Answer:
There are 1.260 moles for every 300.0 grams of Uranium
Explanation:
Answer:
- % Cobalt (II) Nitrate = 30.62%
Explanation:
To calculate mass percent, first we need to <u>calculate the total mass of the mixture</u>:
- Mass Water ⇒ 0.350 kg Water = 350 g water
- Mass Ammonia⇒We use ammonia's molar mass⇒5.4 mol * 17 g/mol = 91.8 g
- Mass cobalt (II) nitrate ⇒ 195.0 g
Total Mass = Mass Water + Mass Ammonia + Mass Cobalt Nitrate
- Total Mass = 350 g+ 91.8 g+ 195 g = 636.8 g
To calculate each component's mass percent, we divide its mass by the total mass and multiply by 100:
- % Water ⇒ 350/636.8 * 100% = 54.96%
- % Ammonia ⇒ 91.8/636.8 * 100% = 0.14%
- % Cobalt (II) Nitrate ⇒ 195/636.8 * 100% = 30.62%
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
