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Answer to Question #91997 in General Chemistry for jimbo
Question #91997
How many milliliters of
11.5 M HCl(aq)
are needed to prepare
490.0 mL
of
1.00 M HCl(aq)
11.5 M HCl(aq)
are needed to prepare
490.0 mL
of
1.00 M HCl(aq)
Expert's answer
"M1*V1=M2*V2"
"V1=M2*V2\/V1=490mL*1M\/11.5M=42.61mL"
Answer:42.61mL
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