What is the total enthalpy change for the reaction? Is the reaction exothermic or endothermic?
1 answer:
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Here we have to calculate the amount of ion present in the sample.
In the sample solution 0.122g of ion is present.
The reaction happens on addition of excess BaCl₂ in a sample solution of potassium sulfate (K₂SO₄) and sodium sulfate [(Na)₂SO₄] can be written as-
K₂SO₄ = 2K⁺ +
(Na)₂SO₄=2Na⁺ +
Thus, BaCl₂+ = BaSO₄↓ + 2Cl⁻ .
(Na)₂SO₄ and K₂SO₄ is highly soluble in water and the precipitation or the filtrate is due to the BaSO₄ only. As a precipitation appears due to addition of excess BaCl₂ thus the total amount of ion is precipitated in this reaction.
The precipitate i.e. barium sulfate (BaSO₄)is formed in the reaction which have the mass 0.298g.
Now the molecular weight of BaSO₄ is 233.3 g/mol.
We know the molecular weight of sulfate ion () is 96.06 g/mol. Thus in 1 mole of BaSO₄ 96.06 g of
ion is present.
Or. we may write in 233.3 g of BaSO₄ 96.06 g of ion is present. So in 1 g of BaSO₄
g of
ion is present.
Or, in 0.298 g of the filtered mass (0.298×0.411)=0.122g of ion is present.
Complete Question
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Answer:
Answer: 760 uM
Explanation:
the addition of solvent to a solution in a such away that the volume of the solution increases and the concentration decreases is a process know as dilution
The concentration and the volume of the dilute and concentrated solution are given below as follows
Concentrated solution Dilute solution
Given that M1 = 97.0uM
V1 = 47.0 mL
V2 = 6.0 mL
M2 = ?
Therefore M1V1 = M2V2
M2 = M1V1/V2
M2 = (97*47)/6
M2 = 760 uM
Answer: 760 uM
