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.
Answer:
The answer is steam of fossil fuels
Answer:
= 29.64 g NaNO3
Explanation:
Molarity is given by the formula;
Molarity = Moles/Volume in liters
Therefore;
Number of moles = Molarity × Volume in liters
= 1.55 M × 0.225 L
= 0.34875 moles NaNO3
Thus; 0.34875 moles of NaNO3 is needed equivalent to;
= 0.34875 moles × 84.99 g/mol
= 29.64 g
<u>Answer:</u> The value of 
for the given chemical reaction is 0.1415
<u>Explanation:</u>
Equilibrium constant in terms of partial pressure is defined as the ratio of partial pressures of the products and the reactants each raised to the power their stoichiometric ratios. It is expressed as
For a general chemical reaction:
The expression for is written as:
For the given chemical equation:
The expression for for the following equation is:
We are given:
Putting values in above equation, we get:
The value of for the given chemical reaction is 0.1415
