Answer:
1 gramo de metano aporta 50.125 kilojoules.
1 gramo de metano aporta 48.246 kilojoules.
Explanation:
La cantidad de energía liberada por la combustión de una unidad de masa del hidrocarburo (
), en kilojoules por mol, es igual a la cantidad de energía liberada por mol de compuesto (
), en kilojoules por mol, dividido por su masa molar (
), en gramos por mol:
(1)
A continuación, analizamos cada caso:
Metano
1 gramo de metano aporta 50.125 kilojoules.
Octano
1 gramo de metano aporta 48.246 kilojoules.
Increasing the concentration of one or more reactants will often increase the rate of reaction. This occurs because a higher concentration of a reactant will lead to more collisions of that reactant in a specific time period.
Reaction rate increases with concentration, as described by the rate law and explained by collision theory. As reactant concentration increases, the frequency of collision increases. The rate of gaseous reactions increases with pressure, which is, in fact, equivalent to an increase in concentration of the gas.
Answer:
Explanation:
<u>1) Convert the mass of water into number of moles</u>
- Molar mass of water: 18.015 g/mol
- Number of moles, n = mass in grams / molar mass
n = 255 g / 18.015 g/mol = 14.15 mol
<u>2) Use the formula E = n × ΔH vap</u>
This is, you have to multiply the molar ΔH vaporization by the number of moles to find the total energy to boil the given amount of water.
- E = 14.15 mol × 40,650 J/mol = 575,395.5 J
<u>3) Round to the correct number of significant figures.</u>
The mass of water is the measurement with the least number of significant figures (3), so you must report the answer with 3 significant figures,
