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Answer:
1.9 × 10² g NaN₃
1.5 g/L
Explanation:
Step 1: Write the balanced decomposition equation
2 NaN₃(s) ⇒ 2 Na(s) + 3 N₂(g)
Step 2: Calculate the moles of N₂ formed
N₂ occupies a 80.0 L bag at 1.3 atm and 27 °C (300 K). We will calculate the moles of N₂ using the ideal gas equation.
P × V = n × R × T
n = P × V / R × T
n = 1.3 atm × 80.0 L / (0.0821 atm.L/mol.K) × 300 K = 4.2 mol
We can also calculate the mass of nitrogen using the molar mass (M) 28.01 g/mol.
4.2 mol × 28.01 g/mol = 1.2 × 10² g
Step 3: Calculate the mass of NaN₃ needed to form 1.2 × 10² g of N₂
The mass ratio of NaN₃ to N₂ is 130.02:84.03.
1.2 × 10² g N₂ × 130.02 g NaN₃/84.03 g N₂ = 1.9 × 10² g NaN₃
Step 4: Calculate the density of N₂
We will use the following expression.
ρ = P × M / R × T
ρ = 1.3 atm × 28.01 g/mol / (0.0821 atm.L/mol.K) × 300 K = 1.5 g/L
Answer:
1.53x10^22 atoms of Au
Explanation:
To find the atoms of gold we need first, to convert the mass of gold to moles using molar mass of gold (196.97g/mol). Then, these moles must be converted to number of atoms based on definition of moles (1 mole = 6.022x10²³ atoms).
<em>Moles Au:</em>
5.00g Au * (1mol / 196.97g) = 0.0254 moles of Au
<em>Atoms of Au:</em>
0.0254 moles * (6.022x10²³ atoms / 1 mole) =
<h3>1.53x10^22 atoms of Au</h3>
When one observes the phase of matter of water, one observes a physical property of matter.
