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: Option (c) is the correct answer.
Explanation:
When an acid or base is added to a solution then any resistance by the solution in changing the pH of the solution is known as a buffer.
This is because a buffer has the ability to not get affected by the addition of small amounts of an acid or a base. As a result, it helps in maintaining the pH of the solution.
In the give case, when we add the HCl then more number of protons will dissociate. This causes the acetate to react with the protons and leads to the formation of acetic acid.
We know that acetic acid is a weak acid and it dissociates partially or feebly. Therefore, no change in pH will take place.
Thus, we can conclude that equation
represents the chemical reaction that accounts for the fact that acid was added but there was no detectable change in pH.
Answer:
47.3 ml
Explanation:
The graduated cylinder is shown in the image attached.
Now we have to take a good look at the cylinder, the lines between 45 and 50 are 46, 47, 48 and 49. Even though the points in between two lines weren't graduated but we can intelligently guess the correct volume by observing the upper meniscus of the liquid. Hence the answer.