Concentration of a solution can be expressed in terms of molarity and molality
Molarity is the number of moles of solute in a liter of a solution.
Molarity (M) = Moles of solute/Volume(litres) of solution
Molality is the number of moles of solute in one kg of the solution
Molality (m) = Moles of solute/Mass (kg) of solution
Therefore if the volume or the mass of the solution is changed this would affect the concentration.
In addition, volume is a quantity which depends on temperature. However, mass is independent of temperature. Therefore any changes in temperature, can also bring about a change in the molarity of the solution.
Answer:
Mass of water produced is 22.86 g.
Explanation:
Given data:
Mass of hydrogen = 2.56 g
Mass of oxygen = 20.32 g
Mass of water = ?
Solution:
Chemical equation:
2H₂ + O₂ → 2H₂O
Number of moles of oxygen:
Number of moles = mass/ molar mass
Number of moles = 20.32 g/ 32 g/mol
Number of moles = 0.635 mol
Number of moles of hydrogen:
Number of moles = mass/ molar mass
Number of moles = 2.56 g/ 2 g/mol
Number of moles = 1.28 mol
Now we will compare the moles of water with oxygen and hydrogen.
O₂ : H₂O
1 : 2
0.635 ; 2×0.635 = 1.27
H₂ : H₂O
2 : 2
1.28 : 1.28
The number of moles of water produced by oxygen are less thus it will be limiting reactant.
Mass of water produced:
Mass = number of moles × molar mass
Mass = 1.27 × 18 g/mol
Mass = 22.86 g
Answer:
0.46 V
Explanation:
The emf for the cell is given by:
Eº cell = Eº oxidation + Eº reduction
From the given balanced chemical equation, we can deduce that Fe²⁺ has been oxidized to Fe³⁺, and O reduced from 0 to negative 2, according to the half cell reactions:
4Fe²⁺ ⇒ Fe³⁺ + 4e⁻ oxidation
O₂ + 4H⁺ + 4 e⁻ ⇒ 2 H₂O reduction
From reference tables for the standard reduction potential, we get
Eº red Fe³⁺ / Fe²⁺ Eºred = 0.77 V
Eº red O₂ / H₂O Eºred = 1.23 V
Now all we need to do is change the sign of Eº reduction for the species being oxidized ( Fe²⁺ ) and add it to Eº reduction O₂:
Eº cell = Eº oxidation + Eº reduction = - (0.77 V ) + 1.23 V = 0.46 V
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