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.
Hypothesis because law don’t make sense and theory is for something that already has data behind it. In this case you don’t do it’s not hypothesis
Because molecules in the air scatter blue light from<span> the sun </span>more than<span> they </span>scatter red light<span>.</span>
Answer:
1.3 × 10²³ Atoms of Mercury
Solution:
Step 1: Calculate Mass of Mercury using following formula,
Density = Mass ÷ Volume
Solving for Mass,
Mass = Density × Volume
Putting values,
Mass = 13.55 g.cm⁻³ × 3.2 cm³ ∴ 1 cm³ = 1 cc
Mass = 43.36 g
Step 2: Calculating number of Moles using following formula;
Moles = Mass ÷ M.mass
Putting values,
Moles = 43.36 g ÷ 200.59 g.mol⁻¹
Moles = 0.216 mol
Step 3: Calculating Number of Atoms using following formula;
Number of atoms = Moles × 6.022 ×10²³
Putting value of moles,
Number of Atoms = 0.216 mol × 6.022 × 10²³
Number of Atoms = 1.3 × 10²³ Atoms of Hg
The answer would be 20% (B)
