How many grams of calcium cyanide (Ca(CN)2) are contained in 0.79 mol of calcium cyanide?
1 answer:
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Answer:
124.91mL
Explanation:
Given parameters:
P₁ = 1.08atm
V₁ = 250mL
T₁ = 24°C
P₂ = 2.25atm
T₂ = 37.2°C
V₂ = ?
Solution:
To solve this problem, we are going to apply the combined gas law;
P, V and T represents pressure, volume and temperature
1 and 2 delineates initial and final states
Convert the temperature to kelvin;
T₁ = 24°C, T₁ = 24 + 273 = 297K
T₂ = 37.2°C , T₂ = 37.2 + 273 = 310.2K
Input the variables and solve for V₂
V₂ = 124.91mL
Check the attached file for the answer.
