Answer:
The amount of heat released by the combustion of 2 moles of methane = 1605.1 kJ
Option 3 is correct.
ΔH°(combustion of 2 moles of methane)
= -1,605.1 kJ
Explanation:
The balanced equation for the combustion of methane is presented as
CH₄ + 2O₂ → CO₂ + 2H₂O
The heat of formation for the reactants and products are
CH₄ (g): ΔHf = –74.6 kJ/mol;
CO₂ (g): ΔHf = –393.5 kJ/mol;
and H₂O(g): ΔHf = –241.82 kJ/mol.
ΔHf for O₂ = 0 kJ/mol
ΔH°rxn for the combustion of methane is given as
ΔH°rxn = ΣnH°(products) - ΣnH°(reactants)
ΣnH°(products) = (1×-393.5) + (2×-241.82)
= -877.14 kJ/mol
ΣnH°(reactants) = (1×-74.6) + (2×0)
= -74.6 kJ/mol
ΔH°rxn = -877.14 - (-74.6) = -802.54 kJ/mol
For 2 moles of methane, the heat of combustion = 2 moles × -802.54 kJ/mol
= -1,605.08 kJ = -1,605.1 kJ
Hope this Helps!!!
A summary of the Law of multiple proportions is that if A and B form more than one compound, and B1 is the amount of element B which reacts with a fixed mass of A in compound 1, and B2 is the amount of B which reacts with the same fixed mass of B to form compound 2, then the ratio B1:B2 will be small whole numbers.
This law is rather simplistic, and given the range of compounds known today the definition of 'small' is now rather large... but, to answer the question:
in compound one 1.14133g of B reacts with 1g of A. (1.14133=53.3/46.7)
Answer:
P.E = 25.48 J
Explanation:
Given data:
Mass = 2 Kg
Height = 1.3 m
Potential energy = ?
Solution:
Formula:
P.E = m . g . h
P. E = potential energy
m = mass in kilogram
g = acceleration due to gravity
h = height
Now we will put the values in formula.
P.E = m . g . h
P.E = 2 Kg . 9.8 m /s² . 1.3 m
P.E = 25.48 Kg. m² / s²
Kg. m² / s² = J
P.E = 25.48 J
