Answer:
1. 0.0154mole of PbS
2. Double displacement reaction
Explanation:
First, let write a balanced equation for the reaction. This is illustrated below:
Pb(CH3COO)2 + H2S —> PbS + 2 CH3COOH
Molar Mass of Pb(CH3COO)2 = 207 + 2(12 + 3 + 12 + 16 +16) = 207 + 2(59) = 207 + 118 = 325g
Mass of Pb(CH3COO)2 = 5g
Number of mole = Mass /Molar Mass
Number of mole of Pb(CH3COO)2 = 5/325 = 0.0154mole
From the equation,
1mole of Pb(CH3COO)2 produced 1mole of PbS.
Therefore, 0.0154mole of Pb(CH3COO)2 will also produce 0.0154mole of PbS
2. The name of the reaction is double displacement reaction since the ions in the two reactants interchange to form two different products
The chemical elements are arranged in order of increasing atomic number. Answer A.
<span>a pure substance that can be separated into different elements by chemical means</span>
The enthalpy change : -196.2 kJ/mol
<h3>Further explanation </h3>
The change in enthalpy in the formation of 1 mole of the elements is called enthalpy of formation
The enthalpy of formation measured in standard conditions (25 ° C, 1 atm) is called the standard enthalpy of formation (ΔHf °)
(ΔH) can be positive (endothermic = requires heat) or negative (exothermic = releasing heat)
The value of ° H ° can be calculated from the change in enthalpy of standard formation:
∆H ° rxn = ∑n ∆Hf ° (product) - ∑n ∆Hf ° (reactants)
Reaction
2 H₂O₂(l)-→ 2 H₂O(l) + O₂(g)
∆H ° rxn = 2. ∆Hf ° H₂O - 2. ∆Hf °H₂O₂

A) Temperature does not affect reaction rate