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
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Answer:
What mass (g) of barium iodide is contained in 188 mL of a barium iodide solution that has an iodide ion concentration of 0.532 M?
A) 19.6
B) 39.1
C) 19,600
D) 39,100
E) 276
The correct answer to the question is
B) 39.1 grams
Explanation:
To solve the question
The molarity ratio is given by
188 ml of 0.532 M solution of iodide.
Therefore we have number of moles = 0.188 × 0.532 M = 0.100016 Moles
To find the mass, we note that the Number of moles = 
from which we have
Mass = Number of moles × molar mass
Where the molar mass of Barium Iodide = 391.136 g/mol
= 0.100016 moles ×391.136 g/mol = 39.12 g
If a sample of gas is a 0.622-gram, volume of 2.4 L at 287 K and 0.850 atm. Then the molar mass of the gas is 7.18 g/mol
<h3>What is an ideal gas equation?</h3>
The ideal gas law (PV = nRT) relates to the macroscopic properties of ideal gases.
An ideal gas is a gas in which the particles (a) do not attract or repel one another and (b) take up no space (have no volume).
Given :
The ideal gas equation is given below.
n = PV/RT
n = 86126.25 x 0.0024 / 8.314 x 287
n = 0.622 / molar mass (n = Avogardos number)
Molar mass = 7.18 g
Hence, the molar mass of a 0.622-gram sample of gas having a volume of 2.4 L at 287 K and 0.850 atm is 7.18 g
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mass of PbI₂ = 27.6606 g
<h3>Further explanation</h3>
Given
Pb(NO₃)₂ + NaI → PbI₂ + NaNO₃
28.0 grams of Pb(NO₃)₂ react with 18.0 grams of NaI
Required
mass of PbI₂
Solution
Balanced equation
Pb(NO₃)₂ + 2NaI → PbI₂ + 2NaNO₃
The principle of a balanced reaction is the number of atoms in the reactants = the number of atoms in the product
mol Pb(NO₃)₂ :
= 28 : 331,2 g/mol
= 0.0845
mol NaI :
= 18 : 149,89 g/mol
= 0.12
Limiting reactant : mol : coefficient
Pb(NO₃)₂ : 0.0845 : 1 = 0.0845
NaI : 0.12 : 2 = 0.06
NaI limiting reactant (smaller ratio)
mol PbI₂ based on NaI
= 1/2 x 0.12 = 0.06
Mass PbI₂ :
= 0.06 x 461,01 g/mol
= 27.6606 g
