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.
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Answer:
The law is given by the following equation: PV = nRT, where P = pressure, V = volume, n = number of moles, R is the universal gas constant, which equals 0.0821 L-atm / mole-K, and T is the temperature in Kelvin.
Explanation:
2ZnS(s)+3O2(g) -> 2Zns(s) + 2SO3(g)
the above given equation is unbalanced as it contains 4 moles of sulphur in the output but in the input there are only two aoms of sulphur so to balance the equation we will write the equation as given under
balanced equation is
2ZnS(s)+3O2(g) -> 2Zn(s) + 2SO3(g)
In words:
When 2 moles of solid zinc sulfide reacts with 3 moles of oxygen gas gives 2 moles of solid zinc and 2 moles of sulphur trioxide gas.
Answer:
0.3 mole
Explanation:
number of moles grams
one 14+(4×1)+14+(3×16)
1 80
? 24
Therefore, 24×1÷80 = 0.3 moles of ammonium nitrate
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