ollowing table summarizes the amounts and concentrations of the reactants that w

Question

ollowing table summarizes the amounts and concentrations of the reactants that will be used in each trial in Parts A and B. Use the dilution equation M,v, = MA, to calculate the concentration M2 of each reactant in each well after mixing but before any reaction occurs. The first one has been worked for you as an example. 1. The f Volume of 0.30 M Na S203 3.0 mL 1.5 mL 1.0 mL 2.0 mL 2.0 mL 2.0 mL Volume of Water Sample Volume of Part AWell 1 Well 2 -Well 3 Part B Well 4 Well 5 Well 6 1.0 M HCI 2.0 mL 2.0 mL 2.0 mL 3.0 mL 1.5 mL 1.0 mL 1.5 mL 2.0 mL .5 mL 2.0 mL M2 concentration of reactant after mixing M, = concentation of reactant before mixing V volume of reactant before mixing V2 volume of reactant after mixing For well # 1: M2(HCI)-(1.0 M)(2.0 mL)/(5.0 mL) = 0.40 M M2(Na S,0 -(0.30 M)(3.0 mL)/(5.0 mL) 0.18 M 2. Enter the results of the calculations in the data table.

Explanation / Answer

Sample M2 HCl (1.0M) M2 Na2S2O3 (0.30M) Part A Well 1 M2 HCl=(1.0M)(2.0mL)/(5.0mL)=0.40M M2 Na2S2O3 =(0.30M)(3.0mL)/(5.0mL)=0.18M Well 2 M2 HCl=(1.0M)(2.0mL)/(5.0mL)=0.40M M2 Na2S2O3 =(0.30M)(1.5mL)/(5.0mL)=0.09M Well 3 M2 HCl=(1.0M)(2.0mL)/(5.0mL)=0.40M M2 Na2S2O3 =(0.30M)(1.0mL)/(5.0mL)=0.06M Part B Well 4 M2 HCl=(1.0M)(3.0mL)/(5.0mL)=0.6M M2 Na2S2O3 =(0.30M)(2.0mL)/(5.0mL)=0.12M Well 5 M2 HCl=(1.0M)(1.5mL)/(5.0mL)=0.3M M2 Na2S2O3 =(0.30M)(2.0mL)/(5.0mL)=0.12M Well 6 M2 HCl=(1.0M)(1.0mL)/(5.0mL)=0.2M M2 Na2S2O3 =(0.30M)(2.0mL)/(5.0mL)=0.12M

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