200 ml of a solutionof bariun hydroxide containa 171.4 mg of solute.the molarity of the solution will be
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unknown tri protic acid has an equivalent weight of 139.42 g/equiv, its average molecular weight is _________g/mol and __________g would be required to make 100.00 mL of a 1.50 N solution. See my response to your first question.
ap chemistry
A 0.456 gram sample of an unknown mono- protic acid (let’s call it HZ) was dissolved in some water (you pick the amount). Then the acidic solution was titrated to the equivalence point with 32.5 mL of 0.174 M KOH. What is the molecular weight of the unknown acid HZ? 1. 103.2...
chemistry
Three samples of an unknown acid are titrated with 0.4901 M NaOH. The average equivalent weight of the acid using the data below is g/equiv 1 2 3 mass of unknown acid (g) 1.020 0.522 0.742 vol NaOH used (ml) 27.72 14.22 20.02
chemistry - (Dr. Bob222)
A student attempted to synthesize malonic acid which has a molecular weight of 104 g/mol by the same method used to synthesize aspirin. A titration was performed to find the equivalent weight. 0.125 g of malonic acid product was used for a titration that
ap chemistry
A 0.456 gram sample of an unknown mono- protic acid (let’s call it HZ) was dissolved in some water (you pick the amount). Then the acidic solution was titrated to the equivalence point with 32.5 mL of 0.174 M KOH. What is the molecular weight of the unknown acid HZ? 1. 103.2...
chemistry
Three samples of an unknown acid are titrated with 0.4901 M NaOH. The average equivalent weight of the acid using the data below is g/equiv 1 2 3 mass of unknown acid (g) 1.020 0.522 0.742 vol NaOH used (ml) 27.72 14.22 20.02
chemistry - (Dr. Bob222)
A student attempted to synthesize malonic acid which has a molecular weight of 104 g/mol by the same method used to synthesize aspirin. A titration was performed to find the equivalent weight. 0.125 g of malonic acid product was used for a titration that
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