Methylene blue indicates the presence of oxidizing agents because it is oxidized itself by these compounds. When electrons are stripped from methylene blue, the resulting molecule imparts a blue color to the solution--giving a clear sign of a chemical change.
The mass of sodium sulphate, Na₂SO₄, required to prepare the solution is 10.65 g
<h3>How to determine the mole of sodium sulphate Na₂SO₄</h3>
- Volume = 250 mL = 250 / 1000 = 0.25 L
- Molarity = 0.3 M
Mole = Molarity x Volume
Mole of Na₂SO₄ = 0.3 × 0.25
Mole of Na₂SO₄ = 0.075 mole
<h3>How to determine the mass of sodium sulphate Na₂SO₄</h3>
- Molar mass of Na₂SO₄ = 142.05 g/mol
- Mole of Na₂SO₄ = 0.075 mole
Mass = mole × molar mass
Mass of Na₂SO₄ = 0.075 × 142.05
Mass of Na₂SO₄ = 10.65 g
Thus, 10.65 g of Na₂SO₄ is needed to prepare the solution.
Learn more about molarity:
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Answer:
Keq = [CO₂]/[O₂]
Explanation:
Step 1: Write the balanced equation for the reaction at equilibrium
C(s) + O₂(g) ⇄ CO₂(g)
Step 2: Write the expression for the equilibrium constant (Keq)
The equilibrium constant is equal to the product of the concentrations of the products raised to their stoichiometric coefficients divided by the product of the concentrations of the reactants raised to their stoichiometric coefficients. It only includes gases and aqueous species. The equilibrium constant for the given system is:
Keq = [CO₂]/[O₂]
Answer:
3.6 grams of NaCl are needed
Explanation:
Percent solution are solutions whose concentrations are expressed in percentages. The amount(either weight or volume) of a solute is expressed as a percentage of the total weight or volume of solution. Percent solutions can either be expressed as weight/volume % (wt/vol % or w/v %), weight/weight % (wt/wt % or w/w %), or volume/volume % (vol/vol % or v/v %).
A 6.0% wt/wt % solution contains 6 g of solute in 100 g of solution
Therefore, a 100 g solution contains 6.0 g of solute.
60 g of 6.0% solution will contain 60 g solution * 6.0 g solute/ 100 g solution
Mass of NaCl present = 3.6 g of NaCl
