The molarity of the resulting solution obtained by diluting the stock solution is 3 M
<h3>Data obtained from the question </h3>
- Molarity of stock solution (M₁) = 15 M
- Volume of stock solution (V₁) = 500 mL
- Volume of diluted solution (V₂) = 2.5 L = 2.5 × 1000 = 2500 mL
- Molarity of diluted solution (M₂) =?
<h3>How to determine the molarity of diluted solution </h3>
M₁V₁ = M₂V₂
15 × 500 = M₂ × 2500
7500 = M₂ × 2500
Divide both side by 2500
M₂ = 7500 / 2500
M₂ = 3 M
Thus, the volume of the resulting solution is 3 M
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Answer:
Just Barely Base/Neutral
Explanation:
a pH of 8.0 is greater than Neutral (7.0) but is still neutral due to it being more neutral than a base
Answer:
200g or 40 teaspoons
Explanation:
An average human, weighing about 50 pounds, has about 200 g or 40 tps of NACl
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.
