The combustion of isooctane (C₈H₁₈) is written below:
2 C₈H₁₈ (l) + 25 O₂ (g) → 16 CO₂ (g) + 18 H₂O (l)
The formula for heat of combustion is:
ΔHc = (∑Stoichiometric coefficient×ΔHf of products) - (∑Stoichiometric coefficient×ΔHf of reactants), where ΔHf is heat of formation.
ΔHf of isooctane = -259.2 kJ/mol
ΔHf of O₂ = 0 kJ/mol
ΔHf of CO₂ = -393.5 kJ/mol
ΔHf of H₂O = <span>-285.8 kJ/mol
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ΔHc = [(16 mol×-393.5 kJ/mol )+(18 mol×-285.8 kJ/mol)] - [(2 mol×-259.2 kJ/mol) + (25 mol*0 kJ/mol)]
ΔHc = -10,922 kJ
Answer:
Add is wires and lights.
Explanation:
Gas laws include the wire lines the produce temeperature and the lights system.
PV=nRT
n=2.4 moles
T=273.15+50=323.15K
P=2*101325=202650 Pa
R=8.31
Solve for V:
V=nRT/P=2.4*8.31*323.15/202650=.032m^3
Answer:
9.1
Explanation:
Step 1: Calculate the basic dissociation constant of propionate ion (Kb)
Sodium propionate is a strong electrolyte that dissociates according to the following equation.
NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻
Propionate is the conjugate base of propionic acid according to the following equation.
C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻
We can calculate Kb for propionate using the following expression.
Ka × Kb = Kw
Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰
Step 2: Calculate the concentration of OH⁻
The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.
[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M
Step 3: Calculate the concentration of H⁺
We will use the following expression.
Kw = [H⁺] × [OH⁻]
[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M
Step 4: Calculate the pH of the solution
We will use the definition of pH.
pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1
Answer:
A. Inheriting a specific gene
Explanation:
I hope this helps! ^^
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