Calculate the freezing point of a 12.25 m 12.25m aqueous solution of propanol. Freezing point constants can be found in the list
1 answer:
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Answer : The value of and K is, -180 kJ/mol and
Explanation :
The balanced cell reaction will be,
The half-cell reactions are:
Oxidation reaction (anode) :
Reduction reaction (cathode) :
Relationship between standard Gibbs free energy and standard electrode potential follows:
where,
= standard Gibbs free energy
F = Faraday constant = 96500 C
n = number of electrons in oxidation-reduction reaction = 2
= standard electrode potential of the cell = 0.93 V
Now put all the given values in the above formula, we get:
Now we have to calculate the value of 'K'.
where,
= standard Gibbs free energy = -180 kJ/mol
R = gas constant =
T = temperature = 298 K
K = equilibrium constant = ?
Now put all the given values in the above formula 1, we get:
Therefore, the value of and K is, -180 kJ/mol and
Answer: A substance that produces hydroxide ions when placed in water is base.
Explanation:
Bases are the substance:
- Which gives negatively charged hydroxide(
) ions in aqueous solution.
- Which have pH value ranging from 7 to 14.
Where as acid gives positively charged hydronium ion() in aqueous solution.