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Answer to Question #67195 in General Chemistry for MoneyManDEV
Question #67195
What volume of 0.25 MHCl must be diluted to prepare 1.4 L of 5.1*10^-2 MHCl ?
Expert's answer
n1(HCl) = n2(HCl)
c1(HCl)×V1 = c2(HCl)×V2
V1 = c2(HCl)×V2/c1(HCl)
V1 = 5.1×10-2×1.4/0.25 = 0.2856 L = 285.6 ml.
Answer:
285.6 ml.
c1(HCl)×V1 = c2(HCl)×V2
V1 = c2(HCl)×V2/c1(HCl)
V1 = 5.1×10-2×1.4/0.25 = 0.2856 L = 285.6 ml.
Answer:
285.6 ml.
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