Respuesta :
Answer:
The sample of lithium occupies the largest volume.
Explanation:
Given the densities for the four elements, we have the expression [tex]d=\frac{m}{V}[/tex] that shows the relationship between mass and Volume to express the density of an element.
For each element we have:
[tex]d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL[/tex]
[tex]d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL[/tex]
[tex]d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL[/tex]
[tex]d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL[/tex]
The problem says that all the samples have the same mass, so:
[tex]m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m[/tex]
it means that m is a constant
Now, solving for the Volume in each element and with m as a constant, we have:
[tex]V_{lithium}=\frac{m}{d_{lithium}}[/tex]
[tex]V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m[/tex]
[tex]V_{lithium}=1.88\frac{mL}{g}*m[/tex]
[tex]V_{gold}=\frac{m}{d_{gold}}[/tex]
[tex]V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m[/tex]
[tex]V_{gold}=5.18*10^{-2}\frac{mL}{g}*m[/tex]
[tex]V_{aluminum}=\frac{m}{d_{aluminum}}[/tex]
[tex]V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m[/tex]
[tex]V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m[/tex]
[tex]V_{lead}=\frac{m}{d_{lead}}[/tex]
[tex]V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m[/tex]
[tex]V_{lead}=8.85*10^{-2}\frac{mL}{g}*m[/tex]
If we assume m = 1g, we find that:
[tex]V_{lithium}=1.88mL[/tex]
[tex]V_{gold}=5.18*10^{-2}mL[/tex]
[tex]V_{aluminum}=3.70*10^{-1}mL[/tex]
[tex]V_{lead}=8.85*10^{-2}mL[/tex]
So we can see that the sample of lithium occupies the largest volume with 1.88mL
Note that m only can take positive values, so if you change the value of m, always will be the lithium which occupies the largest volume.
