Answer:
hello your question lacks the required reaction pairs below are the missing pairs
Reaction system 1 :
A + B ⇒ D ![-r_{1A} = 10exp[-8000K/T]C_{A}C_{B}](https://tex.z-dn.net/?f=-r_%7B1A%7D%20%20%3D%2010exp%5B-8000K%2FT%5DC_%7BA%7DC_%7BB%7D)
A + B ⇒ U 
Reaction system 2
A + B ⇒ D 
B + D ⇒ U 
Answer : reaction 1 : description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1
condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2
reaction 2 :
description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1
condition to maximize selectivity:
to increase selectivity the concentration of D should be minimal
Explanation:
Reaction system 1 :
A + B ⇒ D ![-r_{1A} = 10exp[-8000K/T]C_{A}C_{B}](https://tex.z-dn.net/?f=-r_%7B1A%7D%20%20%3D%2010exp%5B-8000K%2FT%5DC_%7BA%7DC_%7BB%7D)
A + B ⇒ U 
the selectivity of D is represented using the relationship below
hence SDu = 1/10 * 
description of the reactor system : The desired reaction which is the first reaction possess a higher activation energy and higher temperature is required to kickstart reaction 1
condition to maximize selectivity : To maximize selectivity the concentration of reaction 1 should be higher than that of reaction 2
Reaction system 2
A + B ⇒ D 
B + D ⇒ U 
selectivity of D

hence Sdu = 
description of reactor system : The desired reaction i.e. reaction 1 has a lower activation energy and lower temperatures is required to kickstart reaction 1
condition to maximize selectivity:
to increase selectivity the concentration of D should be minimal
Answer:
C
Explanation:
This is a correct answer i think soo
Iodic acid partially dissociates into H+ and IO3-
Assuming that x is the concentration of H+ at equilibrium, and sine the equation says the same amount of IO3- will be released as that of H+, its concentration is also X. The formation of H+ and IO3- results from the loss of HIO3 so its concentration at equilibrium is 0.20 M - x
Ka = [H+] [IO3-] / [HIO3];
<span>Initially, [H+] ≈ [IO3-] = 0 and [HIO3] = 0.20; </span>
<span>At equilibrium [H+] ≈ [IO3-] = x and [HIO3] = 0.20 - x; </span>
<span>so 0.17 = x² / (0.20 - x); </span>
<span>Solving for x using the quadratic formula: </span>
<span>x = [H+] = 0.063 M or pH = - log [H+] = 1.2.</span>