Here we apply the Clausius-Clapeyron equation:
ln(P₁/P₂) = ΔH/R x (1/T₂ - 1/T₁)
The normal vapor pressure is 4.24 kPa (P₁)
The boiling point at this pressure is 293 K (P₂)
The heat of vaporization is 39.9 kJ/mol (ΔH)
We need to find the vapor pressure (P₂) at the given temperature 355.3 K (T₂)
ln(4.24/P₂) = 39.9/0.008314 x (1/355.3 - 1/293)
P₂ = 101.2 kPa
It’s the nerves but I’m not 100% sure if I’m correct
Answer:
3.6 × 10²⁴ atoms of O
Explanation:
Let's consider the molecular formula of silver nitrate: AgNO₃.
We can establish the following relations:
- 1 mol of AgNO₃ has 6.02 × 10²³ molecules of AgNO₃ (Avogadro's number).
- 1 molecule of AgNO₃ has 3 atoms of oxygen.
The atoms of oxygen in 2.0 moles of silver nitrate are:
