<span>Answer:
1/4 is the average bond order for a pâ’o bond (such as the one shown in blue) in a phosphate ion.</span>
(p1)(V1)/(T1) = (p2)(V2)/(T2)
(1.00 atm)(V) / (273 + 25K) = (40.0 atm)(V/10) / (273 + T)
273 + T = (40.0)(1/10)(273 + 25K) / (1.00)
T = 919°C
Answer:
V₂ = 1.92 L
Explanation:
Given data:
Initial volume = 0.500 L
Initial pressure =2911 mmHg (2911/760 = 3.83 atm)
Initial temperature = 0 °C (0 +273 = 273 K)
Final temperature = 273 K
Final volume = ?
Final pressure = 1 atm
Solution:
Formula:
P₁V₁/T₁ = P₂V₂/T₂
P₁ = Initial pressure
V₁ = Initial volume
T₁ = Initial temperature
P₂ = Final pressure
V₂ = Final volume
T₂ = Final temperature
by putting values,
V₂ = P₁V₁ T₂/ T₁ P₂
V₂ = 3.83 atm × 0.500 L × 273 K / 273 K × 1 atm
V₂ = 522.795 atm .L. K / 273 K.atm
V₂ = 1.92 L
