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>
Answer:
i think it will be warm air rises and takes heat with it eventually it cools and sinks
Explanation:
please mark me as brainliest if it helps
Answer: c. Salt and Water
Explanation:
For example;
When an Arrhenius acid such as; Tetraoxosulphate (VI) acid (H2SO4) reacts with an Arrhenius base such as Potassium hydroxide (KOH), the products formed in this neutralization reaction is a salt known as ''Potassium Sulphate'' (K2SO4) and ''Water'' (H2O).
H2SO4 + KOH -------------> K2SO4 + H2O
Answer:
number of energy levels
Explanation:
for example hydrogen and helium are in period 1 and they have 1 energy level
