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Marina CMI [18]
2 years ago
5

Based on the entries in the following table, which element is most commonly bonded to the acidic hydrogen? table some weak acids

in water at 25 ∘c

Chemistry
2 answers:
Studentka2010 [4]2 years ago
7 0
According to picture below, C₆H₅O⁻ is most commonly bonded to the acidic hydrogen, because it has smallest Ka (<span>the acid dissociation constant).
Chemical reaction: C</span>₆H₅OH(aq) ⇄ C₆H₅O⁻(aq) + H⁺(aq).
Ka = [C₆H₅O⁻] · [H⁺] / [C₆H₅OH]. 
Equolibrium concentration of hydrogen ions is very low in water solution of phenol.

saw5 [17]2 years ago
6 0

The Correct is C₆H₅O⁻ : It is the most commonly bonded to the acidic hydrogen , and O2 is the most element bonded to the acidic hydrogen.


The explanation:


According to the table below we can see that the Ka ( the acid dissociation constant) for the Phenol is the smallest one.


when the Ka expression for this chemical reaction:


C₆H₅OH(aq) ⇄ C₆H₅O⁻(aq) + H⁺(aq)


is


Ka = [C₆H₅O⁻] · [H⁺] / [C₆H₅OH]


So, the water solution of the phenol has a very low equilibrium concentration of H ions [H+] , So C₆H₅O⁻ is the element which most commonly bonded to the acidic hydrogen.

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4 0
2 years ago
Percentage yield of sodium peroxide if 5 g of sodium oxide produces 5.5 g of sodium peroxide
Rama09 [41]
<h3>Answer:</h3>

87.40 %

<h3>Explanation:</h3>

Concept being tested: Percent yield of a product

We are given;

Mass of Sodium oxide 5 g

Experimental or Actual yield of sodium peroxide IS 5.5 g

We are required to calculate the percent yield of sodium peroxide;

The equation for the reaction that forms sodium peroxide is

2Na₂O + O₂ → 2Na₂O₂

<h3>Step 1; moles of sodium oxide</h3>

Moles = mass ÷ molar mass

Molar mass of sodium oxide is 61.98 g/mol

Therefore;

Moles = 5 g ÷ 61.98 g/mol

          = 0.0807 moles

<h3>Step 2: Theoretical moles of sodium peroxide produced </h3>

From the equation, 2 moles of sodium oxide produces 1 mole of sodium peroxide.

Thus, moles of sodium peroxide used is 0.0807 moles

<h3>Step 3: Theoretical mass of sodium peroxide used</h3>

Mass = Number of moles × Molar mass

Molar mass of sodium peroxide = 77.98 g/mol

Therefore;

Theoretical mass = 0.0807 moles × 77.98 g/mol

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Theoretical mass of Na₂O₂ is 6.293 g

<h3>Step 4: Percent yield of Na₂O₂</h3>
  • We know that percent yield is given by the ratio of actual yield to theoretical yield expressed as a percentage.

Percent yield=(\frac{Actual yield}{theoretical yield})100

Percent yield(Na_{2}O_{2})=(\frac{5.5g}{6.293g})100

                       = 87.40 %

Therefore, the percentage yield of sodium peroxide is 87.4%

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So: 50/30.8=1.623
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2 years ago
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