Answer:
1.0190 x 10⁻⁵ mol
Explanation:
We know the titration required 10.19 mL of 0.001000 M KIO₃, from this information we can calculate the number of moles KIO₃ reacted and from there the number of moles of ascorbic acid since it is a monoprotic acid ( 1 equivalent of ascorbic acid to one equivalent KIO₃).
Molarity = mol/V
V KIO₃ = 10.19 mL = 10.19 mL x 1 L/1000 mL = 0.01019 L
⇒ mol KIO₃ = V x M = 0.01019 L x 0.0010 mol / L = 1.0190 x 10⁻⁵ mol KIO₃
# mol ascorbic acid = # mol KIO₃ = 1.0190 x 10⁻⁵ mol
Answer:
59.077 kJ/mol.
Explanation:
- From Arrhenius law: <em>K = Ae(-Ea/RT)</em>
where, K is the rate constant of the reaction.
A is the Arrhenius factor.
Ea is the activation energy.
R is the general gas constant.
T is the temperature.
- At different temperatures:
<em>ln(k₂/k₁) = Ea/R [(T₂-T₁)/(T₁T₂)]</em>
k₂ = 3k₁ , Ea = ??? J/mol, R = 8.314 J/mol.K, T₁ = 294.0 K, T₂ = 308.0 K.
ln(3k₁/k₁) = (Ea / 8.314 J/mol.K) [(308.0 K - 294.0 K) / (294.0 K x 308.0 K)]
∴ ln(3) = 1.859 x 10⁻⁵ Ea
∴ Ea = ln(3) / (1.859 x 10⁻⁵) = 59.077 kJ/mol.
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Answer:
Hope this helped :) good luck! ❤️
Explanation:
A <em>coolant solution</em> is a <u><em>homogeneous </em></u>mixture because the coolant particles are not chemically combined with the water (keep their properties) and they are evenly distributed throughout the water.
Answer:
the would test random expirements
