Can someone walk me through these questions legibly? Thanks 2. If you poured 10.

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Can someone walk me through these questions legibly? Thanks

2. If you poured 10.00 mL of 1.30 M HCI into a 50.00-mL volumetric flask and diluted it to the mark what would the final volume and final concentration of the solution be? Questions 3 and 4 are about the procedure of preparing the standard solutions used in this experiment. The standard solutions will be provided directly in the lab. However, their concentrations will not be labelled on the reagent bottles. You have to calculate the concentrations of Fe2+ in each standard solution in ?g/dL based on the given information below 3 0.1756 g of (NHFe(SO)-6HO was dissolved in water and diluted to 250.00mL. The solution was labelled as Original Fe Stock Solution. The molar mass of (NH) Fe(SO)-6H20 is 392.15 g/mol. Fe+ forms when (NH)2Fe(SO4)2-6H2O is dissolved in water: a. Calculate the moles of (NH Fe(SO4)2 6H2O used. b. Calculate the moles of Fe2+ in the stock solution. c. Calculate the mass of Fe in the stock solution in unit of g d. Calculate the concentration of Fe2 in the stock solution in units of g/L e. Calculate the concentration of Fe2+ in the stock solution in units of ?g/dL. 4. 10.00 mL of the Original Fe Solution was pipetted into a 100.00-mL volumetric flask and diluted to the mark. This diluted stock solution is labelled as Intermediate Fe2 Stock Solution. a. Calculate the concentration of the Intermediate Fe, solution (in ?dL).

Explanation / Answer

2) Use the dilution equation:

C1*V1 = C2*V2

where C1 = initial concentration of the supplied HCl solution = 1.30 M; V1 = volume of stock 1.30 M HCl used = 10.00 mL and V2 = final volume of the diluted HCl solution = 50.00 mL. Plug in values and obtain

(1.30 M)*(10.00 mL) = C2*(50.00 mL)

=====> C2 = (1.30 M)*(10.00 mL)/(50.00 mL) = 0.26 M.

The concentration of the diluted HCl solution is 0.26 M (ans).

3a) Moles of (NH4)2Fe(SO4)2.6H2O used = [mass of (NH4)2Fe(SO4)2.6H2O taken]/[molar mass of (NH4)2Fe(SO4)2.6H2O] = (0.1756 g)/(392.15 g/mol) = 4.4779*10-5 mole (ans).

b) As per the stoichiometric dissociation equation, we have

1 mole (NH4)2Fe(SO4)2.6H2O = 1 mole Fe2+.

Therefore, 4.4779*10-5 mole (NH4)2Fe(SO4)2.6H2O = 4.4779*10-5 mole Fe2+ (ans).

c) The atomic mass of Fe is 55.845 g/mol. Therefore, mass of Fe2+ in the prepared stock solution of (NH4)2Fe(SO4)2.6H2O = (moles of Fe2+)*(atomic mass of Fe2+) = (4.4779*10-5 mole)*(55.845 g/mol) = 2.5007*10-3 g (ans).

d) The volume of the stock solution is 250.00 mL = (250.00 mL)*(1 L/1000 mL) = 0.250 L.

Concentration of Fe2+ in the stock solution = (grams of Fe2+ in the stock solution)/(volume of the stock solution in L) = (2.5007*10-3 g)/(0.250 L) = 0.0100028 g/L ? 0.010 g/L (ans).

e) Concentration of Fe2+ in the stock solution in µg/dL = [(grams of Fe2+)*(1.0*106 µg/1 g)]/[(volume of stock solution in L)*(10 dL/1 L)] = [(2.5007*10-3 g)*(1.0*106 µg/1 g)]/[(0.250 L)*(10 dL/1 L)] = 1000.28 µg/dL ? 1000.3 µg/dL (ans).

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