When water is added to a compound in a chemical equation, a new compound can be formed, it is called a hydrolysis reaction.
<h3>What is hydrolysis reaction?</h3>
There are so many kinds of reaction in chemistry we know that a chemical reaction occurs between reactant molecules in order to yield products in the reactants.
The hydrolysis reaction is a breaking up reaction. In a hydrolysis reaction, water is used to break up a molecule. Hence in a hydrolysis reaction water could be regarded as one of the reactants in the reactions going on in the system.
We know that a hydrolysis is a reaction in which water reacts with another reactant molecule to yield products in the reaction. Thus, when water is added to a compound in a chemical equation, a new compound can be formed, it is called a hydrolysis reaction.
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Answer:
a) Kb = 10^-9
b) pH = 3.02
Explanation:
a) pH 5.0 titration with a 100 mL sample containing 500 mL of 0.10 M HCl, or 0.05 moles of HCl. Therefore we have the following:
[NaA] and [A-] = 0.05/0.6 = 0.083 M
Kb = Kw/Ka = 10^-14/[H+] = 10^-14/10^-5 = 10^-9
b) For the stoichiometric point in the titration, 0.100 moles of NaA have to be found in a 1.1L solution, and this is equal to:
[A-] = [H+] = (0.1 L)*(1 M)/1.1 L = 0.091 M
pKb = 10^-9
Ka = 10^-5
HA = H+ + A-
Ka = 10^-5 = ([H+]*[A-])/[HA] = [H+]^2/(0.091 - [H+])
[H+]^2 + 10^5 * [H+] - 10^-5 * 0.091 = 0
Clearing [H+]:
[H+] = 0.00095 M
pH = -log([H+]) = -log(0.00095) = 3.02
<span>density is defined as mass per volume.. so 94 mL makes no sense.</span>
Answer:
pOH = 11.5
[H⁺] = 0.003 M
[OH⁻] = 3 × 10⁻¹² M
Explanation:
The computation is shown below:
Given that
pH = 2.5
Based on the above information
We know that
pH + pOH = 14 ⇒ pOH = 14 - pH
pOH = 14 - 2.5
pOH = 11.5
[H⁺] = 10^(-pH) = 10^(-2.5)
[H⁺] = 0.003 M
[OH⁻] = 10^(-pOH)
= 10^(-11.5)
= 3 × 10⁻¹² M
[OH⁻] = 3 × 10⁻¹² M
Hence, the above represents the answer
1.75 (moles O2) × 6 (moles H2O) ÷ 9 (moles O2) = 1.17 (moles H2O)
You have to convert moles of O2 into moles of H2O so it's the number of moles you start with (1.75 O2) × the number of moles from the element you want (6 H20), then ÷ by the number of moles that the element that you already have (9 O2).
