Combustion of hydrogen releases 142 j/g of hydrogen reacted. How many kj of energy are released by the combustion of 16.0 oz of
hydrogen? (1 lb = 16 oz; 1 kg = 2.2 lb)
2 answers:
64.61 kJ of energy is released by the combustion of 16.0 oz of hydrogen
<h3>Further explanation</h3>Delta H reaction (ΔH) is the amount of heat/heat change between the system and its environment
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The reaction of a substance with oxygen is called a combustion reaction
the combustion of hydrogen is
<h3>2H₂ + O₂ → 2H₂O.</h3>This equation shows that 2 Hydrogen molecules react with 1 molecule of oxygen to produce 2 molecules of water
In this combustion reaction will produce large thermal energy because it involves breaking the hydrogen gas bond, so the reaction is exothermic
The combustion of hydrogen releases 142 j / g of hydrogen reacted.
Combustion of 16.0 oz of hydrogen will release heat of:
16 oz = 1 lb = 0.455 kg = 455 g
For each gram Hydrogen releases 142 J, then for 455 g releases:
142 x 455 J = 64610 J = 64.61 kJ
<h3>Learn more</h3>the complete combustion of methane
grams of o2 are required to burn 17.0 gal of c8h18
moles of oxygen are required to produce 2.33 moles of water
Keywords: Combustion of hydrogen, ΔH reaction, water, oxygen, exothermic
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Answer:
<h2>127.57 moles</h2>Explanation:
To find the number of moles in a substance given it's number of entities we use the formula
where n is the number of moles
N is the number of entities
L is the Avogadro's constant which is
6.02 × 10²³ entities
From the question we have
We have the final answer as
<h3>127.57 moles</h3>Hope this helps you
Answer : The value of equilibrium constant for this reaction at 262.0 K is
Explanation :
As we know that,
where,
= standard Gibbs free energy = ?
= standard enthalpy = -45.6 kJ = -45600 J
= standard entropy = -125.7 J/K
T = temperature of reaction = 262.0 K
Now put all the given values in the above formula, we get:
The relation between the equilibrium constant and standard Gibbs free energy is:
where,
= standard Gibbs free energy = -12666.6 J
R = gas constant = 8.314 J/K.mol
T = temperature = 262.0 K
K = equilibrium constant = ?
Now put all the given values in the above formula, we get:
Therefore, the value of equilibrium constant for this reaction at 262.0 K is