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:
A) 14. 25 × 10²³ Carbon atoms
B) 34.72 grams
Explanation:
1 molecule of Propane has 3 atoms of Carbon and 8 atoms of Hydrogen.
The sample has 3.84 × 10²⁴ H atoms.
If 8 atoms of Hydrogrn are present in 1 molecule of propane.
3.84 × 10²⁴ H atoms are present in
<u>= 4.75 × 10²³ molecules of Propane</u>.
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No. of Carbon atoms in 1 molecule of propane = 3
=> C atoms in 4.75× 10²³ molecules of Propane = 3 × 4.75 × 10²³
<u>= 14.25 × 10²³ </u>
<u>________________________________________</u>
<u>Gram</u><u> </u><u>Molecular</u><u> </u><u>Mass</u><u> </u><u>of</u><u> </u><u>Propane</u><u>(</u><u>C3H8</u><u>)</u>
= 3 × 12 + 8 × 1
= 36 + 8
= 44 g
1 mole of propane weighs 44g and has 6.02× 10²³ molecules of Propane.
=> 6.02 × 10²³ molecules of Propane weigh = 44 g
=> 4. 75 × 10²³ molecules of Propane weigh =
<u>= 34.72 g</u>
<u>Answer:</u> The equilibrium concentration of bromine gas is 0.00135 M
<u>Explanation:</u>
We are given:
Initial concentration of chlorine gas = 0.0300 M
Initial concentration of bromine monochlorine = 0.0200 M
For the given chemical equation:
<u>Initial:</u> 0.02 0.03
<u>At eqllm:</u> 0.02-2x x 0.03+x
The expression of for above equation follows:
We are given:
Putting values in above equation, we get:
Neglecting the value of x = -0.96 because, concentration cannot be negative
So, equilibrium concentration of bromine gas = x = 0.00135 M
Hence, the equilibrium concentration of bromine gas is 0.00135 M