Answer:
Evolution of gas.
Formation of a precipitate.
Change in color.
Explanation:
Answer:
1) 0.0625 g.
2) 0.0125 g.
Explanation:
<em>1) A solution of NaOH has a concentration of 25.00% by mass. What mass of NaOH is present in 0.250 g of this solution?</em>
mass% of NaOH = [(mass of NaOH)/(mass of solution)] x 100.
mass% of NaOH = 25.0%, mass of NaOH = ??? g, mass of solution = 0.250 g.
∴ mass of NaOH = (mass% of NaOH)(mass of solution)/100 = (25.0%)(0.250 g)/100 = 0.0625 g.
<em>2) What mass of NaOH must be added to the solution to increase the concentration to 30.00% by mass?</em>
We can use the relation:
mass% of NaOH = [(mass of NaOH)/(mass of solution)] x 100.
mass% of NaOH = 30.0%, mass of NaOH = ??? g, mass of solution = 0.250 g.
∴ mass of NaOH = (mass% of NaOH)(mass of solution)/100 = (30.0%)(0.250 g)/100 = 0.075 g.
∴ The mass of NaOH should be added = 0.075 - 0.0625 = 0.0125 g.
The enthalpy change of the reaction is <u>-1347.8 kJ.</u>
<h3>What is the enthalpy change, ΔH, of the reaction?</h3>
The enthalpy change, ΔH, of the reaction is calculated from Hess's law of constant heat summation as follows:
Hess's law states that the enthalpy change of a reaction is the sum of the enthalpies of the intermediate reaction.
Given the reactions below and their enthalpy values;
1. X (s) + 12 O₂ (g)⟶ XO (s) ΔH₁ = −850.5 kJ
2. XCO₃ (s) ⟶ XO (s) + CO₂ (g) ΔH₂ = +497.3 kJ
The enthalpy change, ΔH, of the reaction whose equation is given below, will be:
X (s) + 12 O₂ (g) + CO₂ (g) ⟶ XCO₃ (s)
ΔH = ΔH₁ - ΔH₂
ΔH = − 850.5 kJ - (+497.3 kJ)
ΔH = -1347.8 kJ
Learn more about enthalpy change at: brainly.com/question/14047927
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