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:
Mass of water produced is 22.86 g.
Explanation:
Given data:
Mass of hydrogen = 2.56 g
Mass of oxygen = 20.32 g
Mass of water = ?
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 20.32 g/ 32 g/mol
Number of moles = 0.635 mol
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 2.56 g/ 2 g/mol
Number of moles = 1.28 mol
Now we will compare the moles of water with oxygen and hydrogen.
O₂ : H₂O
1 : 2
0.635 ; 2×0.635 = 1.27
H₂ : H₂O
2 : 2
1.28 : 1.28
The number of moles of water produced by oxygen are less thus it will be limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 1.27 × 18 g/mol
Mass = 22.86 g
Formula : BaI₂. <span>barium iodide</span>
Answer:
ΔU = 25.8 J
Explanation:
The gas absorbs 33.3 J of heat, that is, Q = 33.3 J.
The work (W) of expansion can be calculated using the following expression:
W = -P. ΔV
where,
P is the external pressure
ΔV is the change in volume
W = -1.45 × 10⁴ N . m⁻² × (8.40 × 10⁻⁴ m³ - 3.24 × 10⁻⁴ m³) = -7.48 J
The change in the internal energy (ΔU) is:
ΔU = Q + W
ΔU = 33.3 J + (-7.48 J) = 25.8 J
Answer: A
Aloha is the one that loses a helium