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Answer:
V₁ = 228 mL
Explanation:
There is some info missing. I think this is the original question.
<em>A chemist must prepare 575 mL of 50.0 mM aqueous potassium permanganate (KMnO₄) working solution. He'll do this by pouring out some 0.126M aqueous potassium permanganate stock solution into a graduated cylinder and diluting it with distilled water. Calculate the volume in mL of the potassium permanganate stock solution that the chemist should pour out. Round your answer to 3 significant digits.</em>
<em />
We want to prepare 575 mL (V₂) of a 50.0 mM solution (C₂). To do so, we start, from a stock solution of concentration C₁ = 0.126 M. In order to find the volume V₁ that the chemist should pour out, we will use the dilution rule.
C₁ × V₁ = C₂ × V₂
0.126 M × V₁ = 50.0 mM × 575 mL
0.126 M × V₁ = 50.0 × 10⁻³ M × 575 mL
V₁ = 228 mL
Answer:
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Explanation:
Mole=mass/molecular mass (Mr)
mole=volume/ 22.4
Molecular amount: Mole=amount atoms/ Avogadro's number
Concentration: mole=volume x M
