The activation energy of a reaction is the minimum energy that must be overcome in order for the reaction to take place. One way of reaching the activation energy is by manipulating the process conditions like pressure or temperature. But the most common method is by adding an enzyme. An enzyme speeds up the rate of the reaction but does not actively take part in it.
An analogy would be pushing heavy wooden block down a slope. No matter how many people push on it, the block won't move because of friction. But if you spill oil on the floor, the block would effortlessly move down the slope. The oil here is like an enzyme in a reaction.
D only..
A is pure compound
B is mixture of compound and element
C is mixture of compounds
D is mixture of elements
Answer:
ΔS° = -268.13 J/K
Explanation:
Let's consider the following balanced equation.
3 NO₂(g) + H₂O(l) → 2 HNO₃(l) + NO(g)
We can calculate the standard entropy change of a reaction (ΔS°) using the following expression:
ΔS° = ∑np.Sp° - ∑nr.Sr°
where,
ni are the moles of reactants and products
Si are the standard molar entropies of reactants and products
ΔS° = [2 mol × S°(HNO₃(l)) + 1 mol × S°(NO(g))] - [3 mol × S°(NO₂(g)) + 1 mol × S°(H₂O(l))]
ΔS° = [2 mol × 155.6 J/K.mol + 1 mol × 210.76 J/K.mol] - [3 mol × 240.06 J/K.mol + 1 mol × 69.91 J/k.mol]
ΔS° = -268.13 J/K
Despite its appearance, air has a ‘thickness’ so when the sun is high in the sky the light travels through the air on a very much shorter path than when it is low on the horizon.
Imagine that air water and you are below the surface, the light from an overhead sun will be quite sharp and bright, but if lower in the sky it will have to travel through much more water to reach you, so will look less bright and sharp. It ma not seem the same, but the atmosphere is just like very thin water, and a low lying sun will be drastically reduced in strength, so all you will see is a sun with a shift to the red end of the spectrum as all the actinic part will be filtered away by that thicker atmosphere.
If you could explain what “merkels” are, the question would be more clear.
