Question

post - lab questions
show all work for the calculations below, and express your final answer with the correct units and number of significant figures. circle your final answer.

1. indicate the appropriate type of glassware (a 100 - ml beaker, a 50 - ml graduated cylinder, or a buret) to use when measuring out the following:

	type(s) of glassware that can be used
15.00 ml of solution
approximately 50 ml of di water

1. consider the measurements recorded for a rectangular metal sample using an analytical balance and a centimeter ruler similar to those used in this experiment.

	experimental measurements
length of metal sample	4.60 cm
width of metal sample	1.40 cm
thickness of metal sample	1.30 cm
volume of metal sample
density of metal sample

a. calculate the volume and density for the trial to fill in the blanks in the data table above.
b. explain if and how the experimentally determined density would change if a centigram balance (±0.01 g) is used instead of an analytical balance. (hint: round the mass to the correct number of decimal places, and calculate the resulting density with the less accurate mass before answering this question.)

1. compare the density of your unknown metal to the densities of common substances below:

substance	density (in g/ml)
lead	11.4

b. identify your solution by comparing its density with the densities provided above:
c. calculate the percent error of your solution.

Explanation:

Step1: Calculate volume of metal sample

The volume $V$ of a rectangular - shaped object is given by $V = l\times w\times h$, where $l$ is length, $w$ is width and $h$ is thickness.
$V=4.60\ cm\times1.40\ cm\times1.30\ cm$
$V = 8.364\ cm^{3}$

Step2: Calculate density of metal sample

The density $
ho$ is given by the formula $
ho=\frac{m}{V}$, where $m$ is mass and $V$ is volume.
$
ho=\frac{95.524\ g}{8.364\ cm^{3}}\approx11.4\ g/cm^{3}$ (rounded to 3 significant figures)

Step3: Analyze effect of using centigram balance

If a centigram balance ($\pm0.01\ g$) is used, the mass of the metal sample $m = 95.52\ g$ (rounded to two decimal - places).
The new density $
ho_{new}=\frac{95.52\ g}{8.364\ cm^{3}}\approx11.4\ g/cm^{3}$ (rounded to 3 significant figures). The density value remains the same to 3 significant figures because the change in mass due to the less - accurate balance is not significant enough to affect the density value at this level of precision.

Step4: Identify the metal

Comparing the calculated density of approximately $11.4\ g/cm^{3}$ with the given densities, the metal is lead since its density is $11.4\ g/mL$ (and $1\ cm^{3}=1\ mL$).

Step5: Calculate percent error

Assuming the accepted density of lead $
ho_{accepted}=11.4\ g/mL$ and the experimental density $
ho_{exp}=11.4\ g/mL$
The percent - error formula is $\text{Percent Error}=\frac{\vert
ho_{exp}-
ho_{accepted}\vert}{
ho_{accepted}}\times100\%$
$\text{Percent Error}=\frac{\vert11.4 - 11.4\vert}{11.4}\times100\% = 0\%$

Answer:

Volume of Metal Sample: $8.36\ cm^{3}$ (3 significant figures)
Density of Metal Sample: $11.4\ g/cm^{3}$
The density would not change significantly when using a centigram balance.
The metal is lead.
Percent Error: $0\%$