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.
Because ionic compounds' strong bonds form network structures, which have a stronger attraction than the covalent compounds which are molecules.
PV = nRT. Where P = pressure, V = volume, n = number of moles, R = universal gas constant and T = temperature. Hope this helps!
Answer:
1.07 g
Explanation:
Half-life of Pu-234 = 4.98 hours
Initially present = 45 g
mass remains after 27 hours = ?
Solution:
Formula
mass remains = 1/ 2ⁿ (original mass) ……… (1)
Where “n” is the number of half lives
To find "n" for 27 hours
n = time passed / half-life . . . . . . . .(2)
put values in equation 2
n = 27 hr / 4.98 hr
n = 5.4
Mass after 27 hr
Put values in equation 1
mass remains = 1/ 2ⁿ (original mass)
mass remains = 1/ 2^5.4 (45 g)
mass remains = 1/ 42.2 (45 g)
mass remains = 0.0237 x 45 g
mass remains = 1.07 g
