Answer:
These metals are less reactive than the neighbouring alkali metal. Magnesium is less active than sodium; calcium is less active than potassium; and so on. These metals become more active as we go down the column.
Explanation:
Magnesium is more active than beryllium; calcium is more active than magnesium; and so on.
Answer:
The correct option is;
B. -112 kJ
Explanation:
The parameters given are;
N (g) + O₂ (g) → NO (g) ΔH = 90 kJ/mol
N (g) + O₂ (g) → NO₂ (g) ΔH = 34 kJ/mol
The required chemical reaction is given as follows;
2NO (g) + O₂ (g) → 2NO₂ (g)
Therefore, the heat of formation of 2 moles of NO = 2 × 90 = 180 kJ
The heat of formation of 2 moles of NO₂ = 2 × 34 = 68 kJ
Hence, given that the heat of formation of O₂ at room temperature = 0 kJ/mol, we have;
Change in enthalpy of the chemical reaction = Heat of formation of the products - Heat of formation of the reactants
Change in enthalpy of the chemical reaction = 68 kJ - 180 kJ = -112 kJ
Change in enthalpy of the chemical reaction = -112 kJ.
Answer:
0.75M Fe²⁺
Explanation:
First, we need to balance the redox reaction in acidic medium. Then, we can obtain moles of KMnO4 and with the reaction moles and molarity of the Fe²⁺ solution:
<em>Redox Balance:</em>
Fe²⁺ → Fe³⁺ + 1e⁻
5e⁻ + 8H⁺ + MnO₄⁻ → Mn²⁺ + 4H₂O
___________________________
5Fe²⁺ + 5e⁻ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + 5e⁻ + Mn²⁺ + 4H₂O
<h3>5Fe²⁺ + 8H⁺ + MnO₄⁻ → 5Fe³⁺ + Mn²⁺ + 4H₂O</h3>
<em>Moles of KMnO₄:</em>
70.0mL = 0.0700L * (0.150mol / L) = 0.0105 moles KMnO₄
<em>Moles and molarity Fe²⁺:</em>
0.0105 moles KMnO₄ * (5 moles Fe²⁺ / 1mol KMnO₄) = 0.0525 moles Fe²⁺
In 70.0mL = 0.0700L:
0.0525 moles Fe²⁺ / 0.0700L =
<h3>0.75M Fe²⁺</h3>