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.
Answer:
Option B. 3.0 M
Explanation:
From the question given above, the following data were obtained:
Volume of solution = 3.0 L
Mole of MgCl₂ = 9 moles
Molarity =?
Molarity can simply be defined as the mole of solute per unit litre of the solution. Mathematically, it can be expressed as:
Molarity = mole of solute /Volume of solution
With the above formula, we can obtain the molarity of the solution as follow:
Volume of solution = 3.0 L
Mole of MgCl₂ = 9 moles
Molarity =?
Molarity = mole of solute /Volume of solution
Molarity = 9 / 3
Molarity = 3 mol/L = 3.0 M
Thus, the molarity of the solution is 3 M
Answer:
3.59x10⁻⁴ mol
Explanation:
Assuming ideal behaviour we can solve this problem by using the<em> PV=nRT formula</em>, where:
- R = 8314.46 Pa·L·mol⁻¹·K⁻¹
We<u> input the data given by the problem</u>:
- 205 Pa * 5.68 L = n * 8314.46 Pa·L·mol⁻¹·K⁻¹ * 390.4 K
And <u>solve for n</u>:
