Answer:
4.26 %
Explanation:
There is some info missing. I think this is the original question.
<em>Calculate the percent ionization of nitrous acid in a solution that is 0.249 M in nitrous acid. The acid dissociation constant of nitrous acid is 4.50 × 10
⁻⁴.</em>
<em />
Step 1: Given data
Initial concentration of the acid (Ca): 0.249 M
Acid dissociation constant (Ka): 4.50 × 10
⁻⁴
Step 2: Write the ionization reaction for nitrous acid
HNO₂(aq) ⇒ H⁺(aq) + NO₂⁻(aq)
Step 3: Calculate the concentration of nitrite in the equilibrium ([A⁻])
We will use the following expression.
Step 4: Calculate the percent ionization of nitrous acid
We will use the following expression.
Answer: H2 + Cl2 => 2HCl
Explanation: the equation is now balanced. 2 H and 2 Cl atoms both in the reactant and product side.
The volume of HCl = 2.10⁴ mL
<h3>Further explanation</h3>
Given
5.0 M(mol/L) HCl solution
3 g of Mg
Required
Volume of HCl
Solution
Reaction
Mg + 2HCl ⇒ MgCl₂ + H₂
mol of Mg (Ar=24 g/mol) :
mol = mass : Ar
mol = 3 : 24
mol = 0.125
From the equation, mol ratio of Mg : HCl = 1 : 2, so mol HCl :
= 2/1 x mol Mg
= 2 x 0.125
= 0.25
Molarity : mol solute per liters solution
M = m x V
V = M : m
V = 5 mol/L : 0.25
V = 20 L = 2.10⁴ mL
Due to the iron hydrogen and chromium diluted oxygen. the iron become more oxydized.
The answer is b opinions and beliefs