What is the molarity of a solution that contains 200. 0 g of lithium chloride in 450 mL? molar mass of chloride= 42. 394g/mol
1 answer:
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Answer:
The standard cell potential is 1.40 V. The correct option is the option D (+1.40 V)
Explanation:
Oxide-reduction reactions, also called redox, involve the transfer or transfer of electrons between two or more chemical species. In these reactions two substances interact: the reducing agent and the oxidizing agent.
An oxidizing element or oxidizing agent is one that reaches a stable energy state as a result of which the oxidant is reduced and gains electrons. The oxidizing agent causes oxidation of the reducing agent generating the loss of electrons of the substance and, therefore, oxidizes in the process.
In other words, the oxidizing agent is that chemical species that in a redox process accepts electrons released by the reducing agent and, therefore, is reduced in said process. The oxidizing agent is reduced because, upon receiving electrons from the reducing agent, a decrease in the value of the charge or oxidation number of one of the atoms of the oxidizing agent is induced .
Electrochemical cells, galvanic cells or batteries are called devices that are capable of transforming chemical energy originated in a spontaneous redox process into electrical energy.
The cellular potential is generally in standard conditions, that is, 1 M with respect to solute concentrations in solution and 1 atm for gases.
In this case you have the reaction:
3 Cl₂(g) + 2 Fe(s) → 6 Cl⁻(aq) + 2 Fe³⁺(aq)
In this case the following half-reactions occur:
Semi-reaction of oxidation ( an atom or group of atoms loses electrons, or increases its positive charges): Fe³⁺(aq) + 3 e- -->Fe(s); E⁰ = -0.04 V
Semi-reaction of reduction (an atom or group of atoms gains electrons, increasing its negative charges): Cl₂(g) + 2 e- --> 2 Cl-(aq); E⁰=1.36 V
In an electrochemical cell at 25°C the potentials of the semi-reactions are usually measured in the sense of reduction and generally the standard potential between both electrochemical cells will be:
E⁰=1.36 V - (-0.04 V)
E⁰=1.36 V + 0.04 V
<em>E⁰=1.40 V</em>
<em><u>The standard cell potential is 1.40 V. The correct option is the option D (+1.40 V)</u></em>
Answer:
a. n = 6,54 moles.
b. CH₃COOH(aq) + NaHCO₃(aq) → CH₃COONa(aq) + H₂O (l) + CO₂(g)
c. 549g of NaHCO₃
d. 7,85L of CH₃COOH
Explanation:
a. It is possible to know moles of CO₂ required to inflate the air bag using:
n = PV/RT
Where n are moles; P is pressure (1atm at room conditions); V is volume (160L); R is gas constant (0,082 atmL/molK); and T is temperature (298,15K at room conditions.
Replacing:
<em>n = 6,54 moles.</em>
b. The reaction of CH₃COOH with NaHCO₃ produce:
CH₃COOH(aq) + NaHCO₃(aq) → CH₃COONa(aq) + H₂O (l) + CO₂(g)
c. 1 mol of CO₂ is produced from 1 mol of NaHCO₃, that means 6,54moles of CO₂ are produced from <em>6,54 moles of NaHCO₃</em>. In grams:
6,54 moles NaHCO₃× = <em>549g of NaHCO₃</em>
d. Again, you require 6,54 moles of CH₃COOH. If your acetic acid solution is 0,833M you need:
6,54moles× = <em>7,85L of CH₃COOH</em>
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I hope it helps!