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: [N2]₀ = 10M and [H2]₀ = 11M
Explanation: To calculate the initial concentration, you would have to set up an ICE table, which is an organized way of tracking known quantities or the ones you want to find. ICE stands for:
I is initial amount;
C is change in concentration;
E is for equilibrium concentration;
For the mixture,
N2 3H2 2NH3
I [N2]₀ [H2]₀ 0
C - x -3x +2x
E [N2]₀ - x =8 [H2]₀ - 3x =5 2x =4
With the product, we can find "x":
2x=4
x=2M
With x=2, find the concentrations:
[N2]₀ - x = 8
[N2]₀ = 10M
[H2]₀ - 3x = 5
[H2]₀ = 11M
The initial concentrations of nitrogen gas [N2] is 10.0 M and of hydrogen gas [H2] is 11.0 M.
The answer is "elements" :)
Answer:
0.00370 g
Explanation:
From the given information:
To determine the amount of acid remaining using the formula:
where;
v_1 = volume of organic solvent = 20-mL
n = numbers of extractions = 4
v_2 = actual volume of water = 100-mL
k_d = distribution coefficient = 10
∴
Thus, the final amount of acid left in the water = 0.012345 * 0.30
= 0.00370 g
