Answer:
[tex]\boxed {\boxed {\sf A. \ 0.365 \ g\ He}}[/tex]
Explanation:
1. Convert Atoms to Moles
We must use Avogadro's Number: 6.022*10²³. This is the number of particles (atoms, molecules, ions, etc.) in 1 mole of a substance. In this case, the particles are atoms of helium. We can create a ratio.
[tex]\frac {6.022*10^{23} \ atoms \ He}{1 \ mol \ He}[/tex]
Multiply by the given number of helium atoms.
[tex]5.50 *10^{22} \ atoms \ He *\frac {6.022*10^{23} \ atoms \ He}{1 \ mol \ He}[/tex]
Flip the fraction so the atoms of helium cancel.
[tex]5.50 *10^{22} \ atoms \ He *\frac {1 \ mol \ He}{6.022*10^{23} \ atoms \ He}[/tex]
[tex]5.50 *10^{22} *\frac {1 \ mol \ He}{6.022*10^{23} }[/tex]
[tex]\frac {5.50 *10^{22} \ mol \ He}{6.022*10^{23} }= 0.09133178346 \ mol \ He[/tex]
2. Convert Moles to Grams
We must use the molar mass, which is found on the Periodic Table.
Use this as a ratio.
[tex]\frac { 4.00 \ g \ He }{ 1 \ mol \ He}[/tex]
Multiply by the number of moles we calculated. The moles will then cancel.
[tex]0.09133178346 \ mol \ He *\frac { 4.00 \ g \ He }{ 1 \ mol \ He}[/tex]
[tex]0.09133178346*\frac { 4.00 \ g \ He }{ 1 }[/tex]
[tex]0.3653271338 \ g\ He[/tex]
3. Round
The original measurement has 3 significant figures (5, 5, and 0). Our answer must have the same. For the number we calculated, it is thousandth place. The 3 in the ten thousandth place tells us to leave the 5.
[tex]0.365 \ g\ He[/tex]
The mass is 0.365 grams of helium so choice A is correct.
