the time required for the activity of a substance taken into the body to lose one half its initial effectiveness. Informal. a brief period during which something flourishes before dying out.
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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:
brainly.com/question/15370276
Answer:
-372000 J or -372 KJ
Explanation:
We have the electrochemical reaction as;
Mg(s) + Fe^2+(aq)→ Mg^2+(aq) + Fe(s)
We must first calculate the E∘cell from;
E∘cathode - E∘anode
E∘cathode = -0.44 V
E∘anode = -2.37 V
Hence;
E∘cell = -0.44 V -(-2.37 V)
E∘cell = 1.93 V
n= 2 since two electrons were transferred
F=96,500C/(mol e−)
ΔG∘=−nFE∘
ΔG∘= -( 2 * 96,500 * 1.93)
ΔG∘= -372000 J or -372 KJ
