The volume of chlorine molecules produced at STP would be 96 dm³.
<h3>Stoichiometric problem</h3>
Sodium chloride ionizes during electrolysis to produce sodium and chlorine ions as follows:
This means that 1 mole of sodium chloride will produce 1 mole of sodium ion and 1 mole of chlorine ion respectively.
Recall that: mole = mass/molar mass
Hence, 234 g of sodium chloride will give:
234/58.44 = 4.00 moles.
Thus, the equivalent number of moles of chlorine produced by 234 g of sodium chloride will be 4 moles.
Recall that:
1 mole of every gas at Standard Temperature and Pressure = 24 Liters.
Hence:
4 moles of chlorine = 4 x 24 = 96 Liters or 96 dm³.
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The decomposition of ammonia is characterized by the following decomposition equation:
2NH₃<span> → N</span>₂ <span> + 3H</span>₂
The mole ratio of N₂ : H₂ is 1 : 3
If the number of moles of N₂ = 0.0351 mol
Then the number of moles of H₂ = 0.0351 mol × 3
= 0.1053 mol
The number of moles of hydrogen gas produced when 0.0351 mol of Nitrogen gas is produced after the decomposition of Ammonia is 0.105 mol (OPTION 3).
Answer:
9.1
Explanation:
Step 1: Calculate the basic dissociation constant of propionate ion (Kb)
Sodium propionate is a strong electrolyte that dissociates according to the following equation.
NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻
Propionate is the conjugate base of propionic acid according to the following equation.
C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻
We can calculate Kb for propionate using the following expression.
Ka × Kb = Kw
Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰
Step 2: Calculate the concentration of OH⁻
The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.
[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M
Step 3: Calculate the concentration of H⁺
We will use the following expression.
Kw = [H⁺] × [OH⁻]
[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M
Step 4: Calculate the pH of the solution
We will use the definition of pH.
pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1
The electrophilic bromination or chlorination of benzene requires Lewis acid along with the halogen.
<h3>
What is bromination of benzene?</h3>
The bromination or chlorination of benzene is an example of an electrophilic aromatic substitution reaction.
During the reaction, the bromine forms a sigma bond to the benzene ring, yielding an intermediate. Subsequently a a proton is removed from the intermediate to form a substituted benzene ring.
This reaction is achieved with the help of Lewis acid as catalysts.
Thus, the electrophilic bromination or chlorination of benzene requires Lewis acid along with the halogen.
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