The reaction involved in present case is:
Net Reaction: Ca + 1/2 O2 → CaO. ..................(1)
In terms of oxidation and reduction, the reaction can be shown at
Oxidation: Ca → Ca2+ + 2e- .................(2)
Reduction: 1/2O2 + 2e- → O2-...................(3)
From, reaction 1 it can be seen that 1 mol of Ca reacts with 1/2 mol of O2 to form 1 mol of CaO.
From, reaction 2 it can be seen that 1 mol of Ca, generates 2 mol of e-.
Thus, when 1/2 mol of Ca is used in reaction, it will lose 1 mol of electrons.
Answer is: D. Na2SO4.
b(solution) = 0.500 mol ÷ 2.0 L.
b(solution) = 0.250 mol/L.
b(solution) = 0.250 m; molality of the solutions.
ΔT = Kf · b(solution) · i.
Kf - the freezing point depression constant.
i - Van 't Hoff factor.
Dissociation of sodium sulfate in water: Na₂SO₄(aq) → 2Na⁺(aq) + SO₄²⁻(aq).
Sodium sulfate dissociates on sodium cations and sulfate anion, sodium sulfate has approximately i = 3.
Sodium chloride (NaCl) and potassium iodide (KI) have Van 't Hoff factor approximately i = 2.
Carbon dioxide (CO₂) has covalent bonds (i = 1, do not dissociate on ions).
Because molality and the freezing point depression constant are constant, greatest freezing point lowering is solution with highest Van 't Hoff factor.
The hours taken for concentration to decrease from 0 to 74 min. to 0.21 m is 91.7 hours.
<h3>What is the rate law of a reaction?</h3>
Rate law depicts the rate of a chemical reaction depend on the concentration of the reactant.
The given reaction is second order reaction
Thus, the hours taken for concentration to decrease from 0 to 74 min. to 0.21 m is 91.7 hours.
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Answer:
Difference between Vapor and Gas
Usually, a vapour phase consists of a phase with two different substances at room temperature, whereas a gas phase consists of a single substance at a defined thermodynamic range, at room temperature
Explanation:
Answer:
The IUPAC structure only shows bond pairs and lone pairs. In the flouromethane structure above, there is only one bond pair and three lone pairs of electrons. Therefore there is one electron remaining, but since it doesn't not make up a pair, it is ignored in the structure but theoretically it is present.
