Answer:
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
Explanation:
Step 1: Data given
The combustion reaction of octane produces 5104.1 kJ per mol octane
Step 2: The balanced equation
C8H18(g) + 12.5 O2 ⟶ 8CO2 (g) + 9 H2O (g) ∆H°rxn = -5104.1 kJ/mol
Step 3:
∆H°rxn = ∆H°f of products minus the ∆H° of reactants
∆H°rxn = ∆H°f products - [∆H°f reactants]
-5104.1 kJ/mol = (8*∆H°fCO2 + 9*∆H°fH20) - (∆H°fC8H18 + 12.5∆H°fO2)
∆H°f C8H18 = ∆H°f 8CO2 + ∆H°f 9H2O+ 5104.1 kJ/mol
∆H°f C8H18 = 8 * (-393.5 kJ)/mol + 9 * (-241.8 kJ/mol)] + 5104.1 kJ
/mol
∆H°f C8H18 = -220.1 kJ/mol
The standard enthalpy of formation of this isomer of octane is -220.1 kJ/mol
126 grams of H2O is formed.
Explanation:
Data given:
volume of the gas = 88 Liters
pressure = 720 mm Hg or 0.947 atm
temperature T = 22 Degrees or 295.15 K
R = 0.08021 atm L/mole K
n =?
The formula is used is of ideal gas law to know the number of moles of CH4 undergoing combustion.
PV = nRT
n =
putting the values in the equation
= 0.947 X 88/ 0.08021 X 295.15
n = 3.5 moles
balanced reaction for combustion of methane
CH4 + O2 ⇒ CO2 + 2H20
1 mole of CH4 undergoes combustion to form 2 moles of water
3.5 moles will give x moles of water
2/1 = x/3.5
x = 7 moles of water (atomic mass of water = 18 gram/mole)
mass = atomic mass x number of moles
mass = 18 x 7
=126 grams of water is formed.
Answer:
The answer to the question is
<h2>
A. Air, Soil, Sand, Water</h2>
Explanation:
Answer:
Transparency to visible light and microwaves. Very good resistance to ageing, wear and heat. Lightweight, impact and shatter resistant. Good gas and moisture barrier properties
Explanation:
Answer:
True
Explanation:
I used my notes from class today. I could be wrong.