B.) absorb energy as it goes to the ground state
Since most acetic acid molecules do not dissociate when a sample is dissolved in water, acetic acid and water molecule interactions are what determine how soluble a substance is. In its liquid state, water forms a network of hydrogen bonds among its molecules; when a material dissolves in water, this network of hydrogen bonds is broken.
<h3>What is acetic acid such a poor electrolyte?</h3>
Since acetic acid is a weak acid, only a small portion of the acetic acid molecules react to produce ethanoate and hydronium ions, shifting the equilibrium position substantially to the left. Aqueous acetic acid is a weak electrolyte as a result of the existence of these few ions.
<h3>Why does acetic acid have a low conductivity?</h3>
The majority ($> 99%) of the acetic acid molecules remain after dissolution, with only a very tiny portion deprotonating to form acetate anions. Only the latter are charged compounds, and as a result, only these increase the conductivity of the solution. They are few, and conductivity is low.
<h3>What makes acetic acid more potent than water?</h3>
The equilibrium shifts to the left and the concentration of hydrogen ions decreases when a strong acid is added to the buffer solution. A strong base is also added, which causes the equilibrium to move to the left and results in a smaller pH increase. Because of this, acetic acid is a better buffer than water.
Learn more about Acetic acid
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Answer:
B
Explanation:
The last 2 answers can immediately be cancelled out as they're not combustion of ethanol, combustion in this context is not a reversable reaction so cannot be referenced in this way.
The answer is determined to be B just by simply counting the numbers of each of the elements on the reactants (left) side and the products (right) side and ensuring they're the same
In A on the reactants side there is 2C 6H 3O and on the products side there is 1C 3O and 2H its not equal so it is not correct
In B on the reactants side there is 2C 6H 7O and on the products side there is 2C 6H 7O its equal so this answer is correct
Answer:
The freezing point of chloroform (CHCl₃) in presence of benzene = - 64.252 °C.
Explanation:
- We can solve this problem using the relation:
ΔTf = Kf.m,
Where, ΔTf is the elevation of freezing point,
Kf is the freezing point depression constant <em>(Kf of chloroform = 4.7 °C/m)</em>,
m is the molality of the solution.
- We can get the molality of benzene from the relation:
m = (mass / molar mass) solute (benzene) x (1000 / mass of solvent (CHCl₃))
m = (1.25 g / 78.11 g/mol) x (1000 / 100.0 g) = 0.160 m.
ΔTf = Kf.m = (4.70 °C/m) (0.160 m) = 0.752 °C.
∴ The freezing point in presence of benzene = The freezing point of pure CHCl₃ - ΔTf = - 63.50 °C – 0.752 °C = - 64.252 °C.
