This separation technique is a 4-step procedure. First, add H₂SO₄ to the solution. Because of common ion effect, BaSO₄ will not react, only Mg(OH)₂.
Mg(OH)₂ + H₂SO₄ → MgSO₄ + 2 H₂O
The aqueous solution will now contain MgSO₄ and BaSO₄. Unlike BaSO₄, MgSO₄ is soluble in water. So, you filter out the solution. You can set aside the BaSO₄ on the filter paper. To retrieve Mg(OH)₂, add NaOH.
MgSO₄ + 2 NaOH = Mg(OH)₂ + Na₂SO₄
Na₂SO₄ is soluble in water, while Mg(OH)₂ is not. Filter this solution again. The Mg(OH)₂ is retrieved in solid form on the filter paper.
Answer: A material that readily transmits energy is a conductor, while one that resists energy transfer is called an insulator .
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Answer:
ΔS surrounding (entropy change of the reservoir) = -1 J/K
Explanation:
Given:
Change in heat (ΔH) = 150 joules
Temperature (T) = 150 K
Find:
ΔS surrounding (entropy change of the reservoir)
Computation:
ΔS surrounding (entropy change of the reservoir) = - ΔH / T
ΔS surrounding (entropy change of the reservoir) = - 150 / 150
ΔS surrounding (entropy change of the reservoir) = -1 J/K
Answer:
salt and water
Explanation:
they react to neutralize the acid and base properties,producing a salt while the H(+) cation of the acid combines with OH(-) anion of the base to form water. I hope it's helpful
