Answer:
the ability to breed successfully
Explanation:
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An adiabatic process is when the system is insulated that no heat is released to the surroundings. For this type of process, we have a derived formula written below:
(T₂/T₁)^C = (V₁/V₂)
where C = Cv/nR
From the complete problem shown in the attached picture, Cv = (3/2)R. Thus,
C= (3/2)/1 mol = 3/2
(T₂/305 K)^(3/2) = (8.5 L/82 L)
Solving for T₂,
<em>T₂ = 67.3 K</em>
We have to distinguish a solution of sodium sulphite from a solution of sodium sulphate in the laboratory.
Solution of sodium sulphite is acidified with dilute HCl and to that when few drops of barium chloride (BaCl₂) solution is added, white precipitate is formed. The white precipitate is soluble in HCl.
To this solution 2 drops of iodine (I₂) solution is added and brown colour of iodine is discharged as I₂ gets reduced to HI.
The reactions involved in case of sodium sulphite is are:
Na₂SO₃ + BaCl₂ = BaSO₃ ↓ + 2NaCl
(white precipitate)
BaSO₃ + 2HCl = BaCl₂ + H₂SO₃
H₂SO₃ + I₂ + H₂O = H₂SO₄ + 2HI
On the other hand, solution of sodium sulphate is acidified with dilute HCl and to that when few drops of barium chloride (BaCl₂) solution is added, white precipitate is formed. The white precipitate of BaSO₄ is formed which is insoluble in HCl.
Na₂SO₄+ BaCl₂ = BaSO₄ ↓ + 2NaCl
(white precipitate)