It can allow the molecule (like water) to be polar because it has a negative and positive side to it (oxygen holds the negatives tight causing the hydrogens to be positive).
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.
Plants require pH to thrive which in turn gives us food.
There will be 7.5 g of Be-11 remaining after 28 s.
If 14 s = 1 half-life, 28 s = 2 half-lives.
After the first half-life, ½ of the Be-11 (15 g) will disappear, and 15 g will remain.
After the second half-life, ½ of the 15 g (7.5 g) will disappear, and 7.5 g will remain.
In symbols,
<em>N</em> = <em>N</em>₀(½)^<em>n</em>
where
<em>n</em> = the number of half-lives
<em>N</em>₀ = the original amount
<em>N</em> = the amount remaining after <em>n</em> half-lives
