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Ne4ueva [31]
2 years ago
6

A party balloon contains 5.50 x 10 ^22 atoms of helium gas. What is the mass, in grams of the

Chemistry
1 answer:
eduard2 years ago
7 0

Answer:

\boxed {\boxed {\sf A. \ 0.365 \ g\ He}}

Explanation:

<u>1. Convert Atoms to Moles</u>

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.

\frac {6.022*10^{23} \ atoms \ He}{1 \ mol \ He}

Multiply by the given number of helium atoms.

5.50 *10^{22} \ atoms \ He *\frac {6.022*10^{23} \ atoms \ He}{1 \ mol \ He}

Flip the fraction so the atoms of helium cancel.

5.50 *10^{22} \ atoms \ He *\frac {1 \ mol \ He}{6.022*10^{23} \ atoms \ He}

5.50 *10^{22} *\frac {1 \ mol \ He}{6.022*10^{23} }

\frac {5.50 *10^{22} \ mol \ He}{6.022*10^{23} }= 0.09133178346 \ mol \ He

<u>2. Convert Moles to Grams</u>

We must use the molar mass, which is found on the Periodic Table.

  • Helium (He): 4.00 g/mol

Use this as a ratio.

\frac { 4.00 \ g \ He }{ 1 \ mol \ He}

Multiply by the number of moles we calculated. The moles will then cancel.

0.09133178346 \ mol \ He *\frac { 4.00 \ g \ He }{ 1 \ mol \ He}

0.09133178346*\frac { 4.00 \ g \ He }{ 1 }

0.3653271338 \ g\ He

<u>3. Round </u>

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.

0.365 \ g\ He

The mass is <u>0.365 grams of helium</u> so choice A is correct.

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