Given the densities for the four elements, we have the expression $d=\frac{m}{V}$ that shows the relationship between mass and Volume to express the density of an element.

For each element we have:

$d_{lithium}=\frac{m_{lithium}}{V_{lithium}}=0.53g/mL$

$d_{gold}=\frac{m_{gold}}{V_{gold}}=19.3g/mL$

$d_{aluminum}=\frac{m_{aluminum}}{V_{aluminum}}=2.70g/mL$

$d_{lead}=\frac{m_{lead}}{V_{lead}}=11.3g/mL$

The problem says that all the samples have the same mass, so:

$m_{lithium}=m_{gold}=m_{aluminum}=m_{lead}=m$

it means that m is a constant

Now, solving for the Volume in each element and with m as a constant, we have:

$V_{lithium}=\frac{m}{d_{lithium}}$

$V_{lithium}=\frac{1}{0.53\frac{g}{mL}} *m$

$V_{lithium}=1.88\frac{mL}{g}*m$

$V_{gold}=\frac{m}{d_{gold}}$

$V_{gold}=\frac{1}{19.3\frac{g}{mL}} *m$

$V_{gold}=5.18*10^{-2}\frac{mL}{g}*m$

$V_{aluminum}=\frac{m}{d_{aluminum}}$

$V_{aluminum}=\frac{1}{2.70\frac{g}{mL}} *m$

$V_{aluminum}=3.70*10^{-1}\frac{mL}{g}*m$

$V_{lead}=\frac{m}{d_{lead}}$

$V_{lead}=\frac{1}{11.3\frac{g}{mL}} *m$

$V_{lead}=8.85*10^{-2}\frac{mL}{g}*m$

If we assume m = 1g, we find that:

$V_{lithium}=1.88mL$

$V_{gold}=5.18*10^{-2}mL$

$V_{aluminum}=3.70*10^{-1}mL$

$V_{lead}=8.85*10^{-2}mL$

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.

4 0

2 years ago

PLEASE HELP ME !!!!!!!!!!!!!!

sergeinik [125]

Answer:

Explanation:

3 0

3 years ago

Fe3O4(s) + 4H2(g)3Fe(s) + 4H2O(g) Using standard thermodynamic data at 298K, calculate the entropy change for the surroundings w

alexandr402 [8]

Answer:

dS= 1.79*169.504

j/k = 303.41 j/k

Explanation:

Fe3O4(s) + 4H2(g) --> 3Fe (s)+ 4H2O(g)

dS(Fe3O4) =146.4 j/k

dS(H2) =130.684

dS(Fe) =27.78

dS(H2O) =188.825

dSrxn = dS[product]-dS[reactants]

= 3*dS(Fe)+ 4*dS(H2O)-[1*dS(Fe3O4)+ 4dS(H2)]

= [3*27.78 +4*188.825-146.4 -4*130.684] j/k = 169.504 j/k

This is the dS for 1mole Fe3O4

for 1.79 mols Fe3O4

dS= 1.79*169.504 j/k = 303.41 j/k

7 0

3 years ago