Answer:
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Answer:
Option (1) Br– is the catalyst, and the reaction follows a faster pathway with Br– than without
Explanation:
Let us consider the equation below:
Step 1:
H2O2(aq) + Br–(aq) → H2O(l) + BrO–(aq)
Step 2:
BrO–(aq) + H2O2(aq) → H2O(l) + O2(g) + Br–(aq)
From the above equation, we can see that Br– is unchanged.
This implies that Br– is the catalyst as catalyst does not take part in a chemical reaction but they create an alternate pathway to lower the activation energy in order for the reaction to proceed at a much faster rate to arrive at the products.
Answer:
The products are: KCl03 and H20.
Explanation:
The reaction between HC03 (chloric acid) and KOH (potassium hydroxide) is:
HC03 + KOH ----> KCl03 (KCl03 and H20) + H20 (water)
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<em>The reaction is of the double displacement type (in this case parts of the reagents are exchanged, producing two generating new compounds).</em>
Answer : The equilibrium concentration of 
in the solution is, 
Explanation :
The dissociation of acid reaction is:
Initial conc. c 0 0
At eqm. c-x x x
Given:
c = 
The expression of dissociation constant of acid is:
![K_a=\frac{[H_3O^+][C_6H_5COO^-]}{[C_6H_5COOH]}](https://tex.z-dn.net/?f=K_a%3D%5Cfrac%7B%5BH_3O%5E%2B%5D%5BC_6H_5COO%5E-%5D%7D%7B%5BC_6H_5COOH%5D%7D)
Now put all the given values in this expression, we get:
![6.3\times 10^{-5}=\frac{(x)\times (x)}{[(7.0\times 10^{-2})-x]}](https://tex.z-dn.net/?f=6.3%5Ctimes%2010%5E%7B-5%7D%3D%5Cfrac%7B%28x%29%5Ctimes%20%28x%29%7D%7B%5B%287.0%5Ctimes%2010%5E%7B-2%7D%29-x%5D%7D)
Thus, the equilibrium concentration of in the solution is,
Number 9 is A and 10 is A
