Answer is: volume of CO₂ is 0,113 dm³.
Ideal gas law = pV = nRT.
p = 850 PSI = 5860543,6992 Pa.
Psi <span>is the abbreviation of pound per square inch.
T = 21</span>°C = 294,15 K.
n = 0,273 mol.
R = 8,314 J/K·mol.
V = nRT ÷ p
V = 0,273 mol · 8,314 J/K·mol · 294,15 K ÷ 5860543,6992 Pa.
V = 0,00011 m³ = 0,113 dm³.
Answer is: pH of barium hydroxide is 13.935.
Chemical dissociation of barium hydroxide in water:
Ba(OH)₂(aq) → Ba²⁺(aq) + 2OH⁻(aq).
c(Ba(OH)₂) = 0.43 M.
V(Ba(OH)₂) = 100 mL ÷ 1000 mL/L = 0.1 L.
n(Ba(OH)₂) = 0.43 mol/L · 0.1 L.
n(Ba(OH)₂) = 0.043 mol.
From chemical reaction: n(Ba(OH)₂) : n(OH⁻) = 1 : 2.
n(OH⁻) = 0.086 mol.
c(OH⁻) = 0.86 mol/L.
pOH = -logc(OH⁻).
pOH = 0.065.
pH = 14 - 0.065 = 13.935.
1. 1 M , 2 M , 1 M
2. 10 mol , 0.1 mol , 0.5 mol
3. 0.5 L , 6.6 L , 5/21 L
M = mol/L
The answers are in the attached file
Answer:
1. C4H8 + 6O2 -----> 4CO2 + 4H20
2. 3836.77 kcal
Explanation:
1. Balanced equation for the complete combustion of cyclobutane:
C4H8 + 6O2 -----> 4CO2 + 4H20
2. Heat of combustion of cyclobutane = 650.3 kcal/mol
Molecular weight of cyclobutane, C4H8 = 56.1 g/mol
Mole of C4H8 : mass of cyclobutane/Molecular weight of cyclobutane
Mole of C4H8 = 331/56.1 = 5.9 mol
Energy released during combustion = 5.9 mol × 650.3 kcal/mol = 3836.77kcal
Therefore the energythat is released during the complete combustion of 331 grams of cyclobutane is 3836.77kcal
