Answer:
1.01 atm
Step-by-step explanation:
To solve this problem, we can use the <em>Combined Gas Laws</em>:
p₁V₁/T₁ = p₂V₂/T₂ Multiply each side by T₂
p₁V₁T₂ = p₂V₂ Divide each side by V₂
p₂ = p₁ × V₁/V₂ × T₂/T₁
Data:
p₁ = 1.05 atm; V₁ = 285 mL; T₁ = 15.8 °C
p₂ = ?; V₂ = 292 mL; T₂ = 11.2 °C
Calculations:
(a) Convert <em>temperatures to kelvins
</em>
T₁ = (15.8+273.15) K = 288.95 K
T₂ = (11.2+273.15) K = 284.35 K
(b) Calculate the <em>pressure
</em>
p₂ = 1.05 atm× (285/292) × (284.35/288.95)
= 1.05 atm × 0.9760 × 0.9840
= 1.01 atm
Answer:
Explanation:
Step 1: Data given
The equilibrium constant, Kc, for the following reaction is 4.76 * 10^-4 at 431 K
The equilibrium concentration of Cl2(g) is 0.233 M
Step 2: The balanced equation
PCl5(g) ⇄ PCl3(g) + Cl2(g)
Step 3: The initial concentration
[PCl5]= Y M
[PCl] = 0M
[Cl2] = 0M
Step 4: Calculate the concentration at equilibrium
[PCl5] = Y + X M = Y - 0.233 M
[PCl]= XM = 0.233 M
[Cl2]= XM = 0.233 M
Step 5: Define Kc
Kc = [Cl2]* [PCl3] / [PCl5]
4.76 * 10^-4 = 0.233² / (Y -0.233)
0.000476 = 0.05429 / (Y - 0.233)
Y - 0.233 = 0.05429 / 0.000476
Y - 0.233 = 114.05 M
Y = 114.283 M = the initial concentration
The concentration of PCl5 at the equilibrium is 114.05 M
Answer:
Group 1: LiOH, NaOH, KOH
Group 2: Ca(OH)2, Sr(OH)2, Ba(OH)2
Explanation:
Ancient cyanobacteria released hydrogen which assisted in creating the atomsphere as we know it today.