When they ask you for the solution they are usually asking for x. So solve for X.
X=-12 so you only have one solution. so the answer is B
The states in which water occur in are ice liquid and gas
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:
1.99 atm
Explanation:
Step 1:
Data obtained from the question. This include the following:
Initial pressure (P1) = 0.520 atm
Initial temperature (T1) = 26.2°C
Initial volume (V1) = 15.4L
Final temperature (T2) = constant = 26.2°C
Final volume (V2) = 4.02L
Final pressure (P2) =..?
Step 2:
Determination of the new pressure of the gas.
Since the temperature of the gas is constant, it means the gas is obeying Boyle's law. Thus, the new pressure of the gas can be obtained by applying the Boyle's law equation as shown below:
P1V1 = P2V2
0.520 x 15.4 = P2 x 4.02
Divide both side by 4.02
P2 = (0.520 x 15.4) / 4.02
P2 = 1.99 atm
Therefore, the new pressure of the gas is 1.99 atm
