Answer:
A. P₂ / P₁ = 2
B. P₂ / P₁ = 1.1
Explanation:
A. Determination of the ratio P₂/P₁
Volume = constant
Initial temperature (T₁) = 46 K
Final temperature (T₂) = 92 K
Final pressure /Initial pressure (P₂/P₁) =?
P₁/T₁ = P₂/T₂
P₁/46 = P₂/92
Cross multiply
46 × P₂ = P₁ × 92
Divide both side by P₁
46 × P₂ / P₁ = 92
Divide both side by 46
P₂ / P₁ = 92 / 46
P₂ / P₁ = 2
B. Determination of the ratio P₂/P₁
Volume = constant
Initial temperature (T₁) = 35.4 °C = 35.4 + 273 = 308.4 K
Final temperature (T₂) = 69.0 °C = 69 + 273 = 342 K
Final pressure /Initial pressure (P₂/P₁) =?
P₁/T₁ = P₂/T₂
P₁/308.4 = P₂/342
Cross multiply
308.4 × P₂ = P₁ × 342
Divide both side by P₁
308.4 × P₂ / P₁ = 342
Divide both side by 308.4
P₂ / P₁ = 342 / 308.4
P₂ / P₁ = 1.1
Your answer would be D friction.
Answer:
Explanation:
Hello,
In this case, we consider oxygen as an ideal gas, for that reason, we use yhe ideal gas equation to compute the moles based on:
Hence, at 3.50 atm and 25 °C for a volume of 2.00 L we compute the moles considering absolute temperature in Kelvins:
Best regards.
Answer:
101.63° C
Explanation:
Volume expansivity γa = γr - γ g = 18 × 10⁻⁵ - 2.0 × 10⁻⁵ = 16 × 10⁻⁵ /K
v₂ - v₁ / v₁θ = 16 × 10⁻⁵ /K
(500 - 492 ) mL / (492 × 16 × 10⁻⁵) = θ
θ = 101.63° C
PbCl₂(aq) → Pb²⁺(aq) + 2Cl⁻(aq)
NaCl(aq) → Na⁺(aq) + Cl⁻(aq)
Pb²⁺(aq) + 2Cl⁻(aq) ⇄ PbCl₂(s)
At increase the concentration of chloride ions - concentration of lead ions decreases, the lead chloride is formed.
