## anonymous one year ago How many milliliters of 11.5 M HCl(aq) are needed to prepare 935.0 mL of 1.00 M HCl(aq)?

You need to use the dilution equation, where you have two sets of conditions: $$\sf \huge \underbrace{M_1V_1}_{have }=\underbrace{M_2V_2}_{sought}$$ $$\sf 11.5 M *V_1=935.0 mL* 1.00 M \rightarrow V_1=\dfrac{935.0 mL* 1.00 M}{11.5 M }$$