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Select the correct answer.

A person has an unknown element with a volume of 16 cubic inches and a weight of 4.13 pounds.

\begin{tabular}{|l|l|}
\hline \multicolumn{1}{|c|}{Density} & \multicolumn{1}{c|}{[tex]$lb /$[/tex] in [tex]$^3$[/tex]} \\
\hline Titanium & 0.161 \\
\hline Zinc & 0.258 \\
\hline Iron & 0.284 \\
\hline Silver & 0.379 \\
\hline
\end{tabular}

Using the table, what is the element?

A. Silver

B. Zinc

C. Titanium

D. Iron


Sagot :

Let's solve this problem step-by-step:

1. Find the density of the unknown element:

The volume of the unknown element is given as [tex]\( 16 \)[/tex] cubic inches and the weight is [tex]\( 4.13 \)[/tex] pounds.

The density of a material is calculated using the formula:
[tex]\[ \text{Density} = \frac{\text{Weight}}{\text{Volume}} \][/tex]

So, we plug in the given values:
[tex]\[ \text{Density} = \frac{4.13 \, \text{pounds}}{16 \, \text{cubic inches}} = 0.258125 \, \text{lb/in}^3 \][/tex]

2. Compare the calculated density with the given densities in the table:

\begin{tabular}{|l|l|}
\hline \text{Element} & \text{Density} ([tex]$\text{lb/in}^3$[/tex]) \\
\hline Titanium & 0.161 \\
\hline Zinc & 0.258 \\
\hline Iron & 0.284 \\
\hline Silver & 0.379 \\
\hline
\end{tabular}

The calculated density of [tex]\( 0.258125 \, \text{lb/in}^3 \)[/tex] is very close to the density of Zinc, which is [tex]\( 0.258 \, \text{lb/in}^3 \)[/tex].

3. Select the element with the closest matching density:

Since [tex]\( 0.258125 \, \text{lb/in}^3 \)[/tex] is almost identical to [tex]\( 0.258 \, \text{lb/in}^3 \)[/tex], the unknown element is Zinc.

Therefore, the correct answer is:
[tex]\[ \boxed{\text{B. Zinc}} \][/tex]