper? the density for pure copper is 8.96 g/cm3 and the density of pure zinc is 7.13 g/cm3. report your answer to one decimal place.
note: this question is meant to give you practice before your data. the density value generated is random within a certain realistic constraint. it will not be representative of your data in the experiment.
type answer.
enter a number

Explanation:

Step1: Define variables

Let \( x \) be the percent abundance of zinc (in decimal form, so \( 0 \leq x \leq 1 \)) and \( 1 - x \) be the percent abundance of copper. The density of the mixture \(
ho_{mix} \) is given by the weighted average of the densities of zinc (\(
ho_{Zn} = 7.13 \, \text{g/cm}^3 \)) and copper (\(
ho_{Cu} = 8.96 \, \text{g/cm}^3 \)). The formula for the density of the mixture is:

$$ ho_{mix} = x ho_{Zn} + (1 - x) ho_{Cu} $$

We know \(
ho_{mix} = 7.843 \, \text{g/cm}^3 \).

Step2: Substitute values and solve for \( x \)

Substitute the known values into the formula:

$$ 7.843 = x(7.13) + (1 - x)(8.96) $$

Expand the right - hand side:

$$ 7.843 = 7.13x+8.96 - 8.96x $$

Combine like terms:

$$ 7.843=8.96 - 1.83x $$

Subtract \( 8.96 \) from both sides:

$$ 7.843 - 8.96=- 1.83x $$

$$ - 1.117=-1.83x $$

Divide both sides by \( - 1.83 \):

$$ x=\frac{-1.117}{-1.83}\approx0.610 $$

Step3: Convert to percent

To convert \( x \) to a percent, multiply by \( 100 \):
\( 0.610\times100 = 61.0\% \)

Answer:

\( 61.0 \)