Hey there!:
Given the reaction:
2 C2H2 + 5 O2 → 4 CO2 + 2 H2O
5 moles O2 ------------- 4 moles CO2
3.00 moles O2 ---------- ( moles of CO2 ?? )
moles of CO2 = 3.00 * 4 / 5
moles of CO2 = 12 / 5
moles of CO2 = 2.4 moles
So, molar mass CO2 = 44.01 g/mol
Therefore:
1 mole CO2 -------------- 44.01 g
2.4 moles CO2 ---------- ( mass of CO2 )
mass of CO2 = 2.4 * 44.01 / 1
mass of CO2 = 106 g
Answer A
Hope that helps!
Because the water seeped through the bag (? I don’t know the context correct me if I’m wrong)
He was credited with discovering the subatomic particle also known as the electron in 1897.
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
Elements that exceed the octet rule must have an unoccupied d orbital. I believe.
