Ionization of H2CO3
<span>Step-1: </span>
<span>H2CO3 (aq) + H2O (l) <----> HCO3- (aq) + H3O+ (aq) </span>
<span>Step-2: </span>
<span>HCO3- (aq) + H2O (l) <----> CO32- (aq) + H3O+ (aq) </span>
<span>Ionization of H3BO3 </span>
<span>Step 1 is </span>
<span>H3BO3(aq) + H2O(l) <=> H2BO3-(aq) + H3O+(aq) </span>
<span>Step-2: </span>
<span>H2BO3-(aq) + H2O(l) <=> HBO3^-2(aq) + H3O+(aq) </span>
<span>Step 3 is</span>
<span>HBO3^-2(aq) + H2O(l) <=> BO3^-3(aq) + H3O+(aq)
credits go to Mary Nancy :)</span>
Answer:
The pH decreases very quickly, more quickly than earlier or later in the titration.
I think because SA-SB titration have the straight line where the equivalence point is where pH change is very steep and fast. Also pH is gonna decrease because after equivalence point you have more acid then base.
Explanation:
The balanced equation for the above reaction is
2Al + 6H₂O ---> 2Al(OH)₃ + 3H₂
stoichiometry of Al to H₂ is 2:3
number of Al moles reacted - 78.33 g / 27 g/mol = 2.901 mol
according to molar ratio
2 mol of Al forms - 3 mol of H₂
therefore 2.901 mol of Al - forms 3/2 x 2.901 = 4.352 mol
molar volume states that 1 mol of any gas occupies a volume of 22.4 L at STP
if 1 mol occupies 22.4 L
then 4.352 mol occupies - 22.4 L/mol x 4.352 mol = 97.48 L
volume occupied by H₂ is 97.48 L
Potassium iodide will act as the catalyst to decompose hydrogen peroxide into water and oxygen. A catalyst is a substance that participates in a chemical reaction and influences its speed without undergoing permanent change.<span>
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