Problemas

3. Perform the following conversions. a. 1.51times 10^25 atoms of Si to mol of Si b. 4.25times 10^24 mol of H_(2)SO_(4) to molecules of H_(2)SO_(4) C. 8.95times 10^25 molecules of CCl_(4) to mol of CCl_(4)
Solución
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Julio Césarprofessionell · Tutor durante 6 años
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a. To convert atoms to moles, we need to use Avogadro's number, which is $6.022\times 10^{23}$ atoms/mol.<br /><br />Given: $1.51\times 10^{25}$ atoms of Si<br /><br />To find the number of moles of Si, we divide the given number of atoms by Avogadro's number:<br /><br />Number of moles of Si = $\frac{1.51\times 10^{25}}{6.022\times 10^{23}}$<br /><br />Number of moles of Si = 251.5 mol<br /><br />Therefore, $1.51\times 10^{25}$ atoms of Si is equivalent to 251.5 mol of Si.<br /><br />b. To convert moles to molecules, we again use Avogadro's number.<br /><br />Given: $4.25\times 10^{24}$ mol of $H_{2}SO_{4}$<br /><br />To find the number of molecules of $H_{2}SO_{4}$, we multiply the given number of moles by Avogadro's number:<br /><br />Number of molecules of $H_{2}SO_{4}$ = $4.25\times 10^{24} \times 6.022\times 10^{23}$<br /><br />Number of molecules of $H_{2}SO_{4}$ = $2.56\times 10^{48}$<br /><br />Therefore, $4.25\times 10^{24}$ mol of $H_{2}SO_{4}$ is equivalent to $2.56\times 10^{48}$ molecules of $H_{2}SO_{4}$.<br /><br />c. To convert molecules to moles, Avogadro's number.<br /><br />Given: $8.95\times 10^{25}$ molecules of $CCl_{4}$<br /><br />To find the number of moles of $CCl_{4}$, we divide the given number of molecules by Avogadro's number:<br /><br />Number of moles of $CCl_{4}$ = $\frac{8.95\times 10^{25}}{6.022\times 10^{23}}$<br /><br />Number of moles of $CCl_{4}$ = 148.5 mol<br /><br />Therefore, $8.95\times 10^{25}$ molecules of $CCl_{4}$ is equivalent to 148.5 mol of $CCl_{4}$.
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