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 molarity of the stock solution is 1.25 M.
<u>Explanation:</u>
We have to find the molarity of the stock solution using the law of volumetric analysis as,
V1M1 = V2M2
V1 = 150 ml
M1 = 0.5 M
V2 = 60 ml
M2 = ?
The above equation can be rearranged to get M2 as,
M2 = 
Plugin the values as,
M2 = 
= 1.25 M
So the molarity of the stock solution is 1.25 M.
Answer:
P₂ = 28.5 torr
Explanation:
Given data:
Initial pressure = 38 torr
Initial volume = 500 L
Final volume = 677 L
Final pressure = ?
Solution:
P₁V₁ = P₂V₂
P₁ = Initial pressure
V₁ = Initial volume
P₂ = Final pressure
V₂ = Final volume
Now we will put the vales in formula.
P₁V₁ = P₂V₂
P₂ = P₁V₁ /V₂
P₂ = 38 torr × 500 L / 667 L
P₂ = 19000 torr. L / 667 L
P₂ = 28.5 torr
Answer:
Explanation:
An electron in 4s is farther away from nucleus and it has higher energy when compared to electron from 1s.
Answer:
Suppose you added some solid NaCl to a saturated solution of NaCl at 20℃ and warmed the mixture to 40℃. What would happen to the added NaCl?
Explanation:
can you help with this one
