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how to prepare 24 μL of 5% H3BO3 
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I am assuming this is a 1,000g mixture. 42% H3BO3 = 420g H3BO3/(420g H3BO3+580g H2O)= 0.42 by mass. 420g H3BO3/(420g H3BO3+580g H2O+Xg H2O)= 0.06 Solve for X Check my thinking. Check my work. There is an alternative approach. You 
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25.0 mL of an H3BO3 solution were titrated with 29.15 mL of a 0.205 M LiOH solution to reach the equivalence point. what is the molarity of the H3BO3 solution? 
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