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If you want to prepare 80.0 mL of 4.00M acid ,How many mL of 12.4 M HCl are required ?

Sagot :

Medunno13

[tex]M_{A}V_{A}=M_{B}V_{B}\\(80.0)(4.00)=V_{B}(12.4)\\V_{B}=\frac{(80.0)(4.00)}{12.4} \approx \boxed{25.8 \text{ mL}}[/tex]

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