What volume of HCl must be used to prepare 350mL of 2.0M HCl solution using a stock solution containing 5L of 12.0M HCl?

~58.3 mL

Explanation:

m1*v1=m2*v2 --> (12.0 M HCL) * v1 = 350mL * 2.0M HCL

--> 350*2.0 = 700

--> 700/12.0 M HCL = 58.3 mL repeating

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