Answer:
300000Pa or 3×10^5 Pa
Explanation:
Since the problem involves only two parameters of volume and pressure, the formula for Boyle's law is suitably used.
Using Boyle's law
P1V1 = P2V2
P1 is the initial pressure = 1.5×10^5Pa
V1 is the initial volume = 0.08m3
P2 is the final pressure (required)
V2 is the final volume = 0.04 m3
From the formula, P2 = P1V1/V2
P2 = 1.5×10^5 × 0.08 ÷ 0.04
= 300000Pa or 3×10^5 Pa.
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.
the answer is A. I (iodine)
no worries
Explanation:
The more reactive element replaces less reactive element during chemical reaction.
Since, potassium is more reactive than beryllium. When potassium reacts with beryllium choride, it replaces beryllium and forms potassium chloride and produces beryllium.
