Question

post lab questions

1. two strips of metal each weighing 100 g are placed into a cylinder containing water. the volume of water displaced is 8.8 ml in one case and 37.0 ml in the other case. identify the two metal strips.

metal au fe al pb ag
density 19.3 7.9 2.7 11.3 10.5

Explanation:

Step1: Recall density formula

Density ($
ho$) is given by the formula $
ho = \frac{m}{V}$, where $m$ is mass and $V$ is volume. We can rearrange this to find volume $V=\frac{m}{
ho}$ or to find density from mass and volume as $
ho=\frac{m}{V}$. Here, mass $m = 100\space g$ for each metal strip, and we have two volumes $V_1 = 8.8\space mL$ and $V_2=37.0\space mL$. We will calculate the density for each case and match with the given densities.

Step2: Calculate density for first case ($V = 8.8\space mL$)

Using $
ho=\frac{m}{V}$, substitute $m = 100\space g$ and $V = 8.8\space mL$.
$
ho_1=\frac{100\space g}{8.8\space mL}\approx11.36\space g/mL$. Looking at the given densities, $Pb$ has a density of $11.3\space g/mL$, which is close (considering significant figures or possible rounding), so this metal is likely $Pb$.

Step3: Calculate density for second case ($V = 37.0\space mL$)

Using $
ho=\frac{m}{V}$, substitute $m = 100\space g$ and $V = 37.0\space mL$.
$
ho_2=\frac{100\space g}{37.0\space mL}\approx2.70\space g/mL$. Looking at the given densities, $Al$ has a density of $2.7\space g/mL$, so this metal is $Al$.

Answer:

The metal with volume displacement $8.8\space mL$ is $Pb$ (Lead) and the metal with volume displacement $37.0\space mL$ is $Al$ (Aluminum).