post - lab questions\nshow all work for the calculations below, and express your final answer with the…

post - lab questions\nshow all work for the calculations below, and express your final answer with the correct units and number of significant figures. circle your final answer.\n1. 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:\n| |type(s) of glassware that can be used|\n|----|----|\n|8.00 ml of solution| |\n|15.00 ml of solution| |\n|approximately 50 ml of di water| |\n2. consider the measurements recorded for a rectangular metal sample using an analytical balance and a centimeter ruler similar to those used in this experiment.\n| |experimental measurements|\n|----|----|\n|mass of metal sample|95.524 g|\n|length of metal sample|4.60 cm|\n|width of metal sample|1.40 cm|\n|thickness of metal sample|1.30 cm|\n|volume of metal sample| |\n|density of metal sample| |\na. calculate the volume and density for the trial to fill in the blanks in the data table above.\nb. 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.)\n3. compare the density of your unknown metal to the densities of common substances below:\n|substance|density (in g/ml)|\n|----|----|\n|gold|19.3|\n|lead|11.4|\nb. identify your solution by comparing its density with the densities provided above:\nc. calculate the percent error of your solution.
Answer
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 $\rho$ is given by the formula $\rho=\frac{m}{V}$, where $m$ is mass and $V$ is volume. $\rho=\frac{95.524\ g}{8.364\ cm^{3}}\approx11.4\ g/cm^{3}$ (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 $\rho_{new}=\frac{95.52\ g}{8.364\ cm^{3}}\approx11.4\ g/cm^{3}$ (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 $\rho_{accepted}=11.4\ g/mL$ and the experimental density $\rho_{exp}=11.4\ g/mL$ The percent - error formula is $\text{Percent Error}=\frac{\vert\rho_{exp}-\rho_{accepted}\vert}{\rho_{accepted}}\times100%$ $\text{Percent Error}=\frac{\vert11.4 - 11.4\vert}{11.4}\times100% = 0%$
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: $0%$