Answer:
To increase the yield of H₂ we would use a low temperature.
For an exothermic reaction such as this, decreasing temperature increases the value of K and the amount of products at equilibrium. Low temperature increases the value of K and the amount of products at equilibrium.
Explanation:
Let´s consider the following reaction:
CO(g) + H₂O(g) ⇌ CO₂(g) + H₂(g)
When a system at equilibrium is disturbed, the response of the system is explained by Le Chatelier's Principle: <em>If a system at equilibrium suffers a perturbation (in temperature, pressure, concentration), the system will shift its equilibrium position to counteract such perturbation</em>.
In this case, we have an exothermic reaction (ΔH° < 0). We can imagine heat as one of the products. If we decrease the temperature, the system will try to raise it favoring the forward reaction to release heat and, at the same time, increasing the yield of H₂. By having more products, the value of the equilibrium constant K increases.
Answer:
Second Law
Explanation:
Newton's second law states that the acceleration caused in a body is directly proportional to the force applied and inversely proportion to the mass of the body.
This is given by :

In this case the suggestion given to reduce the aircraft's cargo load is the right move as reducing the load on the aircraft will decrease the mass of the whole aircraft. This in turn will help the aircraft to accelerate more as acceleration inversely varies with mass. Thus the aircraft will be able to reach its flying speed even on a short run way.
Hence, Newton's second law is applied.
Answer:
The correct answer is (b)
Explanation:
Charles law describes the behavior of gases when heated. Charles law states that the volume of a given mass of gas would increase as its Kelvin temperature increases provided the pressure is held constant. That is the volume of a given mass of gas is directly proportional to its Kelvin temperature at constant pressure
Answer:
See explanation and image attached
Explanation:
The standard cell potential at 298 K is given by;
E°cathode - E°anode
Hence;
E°cell = 0.34 V - (-0.76 V)
E°cell = 0.34 V + 0.76 V
E°cell = 1.1 V
To reduce Zn^2+ to Zn then Zn must be the cathode, hence;
E°cell = (-0.76 V) - 0.34 V
E°cell = -1.1 V