Twice as much more will the freezing point of water be lowered in beaker a than in beaker b.
<h3>What determines freezing point?</h3>
A liquid's freezing point rises if the intermolecular interactions between its molecules are strong. The freezing point, however, drops if the molecules of inter - molecular are minimal. The process through which a substance transforms from a liquid into a solid is known as freezing.
<h3>How significant is freezing point?</h3>
Freezing points play a big role in occupational safety. A chemical may perhaps turn harmful if held below its freezing point. A critical safety benchmark for assessing the effects of worker exposure to cold environments is the freezing point.
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<span>Answer:
Molecular:
HC2H3O2(aq) + KOH(aq) --> KC2H3O2(aq) + H2O(l)
Complete ionic:
HC2H3O2(aq) + K+(aq) + OH-(aq) --> K+(aq) + C2H3O2-(aq) + H2O(l)
Net Ionic:
HC2H3O2(aq) + OH-(aq) --> C2H3O2-(aq) + H2O(l)</span>
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.
Answer:
A hydrocarbon and oxygen
Explanation:
hydrocarbon is being burned, and oxygen is necessary to burn things
