Answer:
1.57 mol NaN₃
Explanation:
- 2 NaN₃ (s) → 2 Na (s) + 3 N₂ (g)
First we <u>use PV=nRT to calculate the number of N₂ moles that need to be produced</u>:
- R = 0.082 atm·L·mol⁻¹·K⁻¹
- T = 23.7 °C ⇒ 23.7 + 273.16 = 296.86 K
<u>Inputing the data</u>:
- 1.07 atm * 53.4 L = n * 0.082 atm·L·mol⁻¹·K⁻¹ * 296.86
And <u>solving for n</u>:
Finally we <u>convert N₂ moles into NaN₃ moles</u>, using <em>the stoichiometric coefficients of the balanced reaction</em>:
- 2.35 mol N₂ *
= 1.57 mol NaN₃
The specific heat capacity of Titanium is found as 0.522 J / g°C.
<u>Explanation:</u>
We have to find the specific capacity using the formula as,
q = m × c × ΔT
Where,
q is the heat absorbed by the sample = 476 J
m is the mass of the sample = 43.56 g
c is the specific heat capacity of the sample = ?
ΔT is the temperature difference = 41.06° C - 20.13°C = 20.93° C
Now, we have to rewrite the equation to get the specific heat capacity as,
c = 
= 
= 0.522 J / g°C
So the specific heat capacity of Titanium is found as 0.522 J / g°C.
As,
w/w % = weight of Solute / Weight of Solution × 100 --- (1)
Finding weight of Solute:
As,
1 mole of NaOH = 40 g
So,
18 moles = X g
Solving for X,
X = (18 mol × 40 g) ÷ 1 mol
X = 720 g
Finding mass of Solution:
Multiply density of the solution by 100 mL
So,
= 1.48 g/mL × 1000 mL
= 1480 g
Putting values in eq. 1,
w/w % = 720 g / 1480 × 100
w/w % = 48.64 %
