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
I will try my best to sovle this question...
Im Pretty sure it has to be A. it needs to balance it would not stand in place if not balanced.
Halogens (atoms with 7 valence electrons) and Hydrogen
or generally, atoms with their shells almost full
Molar mass ( CuSO₄) = 159.609 g/mol
159.609 g ----------------- 6.02 x 10²³ molecules
? g ------------------ 3.36 x 10²³ molecules
mass = ( 3.36 x10²³) x 159.609 / 6.02 x 10²³
mass = 5.36 x 10²⁴ / 6.02 x 10²³
mass = 8.90 g
hope this helps!
