During the electrolysis of the molten lithium chloride, the Lithium ions (Li⁺) at the cathode undergoes reduction, and the electron configuration of lithium becomes 1s²2s¹.
<h3>What is electrolysis?</h3>
Electrolysis can be described as the process in which the electric current is passed through the chemical compound to break them. In this process, the atoms and ions are interchanged by the addition or removal of electrons.
The ions are allowed to move freely in this process. When an ionic compound is melted or dissolved in water then ions are produced which can move freely.
During the electrolysis of molten lithium chloride, the lithium ions reach the cathode and accept the electrons while chloride ions reach at anode and loss electrons to become chlorine gas.
At anode : 2 Cl⁻ → Cl₂ + 2e⁻
At cathode: 2 Li⁺ + 2e⁻ → Li
Learn more about electrolysis, here:
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Answer:
2Mg + O₂ → 2MgO
Explanation:
Chemical equation:
Mg + O₂ → MgO
Balanced chemical equation:
2Mg + O₂ → 2MgO
The balanced equation s given above and it completely follow the law of conservation of mass.
Law of conservation of mass:
According to the law of conservation mass, mass can neither be created nor destroyed in a chemical equation.
This law was given by french chemist Antoine Lavoisier in 1789. According to this law mass of reactant and mass of product must be equal, because masses are not created or destroyed in a chemical reaction.
Steps to balanced the equation:
Step 1:
Mg + O₂ → MgO
Mg = 1 Mg = 1
O = 2 O = 1
Step 2:
2Mg + O₂ → MgO
Mg = 2 Mg = 1
O = 2 O = 1
Step 3:
2Mg + O₂ → 2MgO
Mg = 2 Mg = 2
O = 2 O = 2
Answer:
[OH⁻] = 3.34x10⁻³M; Percent ionization = 0.54%; pH = 11.52
Explanation:
Kb of the reaction:
NH3 + H2O(l) ⇄ NH4+ + OH-
Is:
Kb = 1.8x10⁻⁵ = [NH₄⁺] [OH⁻] / [NH₃]
<em>As all NH₄⁺ and OH⁻ comes from the same source we can write: </em>
<em>[NH₄⁺] = [OH⁻] = X</em>
<em>And as </em>[NH₃] = 0.619M
1.8x10⁻⁵ = [X] [X] / [0.619M]
1.11x10⁻⁵ = X²
3.34x10⁻³ = X = [NH₄⁺] = [OH⁻]
<h3>[OH⁻] = 3.34x10⁻³M</h3><h3 />
% ionization:
[NH₄⁺] / [NH₃] * 100 = 3.34x10⁻³M / 0.619M * 100 = 0.54%
pH:
As pOH = -log [OH-]
pOH = 2.48
pH = 14 - pOH
<h3>pH = 11.52</h3>
Answer: 0.151
Explanation:
Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.
![Rate=-\frac{1d[HBr]}{2dt}=+\frac{1d[H_2]}{2dt}=+\frac{1d[Br_2]}{dt}](https://tex.z-dn.net/?f=Rate%3D-%5Cfrac%7B1d%5BHBr%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7B1d%5BH_2%5D%7D%7B2dt%7D%3D%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D)
Given: ![-\frac{d[HBr]}{dt}]=0.301](https://tex.z-dn.net/?f=-%5Cfrac%7Bd%5BHBr%5D%7D%7Bdt%7D%5D%3D0.301)
Putting in the values we get:
![\frac{0.301}{2}=+\frac{1d[Br_2]}{dt}](https://tex.z-dn.net/?f=%5Cfrac%7B0.301%7D%7B2%7D%3D%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D)
![+\frac{1d[Br_2]}{dt}=0.151](https://tex.z-dn.net/?f=%2B%5Cfrac%7B1d%5BBr_2%5D%7D%7Bdt%7D%3D0.151)
Thus the rate of appearance of 
is 0.151
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</span><span>
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</span>
