Answer:
P₂ = 0.09 atm
Explanation:
According to general gas equation:
P₁V₁/T₁ = P₂V₂/T₂
Given data:
Initial volume = 0.225 L
Initial pressure = 338 mmHg (338/760 =0.445 atm)
Initial temperature = 72 °C (72 +273 = 345 K)
Final temperature = -15°C (-15+273 = 258 K)
Final volume = 1.50 L
Final pressure = ?
Solution:
P₁V₁/T₁ = P₂V₂/T₂
P₂ = P₁V₁ T₂/ T₁ V₂
P₂ = 0.445 atm × 0.225 L × 258 K / 345 K × 1.50 L
P₂ = 25.83 atm .L. K / 293 K . L
P₂ = 0.09 atm
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:
According to Le Chatelier's principle, increasing the reaction temperature of an exothermic reaction causes a shift to the left and decreasing the reaction temperature causes a shift to the right.
Explanation:
C6H12O6(s) + 6O2(g) ⇌6CO2(g) + 6H2O(g)
We are told that the forward reaction is exothermic, meaning heat is removed from the reacting substance to the surroundings.
According to Le Chatelier's principle,
1. for an exothermic reaction, on increasing the temperature, there is a shift in equilibrium to the left and formation of the product is favoured.
2. if the temperature of the system is decreased, the equilibrium shifts to right and the formation of the reactants is favoured.
3. if the reaction temperature is kept constant, the system is at equilibrium and there is no shift to the right nor to the left.
Answer:
D.Lowering the temperature is the best option.
Explanation:
The value of equilibrium constants aren't changed with change in the pressure or concentrations of reactants and products in equilibrium. The only thing that changes the value of equilibrium constant is a change of temperature.
In the reaction below for example;
A + B <==>C+D
If you have moved the position of the equilibrium to the right (and so increased the amount of C and D), why hasn't the equilibrium constant increased?
Let's assume that the equilibrium constant mustn't change if you decrease the concentration of C - because equilibrium constants are constant at constant temperature. Why does the position of equilibrium move as it does?
If you decrease the concentration or pressure of C, the top of the Kc expression gets smaller. That would change the value of Kc. In order for that not to happen, the concentrations of C and D will have to increase again, and those of A and B must decrease. That happens until a new balance is reached when the value of the equilibrium constant expression reverts to what it was before.
