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
As you can see in the picture we have +ΔH so that means for this reaction we need to GET heat. so the answer is A. endothermic :))
i hope this is helpful
have a nice day
Answer: 41.5 mL
Explanation:
Molarity of a solution is defined as the number of moles of solute dissolved per liter of the solution.
where,
n = moles of solute
= volume of solution in L
Given : 59.4 g of 
in 100 g of solution
moles of 
Volume of solution =
Now put all the given values in the formula of molality, we get
To calculate the volume of acid, we use the equation given by neutralisation reaction:
where,
are the molarity and volume of stock acid which is
are the molarity and volume of dilute acid which is
We are given:
Putting values in above equation, we get:
Thus 41.5 mL of the solution would be required to prepare 1550 mL of a .30M solution of the acid
Answer:
b. One electron state is an anti-bonding orbital, which results in an absence of electron density between atoms.
Explanation:
