Answer:
At equilibrium, the concentration of the reactants will be greater than the concentration of the products. This does not depend on the initial concentrations of the reactants and products.
Explanation:
The value of Kc gives us an idea of the extent of the reaction. A big Kc (Kc > 1) means that in the equilibrium there are more products than reactants, and the opposite happens for a small Kc (Kc < 1). The equilibrium is reached no matter what the initial concentrations are.
The value of the equilibrium constant is relatively SMALL; therefore, the concentration of reactants will be GREATER THAN the concentration of products. This result is INDEPENDENT OF the initial concentration of the reactants and products.
Option A: Clouds
In the morning, air is cool and as sun begins to rise it starts increasing the temperature of air. By time, the air becomes warmer and warmer. Depending upon the surrounding conditions, air in different areas heat up at different rates.
Due to this heating, thermal formation takes place, this is due to uneven heating of surface of earth. The thermal formation at surface causes difference in temperature of surface of the earth and air around it. The warm air has tendency to rise thus, the air in the thermal rise and expand. Due to expansion it cools down, this process continues till the temperature of thermal air reaches equals to the temperature of surrounding air. This results in the formation of cloud.
Thus, when a humid air mass rises into a cooler temperature area, clouds formation takes place
Answer:
<h3>I don't know what is the answer of your question sorry never mind..</h3>
Explanation:
<h3>And please marks me as brainliest... </h3>
T is amount after time t
<span>Ao is initial amount </span>
<span>t is time </span>
<span>HL is half life </span>
<span>log (At) = log [ Ao x (1/2)^(t/HL) ] </span>
<span>log (At) = log Ao + log (1/2)^(t/HL) </span>
<span>log (At) = log Ao + (t/HL) x log (1/2) </span>
<span>( log At - log Ao) / log (1/2) = t / HL </span>
<span>log (At/Ao) / log (1/2) = t / HL </span>
<span>HL = t / [( log (At / Ao)) / log (1/2) ] </span>
<span>HL = 14.4 s / [ ( log (12.5 / 50) / log (1/2) ] </span>
<span>HL = 14.4 s / 2 = 7.2 seconds </span>
